Practice Problems: Resonance in AC Circuits

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about resonance in AC circuits.

Conductance, Susceptance, and Admittance

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance ( \frac{1}{R}. Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B=\frac{1}{X} and admittance, the reciprocal of impedance Y=\frac{1}{Z} ) are very necessary in order to draw phasor diagrams for parallel networks.

Capacitance adding in parallel; capacitive reactance and impedance adding in series

When students first encounter capacitance, they are struck by how this quantity adds when capacitors
are connected in parallel, not in series as it is for resistors and inductors. They are surprised again, though,
when they discover that the opposition to current offered by capacitors (either as scalar reactance or phasor
impedance) adds in series just as resistance adds in series and inductive reactance/impedance adds in series.
Remember: ohms always add in series, no matter what their source(s); only farads add in parallel (omitting
siemens or mhos, the units for conductance and admittance and susceptance, which of course also add in
parallel).

Question 1. (Click on arrow for answer)

If a metal bar is struck against a hard surface, the bar will “ring” with a characteristic frequency. This is the fundamental principle upon which tuning forks work:

Should be image 00600x01

The ability of any physical object to “ring” like this after being struck is dependent upon two complementary properties: mass and elasticity. An object must possess both mass and a certain amount of “springiness” in order to physically resonate.

Describe what would happen to the resonant frequency of a metal bar if it were made of a more elastic (less “stiff”) metal? What would happen to the resonant frequency if an extra amount of mass were added to the end being struck?

File Num: 00600

Answer

In either case, the resonant frequency of the bar would decrease.


Notes

Electrical resonance is so closely related to physical resonance, that I believe questions like this help students grasp the concept better. Everyone knows what resonance is in the context of a vibrating object (tuning fork, bell, wind chime, guitar string, cymbal head), even if they have never heard of the term “resonance” before. Getting them to understand that mechanical resonance depends on the complementary qualities of mass and elasticity primes their minds for understanding that electrical resonance depends on the complementary qualities of inductance and capacitance.





Question 2. (Click on arrow for answer)

This simple electric circuit is capable of resonance, whereby voltage and current oscillate at a frequency characteristic to the circuit:

Should be image 00601x01

In a mechanical resonant system — such as a tuning fork, a bell, or a guitar string — resonance occurs because the complementary properties of mass and elasticity exchange energy back and forth between each other in kinetic and potential forms, respectively. Explain how energy is stored and transferred back and forth between the capacitor and inductor in the resonant circuit shown in the illustration, and identify which of these components stores energy in kinetic form, and which stores energy in potential form.

File Num: 00601

Answer

Capacitors store energy in potential form, while inductors store energy in kinetic form.


Notes

Ask your students to define “potential” and “kinetic” energy. These terms, of course, are central to the question, and I have not bothered to define them. This omission is purposeful, and it is the students’ responsibility to research the definitions of these words in the process of answering the question. If a substantial number of your students stopped trying to answer the question when they encountered new words (instead of taking initiative to find out what the words mean), then it indicates a need to focus on independent learning skills (and attitudes!).

Discuss a typical “cycle” of energy exchange between kinetic and potential forms in a vibrating object, and then relate this exchange process to the oscillations of a tank circuit (capacitor and inductor).





Question 3. (Click on arrow for answer)

Resonant electric circuits are analogous to resonant mechanical systems. They both oscillate, and their oscillation is founded on an exchange of energy between two different forms.

Mechanical engineers studying vibrations in machinery sometimes use capacitors and inductors to model the physical characteristics of mechanical systems. Specifically, capacitors model elasticity, while inductors model mass.

Explain what mechanical quantities in a resonant system are analogous to voltage and current in a resonant circuit.

File Num: 01140

Answer

Mechanical force and velocity are analogous to electrical voltage and current, respectively.


Challenge question: specifically relate voltage and current for an inductor to force and velocity for a mass, and voltage and current for a capacitor to force and velocity for a spring. Illustrate the similarities mathematically, where possible!


Notes

This is a challenging question, and it is one I would reserve for students destined to become engineers. However, once answered, it brings deep insight to the phenomenon of resonance both mechanical and electrical.





Question 4. (Click on arrow for answer)

If an oscilloscope is set up for “single-sweep” triggering and connected to a DC-excited resonant circuit such as the one shown in the following schematic, the resulting oscillation will last just a short time (after momentarily pressing and releasing the pushbutton switch):

Should be image 03290x01

Explain why the oscillations die out, rather than go on forever. Hint: the answer is fundamentally the same as why a swinging pendulum eventually comes to a stop.

File Num: 03290

Answer

No resonant circuit is completely free of dissipative elements, whether resistive or radiative, and so some energy is lost each cycle.


Notes

A circuit such as this is easy to build and demonstrate, but you will need a digital storage oscilloscope to successfully capture the damped oscillations. Also, the results may be tainted by switch “bounce,” so be prepared to address that concept if you plan to demonstrate this to a live audience.

You might want to ask your students how they would suggest building a “tank circuit” that is as free from energy losses as possible. If a perfect tank circuit could be built, how would it act if momentarily energized by a DC source such as in this setup?





Question 5. (Click on arrow for answer)

How may the resonant frequency of this tank circuit be increased?

Should be image 00602x01

File Num: 00602

Answer

The resonant frequency of this tank circuit may be increased by substituting a smaller-value capacitor in for the capacitor value it presently has.


Note: this is not the only way to increase this circuit’s resonant frequency!


Notes

Challenge your students to explain another method for increasing the resonant frequency of this tank circuit, besides decreasing the value of the capacitor. Discuss how any of these alterations to the circuit affect the typical energy “cycle” between kinetic and potential forms, and why they lead to an increased frequency.





Question 6. (Click on arrow for answer)

Very interesting things happen to resonant systems when they are “excited” by external sources of oscillation. For example, a pendulum is a simple example of a mechanically resonant system, and we all know from experience with swings in elementary school that we can make a pendulum achieve very high amplitudes of oscillation if we “oscillate” our legs at just the right times to match the swing’s natural (resonant) frequency.

Identify an example of a mechanically resonant system that is “excited” by an external source of oscillations near its resonant frequency. Hint: research the word “resonance” in engineering textbooks, and you are sure to read about some dramatic examples of resonance in action.

File Num: 00603

Answer

Large buildings have (very low) resonant frequencies, that may be matched by the motion of the ground in an earthquake, so that even a relatively small earthquake can cause major damage to the building.


Challenge question: after researching the behavior of mechanical resonant systems when driven by external oscillations of the same frequency, determine what the effects might be of external oscillations on an electrical resonant system.


Notes

Many, many examples of mechanical resonance exist, some of which are quite dramatic. A famous example of destructive mechanical resonance (of a well-known bridge in Washington state) has been immortalized in video form, and is easily available on the internet. If possible, provide the means within your classroom to display a video clip on computer, for any of the students who happen to find this video file and bring it to discussion.





Question 7. (Click on arrow for answer)

If a capacitor and an inductor are connected in series, and energized by an AC voltage source with such a frequency that the reactances of each component are 125 \Omega and 170 \Omega, respectively, what is the total impedance of the series combination?

File Num: 00606

Answer

45 \Omega \angle 90^{o}


Now, of course, you are wondering: “how can two series-connected components have a total impedance that is less than either of their individual impedances?” Don’t series impedances add to equal the total impedance, just like series resistances? Be prepared to explain what is happening in this circuit, during discussion time with your classmates.


Notes

This question is an exercise in complex number arithmetic, and it is quite counter-intuitive at first. Discuss this problem in depth with your students, so that they are sure to comprehend the phenomenon of series-canceling impedances.





Question 8. (Click on arrow for answer)

Calculate all voltage drops and current in this LC circuit at each of the given frequencies:

Should be image 01873x01$$\vbox{\offinterlineskip \halign{\strut \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil \vrule \cr \noalign{\hrule}

Frequency & V_L & V_C & I_{total} \cr

\noalign{\hrule}

50 Hz & & & \cr

\noalign{\hrule}

60 Hz & & & \cr

\noalign{\hrule}

70 Hz & & & \cr

\noalign{\hrule}

80 Hz & & & \cr

\noalign{\hrule}

90 Hz & & & \cr

\noalign{\hrule}

100 Hz & & & \cr

\noalign{\hrule} } % End of \halign }$$ % End of \vbox

Also, calculate the resonant frequency of this circuit.

File Num: 01873

Answer

$$\vbox{\offinterlineskip \halign{\strut \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil \vrule \cr \noalign{\hrule}

Frequency & V_L & V_C & I_{total} \cr

\noalign{\hrule}

50 Hz & 0.121 V & 0.371 V & 1.16 mA \cr

\noalign{\hrule}

60 Hz & 0.221 V & 0.471 V & 1.77 mA \cr

\noalign{\hrule}

70 Hz & 0.440 V & 0.690 V & 3.03 mA \cr

\noalign{\hrule}

80 Hz & 1.24 V & 1.49 V & 7.48 mA \cr

\noalign{\hrule}

90 Hz & 4.25 V & 4.03 V & 22.8 mA \cr

\noalign{\hrule}

100 Hz & 1.07 V & 0.821 V & 5.16 mA \cr

\noalign{\hrule} } % End of \halign }$$ % End of \vboxf_r = 87.6 \hbox{ Hz}

Notes

This is nothing more than number-crunching, though some students may have found novel ways to speed up their calculations or verify their work.





Question 9. (Click on arrow for answer)

Suppose we were to build a series “LC” circuit and connect it to a function generator, where we could vary the frequency of the AC voltage powering it:

Should be image 00604x01

Calculate the amount of current in the circuit, given the following figures:


  • Power supply voltage = 3 volts RMS
  • Power supply frequency = 100 Hz
  • Capacitor = 4.7 \muF
  • Inductor = 100 mH

Then, describe what happens to the circuit current as the frequency is gradually increased.

File Num: 00604

Answer

Circuit current = 10.88 mA RMS. As the frequency is gradually increased, the circuit current increases as well.


Follow-up question: what do you suppose might happen to the circuit current if the frequency is increased to the point that the reactances of the inductor and capacitor completely cancel each other? What safety concerns might arise from this possibility?


Notes

In order for your students to arrive at the answer of circuit current increasing with frequency, they must perform a few calculations at different frequencies. Do this together, as a group, and note how the circuit’s impedance changes with frequency.





Question 10. (Click on arrow for answer)

Calculate the power supply frequency at which the reactances of a 33 \muF and a 75 mH inductor are exactly equal to each other. Derive a mathematical equation from the individual reactance equations (X_L = 2 \pi f L and X_C = {1 \over {2 \pi f C}}), solving for frequency (f) in terms of L and C in this condition.

Calculate the total impedance of these two components, if they were connected in series, at that frequency.

File Num: 00607

Answer

f_{resonant} = 101.17 Hz. At this frequency, Z_{series} = 0 \Omega.

Notes

The answer gives away the meaning of this question: the determination of an LC circuit’s resonant frequency. Students may be surprised at the total impedance figure of 0 \Omega, but this is really nothing more than an extension of the “impedance cancellation” concept they’ve seen before in other series LC circuit questions. In this case, the cancellation concept has merely been taken to the ultimate level of total cancellation between the two impedances.





Question 11. (Click on arrow for answer)

Calculate all voltages and currents in this circuit, at a power supply frequency near resonance:

Should be image 00608x01

Based on your calculations, what general predictions can you make about series-resonant circuits, in terms of their total impedance, their total current, and their individual component voltage drops?

File Num: 00608

Answer

In a series LC circuit near resonance, Z_{total} is nearly zero, I_{total} is large, and both E_L and E_C are large as well.


Follow-up question: suppose the capacitor were to fail shorted. Identify how this failure would alter the circuit’s current and voltage drops.


Notes

This question is given without a specified source frequency for a very important reason: to encourage students to “experiment” with numbers and explore concepts on their own. Sure, I could have given a power supply frequency as well, but I chose not to because I wanted students to set up part of the problem themselves.

In my experience teaching, students will often choose to remain passive with regard to a concept they do not understand, rather than aggressively pursue an understanding of it. They would rather wait and see if the instructor happens to cover that concept during class time than take initiative to explore it on their own. Passivity is a recipe for failure in life, and this includes intellectual endeavors as much as anything else. The fundamental trait of autonomous learning is the habit of pursuing the answer to a question, without being led to do so. Questions like this, which purposefully omit information, and thus force the student to think creatively and independently, teach them to develop this trait.





Question 12. (Click on arrow for answer)

Calculate all voltages and currents in this circuit, at a power supply frequency near resonance:

Should be image 00609x01

Based on your calculations, what general predictions can you make about parallel-resonant circuits, in terms of their total impedance, their total current, and their individual component currents?

File Num: 00609

Answer

In a parallel LC circuit near resonance, Z_{total} is nearly infinite, I_{total} is small, and both I_L and I_C are large as well.


Follow-up question: suppose the inductor were to fail open. Identify how this failure would alter the circuit’s current and voltage drops.


Notes

This question is given without a specified source frequency for a very important reason: to encourage students to “experiment” with numbers and explore concepts on their own. Sure, I could have given a power supply frequency as well, but I chose not to because I wanted students to set up part of the problem themselves.

In my experience teaching, students will often choose to remain passive with regard to a concept they do not understand, rather than aggressively pursue an understanding of it. They would rather wait and see if the instructor happens to cover that concept during class time than take initiative to explore it on their own. Passivity is a recipe for failure in life, and this includes intellectual endeavors as much as anything else. The fundamental trait of autonomous learning is the habit of pursuing the answer to a question, without being led to do so. Questions like this, which purposefully omit information, and thus force the student to think creatively and independently, teach them to develop this trait.





Question 13. (Click on arrow for answer)

The following schematic shows the workings of a simple AM radio receiver, with transistor amplifier:

Should be image 00611x01

The “tank circuit” formed of a parallel-connected inductor and capacitor network performs a very important filtering function in this circuit. Describe what this filtering function is.

File Num: 00611

Answer

The “tank circuit” filters out all the unwanted radio frequencies, so that the listener hears only one radio station broadcast at a time.


Follow-up question: how might a variable capacitor be constructed, to suit the needs of a circuit such as this? Note that the capacitance range for a tuning capacitor such as this is typically in the pico-Farad range.


Notes

Challenge your students to describe how to change stations on this radio receiver. For example, if we are listening to a station broadcasting at 1000 kHz and we want to change to a station broadcasting at 1150 kHz, what do we have to do to the circuit?

Be sure to discuss with them the construction of an adjustable capacitor (air dielectric).





Question 14. (Click on arrow for answer)

Calculate the resonant frequency of this parallel LC circuit, and qualitatively describe its total impedance (Z_{total}) when operating at resonance:

Should be image 04043x01

File Num: 04043

Answer

f_r = 6.195 kHz
Z_{total} @ f_r = \infty

Notes

Nothing special to note here, just an application of the resonance formula and a review of parallel LC resonance.





Question 15. (Click on arrow for answer)

Does a series LC circuit “appear” capacitive or inductive (from the perspective of the AC source powering it) when the source frequency is greater than the circuit’s resonant frequency? What about a parallel resonant circuit? In each case, explain why.

File Num: 01563

Answer

A series LC circuit will appear inductive when the source frequency exceeds the resonant frequency. A parallel LC circuit will appear capacitive in the same condition.


Notes

Ask your students to explain their answers mathematically.





Question 16. (Click on arrow for answer)

A paradoxical property of resonant circuits is that they have the ability to produce quantities of voltage or current (in series and parallel circuits, respectively) exceeding that output by the power source itself. This is due to the cancellation of inductive and capacitive reactances at resonance.

Not all resonant circuits are equally effective in this regard. One way to quantify the performance of resonant circuits is to assign them a quality factor, or Q rating. This rating is very similar to the one given inductors as a measure of their reactive “purity.”

Suppose we have a resonant circuit operating at its resonant frequency. How may we calculate the Q of this operating circuit, based on empirical measurements of voltage or current? There are two answers to this question: one for series circuits and one for parallel circuits.

File Num: 01390

Answer

Q_{series} = {E_C \over E_{source}} = {E_L \over E_{source}}Q_{parallel} = {I_C \over I_{source}} = {I_L \over I_{source}}

Follow-up question: what unique safety hazards may high-Q resonant circuits pose?


Notes

Ask your students to determine which type of danger(s) are posed by high-Q series and parallel resonant circuits, respectively. The answer to this question may seem paradoxical at first: that series resonant circuits whose overall impedance is nearly zero can manifest large voltage drops, while parallel resonant circuits whose overall impedance is nearly infinite can manifest large currents.





Question 17. (Click on arrow for answer)

The Q factor of a series inductive circuit is given by the following equation:

Q = {X_L \over R_{series}}

Likewise, we know that inductive reactance may be found by the following equation:

X_L = 2 \pi f L

We also know that the resonant frequency of a series LC circuit is given by this equation:

f_r = {1 \over {2 \pi \sqrt{LC}}}

Through algebraic substitution, write an equation that gives the Q factor of a series resonant LC circuit exclusively in terms of L, C, and R, without reference to reactance (X) or frequency (f).

File Num: 01683

Answer

Q = {1 \over R} \sqrt{L \over C}

Notes

This is merely an exercise in algebra. However, knowing how these three component values affects the Q factor of a resonant circuit is a valuable and practical insight!





Question 18. (Click on arrow for answer)

Shown here are two frequency response plots (known as Bode plots) for a pair of series resonant circuits. Each circuit has the same inductance and capacitance values, but different resistance values. The “output” is voltage measured across the resistor of each circuit:

Should be image 01391x01

Which one of these plots represents the response of the circuit with the greatest Q, or quality factor?

File Num: 01391

Answer

The steeper plot corresponds to the circuit with the greatest Q.


Follow-up question: assuming both the inductance and the capacitance values are the same in these two resonant circuits, explain which circuit has the greatest resistance (R_1 or R_2).


Challenge question: what does the word “normalized” mean with respect to the vertical axis scale of the Bode plot?


Notes

When your students study resonant filter circuits, they will better understand the importance of Q. For now, though, it is enough that they comprehend the basic notion of how Q impacts the voltage dropped by any one component in a series resonant circuit across a range of frequencies.





Question 19. (Click on arrow for answer)

The Q, or quality factor, of an inductor circuit is defined by the following equation, where X_s is the series inductive reactance and R_s is the series resistance:

Q = {X_s \over R_{s}}

We also know that we may convert between series and parallel equivalent AC networks with the following conversion equations:

R_{s} R_{p} = Z^2 \hbox{\hskip 40pt} X_{s} X_{p} = Z^2Should be image 02096x01

Series and parallel LR networks, if truly equivalent, should share the same Q factor as well as sharing the same impedance. Develop an equation that solves for the Q factor of a parallel LR circuit.

File Num: 02096

Answer

Q = {R_p \over X_p}

Follow-up question: what condition gives the greatest value for Q, a low parallel resistance or a high parallel resistance? Contrast this against the effects of low versus high resistance in a series LR circuit, and explain both scenarios.


Notes

This is primarily an exercise in algebraic substitution, but it also challenges students to think deeply about the nature of Q and what it means, especially in the follow-up question.





Question 20. (Click on arrow for answer)

There is a direct, mathematical relationship between bandwidth, resonant frequency, and Q in a resonant filter circuit, but imagine for a moment that you forgot exactly what that formula was. You think it must be one of these two, but you are not sure which:

\hbox{Bandwidth} = {Q \over f_r} \hbox{(or possibly)} \hbox{Bandwidth} = {f_r \over Q}

Based on your conceptual knowledge of how a circuit’s quality factor affects its frequency response, determine which of these formulae must be incorrect. In other words, demonstrate which of these must be correct rather than simply looking up the correct formula in a reference book.

File Num: 01870

Answer

Hint: the greater the value of Q, the less bandwidth a resonant circuit will have.


Notes

The purpose of this question is not necessarily to get students to look this formula up in a book, but rather to develop their qualitative reasoning and critical thinking skills. Forgetting the exact form of an equation is not a rare event, and it pays to be able to select between different forms based on a conceptual understanding of what the formula is supposed to predict.

Note that the question asks students to identify the wrong formula, and not to tell which one is right. If all we have are these to formulae to choose from, and a memory too weak to confidently recall the correct form, the best that logic can do is eliminate the wrong formula. The formula making the most sense according to our qualitative analysis may or may not be precisely right, because we could very well be forgetting a multiplicative constant (such as 2 \pi).





Question 21. (Click on arrow for answer)

Suppose you have a 110 mH inductor, and wish to combine it with a capacitor to form a band-stop filter with a “notch” frequency of 1 kHz. Draw a schematic diagram showing what the circuit would look like (complete with input and output terminals) and calculate the necessary capacitor size to do this, showing the equation you used to solve for this value. Also, calculate the bandwidth of this notch filter, assuming the inductor has an internal resistance of 20 ohms, and that there is negligible resistance in the rest of the circuit.

File Num: 01872

Answer

Should be image 01872x01

The bandwidth of this 1 kHz notch filter is approximately 29 Hz.


Follow-up question: suppose you looked around but could not find a capacitor with a value of 0.23 \muF. What could you do to obtain this exact capacitance value? Be as specific and as practical as you can in your answer!


Notes

In my answer I used the series-resonant formula f_r = {1 \over {2 \pi \sqrt{LC}}}, since the series formula gives good approximations for parallel resonant circuits with Q factors in excess of 10.

The follow-up question is very practical, since it is often common to need a component value that is non-standard. Lest any of your students suggest obtaining a variable capacitor for this task, remind them that variable capacitors are typically rated in the pico-Farad range, and would be much too small for this application.





Question 22. (Click on arrow for answer)

Shown here are two frequency response plots (known as Bode plots) for a pair of series resonant circuits with the same resonant frequency. The “output” is voltage measured across the resistor of each circuit:

Should be image 01682x01

Determine which plot is associated with which circuit, and explain your answer.

File Num: 01682

Answer

The steeper plot corresponds to the circuit with the greatest {L \over C} ratio.


Follow-up question: what kind of instrument(s) would you use to plot the response of a real resonant circuit in a lab environment? Would an oscilloscope be helpful with this task? Why or why not?


Notes

Discuss with your students why the LC circuit with the greatest {L \over C} ratio has the steeper response, in terms of reactances of the respective components at the resonant frequency.

The purpose of this question is to get students to realize that not all resonant circuits with identical resonant frequencies are alike! Even with ideal components (no parasitic effects), the frequency response of a simple LC circuit varies with the particular choice of component values. This is not obvious from inspection of the resonant frequency formula:

f_r = {1 \over {2 \pi \sqrt{LC}}}




Question 23. (Click on arrow for answer)

Not only do reactive components unavoidably possess some parasitic (“stray”) resistance, but they also exhibit parasitic reactance of the opposite kind. For instance, inductors are bound to have a small amount of capacitance built-in, and capacitors are bound to have a small amount of inductance built-in. These effects are not intentional, but they exist anyway.

Describe how a small amount of capacitance comes to exist within an inductor, and how a small amount of inductance comes to exist within a capacitor. Explain what it is about the construction of these two reactive components that allows the existence of “opposite” characteristics.

File Num: 00593

Answer

Capacitance exists any time there are two conductors separated by an insulating medium. Inductance exists any time a magnetic field is permitted to exist around a current-carrying conductor. Look for each of these conditions within the respective structures of inductors and capacitors to determine where the parasitic effects originate.


Notes

Once students have identified the mechanisms of parasitic reactances, challenge them with inventing means of minimizing these effects. This is an especially practical exercise for understanding parasitic inductance in capacitors, which is very undesirable in decoupling capacitors used to stabilize power supply voltages near integrated circuit “chips” on printed circuit boards. Fortunately, most of the stray inductance in a decoupling capacitor is due to how it’s mounted to the board, rather than anything within the structure of the capacitor itself.





Question 24. (Click on arrow for answer)

Given the unavoidable presence of parasitic inductance and/or capacitance in any electronic component, what does this mean in terms of resonance for single components in AC circuits?

File Num: 00594

Answer

Parasitic reactance means that any single component is theoretically capable of resonance, all on its own!


Follow-up question: at what frequency would you expect a component to self-resonate? Would this be a very low frequency, a very high frequency, or a frequency within the circuit’s normal operating range? Explain your answer.


Notes

This question grew out of several years’ worth of observations, where students would discover self-resonant effects in large (> 1 Henry) inductors at modest frequencies. Being a recurring theme, I thought it prudent to include this question within my basic electronics curriculum.

One component that tends to be more immune to self-resonance than others is the lowly resistor, especially resistors of large value. Ask your students why they think this might be. A mechanical analogy to self-resonance is the natural frequency of vibration for an object, given the unavoidable presence of both elasticity and mass in any object. The mechanical systems most immune to vibratory resonance, though, are those with a high degree of intrinsic friction.





Question 25. (Click on arrow for answer)

A capacitor has been connected in parallel with the solenoid coil to minimize arcing of the switch contacts when opened:

Should be image 00610x01

The only problem with this solution is, resonance between the capacitor and the solenoid coil’s inductance is causing an oscillating voltage (commonly known as ringing) to appear across the terminals of each. This high-frequency “ringing” is generating bursts of radio interference whenever the switch contacts open. Radio interference is not good.

You know that the underlying cause of this “ringing” is resonance, yet you cannot simply remove the capacitor from the circuit because you know that will result in decreased operating life for the switch contacts, as the solenoid’s inductive “kickback” will cause excessive arcing. How do you overcome this problem without creating another problem?

File Num: 00610

Answer

Like many realistic problems, there is more than one possible solution. One way to approach this problem is to think of an analogous situation, and how the same type of problem was solved by someone else in that context. For example, how do automotive engineers solve the problem of mechanical resonance destabilizing a vehicle after it runs over a bump in the road? What did they invent to dampen the natural “bouncing” of the vehicle’s suspension system, without defeating the purpose of the suspension system altogether? And how might you apply this principle to an electric circuit?


Follow-up question: besides shortening the life of the switch, what other undesirable effects can switch “arcing” have? Can you think of any scenarios where an arcing switch could pose a safety hazard?


Notes

Besides posing a practical problem-solving scenario to students, this question is a good lead-in to the topic of antiresonance. Be sure to allow plenty of class discussion time for this question, as many topics are likely to be covered as students discuss alternative problem-solving strategies.





Question 26. (Click on arrow for answer)

An alternative to “tank circuit” combinations of L and C in many electronic circuits is a small device known as a crystal. Explain how a “crystal” may take the place of a tank circuit, and how it functions.

File Num: 01869

Answer

Crystals are mechanical resonators made of a piezoelectric material (usually quartz).


Notes

My answer here is purposefully vague, to inspire students to research on their own.





Question 27. (Click on arrow for answer)

\centerlineDon’t just sit there! Build something!!

Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.

You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuit-building exercises, follow these steps:


\item{1.} Carefully measure and record all component values prior to circuit construction. \item{2.} Draw the schematic diagram for the circuit to be analyzed. \item{3.} Carefully build this circuit on a breadboard or other convenient medium. \item{4.} Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements one-by-one on the diagram. \item{5.} Mathematically analyze the circuit, solving for all voltage and current values. \item{6.} Carefully measure all voltages and currents, to verify the accuracy of your analysis. \item{7.} If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully re-calculate the values and re-measure.

For AC circuits where inductive and capacitive reactances (impedances) are a significant element in the calculations, I recommend high quality (high-Q) inductors and capacitors, and powering your circuit with low frequency voltage (power-line frequency works well) to minimize parasitic effects. If you are on a restricted budget, I have found that inexpensive electronic musical keyboards serve well as “function generators” for producing a wide range of audio-frequency AC signals. Be sure to choose a keyboard “voice” that closely mimics a sine wave (the “panflute” voice is typically good), if sinusoidal waveforms are an important assumption in your calculations.

As usual, avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading”. I recommend resistor values between 1 k\Omega and 100 k\Omega.

One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another time-saving technique is to re-use the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.

File Num: 00605

Answer

Let the electrons themselves give you the answers to your own “practice problems”!


Notes

It has been my experience that students require much practice with circuit analysis to become proficient. To this end, instructors usually provide their students with lots of practice problems to work through, and provide answers for students to check their work against. While this approach makes students proficient in circuit theory, it fails to fully educate them.

Students don’t just need mathematical practice. They also need real, hands-on practice building circuits and using test equipment. So, I suggest the following alternative approach: students should build their own “practice problems” with real components, and try to mathematically predict the various voltage and current values. This way, the mathematical theory “comes alive,” and students gain practical proficiency they wouldn’t gain merely by solving equations.

Another reason for following this method of practice is to teach students scientific method: the process of testing a hypothesis (in this case, mathematical predictions) by performing a real experiment. Students will also develop real troubleshooting skills as they occasionally make circuit construction errors.

Spend a few moments of time with your class to review some of the “rules” for building circuits before they begin. Discuss these issues with your students in the same Socratic manner you would normally discuss the worksheet questions, rather than simply telling them what they should and should not do. I never cease to be amazed at how poorly students grasp instructions when presented in a typical lecture (instructor monologue) format!

An excellent way to introduce students to the mathematical analysis of real circuits is to have them first determine component values (L and C) from measurements of AC voltage and current. The simplest circuit, of course, is a single component connected to a power source! Not only will this teach students how to set up AC circuits properly and safely, but it will also teach them how to measure capacitance and inductance without specialized test equipment.

A note on reactive components: use high-quality capacitors and inductors, and try to use low frequencies for the power supply. Small step-down power transformers work well for inductors (at least two inductors in one package!), so long as the voltage applied to any transformer winding is less than that transformer’s rated voltage for that winding (in order to avoid saturation of the core).


A note to those instructors who may complain about the “wasted” time required to have students build real circuits instead of just mathematically analyzing theoretical circuits:


\hskip 1in What is the purpose of students taking your course?

If your students will be working with real circuits, then they should learn on real circuits whenever possible. If your goal is to educate theoretical physicists, then stick with abstract analysis, by all means! But most of us plan for our students to do something in the real world with the education we give them. The “wasted” time spent building real circuits will pay huge dividends when it comes time for them to apply their knowledge to practical problems.

Furthermore, having students build their own practice problems teaches them how to perform primary research, thus empowering them to continue their electrical/electronics education autonomously.

In most sciences, realistic experiments are much more difficult and expensive to set up than electrical circuits. Nuclear physics, biology, geology, and chemistry professors would just love to be able to have their students apply advanced mathematics to real experiments posing no safety hazard and costing less than a textbook. They can’t, but you can. Exploit the convenience inherent to your science, and get those students of yours practicing their math on lots of real circuits!





Question 28. (Click on arrow for answer)

Plot the typical frequency responses of four different filter circuits, showing signal output (amplitude) on the vertical axis and frequency on the horizontal axis:

Should be image 02571x01

Also, identify and label the bandwidth of the filter circuit on each plot.

File Num: 02571

Answer

Should be image 02571x02

Notes

Although “bandwidth” is usually applied first to band-pass and band-stop filters, students need to realize that it applies to the other filter types as well. This question, in addition to reviewing the definition of bandwidth, also reviews the definition of cutoff frequency. Ask your students to explain where the 70.7\% figure comes from. Hint: half-power point!





Question 29. (Click on arrow for answer)

Identify each of these filter types, and explain how you were able to positively identify their behaviors:

Should be image 02098x01

File Num: 02098

Answer

Should be image 02098x02

Follow-up question: in each of the circuits shown, identify at least one single component failure that has the ability to prevent any signal voltage from reaching the output terminals.


Notes

Some of these filter designs are resonant in nature, while others are not. Resonant circuits, especially when made with high-Q components, approach ideal band-pass (or -block) characteristics. Discuss with your students the different design strategies between resonant and non-resonant band filters.

The high-pass filter containing both inductors and capacitors may at first appear to be some form of resonant (i.e. band-pass or band-stop) filter. It actually will resonate at some frequency(ies), but its overall behavior is still high-pass. If students ask about this, you may best answer their queries by using computer simulation software to plot the behavior of a similar circuit (or by suggesting they do the simulation themselves).

Regarding the follow-up question, it would be a good exercise to discuss which suggested component failures are more likely than others, given the relatively likelihood for capacitors to fail shorted and inductors and resistors to fail open.





Question 30. (Click on arrow for answer)

Identify the following filter types, and be prepared to explain your answers:

Should be image 00620x01

File Num: 00620

Answer

Should be image 00620x02

Notes

Some of these filter designs are resonant in nature, while others are not. Resonant circuits, especially when made with high-Q components, approach ideal band-pass (or -block) characteristics. Discuss with your students the different design strategies between resonant and non-resonant band filters.

Although resonant band filter designs have nearly ideal (theoretical) characteristics, band filters built with capacitors and resistors only are also popular. Ask your students why this might be. Is there any reason inductors might purposefully be avoided when designing filter circuits?





Question 31. (Click on arrow for answer)

The cutoff frequency, also known as half-power point or -3dB point, of either a low-pass or a high-pass filter is fairly easy to define. But what about band-pass and band-stop filter circuits? Does the concept of a “cutoff frequency” apply to these filter types? Explain your answer.

File Num: 01871

Answer

Unlike low-pass and high-pass filters, band-pass and band-stop filter circuits have two cutoff frequencies (f_{c1} and f_{c2})!


Notes

This question presents a good opportunity to ask students to draw the Bode plot of a typical band-pass or band-stop filter on the board in front of the class to illustrate the concept. Don’t be afraid to let students up to the front of the classroom to present their findings. It’s a great way to build confidence in them and also to help suppress the illusion that you (the teacher) are the Supreme Authority of the classroom!





Question 32. (Click on arrow for answer)

An interesting technology dating back at least as far as the 1940’s, but which is still of interest today is power line carrier: the ability to communicate information as well as electrical power over power line conductors. Hard-wired electronic data communication consists of high-frequency, low voltage AC signals, while electrical power is low-frequency, high-voltage AC. For rather obvious reasons, it is important to be able to separate these two types of AC voltage quantities from entering the wrong equipment (especially the high-voltage AC power from reaching sensitive electronic communications circuitry).

Here is a simplified diagram of a power-line carrier system:

Should be image 01393x01

The communications transmitter is shown in simplified form as an AC voltage source, while the receiver is shown as a resistor. Though each of these components is much more complex than what is suggested by these symbols, the purpose here is to show the transmitter as a source of high-frequency AC, and the receiver as a load of high-frequency AC.

Trace the complete circuit for the high-frequency AC signal generated by the “Transmitter” in the diagram. How many power line conductors are being used in this communications circuit? Explain how the combination of “line trap” LC networks and “coupling” capacitors ensure the communications equipment never becomes exposed to high-voltage electrical power carried by the power lines, and visa-versa.

File Num: 01393

Answer

Should be image 01393x02

Follow-up question \#1: trace the path of line-frequency (50 Hz or 60 Hz) load current in this system, identifying which component of the line trap filters (L or C) is more important to the passage of power to the load. Remember that the line trap filters are tuned to resonate at the frequency of the communication signal (50-150 kHz is typical).


Follow-up question \#2: coupling capacitor units used in power line carrier systems are special-purpose, high-voltage devices. One of the features of a standard coupling capacitor unit is a spark gap intended to “clamp” overvoltages arising from lightning strikes and other transient events on the power line:

Should be image 01393x03

Explain how such a spark gap is supposed to work, and why it functions as an over-voltage protection device.


Notes

Although power line carrier technology is not used as much for communication in high-voltage distribution systems as it used to be — now that microwave, fiber optic, and satellite communications technology has superseded this older technique — it is still used in lower voltage power systems including residential (home) wiring. Ask your students if they have heard of any consumer technology capable of broadcasting any kind of data or information along receptacle wiring. “X10” is a mature technology for doing this, and at this time (2004) there are devices available on the market allowing one to plug telephones into power receptacles to link phones in different rooms together without having to add special telephone cabling.

Even if your students have not yet learned about three-phase power systems or transformers, they should still be able to discern the circuit path of the communications signal, based on what they know of capacitors and inductors, and how they respond to signals of arbitrarily high frequency.

Information on the coupling capacitor units was obtained from page 452 of the Industrial Electronics Reference Book, published by John Wiley \& Sons in 1948 (fourth printing, June 1953). Although power line carrier technology is not as widely used now as it was back then, I believe it holds great educational value to students just learning about filter circuits and the idea of mixing signals of differing frequency in the same circuit.





Question 33. (Click on arrow for answer)

In this power-line carrier system, a pair of coupling capacitors connects a high-frequency “Transmitter” unit to two power line conductors, and a similar pair of coupling capacitors connects a “Receiver” unit to the same two conductors:

Should be image 01394x01

While coupling capacitors alone are adequate to perform the necessary filtering function needed by the communications equipment (to prevent damaged from the high-voltage electrical power also carried by the lines), that signal coupling may be made more efficient by the introduction of two line tuning units:

Should be image 01394x02

Explain why the addition of more components (in series, no less!) provides a better “connection” between the high-frequency Transmitter and Receiver units than coupling capacitors alone. Hint: the operating frequency of the communications equipment is fixed, or at least variable only over a narrow range.

File Num: 01394

Answer

The introduction of the line-tuning units increases the efficiency of signal coupling by exploiting the principle of resonance between series-connected capacitors and inductors.


Challenge question: there are many applications in electronics where we couple high-frequency AC signals by means of capacitors alone. If capacitive reactance is any concern, we just use capacitors of large enough value that the reactance is minimal. Why would this not be a practical option in a power-line carrier system such as this? Why could we not (or why would we not) just choose coupling capacitors with very high capacitances, instead of adding extra components to the system?


Notes

Although power line carrier technology is not used as much for communication in high-voltage distribution systems as it used to be — now that microwave, fiber optic, and satellite communications technology has come of age — it is still used in lower voltage power systems including residential (home) wiring. Ask your students if they have heard of any consumer technology capable of broadcasting any kind of data or information along receptacle wiring. “X10” is a mature technology for doing this, and at this time (2004) there are devices available on the market allowing one to plug telephones into power receptacles to link phones in different rooms together without having to add special telephone cabling.

I think this is a really neat application of resonance: the complementary nature of inductors to capacitors works to overcome the less-than-ideal coupling provided by capacitors alone. Discuss the challenge question with your students, asking them to consider some of the practical limitations of capacitors, and how an inductor/capacitor resonant pair solves the line-coupling problem better than an oversized capacitor.





All files with file num less than 4100 are Copyright 2003, Tony R. Kuphaldt, released under the Creative Commons Attribution License (v 1.0). All other files are Copyright 2022, David Williams, released under the Creative Commons Attribution License (V 4.0) This means you may do almost anything with this work, so long as you give proper credit.

Practice Problems: RLC in AC Circuits

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about inductors in AC circuits.

Resistance vs. Reactance vs. Impedance

These three terms represent different forms of opposition to electric current. Despite the fact that they are measured in the same unit (ohms: Omega), they are not the same. Resistance is best thought of as electrical friction, whereas reactance is best thought of as electrical inertia. Whereas resistance creates a voltage drop by dissipating energy, reactance creates a voltage drop by storing and releasing energy. Impedance is a term encompassing both resistance and reactance, usually a combination of both.

Phasors, used to represent AC amplitude and phase relations.

A powerful tool used for understanding the operation of AC circuits is the phasor diagram, consisting of arrows pointing in different directions: the length of each arrow representing the amplitude of some AC quantity (voltage, current, or impedance), and the angle of each arrow representing the shift in phase relative to the other arrows. By representing each AC quantity thusly, we may more easily calculate their relationships to one another, with the phasors showing us how to apply trigonometry (Pythagorean Theorem, sine, cosine, and tangent functions) to the various calculations. An analytical parallel to the graphic tool of phasor diagrams is complex numbers, where we represent each phasor (arrow) by a pair of numbers: either a magnitude and angle (polar notation), or by “real” and “imaginary” magnitudes (rectangular notation). Where phasor diagrams are helpful is in applications where their respective AC quantities add: the resultant of two or more phasors stacked tip-to-tail being the mathematical sum of the phasors. Complex numbers, on the other hand, may be added, subtracted, multiplied, and divided; the last two operations being difficult to graphically represent with arrows.

Conductance, Susceptance, and Admittance.

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance (1 \over R). Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B = {1 \over X} and admittance, the reciprocal of impedance Y = {1 \over Z}) are very necessary in order to draw phasor diagrams for parallel networks.

Question 1. (Click on arrow for answer)

Capacitors and inductors are complementary components — both conceptually and mathematically, they seem to be almost exact opposites of each other. Calculate the total impedance of this series-connected inductor and capacitor network:

Should be image 00851x01

File Num: 00851

Answer

Z_{total} = 13 \Omega \angle -90^{o}

Follow-up question: does this circuit “appear” to be inductive or capacitive from the source’s point of view?


Notes

Here, the complementary nature of inductive and capacitive reactances is plain to see: they subtract in series. Challenge your students by asking them what the total impedance of this circuit would be if the two reactances were equal.





Question 2. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 00852x01

File Num: 00852

Answer

Z_{total} = \sqrt{R^2 + (X_L - X_C)^2}

Notes

Ask your students why one of the reactance terms under the radicand is positive and the other is negative. The way this equation is written, does it matter which term is negative? As your students if we would obtain the same answer if it were written as Z_{total} = \sqrt{R^2 + (X_C - X_L)^2} instead. Challenge them to answer this question without using a calculator!





Question 3. (Click on arrow for answer)

Write an equation that solves for the admittance of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Should be image 00854x01

File Num: 00854

Answer

Y_{total} = \sqrt{G^2 + (B_L - B_C)^2}

Notes

Ask your students why one of the reactance terms under the radicand is positive and the other is negative. The way this equation is written, does it matter which term is negative? Ask your students if we would obtain the same answer if the equation were written as Y_{total} = \sqrt{G^2 + (B_C - B_L)^2} instead. Challenge them to answer this question without using a calculator!





Question 4. (Click on arrow for answer)

Calculate the total impedance of this parallel network, given a signal frequency of 12 kHz:

Should be image 01541x01

File Num: 01541

Answer

Z_{total} = 8.911 k\Omega \angle 26.98^{o}

Notes

Ask your students how they obtained the phase angle for this circuit. There is more than one way to calculate this!


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 5. (Click on arrow for answer)

Is this circuit’s overall behavior capacitive or inductive? In other words, from the perspective of the AC voltage source, does it “appear” as though a capacitor is being powered, or an inductor?

Should be image 01554x01

Now, suppose we take these same components and re-connect them in parallel rather than series. Does this change the circuit’s overall “appearance” to the source? Does the source now “see” an equivalent capacitor or an equivalent inductor? Explain your answer.

Should be image 01554x02

File Num: 01554

Answer

Overall, the first (series) circuit’s behavior is inductive. The second (parallel) circuit’s behavior, though, is capacitive.


Follow-up question: which component “dominates” the behavior of a series LC circuit, the one with the least reactance or the one with the greatest reactance? Which component “dominates” the behavior of a parallel LC circuit, the one with the least reactance or the one with the greatest reactance?


Notes

As usual, the real point of this question is to get students to think about the analytical procedure(s) they use, and to engage their minds in problem-solving behavior. Ask them why they think the circuits behave inductively or capacitively.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 6. (Click on arrow for answer)

An AC electric motor operating under loaded conditions draws a current of 11 amps (RMS) from the 120 volt (RMS) 60 Hz power lines. The measured phase shift between voltage and current for this motor is 34^{o}, with voltage leading current.

Determine the equivalent parallel combination of resistance (R) and inductance (L) that is electrically equivalent to this operating motor.

File Num: 01542

Answer

R_{parallel} = 13.16 \Omega
L_{parallel} = 51.75 mH

Challenge question: in the parallel LR circuit, the resistor will dissipate a lot of energy in the form of heat. Does this mean that the electric motor, which is electrically equivalent to the LR network, will dissipate the same amount of heat? Explain why or why not.


Notes

If students get stuck on the challenge question, remind them that an electric motor does mechanical work, which requires energy.





Question 7. (Click on arrow for answer)

Suppose you are building a circuit and you need an impedance of 1500 \Omega \angle -41^{o} at a frequency of 600 Hz. What combination of components could you connect together in series to achieve this precise impedance?

File Num: 00644

Answer

A 1132.1 \Omega resistor connected in series with a 269.6 nF capacitor would suffice.


Notes

As usual, the most important part of your students’ answers is not the figures themselves, but rather their methods of solution. Students should be very familiar with how to calculate the impedance of a series-connected group of components, but calculating component values from an impedance figure may be a challenge to some.





Question 8. (Click on arrow for answer)

It is often useful in AC circuit analysis to be able to convert a series combination of resistance and reactance into an equivalent parallel combination of conductance and susceptance, or visa-versa:

Should be image 00856x01

We know that resistance (R), reactance (X), and impedance (Z), as scalar quantities, relate to one another trigonometrically in a series circuit. We also know that conductance (G), susceptance (B), and admittance (Y), as scalar quantities, relate to one another trigonometrically in a parallel circuit:

Should be image 00856x02

If these two circuits are truly equivalent to one another, having the same total impedance, then their representative triangles should be geometrically similar (identical angles, same proportions of side lengths). With equal proportions, {R \over Z} in the series circuit triangle should be the same ratio as {G \over Y} in the parallel circuit triangle, that is {R \over Z} = {G \over Y}.

Building on this proportionality, prove the following equation to be true:

R_{series} R_{parallel} = {Z_{total}}^2

After this, derive a similar equation relating the series and parallel reactances (X_{series} and X_{parallel}) with total impedance (Z_{total}).

File Num: 00856

Answer

I’ll let you figure out how to turn {R \over Z} = {G \over Y} into R_{series} R_{parallel} = {Z_{total}}^2 on your own!


As for the reactance relation equation, here it is:

X_{series} X_{parallel} = {Z_{total}}^2

Notes

Being able to convert between series and parallel AC networks is a valuable skill for analyzing complex series-parallel combination circuits, because it means any series-parallel combination circuit may then be converted into an equivalent simple-series or simple-parallel, which is mush easier to analyze.

Some students might ask why the conductance/susceptance triangle is “upside-down” compared to the resistance/reactance triangle. The reason has to do with the sign reversal of imaginary quantities when inverted: {1 \over j} = -j. The phase angle of a pure inductance’s impedance is +90 degrees, while the phase angle of the same (pure) inductance’s admittance is -90 degrees, due to reciprocation. Thus, while the X leg of the resistance/reactance triangle points up, the B leg of the conductance/susceptance triangle must point down.





Question 9. (Click on arrow for answer)

Determine an equivalent parallel RC network for the series RC network shown on the left:

Should be image 01540x01

Note that I have already provided a value for the capacitor’s reactance (X_C), which of course will be valid only for a particular frequency. Determine what values of resistance (R) and reactance (X_C) in the parallel network will yield the exact same total impedance (Z_T) at the same signal frequency.

File Num: 01540

Answer

R = 150 \Omega
X_C = 200 \Omega

Follow-up question: explain how you could check your conversion calculations, to ensure both networks are truly equivalent to each other.


Notes

This problem just happens to work out with whole numbers. Believe it or not, I chose these numbers entirely by accident one day, when setting up an example problem to show a student how to convert between series and parallel equivalent networks!





Question 10. (Click on arrow for answer)

Determine the equivalent parallel-connected resistor and inductor values for this series circuit:

Should be image 00855x01

Also, express the total impedance of either circuit (since they are electrically equivalent to one another, they should have the same total impedance) in complex form. That is, express Z as a quantity with both a magnitude and an angle.

File Num: 00855

Answer

R_{parallel} = 2092 \Omega
L_{parallel} = 1.325 H
Z_{total} = 1772 \Omega \angle 32.14^{o}

Notes

There are different methods of solving this problem. Use the discussion time to let students expound on how they approached the problem, pooling together their ideas. Their creativity may surprise you!





Question 11. (Click on arrow for answer)

Determine the equivalent series-connected resistor and capacitor values for this parallel circuit:

Should be image 00858x01

Also, express the total impedance of either circuit (since they are electrically equivalent to one another, they should have the same total impedance) in complex form. That is, express Z as a quantity with both a magnitude and an angle.

File Num: 00858

Answer

R_{series} = 454.8 \Omega
C_{series} = 3.3 \muF
Z_{total} = 1066 \Omega \angle -64.75^{o}

Notes

There are different methods of solving this problem. Use the discussion time to let students expound on how they approached the problem, pooling together their ideas. Their creativity may surprise you!





Question 12. (Click on arrow for answer)

Calculate the impedance of a 145 mH inductor connected in series with 750 \Omega resistor at a frequency of 1 kHz, then determine the necessary resistor and inductor values to create the exact same total impedance in a parallel configuration.

File Num: 00645

Answer

Z_{total} = 1.18 k\Omega \angle 50.54^{o}

If connected in parallel: R = 1.857 k\Omega ; L = 243.3 mH.


Hint: if you are having difficulty figuring out where to start in answering this question, consider the fact that these two circuits, if equivalent in total impedance, will draw the exact same amount of current from a common AC source at 1 kHz.


Notes

This is an interesting question, requiring the student to think creatively about how to convert one configuration of circuit into another, while maintaining the same total effect. As usual, the real purpose of a question like this is to develop problem-solving strategies, rather than to simply obtain an answer.





Question 13. (Click on arrow for answer)

It is not uncommon to see impedances represented in AC circuits as boxes, rather than as combinations of R, L, and/or C. This is simply a convenient way to represent what may be complex sub-networks of components in a larger AC circuit:

Should be image 00859x01

We know that any given impedance may be represented by a simple, two-component circuit: either a resistor and a reactive component connected in series, or a resistor and a reactive component connected in parallel. Assuming a circuit frequency of 250 Hz, determine what combination of series-connected components will be equivalent to this “box” impedance, and also what combination of parallel-connected components will be equivalent to this “box” impedance.

File Num: 00859

Answer

Should be image 00859x02

Notes

Once students learn to convert between complex impedances, equivalent series R-X circuits, and equivalent parallel R-X circuits, it becomes possible for them to analyze the most complex series-parallel impedance combinations imaginable without having to do arithmetic with complex numbers (magnitudes and angles at every step). It does, however, require that students have a good working knowledge of resistance, conductance, reactance, susceptance, impedance, and admittance, and how these quantities relate mathematically to one another in scalar form.





Question 14. (Click on arrow for answer)

It is not uncommon to see impedances represented in AC circuits as boxes, rather than as combinations of R, L, and/or C. This is simply a convenient way to represent what may be complex sub-networks of components in a larger AC circuit:

Should be image 02118x01

We know that any given impedance may be represented by a simple, two-component circuit: either a resistor and a reactive component connected in series, or a resistor and a reactive component connected in parallel. Assuming a circuit frequency of 700 Hz, determine what combination of series-connected components will be equivalent to this “box” impedance, and also what combination of parallel-connected components will be equivalent to this “box” impedance.

File Num: 02118

Answer

Should be image 02118x02

Notes

Once students learn to convert between complex impedances, equivalent series R-X circuits, and equivalent parallel R-X circuits, it becomes possible for them to analyze the most complex series-parallel impedance combinations imaginable without having to do arithmetic with complex numbers (magnitudes and angles at every step). It does, however, require that students have a good working knowledge of resistance, conductance, reactance, susceptance, impedance, and admittance, and how these quantities relate mathematically to one another in scalar form.





Question 15. (Click on arrow for answer)

Complex quantities may be expressed in either rectangular or polar form. Mathematically, it does not matter which form of expression you use in your calculations.

However, one of these forms relates better to real-world measurements than the other. Which of these mathematical forms (rectangular or polar) relates more naturally to measurements of voltage or current, taken with meters or other electrical instruments? For instance, which form of AC voltage expression, polar or rectangular, best correlates to the total voltage measurement in the following circuit?

Should be image 01072x01

File Num: 01072

Answer

Polar form relates much better to the voltmeter’s display of 5 volts.


Follow-up question: how would you represent the total voltage in this circuit in rectangular form, given the other two voltmeter readings?


Notes

While rectangular notation is mathematically useful, it does not apply directly to measurements taken with real instruments. Some students might suggest that the 3.000 volt reading and the 4.000 volt reading on the other two voltmeters represent the rectangular components (real and imaginary, respectively) of voltage, but this is a special case. In cases where resistance and reactance are mixed (e.g. a practical inductor with winding resistance), the voltage magnitude will be neither the real nor the imaginary component, but rather the polar magnitude.





Question 16. (Click on arrow for answer)

Calculate the amount of current through this impedance, and express your answer in both polar and rectangular forms:

Should be image 02119x01

File Num: 02119

Answer

I = 545.45 \muA \angle 21^{o}I = 509.23 \muA + j195.47 \muA

Follow-up question: which of these two forms is more meaningful when comparing against the indication of an AC ammeter? Explain why.


Notes

It is important for your students to realize that the two forms given in the answer are really the same quantity, just expressed differently. If it helps, draw a phasor diagram showing how they are equivalent.

This is really nothing more than an exercise in complex number arithmetic. Have your students present their solution methods on the board for all to see, and discuss how Ohm’s Law and complex number formats (rectangular versus polar) relate to one another in this question.





Question 17. (Click on arrow for answer)

Determine the total impedance of this series-parallel network by first converting it into an equivalent network that is either all-series or all-parallel:

Should be image 01864x01

File Num: 01864

Answer

Equivalent series resistance and reactances:

Should be image 01864x02Z_{total} = 2.638 \hbox{ k}\Omega

Notes

Although there are other methods of solving for total impedance in a circuit such as this, I want students to become comfortable with series/parallel equivalents as an analysis tool.





Question 18. (Click on arrow for answer)

Determine the total impedance of this series-parallel network by first converting it into an equivalent network that is either all-series or all-parallel:

Should be image 01865x01

File Num: 01865

Answer

Equivalent parallel resistance and reactances:

Should be image 01865x02Z_{total} = 4.433 \hbox{ k}\Omega

Notes

Although there are other methods of solving for total impedance in a circuit such as this, I want students to become comfortable with series/parallel equivalents as an analysis tool.





Question 19. (Click on arrow for answer)

Determine the voltage dropped between points A and B in this circuit:

Should be image 02115x01

Hint: convert the parallel RC sub-network into a series equivalent first.

File Num: 02115

Answer

V_{AB} = 10.491 volts

Notes

Although there are other ways to calculate this voltage drop, it is good for students to learn the method of series-parallel subcircuit equivalents. If for no other reason, this method has the benefit of requiring less tricky math (no complex numbers needed!).

Have your students explain the procedures they used to find the answer, so that all may benefit from seeing multiple methods of solution and multiple ways of explaining it.





Question 20. (Click on arrow for answer)

Determine the current through the series LR branch in this series-parallel circuit:

Should be image 02116x01

Hint: convert the series LR sub-network into a parallel equivalent first.

File Num: 02116

Answer

I_{LR} = 3.290 mA

Notes

Yes, that is an AC current source shown in the schematic! In circuit analysis, it is quite common to have AC current sources representing idealized portions of an actual component. For instance current transformers (CT’s) act very close to ideal AC current sources. Transistors in amplifier circuits also act as AC current sources, and are often represented as such for the sake of analyzing amplifier circuits.

Although there are other ways to calculate this voltage drop, it is good for students to learn the method of series-parallel subcircuit equivalents. If for no other reason, this method has the benefit of requiring less tricky math (no complex numbers needed!).

Have your students explain the procedures they used to find the answer, so that all may benefit from seeing multiple methods of solution and multiple ways of explaining it.





Question 21. (Click on arrow for answer)

Test leads for DC voltmeters are usually just two individual lengths of wire connecting the meter to a pair of probes. For highly sensitive instruments, a special type of two-conductor cable called coaxial cable is generally used instead of two individual wires. Coaxial cable — where a center conductor is “shielded” by an outer braid or foil that serves as the other conductor — has excellent immunity to induced “noise” from electric and magnetic fields:

Should be image 00641x01

When measuring high-frequency AC voltages, however, the parasitic capacitance and inductance of the coaxial cable may present problems. We may represent these distributed characteristics of the cable as “lumped” parameters: a single capacitor and a single inductor modeling the cable’s behavior:

Should be image 00641x02

Typical parasitic values for a 10-foot cable would be 260 pF of capacitance and 650 \muH of inductance. The voltmeter itself, of course, is not without its own inherent impedances, either. For the sake of this example, let’s consider the meter’s “input impedance” to be a simple resistance of 1 M\Omega.

Calculate what voltage the meter would register when measuring the output of a 20 volt AC source, at these frequencies:


  • f = 1 Hz ; V_{meter} =
  • f = 1 kHz ; V_{meter} =
  • f = 10 kHz ; V_{meter} =
  • f = 100 kHz ; V_{meter} =
  • f = 1 MHz ; V_{meter} =

File Num: 00641

Answer


  • f = 1 Hz ; V_{meter} = 20 V
  • f = 1 kHz ; V_{meter} = 20 V
  • f = 10 kHz ; V_{meter} = 20.01 V
  • f = 100 kHz ; V_{meter} = 21.43 V
  • f = 1 MHz ; V_{meter} = 3.526 V


Follow-up question: explain why we see a “peak” at 100 kHz. How can the meter possibly see a voltage greater than the source voltage (20 V) at this frequency?


Notes

As your students what this indicates about the use of coaxial test cable for AC voltmeters. Does it mean that coaxial test cable is unusable for any measurement application, or may we use it with little or no concern in some applications? If so, which applications are these?





Question 22. (Click on arrow for answer)

The voltage measurement range of a DC instrument may easily be “extended” by connecting an appropriately sized resistor in series with one of its test leads:

Should be image 00642x01

In the example shown here, the multiplication ratio with the 9 M\Omega resistor in place is 10:1, meaning that an indication of 3.5 volts at the instrument corresponds to an actual measured voltage of 35 volts between the probes.

While this technique works very well when measuring DC voltage, it does not do so well when measuring AC voltage, due to the parasitic capacitance of the cable connecting the test probes to the instrument (parasitic cable inductance has been omitted from this diagram for simplicity):

Should be image 00642x02

To see the effects of this capacitance for yourself, calculate the voltage at the instrument input terminals assuming a parasitic capacitance of 180 pF and an AC voltage source of 10 volts, for the following frequencies:


  • f = 10 Hz ; V_{instrument} =
  • f = 1 kHz ; V_{instrument} =
  • f = 10 kHz ; V_{instrument} =
  • f = 100 kHz ; V_{instrument} =
  • f = 1 MHz ; V_{instrument} =

The debilitating effect of cable capacitance may be compensated for with the addition of another capacitor, connected in parallel with the 9 M\Omega range resistor. If we are trying to maintain a voltage division ratio of 10:1, this “compensating” capacitor must be {1 \over 9} the value of the capacitance parallel to the instrument input:

Should be image 00642x03

Re-calculate the voltage at the instrument input terminals with this compensating capacitor in place. You should notice quite a difference in instrument voltages across this frequency range!


  • f = 10 Hz ; V_{instrument} =
  • f = 1 kHz ; V_{instrument} =
  • f = 10 kHz ; V_{instrument} =
  • f = 100 kHz ; V_{instrument} =
  • f = 1 MHz ; V_{instrument} =

Complete your answer by explaining why the compensation capacitor is able to “flatten” the response of the instrument over a wide frequency range.

File Num: 00642

Answer

With no compensating capacitor:


  • f = 10 Hz ; V_{instrument} = 1.00 V
  • f = 1 kHz ; V_{instrument} = 0.701 V
  • f = 10 kHz ; V_{instrument} = 97.8 mV
  • f = 100 kHz ; V_{instrument} = 9.82 mV
  • f = 1 MHz ; V_{instrument} = 0.982 mV


With the 20 pF compensating capacitor in place:


  • f = 10 Hz ; V_{instrument} = 1.00 V
  • f = 1 kHz ; V_{instrument} = 1.00 V
  • f = 10 kHz ; V_{instrument} = 1.00 V
  • f = 100 kHz ; V_{instrument} = 1.00 V
  • f = 1 MHz ; V_{instrument} = 1.00 V


Hint: without the compensating capacitor, the circuit is a resistive voltage divider with a capacitive load. With the compensating capacitor, the circuit is a parallel set of equivalent voltage dividers, effectively eliminating the loading effect.


Follow-up question: as you can see, the presence of a compensation capacitor is not an option for a high-frequency, 10:1 oscilloscope probe. What safety hazard(s) might arise if a probe’s compensation capacitor failed in such a way that the probe behaved as if the capacitor were not there at all?


Notes

Explain to your students that “\times 10” oscilloscope probes are made like this, and that the “compensation” capacitor in these probes is usually made adjustable to create a precise 9:1 match with the combined parasitic capacitance of the cable and oscilloscope.

Ask your students what the usable “bandwidth” of a home-made \times 10 oscilloscope probe would be if it had no compensating capacitor in it.





All files with file num less than 4100 are Copyright 2003, Tony R. Kuphaldt, released under the Creative Commons Attribution License (v 1.0). All other files are Copyright 2022, David Williams, released under the Creative Commons Attribution License (V 4.0) This means you may do almost anything with this work, so long as you give proper credit.

Practice Problems: Complex Numbers and Phasors

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about inductors in AC circuits.

Resistance vs. Reactance vs. Impedance

These three terms represent different forms of opposition to electric current. Despite the fact that they are measured in the same unit (ohms: Omega), they are not the same. Resistance is best thought of as electrical friction, whereas reactance is best thought of as electrical inertia. Whereas resistance creates a voltage drop by dissipating energy, reactance creates a voltage drop by storing and releasing energy. Impedance is a term encompassing both resistance and reactance, usually a combination of both.

Phasors, used to represent AC amplitude and phase relations.

A powerful tool used for understanding the operation of AC circuits is the phasor diagram, consisting of arrows pointing in different directions: the length of each arrow representing the amplitude of some AC quantity (voltage, current, or impedance), and the angle of each arrow representing the shift in phase relative to the other arrows. By representing each AC quantity thusly, we may more easily calculate their relationships to one another, with the phasors showing us how to apply trigonometry (Pythagorean Theorem, sine, cosine, and tangent functions) to the various calculations. An analytical parallel to the graphic tool of phasor diagrams is complex numbers, where we represent each phasor (arrow) by a pair of numbers: either a magnitude and angle (polar notation), or by “real” and “imaginary” magnitudes (rectangular notation). Where phasor diagrams are helpful is in applications where their respective AC quantities add: the resultant of two or more phasors stacked tip-to-tail being the mathematical sum of the phasors. Complex numbers, on the other hand, may be added, subtracted, multiplied, and divided; the last two operations being difficult to graphically represent with arrows.

Conductance, Susceptance, and Admittance.

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance (1 \over R). Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B = {1 \over X} and admittance, the reciprocal of impedance Y = {1 \over Z}) are very necessary in order to draw phasor diagrams for parallel networks.

Question 1. (Click on arrow for answer)

Evaluate the length of side x in this right triangle, given the lengths of the other two sides:

Should be image 03326x01

File Num: 03326

Answer

x = 10

Notes

This question is a straight-forward test of students’ ability to identify and apply the 3-4-5 ratio to a right triangle.





Question 2. (Click on arrow for answer)

Evaluate the length of side x in this right triangle, given the lengths of the other two sides:

Should be image 03327x01

File Num: 03327

Answer

x = 15

Notes

This question is a straight-forward test of students’ ability to identify and apply the 3-4-5 ratio to a right triangle.





Question 3. (Click on arrow for answer)

The Pythagorean Theorem is used to calculate the length of the hypotenuse of a right triangle given the lengths of the other two sides:

Should be image 02102x01

Write the standard form of the Pythagorean Theorem, and give an example of its use.

File Num: 02102

Answer

I’ll let you research this one on your own!


Follow-up question: identify an application in AC circuit analysis where the Pythagorean Theorem would be useful for calculating a circuit quantity such as voltage or current.


Notes

The Pythagorean Theorem is easy enough for students to find on their own that you should not need to show them. A memorable illustration of this theorem are the side lengths of a so-called 3-4-5 triangle. Don’t be surprised if this is the example many students choose to give.





Question 4. (Click on arrow for answer)

The Pythagorean Theorem is used to calculate the length of the hypotenuse of a right triangle given the lengths of the other two sides:

Should be image 03114x01

Manipulate the standard form of the Pythagorean Theorem to produce a version that solves for the length of A given B and C, and also write a version of the equation that solves for the length of B given A and C.

File Num: 03114

Answer

Standard form of the Pythagorean Theorem:

C = \sqrt{A^2 + B^2}

Solving for A:

A = \sqrt{C^2 - B^2}

Solving for B:

B = \sqrt{C^2 - A^2}

Notes

The Pythagorean Theorem is easy enough for students to find on their own that you should not need to show them. A memorable illustration of this theorem are the side lengths of a so-called 3-4-5 triangle. Don’t be surprised if this is the example many students choose to give.





Question 5. (Click on arrow for answer)

Should be image 02084x01

Identify which trigonometric functions (sine, cosine, or tangent) are represented by each of the following ratios, with reference to the angle labeled with the Greek letter “Theta” (\Theta):

{X \over R} = {X \over Z} = {R \over Z} =

File Num: 02084

Answer

Should be image 02084x01{X \over R} = \tan \Theta = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \sin \Theta = {\hbox{Opposite} \over \hbox{Hypotenuse}}{R \over Z} = \cos \Theta = {\hbox{Adjacent} \over \hbox{Hypotenuse}}

Notes

Ask your students to explain what the words “hypotenuse”, “opposite”, and “adjacent” refer to in a right triangle.





Question 6. (Click on arrow for answer)

Should be image 03113x01

Identify which trigonometric functions (sine, cosine, or tangent) are represented by each of the following ratios, with reference to the angle labeled with the Greek letter “Phi” (\phi):

{R \over X} = {X \over Z} = {R \over Z} =

File Num: 03113

Answer

Should be image 03113x01{R \over X} = \tan \phi = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \cos \phi = {\hbox{Adjacent} \over \hbox{Hypotenuse}}{R \over Z} = \sin \phi = {\hbox{Opposite} \over \hbox{Hypotenuse}}

Notes

Ask your students to explain what the words “hypotenuse”, “opposite”, and “adjacent” refer to in a right triangle.





Question 7. (Click on arrow for answer)

Trigonometric functions such as sine, cosine, and tangent are useful for determining the ratio of right-triangle side lengths given the value of an angle. However, they are not very useful for doing the reverse: calculating an angle given the lengths of two sides.

Should be image 02086x01

Suppose we wished to know the value of angle \Theta, and we happened to know the values of Z and R in this impedance triangle. We could write the following equation, but in its present form we could not solve for \Theta:

\cos \Theta = {R \over Z}

The only way we can algebraically isolate the angle \Theta in this equation is if we have some way to “undo” the cosine function. Once we know what function will “undo” cosine, we can apply it to both sides of the equation and have \Theta by itself on the left-hand side.

There is a class of trigonometric functions known as inverse or “arc” functions which will do just that: “undo” a regular trigonometric function so as to leave the angle by itself. Explain how we could apply an “arc-function” to the equation shown above to isolate \Theta.

File Num: 02086

Answer

\cos \Theta = {R \over Z} \hbox{ Original equation}\hbox<i>. . . applying the "arc-cosine" function to both sides . . .</i>\arccos \left( \cos \Theta \right) = \arccos \left( {R \over Z} \right)\Theta = \arccos \left( {R \over Z} \right)

Notes

I like to show the purpose of trigonometric arcfunctions in this manner, using the cardinal rule of algebraic manipulation (do the same thing to both sides of an equation) that students are familiar with by now. This helps eliminate the mystery of arcfunctions for students new to trigonometry.





Question 8. (Click on arrow for answer)

The impedance triangle is often used to graphically relate Z, R, and X in a series circuit:

Should be image 02076x01

Unfortunately, many students do not grasp the significance of this triangle, but rather memorize it as a “trick” used to calculate one of the three variables given the other two. Explain why a right triangle is an appropriate form to relate these variables, and what each side of the triangle actually represents.

File Num: 02076

Answer

Each side of the impedance triangle is actually a phasor (a vector representing impedance with magnitude and direction):

Should be image 02076x02

Since the phasor for resistive impedance (Z_R) has an angle of zero degrees and the phasor for reactive impedance (Z_C or Z_L) either has an angle of +90 or -90 degrees, the phasor sum representing total series impedance will form the hypotenuse of a right triangle when the first to phasors are added (tip-to-tail).


Follow-up question: as a review, explain why resistive impedance phasors always have an angle of zero degrees, and why reactive impedance phasors always have angles of either +90 degrees or -90 degrees.


Notes

The question is sufficiently open-ended that many students may not realize exactly what is being asked until they read the answer. This is okay, as it is difficult to phrase the question in a more specific manner without giving away the answer!





Question 9. (Click on arrow for answer)

Use the “impedance triangle” to calculate the impedance of this series combination of resistance (R) and inductive reactance (X):

Should be image 02081x01

Explain what equation(s) you use to calculate Z.

File Num: 02081

Answer

Z = 625 \Omega, as calculated by the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 10. (Click on arrow for answer)

Students studying AC electrical theory become familiar with the impedance triangle very soon in their studies:

Should be image 02077x01

What these students might not ordinarily discover is that this triangle is also useful for calculating electrical quantities other than impedance. The purpose of this question is to get you to discover some of the triangle’s other uses.

Fundamentally, this right triangle represents phasor addition, where two electrical quantities at right angles to each other (resistive versus reactive) are added together. In series AC circuits, it makes sense to use the impedance triangle to represent how resistance (R) and reactance (X) combine to form a total impedance (Z), since resistance and reactance are special forms of impedance themselves, and we know that impedances add in series.

List all of the electrical quantities you can think of that add (in series or in parallel) and then show how similar triangles may be drawn to relate those quantities together in AC circuits.

File Num: 02077

Answer

Electrical quantities that add:
  • Series impedances
  • Series voltages
  • Parallel admittances
  • Parallel currents
  • Power dissipations

I will show you one graphical example of how a triangle may relate to electrical quantities other than series impedances:

Should be image 02077x02

Notes

It is very important for students to understand that the triangle only works as an analysis tool when applied to quantities that add. Many times I have seen students try to apply the ZRX impedance triangle to parallel circuits and fail because parallel impedances do not add. The purpose of this question is to force students to think about where the triangle is applicable to AC circuit analysis, and not just to use it blindly.

The power triangle is an interesting application of trigonometry applied to electric circuits. You may not want to discuss power with your students in great detail if they are just beginning to study voltage and current in AC circuits, because power is a sufficiently confusing subject on its own.





Question 11. (Click on arrow for answer)

Explain why the “impedance triangle” is not proper to use for relating total impedance, resistance, and reactance in parallel circuits as it is for series circuits:

Should be image 02078x01

File Num: 02078

Answer

Impedances do not add in parallel.


Follow-up question: what kind of a triangle could be properly applied to a parallel AC circuit, and why?


Notes

Trying to apply the ZRX triangle directly to parallel AC circuits is a common mistake many new students make. Key to knowing when and how to use triangles to graphically depict AC quantities is understanding why the triangle works as an analysis tool and what its sides represent.





Question 12. (Click on arrow for answer)

Examine the following circuits, then label the sides of their respective triangles with all the variables that are trigonometrically related in those circuits:

Should be image 03288x01

File Num: 03288

Answer

Should be image 03288x02

Notes

This question asks students to identify those variables in each circuit that vectorially add, discriminating them from those variables which do not add. This is extremely important for students to be able to do if they are to successfully apply “the triangle” to the solution of AC circuit problems.

Note that some of these triangles should be drawn upside-down instead of all the same as they are shown in the question, if we are to properly represent the vertical (imaginary) phasor for capacitive impedance and for inductor admittance. However, the point here is simply to get students to recognize what quantities add and what do not. Attention to the direction (up or down) of the triangle’s opposite side can come later.





Question 13. (Click on arrow for answer)

Use a triangle to calculate the total voltage of the source for this series RC circuit, given the voltage drop across each component:

Should be image 02107x01

Explain what equation(s) you use to calculate V_{total}, as well as why we must geometrically add these voltages together.

File Num: 02107

Answer

V_{total} = 3.672 volts, as calculated by the Pythagorean Theorem

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 14. (Click on arrow for answer)

Use the “impedance triangle” to calculate the necessary resistance of this series combination of resistance (R) and inductive reactance (X) to produce the desired total impedance of 5.2 k\Omega:

Should be image 02082x01

Explain what equation(s) you use to calculate R, and the algebra necessary to achieve this result from a more common formula.

File Num: 02082

Answer

R = 4.979 k\Omega, as calculated by an algebraically manipulated version of the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 15. (Click on arrow for answer)

Use the “impedance triangle” to calculate the necessary reactance of this series combination of resistance (R) and inductive reactance (X) to produce the desired total impedance of 145 \Omega:

Should be image 02083x01

Explain what equation(s) you use to calculate X, and the algebra necessary to achieve this result from a more common formula.

File Num: 02083

Answer

X = 105 \Omega, as calculated by an algebraically manipulated version of the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 16. (Click on arrow for answer)

Use the “impedance triangle” to calculate the necessary reactance of this series combination of resistance (R) and capacitive reactance (X) to produce the desired total impedance of 300 \Omega:

Should be image 02092x01

Explain what equation(s) you use to calculate X, and the algebra necessary to achieve this result from a more common formula.

File Num: 02092

Answer

X = 214.2 \Omega, as calculated by an algebraically manipulated version of the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 17. (Click on arrow for answer)

A series AC circuit contains 1125 ohms of resistance and 1500 ohms of reactance for a total circuit impedance of 1875 ohms. This may be represented graphically in the form of an impedance triangle:

Should be image 02085x01

Since all side lengths on this triangle are known, there is no need to apply the Pythagorean Theorem. However, we may still calculate the two non-perpendicular angles in this triangle using “inverse” trigonometric functions, which are sometimes called arcfunctions.

Identify which arc-function should be used to calculate the angle \Theta given the following pairs of sides:

R \hbox{ and } ZX \hbox{ and } RX \hbox{ and } Z

Show how three different trigonometric arcfunctions may be used to calculate the same angle \Theta.

File Num: 02085

Answer

\arccos {R \over Z} = 53.13^o\arctan {X \over R} = 53.13^o\arcsin {X \over Z} = 53.13^o

Challenge question: identify three more arcfunctions which could be used to calculate the same angle \Theta.


Notes

Some hand calculators identify arc-trig functions by the letter “A” prepending each trigonometric abbreviation (e.g. “ASIN” or “ATAN”). Other hand calculators use the inverse function notation of a -1 exponent, which is not actually an exponent at all (e.g. \sin^{-1} or \tan^{-1}). Be sure to discuss function notation on your students’ calculators, so they know what to invoke when solving problems such as this.





Question 18. (Click on arrow for answer)

A series AC circuit exhibits a total impedance of 10 k\Omega, with a phase shift of 65 degrees between voltage and current. Drawn in an impedance triangle, it looks like this:

Should be image 02088x01

We know that the sine function relates the sides X and Z of this impedance triangle with the 65 degree angle, because the sine of an angle is the ratio of opposite to hypotenuse, with X being opposite the 65 degree angle. Therefore, we know we can set up the following equation relating these quantities together:

\sin 65^o = {X \over Z}

Solve this equation for the value of X, in ohms.

File Num: 02088

Answer

X = 9.063 k\Omega

Notes

Ask your students to show you their algebraic manipulation(s) in setting up the equation for evaluation.





Question 19. (Click on arrow for answer)

A series AC circuit exhibits a total impedance of 2.5 k\Omega, with a phase shift of 30 degrees between voltage and current. Drawn in an impedance triangle, it looks like this:

Should be image 02087x01

Use the appropriate trigonometric functions to calculate the equivalent values of R and X in this series circuit.

File Num: 02087

Answer

R = 2.165 k\OmegaX = 1.25 k\Omega

Notes

There are a few different ways one could solve for R and X in this trigonometry problem. This would be a good opportunity to have your students present problem-solving strategies on the board in front of class so everyone gets an opportunity to see multiple techniques.





Question 20. (Click on arrow for answer)

Provide a definition for phasor, as the term applies to electrical calculations.

File Num: 04034

Answer

A “phasor” is a complex-number representation of an electrical quantity, such as voltage, current, or impedance.


Notes

The ingredient of complex must be present in any definition of a phasor. A phasor, while it may be classified as a type of vector (possessing both magnitude and direction), is not the same as the vectors commonly used in other areas of physics (e.g. force vectors, electric/magnetic field vectors, etc.).





Question 21. (Click on arrow for answer)

If you have studied complex numbers, you know that the same complex quantity may be written in two different forms: rectangular and polar. Take for example the complex quantity {\sqrt{3} \over 2} + j{1 \over 2}. The following illustration shows this point located on the complex plane, along with its rectangular dimensions:

Should be image 04059x01

Next, we see the same point, on the same complex plane, along with its polar coordinates:

Should be image 04059x02

Written out, we might express the equivalence of these two notations as such:

{\sqrt{3} \over 2} + j{1 \over 2} = 1 \angle {\pi \over 6}\vskip 30pt

Expressed in a more general form, the equivalence between rectangular and polar notations would look like this:

a + jb = c \angle \Theta

However, a problem with the “angle” symbol (\angle) is that we have no standardized way to deal with it mathematically. We would have to invent special rules to describe how to add, subtract, multiply, divide, differentiate, integrate, or otherwise manipulate complex quantities expressed using this symbol. A more profitable alternative to using the “angle” symbol is shown here:

a + jb = c e^{j\Theta}

Explain why this equivalence is mathematically sound.

File Num: 04059

Answer

The equivalence shown is based on Euler’s relation, which is left to you as an exercise to prove.


Notes

This question should probably be preceded by \#04058, which asks students to explore the relationship between the infinite series for e^x, \cos x, and \sin x. In any case, your students will need to know Euler’s relation:

e^{jx} = \cos x + j \sin x




Question 22. (Click on arrow for answer)

Electrical engineers usually express the frequency of an AC circuit in terms of angular velocity, measured in units of radians per second rather than cycles per second (Hertz, or Hz).

First, explain what a radian is. Next, write an equation relating frequency (f) in Hertz to angular velocity (\omega) in radians per second. Hint: the relationship between the two is perhaps most easily understood in terms of a two-pole AC generator, or alternator, where each revolution of the rotor generates one full cycle of AC.

File Num: 04060

Answer

A radian is that angle describing a sector of a circle, whose arc length is equal to the radius of the circle:

Should be image 04060x01

Next, the equivalence between angular velocity (\omega) and frequency (f):

\omega = 2 \pi f

Notes

Personally, I find the rotating alternator model the best way to comprehend the relationship between angular velocity and frequency. If each turn of the rotor is one cycle (2 \pi radians), and frequency is cycles per second, then one revolution per second will be 1 Hertz, which will be 2 \pi radians per second.





Question 23. (Click on arrow for answer)

Suppose two people work together to slide a large box across the floor, one pushing with a force of 400 newtons and the other pulling with a force of 300 newtons:

Should be image 03278x01

The resultant force from these two persons’ efforts on the box will, quite obviously, be the sum of their forces: 700 newtons (to the right).


What if the person pulling decides to change position and push sideways on the box in relation to the first person, so the 400 newton force and the 300 newton force will be perpendicular to each other (the 300 newton force facing into the page, away from you)? What will the resultant force on the box be then?

Should be image 03278x02

File Num: 03278

Answer

The resultant force on the box will be 500 newtons.


Notes

This is a non-electrical application of vector summation, to prepare students for the concept of using vectors to add voltages that are out-of-phase. Note how I chose to use multiples of 3, 4, and 5 for the vector magnitudes.





Question 24. (Click on arrow for answer)

Special types of vectors called phasors are often used to depict the magnitude and phase-shifts of sinusoidal AC voltages and currents. Suppose that the following phasors represent the series summation of two AC voltages, one with a magnitude of 3 volts and the other with a magnitude of 4 volts:

Should be image 01559x01

Explain what each of the following phasor diagrams represents, in electrical terms:

Should be image 01559x02

Also explain the significance of these sums: that we may obtain three different values of total voltage (7 volts, 1 volt, or 5 volts) from the same series-connected AC voltages. What does this mean for us as we prepare to analyze AC circuits using the rules we learned for DC circuits?

File Num: 01559

Answer

Each of the phasor diagrams represents two AC voltages being added together. The dotted phasor represents the sum of the 3-volt and 4-volt signals, for different conditions of phase shift between them.

Please note that these three possibilities are not exhaustive! There are a multitude of other possible total voltages that the series-connected 3 volt and 4 volt sources may create.


Follow-up question: in DC circuits, it is permissible to connect multiple voltage sources in parallel, so long as the voltages (magnitudes) and polarities are the same. Is this also true for AC? Why or why not?


Notes

Be sure to discuss with your students that these three conditions shown are not the only conditions possible! I simply chose 0^{o}, 180^{o}, and 90^{o} because they all resulted in round sums for the given quantities.

The follow-up question previews an important subject concerning AC phase: the necessary synchronization or paralleled AC voltage sources.





Question 25. (Click on arrow for answer)

When drawing phasor diagrams, there is a standardized orientation for all angles used to ensure consistency between diagrams. This orientation is usually referenced to a set of perpendicular lines, like the x and y axes commonly seen when graphing algebraic functions:

Should be image 02099x01

The intersection of the two axes is called the origin, and straight horizontal to the right is the definition of zero degrees (0^{o}). Thus, a phasor with a magnitude of 6 and an angle of 0^{o} would look like this on the diagram:

Should be image 02099x02

Draw a phasor with a magnitude of 10 and an angle of 100 degrees on the above diagram, as well as a phasor with a magnitude of 2 and an angle of -45 degrees. Label what directions 90^{o}, 180^{o}, and 270^{o} would indicate on the same diagram.

File Num: 02099

Answer

Should be image 02099x03

Notes

Graph paper, a ruler, and a protractor may be helpful for your students as they begin to draw and interpret phasor diagrams. Even if they have no prior knowledge of trigonometry or phasors, they should still be able to graphically represent simple phasor systems and even solve for resultant phasors.





Question 26. (Click on arrow for answer)

What does it mean to add two or more phasors together, in a geometric sense? How would one draw a phasor diagram showing the following two phasors added together?

Should be image 02100x01

File Num: 02100

Answer

Here are two ways of showing the same addition:

Should be image 02100x02

Follow-up question: how would you verbally explain the process of phasor addition? If you were to describe to someone else how to add phasors together, what would you tell them?


Notes

Discuss with your students that phasors may also be subtracted, multiplied, and divided. Subtraction is not too difficult to visualize, but addition and multiplication defies geometric understanding for many.





Question 27. (Click on arrow for answer)

Phasors may be symbolically described in two different ways: polar notation and rectangular notation. Explain what each of these notations means, and why either one may adequately describe a phasor.

File Num: 02101

Answer

Polar notation describes a phasor in terms of magnitude (length) and angle:

Should be image 02101x01

Rectangular notation describes a phasor in terms of horizontal and vertical displacement:

Should be image 02101x02

Follow-up question: why do we need the letter j in rectangular notation? What purpose does it serve, and what does it mean?


Notes

When discussing the meaning of j, it might be good to explain what imaginary numbers are. Whether or not you choose to do this depends on the mathematical aptitude and background of your students.





Question 28. (Click on arrow for answer)

These two phasors are written in a form known as polar notation. Re-write them in rectangular notation:

4 \> \angle \> 0^o = 3 \> \angle \> 90^o =

File Num: 00497

Answer

These two phasors, written in rectangular notation, would be 4 + j0 and 0 + j3, respectively, although a mathematician would probably write them as 4 + i0 and 0 + i3, respectively.


Challenge question: what does the lower-case j or i represent, in mathematical terms?


Notes

Discuss with your students the two notations commonly used with phasors: polar and rectangular form. They are merely two different ways of “saying” the same thing. A helpful “prop” for this discussion is the complex number plane (as opposed to a number line — a one-dimensional field), showing the “real” and “imaginary” axes, in addition to standard angles (right = 0^{o}, left = 180^{o}, up = 90^{o}, down = 270^{o}). Your students should be familiar with this from their research, so have one of them draw the number plane on the whiteboard for all to view.

The challenge question regards the origin of complex numbers, beginning with the distinction of “imaginary” numbers as being a separate set of quantities from “real” numbers. Electrical engineers, of course, avoid using the lower-case letter i to denote “imaginary” because it would be so easily be confused with the standard notation for instantaneous current i.





Question 29. (Click on arrow for answer)

Determine the sum of these two phasors, and draw a phasor diagram showing their geometric addition:


(4 \angle 0^o) + (3 \angle 90^o)

How might a phasor arithmetic problem such as this relate to an AC circuit?

File Num: 00495

Answer

(4 \angle 0^o) + (3 \angle 90^o) = (5 \angle 36.87^o)Should be image 00495x01

Notes

It is very helpful in a question such as this to graphically depict the phasors. Have one of your students draw a phasor diagram on the whiteboard for the whole class to observe and discuss.

The relation of this arithmetic problem to an AC circuit is a very important one for students to grasp. It is one thing for students to be able to mathematically manipulate and combine phasors, but quite another for them to smoothly transition between a phasor operation and comprehension of voltages and/or currents in an AC circuit. Ask your students to describe what the magnitude of a phasor means (in this example, the number 5), if that phasor represents an AC voltage. Ask your students to describe what the angle of an AC voltage phasor means, as well (in this case, 36.87^{o}), for an AC voltage.





Question 30. (Click on arrow for answer)

Phasors may be symbolically described in two different ways: polar notation and rectangular notation. Explain what each of these notations means, and why either one may adequately describe a phasor.

File Num: 02101

Answer

Polar notation describes a phasor in terms of magnitude (length) and angle:

Should be image 02101x01

Rectangular notation describes a phasor in terms of horizontal and vertical displacement:

Should be image 02101x02

Follow-up question: why do we need the letter j in rectangular notation? What purpose does it serve, and what does it mean?


Notes

When discussing the meaning of j, it might be good to explain what imaginary numbers are. Whether or not you choose to do this depends on the mathematical aptitude and background of your students.





Question 31. (Click on arrow for answer)

These two phasors are written in a form known as polar notation. Re-write them in rectangular notation:

4 \> \angle \> 0^o = 3 \> \angle \> 90^o =

File Num: 00497

Answer

These two phasors, written in rectangular notation, would be 4 + j0 and 0 + j3, respectively, although a mathematician would probably write them as 4 + i0 and 0 + i3, respectively.


Challenge question: what does the lower-case j or i represent, in mathematical terms?


Notes

Discuss with your students the two notations commonly used with phasors: polar and rectangular form. They are merely two different ways of “saying” the same thing. A helpful “prop” for this discussion is the complex number plane (as opposed to a number line — a one-dimensional field), showing the “real” and “imaginary” axes, in addition to standard angles (right = 0^{o}, left = 180^{o}, up = 90^{o}, down = 270^{o}). Your students should be familiar with this from their research, so have one of them draw the number plane on the whiteboard for all to view.

The challenge question regards the origin of complex numbers, beginning with the distinction of “imaginary” numbers as being a separate set of quantities from “real” numbers. Electrical engineers, of course, avoid using the lower-case letter i to denote “imaginary” because it would be so easily be confused with the standard notation for instantaneous current i.





Question 32. (Click on arrow for answer)

In this graph of two AC voltages, which one is leading and which one is lagging?

Should be image 00499x01

If the 4-volt (peak) sine wave is denoted in phasor notation as 4 \hbox{ V} \angle \> 0^o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.


If the 4-volt (peak) sine wave is denoted in phasor notation as 4 \hbox{ V} \angle \> 90^o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.

File Num: 00499

Answer

The 4-volt (peak) waveform leads the 3-volt (peak) waveform. Conversely, the 3-volt waveform lags behind the 4-volt waveform.


If the 4-volt waveform is denoted as 4 V \angle 0^o, then the 3-volt waveform should be denoted as 3 V \angle -90^o, or 0 - j3 V.


If the 4-volt waveform is denoted as 4 V \angle 90^o (0 + j4 V in rectangular form), then the 3-volt waveform should be denoted as 3 V \angle 0^o, or 3 + j0 V.


Notes

In my years of teaching, I have been surprised at how many students struggle with identifying the “leading” and “lagging” waveforms on a time-domain graph. Be sure to discuss this topic well with your students, identifying methods for correctly distinguishing “leading” waves from “lagging” waves.

This question also provides students with good practice expressing leading and lagging waves in phasor notation. One of the characteristics of phasors made evident in the answer is the relative nature of angles. Be sure to point this out to your students.





Question 33. (Click on arrow for answer)

In this phasor diagram, determine which phasor is leading and which is lagging the other:

Should be image 03286x01

File Num: 03286

Answer

In this diagram, phasor B is leading phasor A.


Follow-up question: using a protractor, estimate the amount of phase shift between these two phasors.


Notes

It may be helpful to your students to remind them of the standard orientation for phase angles in phasor diagrams (0 degrees to the right, 90 degrees up, etc.).





Question 34. (Click on arrow for answer)

Is it appropriate to assign a phasor angle to a single AC voltage, all by itself in a circuit?

Should be image 00496x01

What if there is more than one AC voltage source in a circuit?

Should be image 00496x02

File Num: 00496

Answer

Phasor angles are relative, not absolute. They have meaning only where there is another phasor to compare against.

Angles may be associated with multiple AC voltage sources in the same circuit, but only if those voltages are all at the same frequency.


Notes

Discuss with your students the notion of “phase angle” in relation to AC quantities. What does it mean, exactly, if a voltage is “3 volts at an angle of 90 degrees”? You will find that such a description only makes sense where there is another voltage (i.e., “4 volts at 0 degrees”) to compare to. Without a frame of reference, phasor angles are meaningless.

Also discuss with your students the nature of phase shifts between different AC voltage sources, if the sources are all at different frequencies. Would the phase angles be fixed, or vary over time? Why? In light of this, why do we not assign phase angles when different frequencies are involved?





Question 35. (Click on arrow for answer)

A parallel AC circuit draws 8 amps of current through a purely resistive branch and 14 amps of current through a purely inductive branch:

Should be image 02089x01

Calculate the total current and the angle \Theta of the total current, explaining your trigonometric method(s) of solution.

File Num: 02089

Answer

I_{total} = 16.12 amps\Theta = 60.26^{o} (negative, if you wish to represent the angle according to the standard coordinate system for phasors).

Follow-up question: in calculating \Theta, it is recommended to use the arctangent function instead of either the arcsine or arc-cosine functions. The reason for doing this is accuracy: less possibility of compounded error, due to either rounding and/or calculator-related (keystroke) errors. Explain why the use of the arctangent function to calculate \Theta incurs less chance of error than either of the other two arcfunctions.


Notes

The follow-up question illustrates an important principle in many different disciplines: avoidance of unnecessary risk by choosing calculation techniques using given quantities instead of derived quantities. This is a good topic to discuss with your students, so make sure you do so.





Question 36. (Click on arrow for answer)

A parallel AC circuit draws 100 mA of current through a purely resistive branch and 85 mA of current through a purely capacitive branch:

Should be image 02091x01

Calculate the total current and the angle \Theta of the total current, explaining your trigonometric method(s) of solution.

File Num: 02091

Answer

I_{total} = 131.2 mA\Theta = 40.36^{o}

Follow-up question: in calculating \Theta, it is recommended to use the arctangent function instead of either the arcsine or arc-cosine functions. The reason for doing this is accuracy: less possibility of compounded error, due to either rounding and/or calculator-related (keystroke) errors. Explain why the use of the arctangent function to calculate \Theta incurs less chance of error than either of the other two arcfunctions.


Notes

The follow-up question illustrates an important principle in many different disciplines: avoidance of unnecessary risk by choosing calculation techniques using given quantities instead of derived quantities. This is a good topic to discuss with your students, so make sure you do so.





Question 37. (Click on arrow for answer)

A parallel RC circuit has 10 \muS of susceptance (B). How much conductance (G) is necessary to give the circuit a (total) phase angle of 22 degrees?

Should be image 02090x01

File Num: 02090

Answer

G = 24.75 \muS

Follow-up question: how much resistance is this, in ohms?


Notes

Ask your students to explain their method(s) of solution, including any ways to double-check the correctness of the answer.





Question 38. (Click on arrow for answer)

Determine the total voltage in each of these examples, drawing a phasor diagram to show how the total (resultant) voltage geometrically relates to the source voltages in each scenario:

Should be image 00498x01

File Num: 00498

Answer

Should be image 00498x02

Notes

At first it may confuse students to use polarity marks (+ and -) for AC voltages. After all, doesn’t the polarity of AC alternate back and forth, so as to be continuously changing? However, when analyzing AC circuits, polarity marks are essential for giving a frame of reference to phasor voltages, which like all voltages are measured between two points, and thus may be measured two different ways.





All files with file num less than 4100 are Copyright 2003, Tony R. Kuphaldt, released under the Creative Commons Attribution License (v 1.0). All other files are Copyright 2022, David Williams, released under the Creative Commons Attribution License (V 4.0) This means you may do almost anything with this work, so long as you give proper credit.

To view a copy of the license, visit https://creativecommons.org/licenses/by/1.0/, or https://creativecommons.org/licenses/by/4.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.

Practice Problems: Basic AC Theory

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about inductors in AC circuits.

Resistance vs. Reactance vs. Impedance

These three terms represent different forms of opposition to electric current. Despite the fact that they are measured in the same unit (ohms: Omega), they are not the same. Resistance is best thought of as electrical friction, whereas reactance is best thought of as electrical inertia. Whereas resistance creates a voltage drop by dissipating energy, reactance creates a voltage drop by storing and releasing energy. Impedance is a term encompassing both resistance and reactance, usually a combination of both.

Phasors, used to represent AC amplitude and phase relations.

A powerful tool used for understanding the operation of AC circuits is the phasor diagram, consisting of arrows pointing in different directions: the length of each arrow representing the amplitude of some AC quantity (voltage, current, or impedance), and the angle of each arrow representing the shift in phase relative to the other arrows. By representing each AC quantity thusly, we may more easily calculate their relationships to one another, with the phasors showing us how to apply trigonometry (Pythagorean Theorem, sine, cosine, and tangent functions) to the various calculations. An analytical parallel to the graphic tool of phasor diagrams is complex numbers, where we represent each phasor (arrow) by a pair of numbers: either a magnitude and angle (polar notation), or by “real” and “imaginary” magnitudes (rectangular notation). Where phasor diagrams are helpful is in applications where their respective AC quantities add: the resultant of two or more phasors stacked tip-to-tail being the mathematical sum of the phasors. Complex numbers, on the other hand, may be added, subtracted, multiplied, and divided; the last two operations being difficult to graphically represent with arrows.

Conductance, Susceptance, and Admittance.

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance (1 \over R). Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B = {1 \over X} and admittance, the reciprocal of impedance Y = {1 \over Z}) are very necessary in order to draw phasor diagrams for parallel networks.

Common ground connections on oscilloscope inputs

Oscilloscopes having more than one input ”channel” share common ground connections between these
channels. That is to say, with two or more input cables plugged into an oscilloscope, the ”ground” clip of
each input cable is electrically common with the ground clip of every other input cable. This can easily cause
problems, as points in a circuit connected by multiple input cable ground clips will be made common with
each other (as well as common with the oscilloscope case, which itself is connected to earth ground). One
way to avoid unintentional short-circuits through these ground connections is to only connect one ground
clip of the oscilloscope to the circuit ground, removing or tying back all the other inputs’ ground clips since
they are redundant.

Question 1. (Click on arrow for answer)

What is the difference between DC and AC electricity? Identify some common sources of each type of electricity.

File Num: 00028

Answer

DC is an acronym meaning Direct Current: that is, electrical current that moves in one direction only. AC is an acronym meaning Alternating Current: that is, electrical current that periodically reverses direction (“alternates”).

Electrochemical batteries generate DC, as do solar cells. Microphones generate AC when sensing sound waves (vibrations of air molecules). There are many, many other sources of DC and AC electricity than what I have mentioned here!


Notes

Discuss a bit of the history of AC versus DC in early power systems. In the early days of electric power in the United States of America, there was a heated debate between the use of DC versus AC. Thomas Edison championed DC, while George Westinghouse and Nikola Tesla advocated AC.

It might be worthwhile to mention that almost all the electric power in the world is generated and distributed as AC (Alternating Current), and not as DC (in other words, Thomas Edison lost the AC/DC battle!). Depending on the level of the class you are teaching, this may or may not be a good time to explain why most power systems use AC. Either way, your students will probably ask why, so you should be prepared to address this question in some way (or have them report any findings of their own!).





Question 2. (Click on arrow for answer)

Alternating current produced by electromechanical generators (or alternators as they are sometimes designated) typically follows a sine-wave pattern over time. Plot a sine wave on the following graph, by tracing the height of a rotating vector inside the circle to the left of the graph:

Should be image 00093x01

To illustrate the principle here, I will show how the point is plotted for a rotation of 45^{o}:

Should be image 00093x02

You may wish to use a protractor to precisely mark the angles along the rotation of the circle, in making your sine-wave plot.

File Num: 00093

Answer

Should be image 00093x03

Notes

For many students, this might be the first time they realize trigonometry functions have anything to do with electricity! That voltage and current in an AC circuit might alternate according to a mathematical function available in their calculators is something of a revelation. Be prepared to discuss why rotating electromagnetic machines naturally produce such waveforms. Also, encourage students to make the cognitive connection between the independent variable of a sine function (angle, expressed in units of degrees in this question) to actual shaft rotation in a real generator.





Question 3. (Click on arrow for answer)

All other factors being equal, which possesses a greater potential for inducing harmful electric shock, DC electricity or AC electricity at a frequency of 60 Hertz? Be sure to back up your answer with research data!

File Num: 03289

Answer

From a perspective of inducing electric shock, AC has been experimentally proven to possess greater hazard than DC (all other factors being equal). See the research of Charles Dalziel for supporting data.


Notes

A common misconception is that DC is more capable of delivering a harmful electric shock than AC, all other factors being equal. In fact, this is something I used to teach myself (because I had heard it numerous times from others) before I discovered the research of Charles Dalziel. One of the explanations used to support the myth of DC being more dangerous is that DC has the ability to cause muscle tetanus more readily than AC. However, at 60 Hertz, the reversals of polarity occur so quickly that no human muscle could relax fast enough to enable a shock victim to release a “hot” wire anyway, so that fact that AC stops multiple times per second is of no benefit to the victim.

Do not be surprised if some students react unfavorably to the answer given here! The myth that DC is more dangerous than AC is so prevalent, especially among people who have a little background knowledge of the subject, that to counter it is to invite dispute. This is why I included the condition of supporting any answer by research data in the question.

This just goes to show that there are many misconceptions about electricity that are passed from person to person as “common knowledge” which have little or no grounding in fact (lightning never strikes twice in the same spot, electricity takes the least path of resistance, high current is more dangerous than high voltage, etc., etc.). The study of electricity and electronics is science, and in science experimental data is our sole authority. One of the most important lessons to be learned in science is that human beings have a propensity to believe things which are not true, and some will continue to defend false beliefs even in the face of conclusive evidence.





Question 4. (Click on arrow for answer)

Apply the following terms to this graph of an AC voltage measured over time:


  • Frequency
  • Period
  • Hertz
  • Amplitude

Should be image 00054x01

File Num: 00054

Answer

Should be image 00054x02

Notes

As always, it is more important to be able to apply a term to a real-life example than it is to memorize a definition for that term. In my experience, many students prefer to memorize definitions for terms rather than to go through the trouble of understanding how those terms apply to real life. Make sure students realize just how and why these AC terms apply to a waveform such as this.





Question 5. (Click on arrow for answer)

Frequency used to be expressed in units of cycles per second, abbreviated as CPS. Now, the standardized unit is Hertz. Explain the meaning of the obsolete frequency unit: what, exactly, does it mean for an AC voltage or current to have x number of “cycles per second?”

File Num: 00053

Answer

Each time an AC voltage or current repeats itself, that interval is called a cycle. Frequency, being the rate at which an AC voltage or current repeats itself over time, may be represented in terms of cycles (repetitions) per second.


Notes

Encourage your students to discuss the origins of the new unit (Hertz), and how it actually communicates less information about the thing being measured than the old unit (CPS).





Question 6. (Click on arrow for answer)

If an AC voltage has a frequency of 350 Hz, how long (in time) is its period?

File Num: 00055

Answer

Period = 2.8571 milliseconds


Notes

It is important for students to realize the reciprocal relationship between frequency and period. One is cycles per second while the other is seconds per cycle.





Question 7. (Click on arrow for answer)

Radio waves are comprised of oscillating electric and magnetic fields, which radiate away from sources of high-frequency AC at (nearly) the speed of light. An important measure of a radio wave is its wavelength, defined as the distance the wave travels in one complete cycle.

Suppose a radio transmitter operates at a fixed frequency of 950 kHz. Calculate the approximate wavelength (\lambda) of the radio waves emanating from the transmitter tower, in the metric distance unit of meters. Also, write the equation you used to solve for \lambda.

File Num: 01819

Answer

\lambda \approx 316 meters

I’ll let you find the equation on your own!


Notes

I purposely omit the velocity of light, as well as the time/distance/velocity equation, so that students will have to do some simple research this calculate this value. Neither of these concepts is beyond high-school level science students, and should pose no difficulty at all for college-level students to find on their own.





Question 8. (Click on arrow for answer)

If the only instrument you had in your possession to detect AC voltage signals was an audio speaker, how could you use it to determine which of two AC voltage waveforms has the greatest period?

Should be image 00387x01

File Num: 00387

Answer

Connecting the speaker to each AC voltage source, one at a time, will result in two different audio tones output by the speaker. Whichever tone is lower in pitch is the waveform with the greatest period.


Notes

An audio speaker is an outstanding instrument to use in teaching AC theory, because it makes use of a human sense that most instruments do not. I have constructed a simple headphone-based listening instrument for my own lab use, and have found it invaluable, especially in the absence of an oscilloscope. There is so much the trained ear may discern about an AC waveform based on volume and tone!





Question 9. (Click on arrow for answer)

An oscilloscope is a very useful piece of electronic test equipment. Most everyone has seen an oscilloscope in use, in the form of a heart-rate monitor (electrocardiogram, or EKG) of the type seen in doctor’s offices and hospitals.

When monitoring heart beats, what do the two axes (horizontal and vertical) of the oscilloscope screen represent?

Should be image 00530x01

In general electronics use, when measuring AC voltage signals, what do the two axes (horizontal and vertical) of the oscilloscope screen represent?

Should be image 00530x02

File Num: 00530

Answer

EKG vertical = heart muscle contraction ; EKG horizontal = time

General-purpose vertical = voltage ; General-purpose horizontal = time


Notes

Oscilloscope function is often best learned through interaction. Be sure to have at least one oscilloscope operational in the classroom for student interaction during discussion time.





Question 10. (Click on arrow for answer)

The core of an analog oscilloscope is a special type of vacuum tube known as a Cathode Ray Tube, or CRT. While similar in function to the CRT used in televisions, oscilloscope display tubes are specially built for the purpose of serving an a measuring instrument.

Explain how a CRT functions. What goes on inside the tube to produce waveform displays on the screen?

File Num: 00536

Answer

There are many tutorials and excellent reference books on CRT function — go read a few of them!


Notes

Some of your students may come across photographs and illustrations of CRTs for use in their presentation. If at all possible, provide a way for individual students to share their visual findings with their classmates, through the use of an overhead projector, computer monitor, or computer projector. Discuss in detail the operation of a CRT with your students, especially noting the electrostatic method of electron beam deflection used to “steer” the beam to specific areas on the screen.





Question 11. (Click on arrow for answer)

When the vertical (“Y”) axis of an oscilloscope is shorted, the result should be a straight line in the middle of the screen:

Should be image 00531x01

Determine the DC polarity of the voltage source, based on this illustration:

Should be image 00531x02

File Num: 00531

Answer

Should be image 00531x03

Notes

This question challenges students to figure out both the polarization of the probe (and ground clip), as well as the orientation of the Y axis. It is very important, of course, that the coupling control be set on “DC” in order to successfully measure a DC signal.





Question 12. (Click on arrow for answer)

An oscilloscope is connected to a battery of unknown voltage. The result is a straight line on the display:

Should be image 01672x01

Assuming the oscilloscope display has been properly “zeroed” and the vertical sensitivity is set to 5 volts per division, determine the voltage of the battery.

File Num: 01672

Answer

The battery voltage is slightly greater than 6.5 volts.


Notes

Measuring voltage on an oscilloscope display is very similar to measuring voltage on an analog voltmeter. The mathematical relationship between scale divisions and range is much the same. This is one reason I encourage students to use analog multimeters occasionally in their labwork, if for no other reason than to preview the principles of oscilloscope scale interpretation.





Question 13. (Click on arrow for answer)

A technician prepares to use an oscilloscope to display an AC voltage signal. After turning the oscilloscope on and connecting the Y input probe to the signal source test points, this display appears:

Should be image 00532x01

What display control(s) need to be adjusted on the oscilloscope in order to show fewer cycles of this signal on the screen, with a greater height (amplitude)?

File Num: 00532

Answer

The “timebase” control needs to be adjusted for fewer seconds per division, while the “vertical” control needs to be adjusted for fewer volts per division.


Notes

Discuss the function of both these controls with your students. If possible, demonstrate this scenario using a real oscilloscope and function generator, and have students adjust the controls to get the waveform to display optimally. Challenge your students to think of ways the signal source (function generator) may be adjusted to produce the display, then have them think of ways the oscilloscope controls could be adjusted to fit.





Question 14. (Click on arrow for answer)

A technician prepares to use an oscilloscope to display an AC voltage signal. After turning the oscilloscope on and connecting the Y input probe to the signal source test points, this display appears:

Should be image 00534x01

What display control(s) need to be adjusted on the oscilloscope in order to show a normal-looking wave on the screen?

File Num: 00534

Answer

The “vertical” control needs to be adjusted for a greater number of volts per division.


Notes

Discuss the function of both these controls with your students. If possible, demonstrate this scenario using a real oscilloscope and function generator, and have students adjust the controls to get the waveform to display optimally. Challenge your students to think of ways the signal source (function generator) may be adjusted to produce the display, then have them think of ways the oscilloscope controls could be adjusted to fit.





Question 15. (Click on arrow for answer)

A technician prepares to use an oscilloscope to display an AC voltage signal. After turning the oscilloscope on and connecting the Y input probe to the signal source test points, this display appears:

Should be image 00533x01

What appears on the oscilloscope screen is a vertical line that moves slowly from left to right. What display control(s) need to be adjusted on the oscilloscope in order to show a normal-looking wave on the screen?

File Num: 00533

Answer

The “timebase” control needs to be adjusted for fewer seconds per division.


Notes

Discuss the function of both these controls with your students. If possible, demonstrate this scenario using a real oscilloscope and function generator, and have students adjust the controls to get the waveform to display optimally.





Question 16. (Click on arrow for answer)

A technician prepares to use an oscilloscope to display an AC voltage signal. After turning the oscilloscope on and connecting the Y input probe to the signal source test points, this display appears:

Should be image 00535x01

What display control(s) need to be adjusted on the oscilloscope in order to show a normal-looking wave on the screen?

File Num: 00535

Answer

The “timebase” control needs to be adjusted for a greater number of seconds per division.


Notes

Discuss the function of both these controls with your students. If possible, demonstrate this scenario using a real oscilloscope and function generator, and have students adjust the controls to get the waveform to display optimally. Challenge your students to think of ways the signal source (function generator) may be adjusted to produce the display, then have them think of ways the oscilloscope controls could be adjusted to fit.





Question 17. (Click on arrow for answer)

Determine the frequency of this waveform, as displayed by an oscilloscope with a vertical sensitivity of 2 volts per division and a timebase of 0.5 milliseconds per division:

Should be image 01668x01

File Num: 01668

Answer

400 Hz


Notes

This is just a straightforward exercise in determining period and translating that value into frequency.





Question 18. (Click on arrow for answer)

Assuming the vertical sensitivity control is set to 2 volts per division, and the timebase control is set to 10 \mus per division, calculate the amplitude of this “sawtooth” wave (in volts peak and volts peak-to-peak) as well as its frequency.

Should be image 00541x01

File Num: 00541

Answer


\item{} E_{peak} = 8 V \item{} E_{peak-to-peak} = 16 V \item{} f = 6.67 kHz

Notes

This question is not only good for introducing basic oscilloscope principles, but it is also excellent for review of AC waveform measurements.





Question 19. (Click on arrow for answer)

Most oscilloscopes can only directly measure voltage, not current. One way to measure AC current with an oscilloscope is to measure the voltage dropped across a shunt resistor. Since the voltage dropped across a resistor is proportional to the current through that resistor, whatever wave-shape the current is will be translated into a voltage drop with the exact same wave-shape.

However, one must be very careful when connecting an oscilloscope to any part of a grounded system, as many electric power systems are. Note what happens here when a technician attempts to connect the oscilloscope across a shunt resistor located on the “hot” side of a grounded 120 VAC motor circuit:

Should be image 01820x01

Here, the reference lead of the oscilloscope (the small alligator clip, not the sharp-tipped probe) creates a short-circuit in the power system. Explain why this happens.

File Num: 01820

Answer

The “ground” clip on an oscilloscope probe is electrically common with the metal chassis of the oscilloscope, which in turn is connected to earth ground by the three-prong (grounded) power plug.


Notes

This is a very important lesson for students to learn about line-powered oscilloscopes. If necessary, discuss the wiring of the power system, drawing a schematic showing the complete short-circuit fault current path, from AC voltage source to “hot” lead to ground clip to chassis to ground prong to ground wire to neutral wire to AC voltage source.





Question 20. (Click on arrow for answer)

Most oscilloscopes have at least two vertical inputs, used to display more than one waveform simultaneously:

Should be image 01821x01

While this feature is extremely useful, one must be careful in connecting two sources of AC voltage to an oscilloscope. Since the “reference” or “ground” clips of each probe are electrically common with the oscilloscope’s metal chassis, they are electrically common with each other as well.

Explain what sort of problem would be caused by connecting a dual-trace oscilloscope to a circuit in the following manner:

Should be image 01821x02

File Num: 01821

Answer

The oscilloscope will create an earth-grounded short circuit in this series resistor circuit:

Should be image 01821x03

If the signal generator is earth-grounded through its power cord as well, the problem could even be worse:

Should be image 01821x04

Follow-up question: explain why the second scenario is potentially more hazardous than the first.


Notes

Failing to consider that the “ground” leads on all probes are common to each other (as well as common to the safety ground conductor of the line power system) is a very common mistake among students first learning how to use oscilloscopes. Hopefully, discussing scenarios such as this will help students avoid this problem in their labwork.


Note to Socratic Electronics developers: the oscilloscope shown in figure {\tt 01821×01.eps is made up of individual lines, circles, text elements, etc., rather than a single object as is contained in the Xcircuit library file ({\tt scope.lps}). If you wish to edit the features of this scope, start with the {\tt 01821×01.eps} image file rather than the library object!} Then you may save your modified oscilloscope as a complete object in your own image library for future use.




Question 21. (Click on arrow for answer)

How is it possible to assign a fixed value of voltage or current (such as “120 volts”) to an AC electrical quantity that is constantly changing, crossing 0 volts, and reversing polarity?

File Num: 00051

Answer

We may express quantities of AC voltage and current in terms of peak, peak-to-peak, average, or RMS.


Notes

Before you discuss “RMS” values with your students, it is important to cover the basic idea of how to assign fixed values to quantities that change over time. Since AC waveforms are cyclic (repeating), this is not as difficult to do as one might think.





Question 22. (Click on arrow for answer)

Suppose a DC power source with a voltage of 50 volts is connected to a 10 \Omega load. How much power will this load dissipate?

Now suppose the same 10 \Omega load is connected to a sinusoidal AC power source with a peak voltage of 50 volts. Will the load dissipate the same amount of power, more power, or less power? Explain your answer.

File Num: 00401

Answer

50 volts DC applied to a 10 \Omega load will dissipate 250 watts of power. 50 volts (peak, sinusoidal) AC will deliver less than 250 watts to the same load.


Notes

There are many analogies to explain this discrepancy between the two “50 volt” sources. One is to compare the physical effort of a person pushing with a constant force of 50 pounds, versus someone who pushes intermittently with only a peak force of 50 pounds.





Question 23. (Click on arrow for answer)

Suppose that a variable-voltage AC source is adjusted until it dissipates the exact same amount of power in a standard load resistance as a DC voltage source with an output of 120 volts:

Should be image 00402x01

In this condition of equal power dissipation, how much voltage is the AC power supply outputting? Be as specific as you can in your answer.

File Num: 00402

Answer

120 volts AC RMS, by definition.


Notes

Ask your students, “how much peak voltage is the AC power source outputting? More or less than 120 volts?”

If one of your students claims to have calculated the peak voltage as 169.7 volts, ask them how they arrived at that answer. Then ask if that answer depends on the shape of the waveform (it does!). Note that the question did not specify a “sinusoidal” wave shape. Realistically, an adjustable-voltage AC power supply of substantial power output will likely be sinusoidal, being powered from utility AC power, but it could be a different wave-shape, depending on the nature of the source!





Question 24. (Click on arrow for answer)

Determine the RMS amplitude of this sinusoidal waveform, as displayed by an oscilloscope with a vertical sensitivity of 0.2 volts per division:

Should be image 01818x01

File Num: 01818

Answer

The RMS amplitude of this waveform is approximately 0.32 volts.


Notes

Students must properly interpret the oscilloscope’s display, then correctly convert to RMS units, in order to obtain the correct answer for this question.





Question 25. (Click on arrow for answer)

Determine the RMS amplitude of this square-wave signal, as displayed by an oscilloscope with a vertical sensitivity of 0.5 volts per division:

Should be image 01824x01

File Num: 01824

Answer

The RMS amplitude of this waveform is 0.5 volt.


Notes

Many electronics students I’ve talked to seem to think that the RMS value of a waveform is always {\sqrt{2} \over 2}, no matter what the waveshape. Not true, as evidenced by the answer for this question!

Students must properly interpret the oscilloscope’s display in order to obtain the correct answer for this question. The “conversion” to RMS units is really non-existent, but I want students to be able to explain why it is and not just memorize this fact.





Question 26. (Click on arrow for answer)

Suppose two voltmeters are connected to source of “mains” AC power in a residence, one meter is analog (D’Arsonval PMMC meter movement) while the other is true-RMS digital. They both register 117 volts while connected to this AC source.

Suddenly, a large electrical load is turned on somewhere in the system. This load both reduces the mains voltage and slightly distorts the shape of the waveform. The overall effect of this is average AC voltage has decreased by 4.5\% from where it was, while RMS AC voltage has decreased by 6\% from where it was. How much voltage does each voltmeter register now?

File Num: 02790

Answer

Analog voltmeter now registers: 111.7 volts


True-RMS digital voltmeter now registers: 110 volts


Notes

Students sometimes have difficulty grasping the significance of PMMC meter movements being “average-responding” rather than RMS-responding. Hopefully, the answer to this question will help illuminate this subject more.





Question 27. (Click on arrow for answer)

If we were to express the series-connected DC voltages as phasors (arrows pointing with a particular length and a particular direction, graphically expressing magnitude and polarity of an electrical signal), how would we draw them in such a way that the total (or resultant) phasors accurately expressed the total voltage of each series-connected pair?

Should be image 00493x01

If we were to assign angle values to each of these phasors, what would you suggest?

File Num: 00493

Answer

Should be image 00493x02

In the right-hand circuit, where the two voltage sources are opposing, one of the phasors will have an angle of 0^{o}, while the other will have an angle of 180^{o}.


Notes

Phasors are really nothing more than an extension of the familiar “number line” most students see during their primary education years. The important difference here is that phasors are two-dimensional magnitudes, not one-dimensional, as scalar numbers are.

The use of degrees to measure angles should be familiar as well, even to those students without a strong mathematics background. For example, what does it mean when a skateboarder or stunt bicyclist “does a 180“? It means they turn around so as to face the opposite direction (180 degrees away from) their previous direction.





Question 28. (Click on arrow for answer)

Calculate the total voltage of these series-connected voltage sources:

Should be image 00492x01

File Num: 00492

Answer

This is a “trick” question, because only the total voltage of the DC sources may be predicted with certainty. There is insufficient information to calculate the total AC voltage for the two series-connected AC sources!

Should be image 00492x02

Notes

Discuss with your students exactly why the total voltage of the two series-connected AC sources cannot be determined, given the little information we have about them. Is it possible for their total voltage to be 8 VAC, just like the series-aiding DC sources? Is it possible for their total voltage to be 2 VAC, just like the series-opposing DC sources? Why or why not?





Question 29. (Click on arrow for answer)

Using a computer or graphing calculator, plot the sum of these two sine waves:

Should be image 01557x01

What do you suppose the sum of a 1-volt (peak) sine wave and a 2-volt (peak) sine wave will be, if both waves are perfectly in-phase with each other?


Hint: you will need to enter equations into your plotting device that look something like this:


{\tt y1 = sin x}
{\tt y2 = 2 * sin x}
{\tt y3 = y1 + y2}

File Num: 01557

Answer

Should be image 01557x02

Notes

Graphing calculators are excellent tools to use for learning experiences such as this. In far less time than it would take to plot a third sine wave by hand, students may see the sinusoidal sum for themselves.





Question 30. (Click on arrow for answer)

Using a computer or graphing calculator, plot the sum of these two sine waves:

Should be image 01558x01

Hint: you will need to enter equations into your plotting device that look something like this:


{\tt y1 = sin x}
{\tt y2 = 2 * sin (x + 90)}
{\tt y3 = y1 + y2}

Note: the second equation assumes your calculator has been set up to calculate trigonometric functions in angle units of degrees rather than radians. If you wish to plot these same waveforms (with the same phase shift shown) using radians as the unit of angle measurement, you must enter the second equation as follows:


{\tt y2 = 2 * sin (x + 1.5708)}

File Num: 01558

Answer

Should be image 01558x02

Follow-up question: note that the sum of the 1-volt wave and the 2-volt wave does not equate to a 3-volt wave! Explain why.


Notes

Graphing calculators are excellent tools to use for learning experiences such as this. In far less time than it would take to plot a third sine wave by hand, students may see the sinusoidal sum for themselves.

The point of this question is to get students thinking about how it is possible for sinusoidal voltages to not add up as one might expect. This is very important, because it indicates simple arithmetic processes like addition will not be as simple in AC circuits as it was in DC circuits, due to phase shift. Be sure to emphasize this point to your students.





Question 31. (Click on arrow for answer)

Special types of vectors called phasors are often used to depict the magnitude and phase-shifts of sinusoidal AC voltages and currents. Suppose that the following phasors represent the series summation of two AC voltages, one with a magnitude of 3 volts and the other with a magnitude of 4 volts:

Should be image 01559x01

Explain what each of the following phasor diagrams represents, in electrical terms:

Should be image 01559x02

Also explain the significance of these sums: that we may obtain three different values of total voltage (7 volts, 1 volt, or 5 volts) from the same series-connected AC voltages. What does this mean for us as we prepare to analyze AC circuits using the rules we learned for DC circuits?

File Num: 01559

Answer

Each of the phasor diagrams represents two AC voltages being added together. The dotted phasor represents the sum of the 3-volt and 4-volt signals, for different conditions of phase shift between them.

Please note that these three possibilities are not exhaustive! There are a multitude of other possible total voltages that the series-connected 3 volt and 4 volt sources may create.


Follow-up question: in DC circuits, it is permissible to connect multiple voltage sources in parallel, so long as the voltages (magnitudes) and polarities are the same. Is this also true for AC? Why or why not?


Notes

Be sure to discuss with your students that these three conditions shown are not the only conditions possible! I simply chose 0^{o}, 180^{o}, and 90^{o} because they all resulted in round sums for the given quantities.

The follow-up question previews an important subject concerning AC phase: the necessary synchronization or paralleled AC voltage sources.





Question 32. (Click on arrow for answer)

When drawing phasor diagrams, there is a standardized orientation for all angles used to ensure consistency between diagrams. This orientation is usually referenced to a set of perpendicular lines, like the x and y axes commonly seen when graphing algebraic functions:

Should be image 02099x01

The intersection of the two axes is called the origin, and straight horizontal to the right is the definition of zero degrees (0^{o}). Thus, a phasor with a magnitude of 6 and an angle of 0^{o} would look like this on the diagram:

Should be image 02099x02

Draw a phasor with a magnitude of 10 and an angle of 100 degrees on the above diagram, as well as a phasor with a magnitude of 2 and an angle of -45 degrees. Label what directions 90^{o}, 180^{o}, and 270^{o} would indicate on the same diagram.

File Num: 02099

Answer

Should be image 02099x03

Notes

Graph paper, a ruler, and a protractor may be helpful for your students as they begin to draw and interpret phasor diagrams. Even if they have no prior knowledge of trigonometry or phasors, they should still be able to graphically represent simple phasor systems and even solve for resultant phasors.





Question 33. (Click on arrow for answer)

What does it mean to add two or more phasors together, in a geometric sense? How would one draw a phasor diagram showing the following two phasors added together?

Should be image 02100x01

File Num: 02100

Answer

Here are two ways of showing the same addition:

Should be image 02100x02

Follow-up question: how would you verbally explain the process of phasor addition? If you were to describe to someone else how to add phasors together, what would you tell them?


Notes

Discuss with your students that phasors may also be subtracted, multiplied, and divided. Subtraction is not too difficult to visualize, but addition and multiplication defies geometric understanding for many.





Question 34. (Click on arrow for answer)

The Pythagorean Theorem is used to calculate the length of the hypotenuse of a right triangle given the lengths of the other two sides:

Should be image 02102x01

Write the standard form of the Pythagorean Theorem, and give an example of its use.

File Num: 02102

Answer

I’ll let you research this one on your own!


Follow-up question: identify an application in AC circuit analysis where the Pythagorean Theorem would be useful for calculating a circuit quantity such as voltage or current.


Notes

The Pythagorean Theorem is easy enough for students to find on their own that you should not need to show them. A memorable illustration of this theorem are the side lengths of a so-called 3-4-5 triangle. Don’t be surprised if this is the example many students choose to give.





Question 35. (Click on arrow for answer)

Determine the sum of these two phasors, and draw a phasor diagram showing their geometric addition:


(4 \angle 0^o) + (3 \angle 90^o)

How might a phasor arithmetic problem such as this relate to an AC circuit?

File Num: 00495

Answer

(4 \angle 0^o) + (3 \angle 90^o) = (5 \angle 36.87^o)Should be image 00495x01

Notes

It is very helpful in a question such as this to graphically depict the phasors. Have one of your students draw a phasor diagram on the whiteboard for the whole class to observe and discuss.

The relation of this arithmetic problem to an AC circuit is a very important one for students to grasp. It is one thing for students to be able to mathematically manipulate and combine phasors, but quite another for them to smoothly transition between a phasor operation and comprehension of voltages and/or currents in an AC circuit. Ask your students to describe what the magnitude of a phasor means (in this example, the number 5), if that phasor represents an AC voltage. Ask your students to describe what the angle of an AC voltage phasor means, as well (in this case, 36.87^{o}), for an AC voltage.





Question 36. (Click on arrow for answer)

Phasors may be symbolically described in two different ways: polar notation and rectangular notation. Explain what each of these notations means, and why either one may adequately describe a phasor.

File Num: 02101

Answer

Polar notation describes a phasor in terms of magnitude (length) and angle:

Should be image 02101x01

Rectangular notation describes a phasor in terms of horizontal and vertical displacement:

Should be image 02101x02

Follow-up question: why do we need the letter j in rectangular notation? What purpose does it serve, and what does it mean?


Notes

When discussing the meaning of j, it might be good to explain what imaginary numbers are. Whether or not you choose to do this depends on the mathematical aptitude and background of your students.





Question 37. (Click on arrow for answer)

These two phasors are written in a form known as polar notation. Re-write them in rectangular notation:

4 \> \angle \> 0^o = 3 \> \angle \> 90^o =

File Num: 00497

Answer

These two phasors, written in rectangular notation, would be 4 + j0 and 0 + j3, respectively, although a mathematician would probably write them as 4 + i0 and 0 + i3, respectively.


Challenge question: what does the lower-case j or i represent, in mathematical terms?


Notes

Discuss with your students the two notations commonly used with phasors: polar and rectangular form. They are merely two different ways of “saying” the same thing. A helpful “prop” for this discussion is the complex number plane (as opposed to a number line — a one-dimensional field), showing the “real” and “imaginary” axes, in addition to standard angles (right = 0^{o}, left = 180^{o}, up = 90^{o}, down = 270^{o}). Your students should be familiar with this from their research, so have one of them draw the number plane on the whiteboard for all to view.

The challenge question regards the origin of complex numbers, beginning with the distinction of “imaginary” numbers as being a separate set of quantities from “real” numbers. Electrical engineers, of course, avoid using the lower-case letter i to denote “imaginary” because it would be so easily be confused with the standard notation for instantaneous current i.





Question 38. (Click on arrow for answer)

In this graph of two AC voltages, which one is leading and which one is lagging?

Should be image 00499x01

If the 4-volt (peak) sine wave is denoted in phasor notation as 4 \hbox{ V} \angle \> 0^o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.


If the 4-volt (peak) sine wave is denoted in phasor notation as 4 \hbox{ V} \angle \> 90^o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.

File Num: 00499

Answer

The 4-volt (peak) waveform leads the 3-volt (peak) waveform. Conversely, the 3-volt waveform lags behind the 4-volt waveform.


If the 4-volt waveform is denoted as 4 V \angle 0^o, then the 3-volt waveform should be denoted as 3 V \angle -90^o, or 0 - j3 V.


If the 4-volt waveform is denoted as 4 V \angle 90^o (0 + j4 V in rectangular form), then the 3-volt waveform should be denoted as 3 V \angle 0^o, or 3 + j0 V.


Notes

In my years of teaching, I have been surprised at how many students struggle with identifying the “leading” and “lagging” waveforms on a time-domain graph. Be sure to discuss this topic well with your students, identifying methods for correctly distinguishing “leading” waves from “lagging” waves.

This question also provides students with good practice expressing leading and lagging waves in phasor notation. One of the characteristics of phasors made evident in the answer is the relative nature of angles. Be sure to point this out to your students.





Question 39. (Click on arrow for answer)

A common feature of oscilloscopes is the X-Y mode, where the vertical and horizontal plot directions are driven by external signals, rather than only the vertical direction being driven by a measured signal and the horizontal being driven by the oscilloscope’s internal sweep circuitry:

Should be image 01480x01

The oval pattern shown in the right-hand oscilloscope display of the above illustration is typical for two sinusoidal waveforms of the same frequency, but slightly out of phase with one another. The technical name for this type of X-Y plot is a Lissajous figure.

What should the Lissajous figure look like for two sinusoidal waveforms that are at exactly the same frequency, and exactly the same phase (0 degrees phase shift between the two)? What should the Lissajous figure look like for two sinusoidal waveforms that are exactly 90 degrees out of phase?

A good way to answer each of these questions is to plot the specified waveforms over time on graph paper, then determine their instantaneous amplitudes at equal time intervals, and then determine where that would place the “dot” on the oscilloscope screen at those points in time, in X-Y mode. To help you, I’ll provide two blank oscilloscope displays for you to draw the Lissajous figures on:

Should be image 01480x02

File Num: 01480

Answer

Should be image 01480x03

Challenge question: what kind of Lissajous figures would be plotted by the oscilloscope if the signals were non-sinusoidal? Perhaps the simplest example of this would be two square waves instead of two sine waves.


Notes

Many students seem to have trouble grasping how Lissajous figures are formed. One of the demonstrations I use to overcome this conceptual barrier is an analog oscilloscope and two signal generators set to very low frequencies, so students can see the “dot” being swept across the screen by both waveforms in slow-motion. Then, I speed up the signals and let them see how the Lissajous pattern becomes more “solid” with persistence of vision and the inherent phosphor delay of the screen.





Question 40. (Click on arrow for answer)

Lissajous figures, drawn by an oscilloscope, are a powerful tool for visualizing the phase relationship between two waveforms. In fact, there is a mathematical formula for calculating the amount of phase shift between two sinusoidal signals, given a couple of dimensional measurements of the figure on the oscilloscope screen.

The procedure begins with adjusting the vertical and horizontal amplitude controls so that the Lissajous figure is proportional: just as tall as it is wide on the screen (n). Then, we make sure the figure is centered on the screen and we take a measurement of the distance between the x-axis intercept points (m), as such:

Should be image 01481x01

Determine what the formula is for calculating the phase shift angle for this circuit, given these dimensions. Hint: the formula is trigonometric! If you don’t know where to begin, recall what the respective Lissajous figures look like for a 0^o phase shift and for a 90^o phase shift, and work from there.

File Num: 01481

Answer

\Theta = \sin^{-1} \left({m \over n}\right)

Challenge question: what kind of Lissajous figure would be drawn by two sinusoidal waveforms at slightly different frequencies?


Notes

This is a great exercise in teaching students how to derive an equation from physical measurements when the fundamental nature of that equation (trigonometric) is already known. They should already know what the Lissajous figures for both 0^o and 90^o look like, and should have no trouble figuring out what a and b values these two scenarios would yield if measured similarly on the oscilloscope display. The rest is just fitting the pieces together so that the trigonometric function yields the correct angle(s).





Question 41. (Click on arrow for answer)

As a general rule, inductors oppose change in (choose: voltage or current), and they do so by . . . (complete the sentence).


Based on this rule, determine how an inductor would react to a constant AC current that increases in frequency. Would an inductor drop more or less voltage, given a greater frequency? Explain your answer.

File Num: 00578

Answer

As a general rule, inductors oppose change in current, and they do so by producing a voltage.


An inductor will drop a greater amount of AC voltage, given the same AC current, at a greater frequency.


Notes

This question is an exercise in qualitative thinking: relating rates of change to other variables, without the use of numerical quantities. The general rule stated here is very, very important for students to master, and be able to apply to a variety of circumstances. If they learn nothing about inductors except for this rule, they will be able to grasp the function of a great many inductor circuits.





Question 42. (Click on arrow for answer)

\int f(x) dx Calculus alert!

We know that the formula relating instantaneous voltage and current in an inductor is this:

e = L{di \over dt}

Knowing this, determine at what points on this sine wave plot for inductor current is the inductor voltage equal to zero, and where the voltage is at its positive and negative peaks. Then, connect these points to draw the waveform for inductor voltage:

Should be image 00576x01

How much phase shift (in degrees) is there between the voltage and current waveforms? Which waveform is leading and which waveform is lagging?

File Num: 00576

Answer

Should be image 00576x02

For an inductor, voltage is leading and current is lagging, by a phase shift of 90^{o}.


Notes

This question is an excellent application of the calculus concept of the derivative: relating one function (instantaneous voltage, e) with the instantaneous rate-of-change of another function (current, di \over dt).





Question 43. (Click on arrow for answer)

Does an inductor’s opposition to alternating current increase or decrease as the frequency of that current increases? Also, explain why we refer to this opposition of AC current in an inductor as reactance instead of resistance.

File Num: 00580

Answer

The opposition to AC current (“reactance”) of an inductor increases as frequency increases. We refer to this opposition as “reactance” rather than “resistance” because it is non-dissipative in nature. In other words, reactance causes no power to leave the circuit.


Notes

Ask your students to define the relationship between inductor reactance and frequency as either “directly proportional” or “inversely proportional”. These are two phrases used often in science and engineering to describe whether one quantity increases or decreases as another quantity increases. Your students definitely need to be familiar with both these phrases, and be able to interpret and use them in their technical discussions.

Also, discuss the meaning of the word “non-dissipative” in this context. How could we prove that the opposition to current expressed by an inductor is non-dissipative? What would be the ultimate test of this?





Question 44. (Click on arrow for answer)

What will happen to the brightness of the light bulb as the iron core is moved away from the wire coil in this circuit? Explain why this happens.

Should be image 00095x01

File Num: 00095

Answer

The light bulb will glow brighter when the iron core is moved away from the wire coil, due to the change in inductive reactance (X_{L}).


Follow-up question: what circuit failure(s) could cause the light bulb to glow brighter than it should?


Notes

One direction you might want to lead your students in with this question is how AC power may be controlled using this principle. Controlling AC power with a variable reactance has a definite advantage over controlling AC power with a variable resistance: less wasted energy in the form of heat.





Question 45. (Click on arrow for answer)

An inductor rated at 4 Henrys is subjected to a sinusoidal AC voltage of 24 volts RMS, at a frequency of 60 hertz. Write the formula for calculating inductive reactance (X_L), and solve for current through the inductor.

File Num: 00582

Answer

X_L = 2 \pi f L

The current through this inductor is 15.92 mA RMS.


Notes

I have consistently found that qualitative (greater than, less than, or equal) analysis is much more difficult for students to perform than quantitative (punch the numbers on a calculator) analysis. Yet, I have consistently found on the job that people lacking qualitative skills make more “silly” quantitative errors because they cannot validate their calculations by estimation.

In light of this, I always challenge my students to qualitatively analyze formulae when they are first introduced to them. Ask your students to identify what will happen to one term of an equation if another term were to either increase, or decrease (you choose the direction of change). Use up and down arrow symbols if necessary to communicate these changes graphically. Your students will greatly benefit in their conceptual understanding of applied mathematics from this kind of practice!





Question 46. (Click on arrow for answer)

At what frequency does a 350 mH inductor have 4.7 k\Omega of reactance? Write the formula for solving this, in addition to calculating the frequency.

File Num: 00586

Answer

f = 2.137 kHz

Notes

Be sure to ask your students to demonstrate the algebraic manipulation of the original formula, in providing the answer to this question. Algebraic manipulation of equations is a very important skill to have, and it comes only by study and practice.





Question 47. (Click on arrow for answer)

How much inductance would an inductor have to possess in order to provide 540 \Omega of reactance at a frequency of 400 Hz? Write the formula for solving this, in addition to calculating the frequency.

File Num: 03277

Answer

L = 214.9 mH

Notes

Be sure to ask your students to demonstrate the algebraic manipulation of the original formula, in providing the answer to this question. Algebraic manipulation of equations is a very important skill to have, and it comes only by study and practice.





Question 48. (Click on arrow for answer)

Explain all the steps necessary to calculate the amount of current in this inductive AC circuit:

Should be image 01552x01

File Num: 01552

Answer

I = 15.6 mA

Notes

The current is not difficult to calculate, so obviously the most important aspect of this question is not the math. Rather, it is the procedure of calculation: what to do first, second, third, etc., in obtaining the final answer.





Question 49. (Click on arrow for answer)

In this AC circuit, the resistor offers 300 \Omega of resistance, and the inductor offers 400 \Omega of reactance. Together, their series opposition to alternating current results in a current of 10 mA from the 5 volt source:

Should be image 00584x01

How many ohms of opposition does the series combination of resistor and inductor offer? What name do we give to this quantity, and how do we symbolize it, being that it is composed of both resistance (R) and reactance (X)?

File Num: 00584

Answer

Z_{total} = 500 \Omega.

Follow-up question: suppose that the inductor suffers a failure in its wire winding, causing it to “open.” Explain what effect this would have on circuit current and voltage drops.


Notes

Students may experience difficulty arriving at the same quantity for impedance shown in the answer. If this is the case, help them problem-solve by suggesting they simplify the problem: short past one of the load components and calculate the new circuit current. Soon they will understand the relationship between total circuit opposition and total circuit current, and be able to apply this concept to the original problem.

Ask your students why the quantities of 300 \Omega and 400 \Omega do not add up to 700 \Omega like they would if they were both resistors. Does this scenario remind them of another mathematical problem where 3 + 4 = 5? Where have we seen this before, especially in the context of electric circuits?

Once your students make the cognitive connection to trigonometry, ask them the significance of these numbers’ addition. Is it enough that we say a component has an opposition to AC of 400 \Omega, or is there more to this quantity than a single, scalar value? What type of number would be suitable for representing such a quantity, and how might it be written?





Question 50. (Click on arrow for answer)

While studying DC circuit theory, you learned that resistance was an expression of a component’s opposition to electric current. Then, when studying AC circuit theory, you learned that reactance was another type of opposition to current. Now, a third term is introduced: impedance. Like resistance and reactance, impedance is also a form of opposition to electric current.

Explain the difference between these three quantities (resistance, reactance, and impedance) using your own words.

File Num: 01567

Answer

The fundamental distinction between these terms is one of abstraction: impedance is the most general term, encompassing both resistance and reactance. Here is an explanation given in terms of logical sets (using a Venn diagram), along with an analogy from animal taxonomy:

Should be image 01567x01

Resistance is a type of impedance, and so is reactance. The difference between the two has to do with energy exchange.


Notes

The given answer is far from complete. I’ve shown the semantic relationship between the terms resistance, reactance, and impedance, but I have only hinted at the conceptual distinctions between them. Be sure to discuss with your students what the fundamental difference is between resistance and reactance, in terms of electrical energy exchange.





Question 51. (Click on arrow for answer)

In DC circuits, we have Ohm’s Law to relate voltage, current, and resistance together:

E = I R

In AC circuits, we similarly need a formula to relate voltage, current, and impedance together. Write three equations, one solving for each of these three variables: a set of Ohm’s Law formulae for AC circuits. Be prepared to show how you may use algebra to manipulate one of these equations into the other two forms.

File Num: 00590

Answer

E = I ZI = {E \over Z}Z = {E \over I}

If using phasor quantities (complex numbers) for voltage, current, and impedance, the proper way to write these equations is as follows:

E = IZI = {E \over Z}Z = {E \over I}

Bold-faced type is a common way of denoting vector quantities in mathematics.


Notes

Although the use of phasor quantities for voltage, current, and impedance in the AC form of Ohm’s Law yields certain distinct advantages over scalar calculations, this does not mean one cannot use scalar quantities. Often it is appropriate to express an AC voltage, current, or impedance as a simple scalar number.





Question 52. (Click on arrow for answer)

It is often necessary to represent AC circuit quantities as complex numbers rather than as scalar numbers, because both magnitude and phase angle are necessary to consider in certain calculations.

When representing AC voltages and currents in polar form, the angle given refers to the phase shift between the given voltage or current, and a “reference” voltage or current at the same frequency somewhere else in the circuit. So, a voltage of 3.5 \hbox{ V} \angle -45^o means a voltage of 3.5 volts magnitude, phase-shifted 45 degrees behind (lagging) the reference voltage (or current), which is defined to be at an angle of 0 degrees.

But what about impedance (Z)? Does impedance have a phase angle, too, or is it a simple scalar number like resistance or reactance?


Calculate the amount of current that would go through a 100 mH inductor with 36 volts RMS applied to it at a frequency of 400 Hz. Then, based on Ohm’s Law for AC circuits and what you know of the phase relationship between voltage and current for an inductor, calculate the impedance of this inductor in polar form. Does a definite angle emerge from this calculation for the inductor’s impedance? Explain why or why not.

File Num: 00588

Answer

Z_L = 251.33 \Omega \angle 90^{o}

Notes

This is a challenging question, because it asks the student to defend the application of phase angles to a type of quantity that does not really possess a wave-shape like AC voltages and currents do. Conceptually, this is difficult to grasp. However, the answer is quite clear through the Ohm’s Law calculation (Z = {E \over I}).

Although it is natural to assign a phase angle of 0^{o} to the 36 volt supply, making it the reference waveform, this is not actually necessary. Work through this calculation with your students, assuming different angles for the voltage in each instance. You should find that the impedance computes to be the same exact quantity every time.





Question 53. (Click on arrow for answer)

If a sinusoidal voltage is applied to an impedance with a phase angle of 0^{o}, the resulting voltage and current waveforms will look like this:

Should be image 00631x01

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit.

File Num: 00631

Answer

Should be image 00631x02

Notes

Ask your students to observe the waveform shown in the answer closely, and determine what sign the power values always are. Note how the voltage and current waveforms alternate between positive and negative, but power does not. Of what significance is this to us? What does this indicate about the nature of a load with an impedance phase angle of 0^{o}?





Question 54. (Click on arrow for answer)

If a sinusoidal voltage is applied to an impedance with a phase angle of 90^{o}, the resulting voltage and current waveforms will look like this:

Should be image 00632x01

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit. Also, explain how the mnemonic phrase “ELI the ICE man” applies to these waveforms.

File Num: 00632

Answer

Should be image 00632x02

The mnemonic phrase, “ELI the ICE man” indicates that this phase shift is due to an inductance rather than a capacitance.


Notes

Ask your students to observe the waveform shown in the answer closely, and determine what sign the power values are. Note how the power waveform alternates between positive and negative values, just as the voltage and current waveforms do. Ask your students to explain what negative power could possibly mean.

Of what significance is this to us? What does this indicate about the nature of a load with an impedance phase angle of 90^{o}?


The phrase, “ELI the ICE man” has been used be generations of technicians to remember the phase relationships between voltage and current for inductors and capacitors, respectively. One area of trouble I’ve noted with students, though, is being able to interpret which waveform is leading and which one is lagging, from a time-domain plot such as this.





Question 55. (Click on arrow for answer)

The impedance triangle is often used to graphically relate Z, R, and X in a series circuit:

Should be image 02076x01

Unfortunately, many students do not grasp the significance of this triangle, but rather memorize it as a “trick” used to calculate one of the three variables given the other two. Explain why a right triangle is an appropriate form to relate these variables, and what each side of the triangle actually represents.

File Num: 02076

Answer

Each side of the impedance triangle is actually a phasor (a vector representing impedance with magnitude and direction):

Should be image 02076x02

Since the phasor for resistive impedance (Z_R) has an angle of zero degrees and the phasor for reactive impedance (Z_C or Z_L) either has an angle of +90 or -90 degrees, the phasor sum representing total series impedance will form the hypotenuse of a right triangle when the first to phasors are added (tip-to-tail).


Follow-up question: as a review, explain why resistive impedance phasors always have an angle of zero degrees, and why reactive impedance phasors always have angles of either +90 degrees or -90 degrees.


Notes

The question is sufficiently open-ended that many students may not realize exactly what is being asked until they read the answer. This is okay, as it is difficult to phrase the question in a more specific manner without giving away the answer!





Question 56. (Click on arrow for answer)

Use the “impedance triangle” to calculate the impedance of this series combination of resistance (R) and inductive reactance (X):

Should be image 02081x01

Explain what equation(s) you use to calculate Z.

File Num: 02081

Answer

Z = 625 \Omega, as calculated by the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 57. (Click on arrow for answer)

Use the “impedance triangle” to calculate the necessary reactance of this series combination of resistance (R) and inductive reactance (X) to produce the desired total impedance of 145 \Omega:

Should be image 02083x01

Explain what equation(s) you use to calculate X, and the algebra necessary to achieve this result from a more common formula.

File Num: 02083

Answer

X = 105 \Omega, as calculated by an algebraically manipulated version of the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 58. (Click on arrow for answer)

Should be image 02084x01

Identify which trigonometric functions (sine, cosine, or tangent) are represented by each of the following ratios, with reference to the angle labeled with the Greek letter “Theta” (\Theta):

{X \over R} = {X \over Z} = {R \over Z} =

File Num: 02084

Answer

Should be image 02084x01{X \over R} = \tan \Theta = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \sin \Theta = {\hbox{Opposite} \over \hbox{Hypotenuse}}{R \over Z} = \cos \Theta = {\hbox{Adjacent} \over \hbox{Hypotenuse}}

Notes

Ask your students to explain what the words “hypotenuse”, “opposite”, and “adjacent” refer to in a right triangle.





Question 59. (Click on arrow for answer)

Trigonometric functions such as sine, cosine, and tangent are useful for determining the ratio of right-triangle side lengths given the value of an angle. However, they are not very useful for doing the reverse: calculating an angle given the lengths of two sides.

Should be image 02086x01

Suppose we wished to know the value of angle \Theta, and we happened to know the values of Z and R in this impedance triangle. We could write the following equation, but in its present form we could not solve for \Theta:

\cos \Theta = {R \over Z}

The only way we can algebraically isolate the angle \Theta in this equation is if we have some way to “undo” the cosine function. Once we know what function will “undo” cosine, we can apply it to both sides of the equation and have \Theta by itself on the left-hand side.

There is a class of trigonometric functions known as inverse or “arc” functions which will do just that: “undo” a regular trigonometric function so as to leave the angle by itself. Explain how we could apply an “arc-function” to the equation shown above to isolate \Theta.

File Num: 02086

Answer

\cos \Theta = {R \over Z} \hbox{ Original equation}\hbox<i>. . . applying the "arc-cosine" function to both sides . . .</i>\arccos \left( \cos \Theta \right) = \arccos \left( {R \over Z} \right)\Theta = \arccos \left( {R \over Z} \right)

Notes

I like to show the purpose of trigonometric arcfunctions in this manner, using the cardinal rule of algebraic manipulation (do the same thing to both sides of an equation) that students are familiar with by now. This helps eliminate the mystery of arcfunctions for students new to trigonometry.





Question 60. (Click on arrow for answer)

A series AC circuit contains 1125 ohms of resistance and 1500 ohms of reactance for a total circuit impedance of 1875 ohms. This may be represented graphically in the form of an impedance triangle:

Should be image 02085x01

Since all side lengths on this triangle are known, there is no need to apply the Pythagorean Theorem. However, we may still calculate the two non-perpendicular angles in this triangle using “inverse” trigonometric functions, which are sometimes called arcfunctions.

Identify which arc-function should be used to calculate the angle \Theta given the following pairs of sides:

R \hbox{ and } ZX \hbox{ and } RX \hbox{ and } Z

Show how three different trigonometric arcfunctions may be used to calculate the same angle \Theta.

File Num: 02085

Answer

\arccos {R \over Z} = 53.13^o\arctan {X \over R} = 53.13^o\arcsin {X \over Z} = 53.13^o

Challenge question: identify three more arcfunctions which could be used to calculate the same angle \Theta.


Notes

Some hand calculators identify arc-trig functions by the letter “A” prepending each trigonometric abbreviation (e.g. “ASIN” or “ATAN”). Other hand calculators use the inverse function notation of a -1 exponent, which is not actually an exponent at all (e.g. \sin^{-1} or \tan^{-1}). Be sure to discuss function notation on your students’ calculators, so they know what to invoke when solving problems such as this.





Question 61. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 00850x01

File Num: 00850

Answer

Z_{total} = \sqrt{R^2 + X^2}

Follow-up question: algebraically manipulate this equation to produce two more; one solving for R and the other solving for X.


Notes

Ask your students if this equation looks similar to any other mathematical equations they’ve seen before. If not, square both sides of the equation so it looks like Z^2 = R^2 + X^2 and ask them again.





Question 62. (Click on arrow for answer)

Draw a phasor diagram showing the trigonometric relationship between resistance, reactance, and impedance in this series circuit:

Should be image 01827x01

Show mathematically how the resistance and reactance combine in series to produce a total impedance (scalar quantities, all). Then, show how to analyze this same circuit using complex numbers: regarding component as having its own impedance, demonstrating mathematically how these impedances add up to comprise the total impedance (in both polar and rectangular forms).

File Num: 01827

Answer

Should be image 01827x02Scalar calculationsR = 2.2 \hbox{ k}\Omega X_L = 1.495 \hbox{ k}\OmegaZ_{series} = \sqrt{R^2 + {X_L}^2}Z_{series} = \sqrt{2200^2 + 1495^2} = 2660 \> \Omega
Complex number calculationsZ_R = 2.2 \hbox{ k}\Omega \> \angle \> 0^o Z_L = 1.495 \hbox{ k}\Omega \> \angle \> 90^o (Polar form)Z_R = 2.2 \hbox{ k}\Omega + j0 \> \Omega Z_L = 0 \> \Omega + j1.495 \hbox{ k}\Omega (Rectangular form)
Z_{series} = Z_1 + Z_2 + \cdots Z_n (General rule of series impedances)Z_{series} = Z_R + Z_L (Specific application to this circuit)
Z_{series} = 2.2 \hbox{ k}\Omega \> \angle \> 0^o + 1.495 \hbox{ k}\Omega \> \angle \> 90^o = 2.66 \hbox{ k}\Omega \> \angle \> 34.2^oZ_{series} = (2.2 \hbox{ k}\Omega + j0 \> \Omega) + (0 \> \Omega + j1.495 \hbox{ k}\Omega) = 2.2 \hbox{ k}\Omega + j1.495 \hbox{ k}\Omega

Notes

I want students to see that there are two different ways of approaching a problem such as this: with scalar math and with complex number math. If students have access to calculators that can do complex-number arithmetic, the “complex” approach is actually simpler for series-parallel combination circuits, and it yields richer (more informative) results.

Ask your students to determine which of the approaches most resembles DC circuit calculations. Incidentally, this is why I tend to prefer complex-number AC circuit calculations over scalar calculations: because of the conceptual continuity between AC and DC. When you use complex numbers to represent AC voltages, currents, and impedances, almost all the rules of DC circuits still apply. The big exception, of course, is calculations involving power.





Question 63. (Click on arrow for answer)

Calculate the total impedance for these two 100 mH inductors at 2.3 kHz, and draw a phasor diagram showing circuit impedances (Z_{total}, R, and X):

Should be image 02080x01

Now, re-calculate impedance and re-draw the phasor impedance diagram supposing the second inductor is replaced by a 1.5 k\Omega resistor:

Should be image 02080x02

File Num: 02080

Answer

Should be image 02080x03Should be image 02080x04

Notes

Phasor diagrams are powerful analytical tools, if one knows how to draw and interpret them. With hand calculators being so powerful and readily able to handle complex numbers in either polar or rectangular form, there is temptation to avoid phasor diagrams and let the calculator handle all the angle manipulation. However, students will have a much better understanding of phasors and complex numbers in AC circuits if you hold them accountable to representing quantities in that form.





Question 64. (Click on arrow for answer)

Calculate the total impedance of this series LR circuit and then calculate the total circuit current:

Should be image 02103x01

Also, draw a phasor diagram showing how the individual component impedances relate to the total impedance.

File Num: 02103

Answer

Z_{total} = 6.944 k\OmegaI = 4.896 mA RMS

Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 65. (Click on arrow for answer)

Calculate the magnitude and phase shift of the current through this inductor, taking into consideration its intrinsic winding resistance:

Should be image 00639x01

File Num: 00639

Answer

I = 7.849 mA \angle -87.08^{o}

Notes

Inductors are the least “pure” of any reactive component, due to significant quantities of resistance in the windings. Discuss this fact with your students, and what it means with reference to choosing inductors versus capacitors in circuit designs that could use either.





Question 66. (Click on arrow for answer)

Solve for all voltages and currents in this series LR circuit:

Should be image 01830x01

File Num: 01830

Answer

V_L = 12.60 \hbox{ volts RMS}V_R = 8.137 \hbox{ volts RMS}I = 11.46 \hbox{ milliamps RMS}

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 67. (Click on arrow for answer)

Solve for all voltages and currents in this series LR circuit, and also calculate the phase angle of the total impedance:

Should be image 01831x01

File Num: 01831

Answer

V_L = 13.04 \hbox{ volts RMS}V_R = 20.15 \hbox{ volts RMS}I = 4.030 \hbox{ milliamps RMS}\Theta_Z = 32.91^o

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 68. (Click on arrow for answer)

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

Should be image 01841x01
  • L_1 = 250 \hbox{ mH}
  • L_2 = 60 \hbox{ mH}
  • R_1 = 6.8 \hbox{ k}\Omega
  • R_2 = 1.2 \hbox{ k}\Omega
  • V_{supply} = 13.4 \hbox{ V RMS}
  • f_{supply} = 6.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01841

Answer

  • I_{total} = 0.895 \hbox{ mA}
  • V_{L1} = 9.14 \hbox{ V}
  • V_{L2} = 2.19 \hbox{ V}
  • V_{R1} = 6.08 \hbox{ V}
  • V_{R2} = 1.07 \hbox{ V}
  • \Theta = 57.71^o

I suggest using a dual-trace oscilloscope to measure total voltage (across the supply terminals) and voltage drop across resistor R_2. Theoretically, measuring the voltage dropped by either resistor would be fine, but R_2 works better for practical reasons (oscilloscope input lead grounding). Phase shift then could be measured either in the time domain or by a Lissajous figure analysis.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 69. (Click on arrow for answer)

One way to vary the amount of power delivered to a resistive AC load is by varying another resistance connected in series:

Should be image 01829x01

A problem with this power control strategy is that power is wasted in the series resistance (I^2R_{series}). A different strategy for controlling power is shown here, using a series inductance rather than resistance:

Should be image 01829x02

Explain why the latter circuit is more power-efficient than the former, and draw a phasor diagram showing how changes in L_{series} affect Z_{total}.

File Num: 01829

Answer

Inductors are reactive rather than resistive components, and therefore do not dissipate power (ideally).

Should be image 01829x03

Follow-up question: the inductive circuit is not just more energy-efficient — it is safer as well. Identify a potential safety hazard that the resistive power-control circuit poses due to the energy dissipation of its variable resistor.


Notes

If appropriate, you may want to mention devices called saturable reactors, which are used to control power in AC circuits by the exact same principle: varying a series inductance.





Question 70. (Click on arrow for answer)

A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = {1 \over R}), and it is measured in the unit of siemens.

Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

Should be image 00853x01

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = {1 \over X}), and the reciprocal of impedance is called admittance (Y = {1 \over Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.

Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Should be image 00853x02

File Num: 00853

Answer

Y_{total} = \sqrt{G^2 + B^2}

Follow-up question \#1: draw a phasor diagram showing how Y, G, and B relate.


Follow-up question \#2: re-write this equation using quantities of resistance (R), reactance (X), and impedance (Z), instead of conductance (G), susceptance (B), and admittance (Y).


Notes

Ask your students if this equation looks familiar to them. It should!


The answer to the second follow-up question is a matter of algebraic substitution. Work through this process with your students, and then ask them to compare the resulting equation with other equations they’ve seen before. Does its form look familiar to them in any way?





Question 71. (Click on arrow for answer)

Students studying AC electrical theory become familiar with the impedance triangle very soon in their studies:

Should be image 02077x01

What these students might not ordinarily discover is that this triangle is also useful for calculating electrical quantities other than impedance. The purpose of this question is to get you to discover some of the triangle’s other uses.

Fundamentally, this right triangle represents phasor addition, where two electrical quantities at right angles to each other (resistive versus reactive) are added together. In series AC circuits, it makes sense to use the impedance triangle to represent how resistance (R) and reactance (X) combine to form a total impedance (Z), since resistance and reactance are special forms of impedance themselves, and we know that impedances add in series.

List all of the electrical quantities you can think of that add (in series or in parallel) and then show how similar triangles may be drawn to relate those quantities together in AC circuits.

File Num: 02077

Answer

Electrical quantities that add:
  • Series impedances
  • Series voltages
  • Parallel admittances
  • Parallel currents
  • Power dissipations

I will show you one graphical example of how a triangle may relate to electrical quantities other than series impedances:

Should be image 02077x02

Notes

It is very important for students to understand that the triangle only works as an analysis tool when applied to quantities that add. Many times I have seen students try to apply the ZRX impedance triangle to parallel circuits and fail because parallel impedances do not add. The purpose of this question is to force students to think about where the triangle is applicable to AC circuit analysis, and not just to use it blindly.

The power triangle is an interesting application of trigonometry applied to electric circuits. You may not want to discuss power with your students in great detail if they are just beginning to study voltage and current in AC circuits, because power is a sufficiently confusing subject on its own.





Question 72. (Click on arrow for answer)

Explain why the “impedance triangle” is not proper to use for relating total impedance, resistance, and reactance in parallel circuits as it is for series circuits:

Should be image 02078x01

File Num: 02078

Answer

Impedances do not add in parallel.


Follow-up question: what kind of a triangle could be properly applied to a parallel AC circuit, and why?


Notes

Trying to apply the ZRX triangle directly to parallel AC circuits is a common mistake many new students make. Key to knowing when and how to use triangles to graphically depict AC quantities is understanding why the triangle works as an analysis tool and what its sides represent.





Question 73. (Click on arrow for answer)

Calculate the total impedance for these two 100 mH inductors at 2.3 kHz, and draw a phasor diagram showing circuit admittances (Y_{total}, G, and B):

Should be image 02079x01

Now, re-calculate impedance and re-draw the phasor admittance diagram supposing the second inductor is replaced by a 1.5 k\Omega resistor:

Should be image 02079x02

File Num: 02079

Answer

Should be image 02079x03Should be image 02079x04

Challenge question: why are the susceptance vectors (B_{L1} and B_{L2}) pointed down instead of up as impedance vectors for inductances typically are?


Notes

Phasor diagrams are powerful analytical tools, if one knows how to draw and interpret them. With hand calculators being so powerful and readily able to handle complex numbers in either polar or rectangular form, there is temptation to avoid phasor diagrams and let the calculator handle all the angle manipulation. However, students will have a much better understanding of phasors and complex numbers in AC circuits if you hold them accountable to representing quantities in that form.





Question 74. (Click on arrow for answer)

Calculate the individual currents through the inductor and through the resistor, the total current, and the total circuit impedance:

Should be image 02104x01

Also, draw a phasor diagram showing how the individual component currents relate to the total current.

File Num: 02104

Answer

I_L = 530.5 \muA RMSI_R = 490.2 \muA RMSI_{total} = 722.3 \muA RMSZ_{total} = 3.461 k\Omega

Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 75. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01839x01

Calculate the current necessary to power this motor if the equivalent resistance and inductance is 20 \Omega and 238 mH, respectively.

File Num: 01839

Answer

I_{supply} = 12.29 \hbox{ A}

Notes

This is a practical example of a parallel LR circuit, as well as an example of how complex electrical devices may be “modeled” by collections of ideal components. To be honest, a loaded AC motor’s characteristics are quite a bit more complex than what the parallel LR model would suggest, but at least it’s a start!


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 76. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01840x01

Calculate the equivalent inductance (L_{eq}) if the measured source current is 27.5 amps and the motor’s equivalent resistance (R_{eq}) is 11.2 \Omega.

File Num: 01840

Answer

L_{eq} = 61.11 \hbox{ mH}

Notes

Here is a case where scalar calculations (R, G, X, B, Y) are much easier than complex number calculations (all Z) would be.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 77. (Click on arrow for answer)

Determine the total current and all component currents in this circuit, stating your answers the way a multimeter would register them:

Should be image 01842x01
  • L_1 = 1.2 \hbox{ H}
  • L_2 = 650 \hbox{ mH}
  • R_1 = 33 \hbox{ k}\Omega
  • R_2 = 27 \hbox{ k}\Omega
  • V_{supply} = 19.7 \hbox{ V RMS}
  • f_{supply} = 4.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01842

Answer

  • I_{total} = 2.12 \hbox{ mA}
  • I_{L1} = 581 \> \mu \hbox{A}
  • I_{L2} = 1.07 \hbox{ mA}
  • I_{R1} = 597 \> \mu \hbox{A}
  • I_{R2} = 730 \> \mu \hbox{A}
  • \Theta = 51.24^o

Measuring \Theta with an oscilloscope requires the addition of a shunt resistor into this circuit, because oscilloscopes are (normally) only able to measure voltage, and there is no phase shift between any voltages in this circuit because all components are in parallel. I leave it to you to suggest where to insert the shunt resistor, what resistance value to select for the task, and how to connect the oscilloscope to the modified circuit.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 78. (Click on arrow for answer)

Calculate the total impedances (complete with phase angles) for each of the following inductor-resistor circuits:

Should be image 02106x01

File Num: 02106

Answer

Should be image 02106x02

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 79. (Click on arrow for answer)

A doorbell ringer has a solenoid with an inductance of 63 mH connected in parallel with a lamp (for visual indication) having a resistance of 150 ohms:

Should be image 02105x01

Calculate the phase shift of the total current (in units of degrees) in relation to the total supply voltage, when the doorbell switch is actuated.

File Num: 02105

Answer

\Theta = 81 degrees

Suppose the lamp turned on whenever the pushbutton switch was actuated, but the doorbell refused to ring. Identify what you think to be the most likely fault which could account for this problem.


Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 80. (Click on arrow for answer)

An AC electric motor operating under loaded conditions draws a current of 11 amps (RMS) from the 120 volt (RMS) 60 Hz power lines. The measured phase shift between voltage and current for this motor is 34^{o}, with voltage leading current.

Determine the equivalent parallel combination of resistance (R) and inductance (L) that is electrically equivalent to this operating motor.

File Num: 01542

Answer

R_{parallel} = 13.16 \Omega
L_{parallel} = 51.75 mH

Challenge question: in the parallel LR circuit, the resistor will dissipate a lot of energy in the form of heat. Does this mean that the electric motor, which is electrically equivalent to the LR network, will dissipate the same amount of heat? Explain why or why not.


Notes

If students get stuck on the challenge question, remind them that an electric motor does mechanical work, which requires energy.





Question 81. (Click on arrow for answer)

Should be image 01674x01

File Num: 01674

\vfil \eject

Answer

You may use circuit simulation software to set up similar oscilloscope display interpretation scenarios, for practice or for verification of what you see in this exercise.


Notes

Use a sine-wave function generator for the AC voltage source, and be sure set the frequency to some reasonable value (well within the capability of both the oscilloscope and counter to measure).

If this is not the first time students have done this, be sure to “mess up” the oscilloscope controls prior to them making adjustments. Students must learn how to quickly configure an oscilloscope’s controls to display any arbitrary waveform, if they are to be proficient in using an oscilloscope as a diagnostic tool.





Question 82. (Click on arrow for answer)

Should be image 01693x01

File Num: 01693

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Answer

You may use circuit simulation software to set up similar oscilloscope display interpretation scenarios, for practice or for verification of what you see in this exercise.


Notes

Use a sine-wave function generator for the AC voltage source, and be sure set the frequency to some reasonable value (well within the capability of a multimeter to measure). It is very important that students learn to convert between peak and RMS measurements for sine waves, but you might want to mix things up a bit by having them do the same with triangle waves and square waves as well! It is vital that students realize the rule of V_{RMS} = {V_{peak} \over \sqrt{2}} only holds for sinusoidal signals.

If you do choose to challenge students with non-sinusoidal waveshapes, be very sure that they do their voltmeter measurements using true-RMS meters! This means no analog voltmeters, which are “miscalibrated” so their inherently average-responding movements register (sinusoidal) RMS accurately. Your students must use true-RMS digital voltmeters in order for their non-sinusoidal RMS measurements to correlate with their calculations.

Incidentally, this lab exercise also works well as a demonstration of the importance of true-RMS indicating meters, comparing the indications of analog, non-true-RMS digital, and true-RMS digital on the same non-sinusoidal waveform!





Question 83. (Click on arrow for answer)

Should be image 01660x01

File Num: 01660

\vfil \eject

Answer

You may use circuit simulation software to set up similar oscilloscope display interpretation scenarios, for practice or for verification of what you see in this exercise.


Notes

Use a sine-wave function generator for the AC voltage source, and be sure set the frequency to some reasonable value (well within the capability of both the oscilloscope and counter to measure).





Question 84. (Click on arrow for answer)

Should be image 01616x01

File Num: 01616

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Answer

Use circuit simulation software to verify your predicted and measured parameter values.


Notes

Use a sine-wave function generator for the AC voltage source. I recommend against using line-power AC because of strong harmonic frequencies which may be present (due to nonlinear loads operating on the same power circuit). Specify a standard inductor value.

If students are to use a multimeter to make their current and voltage measurements, be sure it is capable of accurate measurement at the circuit frequency! Inexpensive digital multimeters often experience difficulty measuring AC voltage and current toward the high end of the audio-frequency range.





Question 85. (Click on arrow for answer)

Should be image 01665x01

File Num: 01665

\vfil \eject

Answer

Use circuit simulation software to verify your predicted and measured parameter values.


Notes

Use a sine-wave function generator for the AC voltage source. I recommend against using line-power AC because of strong harmonic frequencies which may be present (due to nonlinear loads operating on the same power circuit). Specify standard resistor and inductor values.

If students are to use a multimeter to make their current and voltage measurements, be sure it is capable of accurate measurement at the circuit frequency! Inexpensive digital multimeters often experience difficulty measuring AC voltage and current toward the high end of the audio-frequency range.

An extension of this exercise is to incorporate troubleshooting questions. Whether using this exercise as a performance assessment or simply as a concept-building lab, you might want to follow up your students’ results by asking them to predict the consequences of certain circuit faults.





Question 86. (Click on arrow for answer)

Should be image 01823x01

File Num: 01823

\vfil \eject

Answer

There really isn’t much you can do to verify your experimental results. That’s okay, though, because the results are qualitative anyway.


Notes

If the oscilloscope does not have its own internal square-wave signal source, use a function generator set up to output square waves at 1 volt peak-to-peak at a frequency of 1 kHz.

If this is not the first time students have done this, be sure to “mess up” the oscilloscope controls prior to them making adjustments. Students must learn how to quickly configure an oscilloscope’s controls to display any arbitrary waveform, if they are to be proficient in using an oscilloscope as a diagnostic tool.





Question 87. (Click on arrow for answer)

Should be image 03933x01

File Num: 03933

\vfil \eject

Answer

I do not provide a grading rubric here, but elsewhere.


Notes

The idea of a troubleshooting log is three-fold. First, it gets students in the habit of documenting their troubleshooting procedure and thought process. This is a valuable habit to get into, as it translates to more efficient (and easier-followed) troubleshooting on the job. Second, it provides a way to document student steps for the assessment process, making your job as an instructor easier. Third, it reinforces the notion that each and every measurement or action should be followed by reflection (conclusion), making the troubleshooting process more efficient.





Question 88. (Click on arrow for answer)

\centerline{NAME: \hskip 80pt \hskip 40pt Troubleshooting Grading Criteria }

You will receive the highest score for which all criteria are met.


100 \% (Must meet or exceed all criteria listed) \item{A.} Absolutely flawless procedure \item{B.} No unnecessary actions or measurements taken \bigskip90 \% (Must meet or exceed these criteria in addition to all criteria for 85\% and below) \item{A.} No reversals in procedure (i.e. changing mind without sufficient evidence) \item{B.} Every single action, measurement, and relevant observation properly documented \bigskip80 \% (Must meet or exceed these criteria in addition to all criteria for 75\% and below) \item{A.} No more than one unnecessary action or measurement \item{B.} No false conclusions or conceptual errors \item{C.} No missing conclusions (i.e. at least one documented conclusion for action / measurement / observation) \bigskip70 \% (Must meet or exceed these criteria in addition to all criteria for 65\%) \item{A.} No more than one false conclusion or conceptual error \item{B.} No more than one conclusion missing (i.e. an action, measurement, or relevant observation without a corresponding conclusion) \bigskip65 \% (Must meet or exceed these criteria in addition to all criteria for 60\%) \item{A.} No more than two false conclusions or conceptual errors \item{B.} No more than two unnecessary actions or measurements \item{C.} No more than one undocumented action, measurement, or relevant observation \item{D.} Proper use of all test equipment \bigskip60 \% (Must meet or exceed these criteria) \item{A.} Fault accurately identified \item{B.} Safe procedures used at all times \bigskip50 \% (Only applicable where students performed significant development/design work — i.e. not a proven circuit provided with all component values) \item{A.} Working prototype circuit built and demonstrated \bigskip0 \% (If any of the following conditions are true) \item{A.} Unsafe procedure(s) used at any point \bigskip

File Num: 03932

Answer

Be sure to document all steps taken and conclusions made in your troubleshooting!


Notes

The purpose of this assessment rubric is to act as a sort of “contract” between you (the instructor) and your student. This way, the expectations are all clearly known in advance, which goes a long way toward disarming problems later when it is time to grade.





Question 89. (Click on arrow for answer)

Determine the frequency of a waveform having a period of 1.4 milliseconds (1.4 ms).

File Num: 03276

Answer

f = 714.29 Hz

Notes

It is important for students to realize the reciprocal relationship between frequency and period. One is cycles per second while the other is seconds per cycle.





Question 90. (Click on arrow for answer)

Assuming the vertical sensitivity control is set to 0.5 volts per division, and the timebase control is set to 2.5 ms per division, calculate the amplitude of this sine wave (in volts peak, volts peak-to-peak, and volts RMS) as well as its frequency.

Should be image 00540x01

File Num: 00540

Answer


\item{} E_{peak} = 2.25 V \item{} E_{peak-to-peak} = 4.50 V \item{} E_{RMS} = 1.59 V \item{} f = 40 Hz

Notes

This question is not only good for introducing basic oscilloscope principles, but it is also excellent for review of AC waveform measurements.





Question 91. (Click on arrow for answer)

Something is wrong with this circuit. Based on the oscilloscope’s display, determine whether the battery or the function generator is faulty:

Should be image 03448x01

File Num: 03448

Answer

The battery is faulty.


Follow-up question: discuss how accidently setting the coupling control on the oscilloscope to “AC” instead of “DC” would also cause this waveform to show on the screen (even with a good battery).


Notes

This question challenges students to apply their knowledge of AC+DC mixed signals to oscilloscope display patterns, in order to determine whether it is the battery or the function generator which has failed.





Question 92. (Click on arrow for answer)

Something is wrong with this circuit. Based on the oscilloscope’s display, determine whether the battery or the function generator is faulty:

Should be image 03449x01

File Num: 03449

Answer

The function generator is faulty.


Follow-up question: explain how this problem could be created simply by connecting the function generator to the circuit with the ground on the left-hand clip instead of the right-hand clip where it should be.

Should be image 03449x02

Notes

This question challenges students to apply their knowledge of AC+DC mixed signals to oscilloscope display patterns, in order to determine whether it is the battery or the function generator which has failed.





Question 93. (Click on arrow for answer)

Shunt resistors are low-value, precision resistors used as current-measuring elements in high-current circuits. The idea is to measure the voltage dropped across this precision resistance and use Ohm’s Law (I = {V \over R}) to infer the amount of current in the circuit:Should be image 03504x01

Since the schematic shows a shunt resistor being used to measure current in an AC circuit, it would be equally appropriate to use an oscilloscope instead of a voltmeter to measure the voltage drop produced by the shunt. However, we must be careful in connecting the oscilloscope to the shunt because of the inherent ground reference of the oscilloscope’s metal case and probe assembly.

Explain why connecting an oscilloscope to the shunt as shown in this second diagram would be a bad idea:

Should be image 03504x02

File Num: 03504

Answer

This would be a bad idea because the oscilloscope’s ground clip would attempt to bypass current around the shunt resistor, through the oscilloscope’s safety ground wire, and back to the grounded terminal of the AC source. Not only would this induce measurement errors, but it could damage the oscilloscope as well.


Follow-up question: identify a better way of connecting this oscilloscope to the shunt resistor.


Notes

The ground-referenced clip on an oscilloscope probe is a constant source of potential trouble for those who do not fully understand it! Even in scenarios where there is little or no potential for equipment damage, placing an earth ground reference on a circuit via the probe clip can make for very strange circuit behavior and erroneous measurements. Problems like this frequently occur when new students attempt to connect their oscilloscopes to circuits powered by signal generators whose outputs are also earth-ground referenced.

In response to the follow-up question, the most obvious answer is to reverse the probe connections: ground clip on the left-hand terminal and probe tip on the right-hand terminal. However, even this might not be the best idea, since it creates a “ground loop” between the oscilloscope and the ground connection at the AC source:

Should be image 03504x03

Ground loops are to be avoided in measurement circuits because they may be the source of some very strange effects, including the coupling of noise voltage from entirely unrelated circuits to the one being measured. To avoid this problem, the best solution for measuring the voltage dropped across the shunt resistor is to use two scope probes and set the scope up for differential voltage measurement:

Should be image 03504x04




Question 94. (Click on arrow for answer)

An electromechanical alternator (AC generator) and a DC-DC inverter both output the same RMS voltage, and deliver the same amount of electrical power to two identical loads:

Should be image 00404x01

However, when measured by an analog voltmeter, the inverter’s output voltage is slightly greater than the alternator’s output voltage. Explain this discrepancy in measurements.

File Num: 00404

Answer

Electromechanical alternators naturally output sinusoidal waveforms. Many DC-AC inverters do not.


Notes

Remember, most analog meter movement designs respond to the average value of a waveform, not its RMS value. If the proportionality between a waveform’s average and RMS values ever change, the relative indications of a true-RMS instrument and an average-based (calibrated to read RMS) instrument will change as well.





Question 95. (Click on arrow for answer)

Is the deflection of an analog AC meter movement proportional to the peak, average, or RMS value of the waveform measured? Explain your answer.

File Num: 00403

Answer

Analog meter deflection is proportional to the average value of the AC waveform measured, for most AC meter movement types. There are some meter movement designs, however, that give indications proportional to the RMS value of the waveform: hot-wire and electrodynamometer movements are of this nature.


Follow-up question: does this mean an average-responding meter movement cannot be calibrated to indicate in RMS units?


Challenge question: why do hot-wire and electrodynamometer meter movements provide true RMS indications, while most other movement designs indicate based on the signal’s average value?


Notes

Students often confuse the terms “average” and “RMS”, thinking they are interchangeable. Discuss the difference between these two terms, both mathematically and practically. While the concepts may seem similar at first, the details are actually quite different.

The question of whether an average-responding instrument can be calibrated to register in RMS units is very practical, since the vast majority of multimeters are calibrated this way. Because the proportionality between the average and RMS values of an AC waveform are dependent on the shape of the waveform, a certain wave-shape must be assumed in order to accurately calibrate an average-responding meter movement for RMS measurement. The assumed wave-shape, of course, is sinusoidal.





Question 96. (Click on arrow for answer)

In calculating the size of wire necessary to carry alternating current to a high-power load, which type of measurement is the best to use for current: peak, average, or RMS? Explain why.

File Num: 00162

Answer

RMS current is the most appropriate type of measurement for calculating wire size.


Notes

A clue to answering this question is this: what actually happens when the ampacity rating of a conductor is exceeded? Why, exactly, is overcurrent a bad thing for conductors in general?

It is important for students to recognize the value of RMS measurements: why do we use them, and in what applications are they the most appropriate type of measurement to use in certain calculations? Ask your students what other applications might best use RMS voltage and current measurements as opposed to peak or average.





Question 97. (Click on arrow for answer)

In calculating the thickness of insulators for high-voltage AC power lines, which type of measurement is the best to use for voltage: peak, average, or RMS? Explain why.

File Num: 00163

Answer

Peak voltage is the most appropriate type of measurement for calculating insulator thickness. The reason why has to do with the time required for an insulator to “flash over.”


Notes

A closely related subject is insulator breakdown, or dielectric strength. What actually happens when the dielectric strength rating of an insulator is exceeded?





Question 98. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03578x01

File Num: 03578

Answer

\Theta \approx 25.9^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 99. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03575x01

File Num: 03575

Answer

\Theta \approx 64.2^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 100. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03577x01

File Num: 03577

Answer

\Theta \approx 34.5^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 101. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03576x01

File Num: 03576

Answer

\Theta \approx 44.4^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 102. (Click on arrow for answer)

Determine which way the movable iron piece needs to go in order to brighten the light bulb in this circuit:

Should be image 03446x01

File Num: 03446

Answer

The movable piece needs to move to the left, creating a larger air gap between the poles of the U-shaped inductor core.


Notes

This question is an exercise in applying practical electromagnetic theory (namely, reluctance of an iron/air flux path) to inductance and inductive reactance. Ask your students to explain their reasoning step-by-step as they give their answers to this question.





Question 103. (Click on arrow for answer)

A solenoid valve is a mechanical shutoff device actuated by electricity. An electromagnet coil produces an attractive force on an iron “armature” which then either opens or closes a valve mechanism to control the flow of some fluid. Shown here are two different types of illustrations, both showing a solenoid valve:

Should be image 03445x01

Some solenoid valves are constructed in such a way that the coil assembly may be removed from the valve body, separating these two pieces so that maintenance work may be done on one without interfering with the other. Of course, this means the valve mechanism will no longer be actuated by the magnetic field, but at least one piece may be worked upon without having to remove the other piece from whatever it may be connected to:

Should be image 03445x02

This is commonly done when replacement of the valve mechanism is needed. First, the coil is lifted off the valve mechanism, then the maintenance technician is free to remove the valve body from the pipes and replace it with a new valve body. Lastly, the coil is re-installed on the new valve body and the solenoid is once more ready for service, all without having to electrically disconnect the coil from its power source.

However, if this is done while the coil is energized, it will overheat and burn up in just a few minutes. To prevent this from happening, the maintenance technicians have learned to insert a steel screwdriver through the center hole of the coil while it is removed from the valve body, like this:

Should be image 03445x03

With the steel screwdriver shank taking the place of the iron armature inside the valve body, the coil will not overheat and burn up even if continually powered. Explain the nature of the problem (why the coil tends to burn up when separated from the valve body) and also why a screwdriver put in place of the iron armature works to prevent this from happening.

File Num: 03445

Answer

With the iron armature no longer in the center of the solenoid coil, the coil’s inductance — and therefore its inductive reactance to AC — dramatically diminishes unless the armature is replaced by something else ferromagnetic.


Notes

When I first saw this practice in action, I almost fell over laughing. It is both practical and ingenious, as well as being an excellent example of variable inductance (and inductive reactance) arising from varying reluctance.





Question 104. (Click on arrow for answer)

Doorbell circuits connect a small lamp in parallel with the doorbell pushbutton so that there is light at the button when it is not being pressed. The lamp’s filament resistance is such that there is not enough current going through it to energize the solenoid coil when lit, which means the doorbell will ring only when the pushbutton switch shorts past the lamp:

Should be image 03447x01

Suppose that such a doorbell circuit suddenly stops working one day, and the home owner assumes the power source has quit since the bell will not ring when the button is pressed and the lamp never lights. Although a dead power source is certainly possible, it is not the only possibility. Identify another possible failure in this circuit which would result in no doorbell action (no sound) and no light at the lamp.

File Num: 03447

Answer


  • Solenoid coil failed open
  • Wire broken anywhere in circuit


Notes

After discussing alternative possibilities with your students, shift the discussion to one on how likely any of these failures are. For instance, how likely is it that the solenoid coil has developed an “open” fault compared to the likelihood of a regular wire connection going bad in the circuit? How do either of these possibilities compare with the likelihood of the source failing as a result of a tripped circuit breaker or other power outage?





Question 105. (Click on arrow for answer)

Suppose someone were to ask you to differentiate electrical reactance (X) from electrical resistance (R). How would you distinguish these two similar concepts from one another, using your own words?

File Num: 03301

Answer

It is really important for you to frame this concept in your own words, so be sure to check with your instructor on the accuracy of your answer to this question! To give you a place to start, I offer this distinction: resistance is electrical friction, whereas reactance is electrical energy storage. Fundamentally, the difference between X and R is a matter of energy exchange, and it is understood most accurately in those terms.


Notes

This is an excellent point of crossover with your students’ studies in elementary physics, if they are studying physics now or have studied physics in the past. The energy-storing actions of inductors and capacitors are quite analogous to the energy-storing actions of masses and springs (respectively, if you associate velocity with current and force with voltage). In the same vein, resistance is analogous to kinetic friction between a moving object and a stationary surface. The parallels are so accurate, in fact, that the electrical properties of R, L, and C have been exploited to model mechanical systems of friction, mass, and resilience in circuits known as analog computers.





Question 106. (Click on arrow for answer)

The Pythagorean Theorem is used to calculate the length of the hypotenuse of a right triangle given the lengths of the other two sides:

Should be image 03114x01

Manipulate the standard form of the Pythagorean Theorem to produce a version that solves for the length of A given B and C, and also write a version of the equation that solves for the length of B given A and C.

File Num: 03114

Answer

Standard form of the Pythagorean Theorem:

C = \sqrt{A^2 + B^2}

Solving for A:

A = \sqrt{C^2 - B^2}

Solving for B:

B = \sqrt{C^2 - A^2}

Notes

The Pythagorean Theorem is easy enough for students to find on their own that you should not need to show them. A memorable illustration of this theorem are the side lengths of a so-called 3-4-5 triangle. Don’t be surprised if this is the example many students choose to give.





Question 107. (Click on arrow for answer)

Suppose two people work together to slide a large box across the floor, one pushing with a force of 400 newtons and the other pulling with a force of 300 newtons:

Should be image 03278x01

The resultant force from these two persons’ efforts on the box will, quite obviously, be the sum of their forces: 700 newtons (to the right).


What if the person pulling decides to change position and push sideways on the box in relation to the first person, so the 400 newton force and the 300 newton force will be perpendicular to each other (the 300 newton force facing into the page, away from you)? What will the resultant force on the box be then?

Should be image 03278x02

File Num: 03278

Answer

The resultant force on the box will be 500 newtons.


Notes

This is a non-electrical application of vector summation, to prepare students for the concept of using vectors to add voltages that are out-of-phase. Note how I chose to use multiples of 3, 4, and 5 for the vector magnitudes.





Question 108. (Click on arrow for answer)

A rectangular building foundation with an area of 18,500 square feet measures 100 feet along one side. You need to lay in a diagonal run of conduit from one corner of the foundation to the other. Calculate how much conduit you will need to make the run:

Should be image 03275x01

Also, write an equation for calculating this conduit run length (L) given the rectangular area (A) and the length of one side (x).

File Num: 03275

Answer

Conduit run = 210 feet, 3.6 inches from corner to corner.


Note: the following equation is not the only form possible for calculating the diagonal length. Do not be worried if your equation does not look exactly like this!

L = {{\sqrt{x^4 + A^2}} \over x}

Notes

Determining the necessary length of conduit for this question involves both the Pythagorean theorem and simple geometry.

Most students will probably arrive at this form for their diagonal length equation:

L = \sqrt{x^2 + \left({A \over x}\right)^2}

While this is perfectly correct, it is an interesting exercise to have students convert the equation from this (simple) form to that given in the answer. It is also a very practical question, as equations given in reference books do not always follow the most direct form, but rather are often written in such a way as to look more esthetically pleasing. The simple and direct form of the equation shown here (in the Notes section) looks “ugly” due to the fraction inside the radicand.





Question 109. (Click on arrow for answer)

Should be image 03113x01

Identify which trigonometric functions (sine, cosine, or tangent) are represented by each of the following ratios, with reference to the angle labeled with the Greek letter “Phi” (\phi):

{R \over X} = {X \over Z} = {R \over Z} =

File Num: 03113

Answer

Should be image 03113x01{R \over X} = \tan \phi = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \cos \phi = {\hbox{Adjacent} \over \hbox{Hypotenuse}}{R \over Z} = \sin \phi = {\hbox{Opposite} \over \hbox{Hypotenuse}}

Notes

Ask your students to explain what the words “hypotenuse”, “opposite”, and “adjacent” refer to in a right triangle.





Question 110. (Click on arrow for answer)

In this phasor diagram, determine which phasor is leading and which is lagging the other:

Should be image 03286x01

File Num: 03286

Answer

In this diagram, phasor B is leading phasor A.


Follow-up question: using a protractor, estimate the amount of phase shift between these two phasors.


Notes

It may be helpful to your students to remind them of the standard orientation for phase angles in phasor diagrams (0 degrees to the right, 90 degrees up, etc.).





Question 111. (Click on arrow for answer)

Is it appropriate to assign a phasor angle to a single AC voltage, all by itself in a circuit?

Should be image 00496x01

What if there is more than one AC voltage source in a circuit?

Should be image 00496x02

File Num: 00496

Answer

Phasor angles are relative, not absolute. They have meaning only where there is another phasor to compare against.

Angles may be associated with multiple AC voltage sources in the same circuit, but only if those voltages are all at the same frequency.


Notes

Discuss with your students the notion of “phase angle” in relation to AC quantities. What does it mean, exactly, if a voltage is “3 volts at an angle of 90 degrees”? You will find that such a description only makes sense where there is another voltage (i.e., “4 volts at 0 degrees”) to compare to. Without a frame of reference, phasor angles are meaningless.

Also discuss with your students the nature of phase shifts between different AC voltage sources, if the sources are all at different frequencies. Would the phase angles be fixed, or vary over time? Why? In light of this, why do we not assign phase angles when different frequencies are involved?





Question 112. (Click on arrow for answer)

Determine the total voltage in each of these examples, drawing a phasor diagram to show how the total (resultant) voltage geometrically relates to the source voltages in each scenario:

Should be image 00498x01

File Num: 00498

Answer

Should be image 00498x02

Notes

At first it may confuse students to use polarity marks (+ and -) for AC voltages. After all, doesn’t the polarity of AC alternate back and forth, so as to be continuously changing? However, when analyzing AC circuits, polarity marks are essential for giving a frame of reference to phasor voltages, which like all voltages are measured between two points, and thus may be measured two different ways.





Question 113. (Click on arrow for answer)

Before two or more operating alternators (AC generators) may be electrically coupled, they must be brought into synchronization with each other. If two alternators are out of “sync” (or out of phase) with each other, the result will be a large fault current when the disconnect switch is closed.

A simple and effective means of checking for “sync” prior to closing the disconnect switch for an alternator is to have light bulbs connected in parallel with the disconnect switch contacts, like this:

Should be image 00491x01

What should the alternator operator look for before closing the alternator switch? Do bright lights indicate a condition of being “in-phase” with the bus, or do dim lights indicate this? What does the operator have to do in order to bring an alternator into “phase” with the bus voltage?

Also, describe what the light bulbs would do if the two alternators were spinning at slightly different speeds.

File Num: 00491

Answer

Dim lights indicate a condition of being “in-phase” with the bus. If the two alternators are spinning at slightly different speeds, there will be a heterodyne effect to the light bulbs’ brightness: alternately growing brighter, then dimmer, then brighter again.


Notes

Proper synchronization of alternators with bus voltage is a task that used to be performed exclusively by human operators, but may now be accomplished by automatic controls. It is still important, though, for students of electricity to understand the principles involved in alternator synchronization, and the simple light bulb technique of sync-indication is an excellent means of clarifying the concept.

Discuss with your students the means of bringing an alternator into phase with an AC bus. If the light bulbs are glowing brightly, what should the operator do to make them dim?

It might also be a good idea to discuss with your students what happens once two synchronized alternators become electrically coupled: the two machines become “locked” together as though they were mechanically coupled, thus maintaining synchronization from that point onward.





Question 114. (Click on arrow for answer)

Suppose a power plant operator was about to bring this alternator on-line (connect it to the AC bus), and happened to notice that neither one of the synchronizing lights was lit at all. Thinking this to be unusual, the operator calls you to determine if something is wrong with the system. Describe what you would do to troubleshoot this system.

Should be image 00500x01

File Num: 00500

Answer

Before you proceed with troubleshooting steps, first try to determine if there is anything wrong with this system at all. Could it be that the operator is just overly cautious, or is their caution justified?


Notes

There may be some students who suggest there is nothing wrong at all with the system. Indeed, since dim (or dark) lights normally indicate synchronization, would not the presence of two dim lights indicate that perfect synchronization had already been achieved? Discuss the likelihood of this scenario with your students, that two independent alternators could be maintaining perfect synchronization without being coupled together.

In regard to troubleshooting, this scenario has great potential for group discussion. Despite there being a simple, single, probable condition that could cause this problem, there are several possible component failures that could have created the condition. Different students will undoubtedly have different methods of approaching the problem. Let each one share their views, and discuss together what would be the best approach.





Question 115. (Click on arrow for answer)

When AC power is initially applied to an electric motor (before the motor shaft has an opportunity to start moving), the motor “appears” to the AC power source to be a large inductor:

Should be image 01826x01

If the voltage of the 60 Hz AC power source is 480 volts RMS, and the motor initially draws 75 amps RMS when the double-pole single-throw switch closes, how much inductance (L) must the motor windings have? Ignore any wire resistance, and assume the motor’s only opposition to current in a locked-rotor condition is inductive reactance (X_L).

File Num: 01826

Answer

X_L = 16.98 \hbox{ mH}

Notes

In reality, motor winding resistance plays a substantial part in this sort of calculation, but I simplified things a bit just to give students a practical context for their introductory knowledge of inductive reactance.





Question 116. (Click on arrow for answer)

In analyzing circuits with inductors, we often take the luxury of assuming the constituent inductors to be perfect; i.e., purely inductive, with no “stray” properties such as winding resistance or inter-winding capacitance.

Real life is not so generous. With real inductors, we have to consider these factors. One measure often used to express the “purity” of an inductor is its so-called Q rating, or quality factor.

Write the formula for calculating the quality factor (Q) of a coil, and describe some of the operational parameters that may affect this number.

File Num: 01389

Answer

Q_{coil} = {X_L \over R}

Notes

Your students should be able to immediately understand that Q is not a static property of an inductor. Let them explain what makes Q vary, based on their knowledge of inductive reactance.





Question 117. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 60 Hz:

Should be image 01832x01

Show your work using two different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01832

Answer

First strategy:L_{total} = 1.1 \hbox{ H}X_{total} = 414.7 \> \OmegaZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega
Second strategy:X_{L1} = 282.7 \> \Omega Z_{L1} = 282.7 \> \Omega \> \angle \> 90^oX_{L2} = 131.9 \> \Omega Z_{L2} = 131.9 \> \Omega \> \angle \> 90^oZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ impedance phasors geometrically add to equal the total impedance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 118. (Click on arrow for answer)

Calculate the total impedance of this LR circuit, once using nothing but scalar numbers, and again using complex numbers:

Should be image 01837x01

File Num: 01837

Answer

Scalar calculationsR_1 = 1.5 \hbox{ k}\Omega G_{R1} = 666.7 \> \mu \hbox{S}X_{L1} = 2.513 \hbox{ k}\Omega B_{L1} = 397.9 \> \mu \hbox{S}Y_{total} = \sqrt{G^2 + B^2} = 776.4 \> \mu \hbox{S}Z_{total} = {1 \over Y_{total}} = 1.288 \hbox{ k}\Omega
Complex number calculationsR_1 = 1.5 \hbox{ k}\Omega Z_{R1} = 1.5 \hbox{ k}\Omega \> \angle \> 0^oX_{L1} = 2.513 \hbox{ k}\Omega Z_{L1} = 2.513 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = { 1 \over {{1 \over Z_{R1}} + {1 \over Z_{L1}}}} = 1.288 \hbox{ k}\Omega \> \angle \> 30.83^o

Notes

Some electronics textbooks (and courses) tend to emphasize scalar impedance calculations, while others emphasize complex number calculations. While complex number calculations provide more informative results (a phase shift given in every variable!) and exhibit conceptual continuity with DC circuit analysis (same rules, similar formulae), the scalar approach lends itself better to conditions where students do not have access to calculators capable of performing complex number arithmetic. Yes, of course, you can do complex number arithmetic without a powerful calculator, but it’s a lot more tedious and prone to errors than calculating with admittances, susceptances, and conductances (primarily because the phase shift angle is omitted for each of the variables).





Question 119. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 120 Hz:

Should be image 01833x01

Show your work using three different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual admittances first (Y_{L1} and Y_{L2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
  • Using complex numbers: calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01833

Answer

First strategy:L_{total} = 391.3 \hbox{ mH}X_{total} = 295.0 \> \OmegaZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega
Second strategy:Z_{L1} = X_{L1} = 377.0 \> \Omega
Y_{L1} = {1 \over Z_{L1}} = 2.653 \> \hbox{mS}
Z_{L1} = X_{L2} = 1.357 \hbox{ k}\Omega
Y_{L2} = {1 \over Z_{L2}} = 736.8 \> \mu \hbox{S}
Y_{total} = 3.389 \> \hbox{mS}
Z_{total} = {1 \over Y_{total}} = 295 \> \Omega
Third strategy: (using complex numbers)X_{L1} = 377.0 \> \Omega Z_{L1} = 377.0 \> \Omega \> \angle \> 90^oX_{L2} = 1.357 \hbox{ k}\Omega Z_{L2} = 1.357 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ admittance phasors geometrically add to equal the total admittance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 120. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 75^{o}:

Should be image 03282x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03282

Answer

f = 11.342 kHzf = {R \over {2 \pi L \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 121. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of 44^{o}:

Should be image 03283x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03283

Answer

R = 6.826 k\OmegaR = 2 \pi f L \tan \theta

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 122. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -40^{o}:

Should be image 03280x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03280

Answer

f = 2.804 kHzf = - {{R \tan \theta} \over {2 \pi L}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 123. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of -60^{o}:

Should be image 03281x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03281

Answer

R = 2.902 k\OmegaR = - {{2 \pi f L} \over {\tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





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To view a copy of the license, visit https://creativecommons.org/licenses/by/1.0/, or https://creativecommons.org/licenses/by/4.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.

Practice Problems: Inductors in AC Circuits

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about inductors in AC circuits.

Resistance vs. Reactance vs. Impedance

These three terms represent different forms of opposition to electric current. Despite the fact that they are measured in the same unit (ohms: Omega), they are not the same. Resistance is best thought of as electrical friction, whereas reactance is best thought of as electrical inertia. Whereas resistance creates a voltage drop by dissipating energy, reactance creates a voltage drop by storing and releasing energy. Impedance is a term encompassing both resistance and reactance, usually a combination of both.

Phasors, used to represent AC amplitude and phase relations.

A powerful tool used for understanding the operation of AC circuits is the phasor diagram, consisting of arrows pointing in different directions: the length of each arrow representing the amplitude of some AC quantity (voltage, current, or impedance), and the angle of each arrow representing the shift in phase relative to the other arrows. By representing each AC quantity thusly, we may more easily calculate their relationships to one another, with the phasors showing us how to apply trigonometry (Pythagorean Theorem, sine, cosine, and tangent functions) to the various calculations. An analytical parallel to the graphic tool of phasor diagrams is complex numbers, where we represent each phasor (arrow) by a pair of numbers: either a magnitude and angle (polar notation), or by “real” and “imaginary” magnitudes (rectangular notation). Where phasor diagrams are helpful is in applications where their respective AC quantities add: the resultant of two or more phasors stacked tip-to-tail being the mathematical sum of the phasors. Complex numbers, on the other hand, may be added, subtracted, multiplied, and divided; the last two operations being difficult to graphically represent with arrows.

Conductance, Susceptance, and Admittance.

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance (1 \over R). Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B = {1 \over X} and admittance, the reciprocal of impedance Y = {1 \over Z}) are very necessary in order to draw phasor diagrams for parallel networks.

Question 1. (Click on arrow for answer)

Voltage divider circuits may be constructed from reactive components just as easily as they may be constructed from resistors. Take this capacitive voltage divider, for instance:

Should be image 00638x01

Calculate the magnitude and phase shift of V_{out}. Also, describe what advantages a capacitive voltage divider might have over a resistive voltage divider.

File Num: 00638

Answer

V_{out} = 1.754 V \angle 0^{o}

Follow-up question \#1: explain why the division ratio of a capacitive voltage divider remains constant with changes in signal frequency, even though we know that the reactance of the capacitors (X_{C1} and X_{C2}) will change.


Follow-up question \#2: one interesting feature of capacitive voltage dividers is that they harbor the possibility of electric shock after being disconnected from the voltage source, if the source voltage is high enough and if the disconnection happens at just the right time. Explain why a capacitive voltage divider poses this threat whereas a resistive voltage divider does not. Also, identify what the time of disconnection from the AC voltage source has to do with shock hazard.


Notes

Capacitive voltage dividers find use in high-voltage AC instrumentation, due to some of the advantages they exhibit over resistive voltage dividers. Your students should take special note of the phase angle for the capacitor’s voltage drop. Why it is 0 degrees, and not some other angle?





Question 2. (Click on arrow for answer)

A technician needs to know the value of a capacitor, but does not have a capacitance meter nearby. In lieu of this, the technician sets up the following circuit to measure capacitance:

Should be image 02114x01

You happen to walk by this technician’s workbench and ask, “How does this measurement setup work?” The technician responds, “You connect a resistor of known value (R) in series with the capacitor of unknown value (C_x), then adjust the generator frequency until the oscilloscope shows the two voltage drops to be equal, and then you calculate C_x.”

Explain how this system works, in your own words. Also, write the formula you would use to calculate the value of C_x given f and R.

File Num: 02114

Answer

I’ll let you figure out how to explain the operation of this test setup. The formula you would use looks like this:

C_x = {1 \over {2 \pi f R}}

Follow-up question: could you use a similar setup to measure the inductance of an unknown inductor L_x? Why or why not?


Challenge question: astute observers will note that this setup might not work in real life because the ground connection of the oscilloscope is not common with one of the function generator’s leads. Explain why this might be a problem, and suggest a practical solution for it.


Notes

This method of measuring capacitance (or inductance for that matter) is fairly old, and works well if the unknown component has a high Q value.





Question 3. (Click on arrow for answer)

A student measures voltage drops in an AC circuit using three voltmeters and arrives at the following measurements:

Should be image 01566x01

Upon viewing these measurements, the student becomes very perplexed. Aren’t voltage drops supposed to add in series, just as in DC circuits? Why, then, is the total voltage in this circuit only 10.8 volts and not 15.74 volts? How is it possible for the total voltage in an AC circuit to be substantially less than the simple sum of the components’ voltage drops?

Another student, trying to be helpful, suggests that the answer to this question might have something to do with RMS versus peak measurements. A third student disagrees, proposing instead that at least one of the meters is badly out of calibration and thus not reading correctly.

When you are asked for your thoughts on this problem, you realize that neither of the answers proposed thus far are correct. Explain the real reason for the “discrepancy” in voltage measurements, and also explain how you could experimentally disprove the other answers (RMS vs. peak, and bad calibration).

File Num: 01566

Answer

AC voltages still add in series, but phase must also be accounted for when doing so. Unfortunately, multimeters provide no indication of phase whatsoever, and thus do not provide us with all the information we need. (Note: just by looking at this circuit’s components, though, you should still be able to calculate the correct result for total voltage and validate the measurements.)

I’ll let you determine how to disprove the two incorrect explanations offered by the other students!


Challenge question: calculate a set of possible values for the capacitor and resistor that would generate these same voltage drops in a real circuit. Hint: you must also decide on a value of frequency for the power source.


Notes

This question has two different layers: first, how to reconcile the “strange” voltage readings with Kirchhoff’s Voltage Law; and second, how to experimentally validate the accuracy of the voltmeters and the fact that they are all registering the same type of voltage (RMS, peak, or otherwise, it doesn’t matter). The first layer of this question regards the basic concepts of AC phase, while the second exercises troubleshooting and critical thinking skills. Be sure to discuss both of these topics in class with your students.





Question 4. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 01844x01

File Num: 01844

Answer

Z_{total} = \sqrt{R^2 + X^2}

Notes

Ask your students if this equation looks similar to any other mathematical equations they’ve seen before. If not, square both sides of the equation so it looks like Z^2 = R^2 + X^2 and ask them again.





Question 5. (Click on arrow for answer)

Use a triangle to calculate the total voltage of the source for this series RC circuit, given the voltage drop across each component:

Should be image 02107x01

Explain what equation(s) you use to calculate V_{total}, as well as why we must geometrically add these voltages together.

File Num: 02107

Answer

V_{total} = 3.672 volts, as calculated by the Pythagorean Theorem

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 6. (Click on arrow for answer)

Determine the phase angle (\Theta) of the current in this circuit, with respect to the supply voltage:

Should be image 01853x01

File Num: 01853

Answer

\Theta = 26.51^o

Challenge question: explain how the following phasor diagram was determined for this problem:

Should be image 01853x02

Notes

This is an interesting question for a couple of reasons. First, students must determine how they will measure phase shift with just the two voltage indications shown by the meters. This may present a significant challenge for some. Discuss problem-solving strategies in class so that students understand how and why it is possible to determine \Theta.

Secondly, this is an interesting question because it shows how something as abstract as phase angle can be measured with just a voltmeter — no oscilloscope required! Not only that, but we don’t even have to know the component values either! Note that this technique works only for simple circuits.

A practical point to mention here is that multimeters have frequency limits which must be considered when taking measurements on electronic circuits. Some high-quality handheld digital meters have frequency limits of hundred of kilohertz, while others fail to register accurately at only a few thousand hertz. Unless we knew these two digital voltmeters were sufficient for measuring at the signal frequency, their indications would be useless to us.





Question 7. (Click on arrow for answer)

Due to the effects of a changing electric field on the dielectric of a capacitor, some energy is dissipated in capacitors subjected to AC. Generally, this is not very much, but it is there. This dissipative behavior is typically modeled as a series-connected resistance:

Should be image 01847x01

Calculate the magnitude and phase shift of the current through this capacitor, taking into consideration its equivalent series resistance (ESR):

Should be image 01847x02

Compare this against the magnitude and phase shift of the current for an ideal 0.22 \muF capacitor.

File Num: 01847

Answer

I = 3.732206 mA \angle 89.89^{o} for the real capacitor with ESR.
I = 3.732212 mA \angle 90.00^{o} for the ideal capacitor.

Follow-up question \#1: can this ESR be detected by a DC meter check of the capacitor? Why or why not?


Follow-up question \#2: explain how the ESR of a capacitor can lead to physical heating of the component, especially under high-voltage, high-frequency conditions. What safety concerns might arise as a result of this?


Notes

Although capacitors do contain their own parasitic effects, ESR being one of them, they still tend to be much “purer” components than inductors for general use. This is another reason why capacitors are generally favored over inductors in applications where either will suffice.





Question 8. (Click on arrow for answer)

Solve for all voltages and currents in this series RC circuit:

Should be image 01848x01

File Num: 01848

Answer

V_C = 14.39 \hbox{ volts RMS}V_R = 4.248 \hbox{ volts RMS}I = 903.9 \> \mu \hbox{A RMS}

Follow-up question: identify the consequences of a shorted capacitor in this circuit, with regard to circuit current and component voltage drops.


Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 9. (Click on arrow for answer)

Solve for all voltages and currents in this series RC circuit, and also calculate the phase angle of the total impedance:

Should be image 01849x01

File Num: 01849

Answer

V_C = 47.56 \hbox{ volts peak}V_R = 6.508 \hbox{ volts peak}I = 1.972 \hbox{ milliamps peak}\Theta_Z = -82.21^o

Follow-up question: what would we have to do to get these answers in units RMS instead of units “peak”?


Notes

Bring to your students’ attention the fact that total voltage in this circuit is given in “peak” units rather than RMS, and what effect this has on our answers.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 10. (Click on arrow for answer)

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

Should be image 01851x01
  • C_1 = 125 \hbox{ pF}
  • C_2 = 71 \hbox{ pF}
  • R_1 = 6.8 \hbox{ k}\Omega
  • R_2 = 1.2 \hbox{ k}\Omega
  • V_{supply} = 20 \hbox{ V RMS}
  • f_{supply} = 950 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01851

Answer

  • I_{total} = 2.269 \hbox{ mA}
  • V_{C1} = 3.041 \hbox{ V}
  • V_{C2} = 5.354 \hbox{ V}
  • V_{R1} = 15.43 \hbox{ V}
  • V_{R2} = 2.723 \hbox{ V}
  • \Theta = -24.82^o (voltage lagging current)

I suggest using a dual-trace oscilloscope to measure total voltage (across the supply terminals) and voltage drop across resistor R_2. Theoretically, measuring the voltage dropped by either resistor would be fine, but R_2 works better for practical reasons (oscilloscope input lead grounding). Phase shift then could be measured either in the time domain or by a Lissajous figure analysis.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.

It is noteworthy that the low capacitances shown here approach parasitic capacitances between circuit board traces. In other words, whoever designs a circuit to operate at 950 kHz cannot simply place components at will on the board, but must consider the traces themselves to be circuit elements (both capacitive and inductive in nature!). The calculations used to obtain the given answers, of course, assume ideal conditions where the PC board is not considered to possess capacitance or inductance.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 11. (Click on arrow for answer)

Calculate the voltage drops across all components in this circuit, expressing them in complex (polar) form with magnitudes and phase angles each:

Should be image 01852x01

File Num: 01852

Answer

V_{C1} = 0.921 \hbox { V} \> \angle -52.11^oV_{C2} = 0.921 \hbox { V} \> \angle -52.11^oV_{R1} = 1.184 \hbox { V} \> \angle \> 37.90^o

Follow-up question: how much phase shift is there between the capacitors’ voltage drop and the resistor’s voltage drop? Explain why this value is what it is.


Notes

The first challenge of this question is for students to figure out how to reduce this series-parallel combination to something simpler. Fortunately, this is very easy to do if one remembers the properties of parallel capacitances.

Students may be surprised to discover the phase shift between V_C and V_R is the value it is. However, this should not remain a mystery. Discuss this with your class, taking time for all of them to understand why the voltage phasors of a resistor and a capacitor in a simple series circuit will always be orthogonal.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 12. (Click on arrow for answer)

In this circuit, a series resistor-capacitor network creates a phase-shifted voltage for the “gate” terminal of a power-control device known as a TRIAC. All portions of the circuit except for the RC network are “shaded” for de-emphasis:

Should be image 00637x01

Calculate how many degrees of phase shift the capacitor’s voltage is, compared to the total voltage across the series RC network, assuming a frequency of 60 Hz, and a 50\% potentiometer setting.

File Num: 00637

Answer

E_C phase shift = -76.7^{o}

Challenge question: what effect will a change in potentiometer setting have on this phase angle? Specifically, will increasing the resistance make the phase shift approach -90^{o} or approach 0^{o}?


Notes

In this question, I purposely omitted any reference to voltage levels, so the students would have to set up part of the problem themselves. The goal here is to build problem-solving skills.





Question 13. (Click on arrow for answer)

A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = {1 \over R}), and it is measured in the unit of siemens.

Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

Should be image 01845x01

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = {1 \over X}), and the reciprocal of impedance is called admittance (Y = {1 \over Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.

Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Should be image 01845x02

File Num: 01845

Answer

Y_{total} = \sqrt{G^2 + B^2}

Follow-up question \#1: draw a phasor diagram showing how Y, G, and B relate.


Follow-up question \#2: re-write this equation using quantities of resistance (R), reactance (X), and impedance (Z), instead of conductance (G), susceptance (B), and admittance (Y).


Notes

Ask your students if this equation looks familiar to them. It should!


The answer to the challenge question is a matter of algebraic substitution. Work through this process with your students, and then ask them to compare the resulting equation with other equations they’ve seen before. Does its form look familiar to them in any way?





Question 14. (Click on arrow for answer)

Calculate the total impedance offered by these three capacitors to a sinusoidal signal with a frequency of 4 kHz:


  • C_1 = 0.1 \> \mu \hbox{F}
  • C_2 = 0.047 \> \mu \hbox{F}
  • C_3 = 0.033 \> \mu \hbox{F}

Should be image 01846x01

State your answer in the form of a scalar number (not complex), but calculate it using two different strategies:


  • Calculate total capacitance (C_{total}) first, then total impedance (Z_{total}).
  • Calculate individual admittances first (Y_{C1}, Y_{C2}, and Y_{C3}), then total impedance (Z_{total}).

File Num: 01846

Answer

First strategy:C_{total} = 0.18 \> \mu \hbox{F}Z_{total} = 221 \> \Omega
Second strategy:Y_{C1} = 2.51 \hbox{ mS}Y_{C2} = 1.18 \> \hbox{ mS}Y_{C3} = 829 \> \mu \hbox{S}Y_{total} = 4.52 \hbox{ mS}Z_{total} = 221 \> \Omega

Notes

This question is another example of how multiple means of calculation will give you the same answer (if done correctly!). Make note to your students that this indicates an answer-checking strategy!





Question 15. (Click on arrow for answer)

Calculate the total impedance of these parallel-connected components, expressing it in polar form (magnitude and phase angle):

Should be image 02108x01

Also, draw an admittance triangle for this circuit.

File Num: 02108

Answer

Z_{total} = 391.4 \Omega \angle -39.9^{o}Should be image 02108x02

Notes

Some students may wonder why every side of the triangle is represented by a Y term, rather than Y for the hypotenuse, G for the adjacent, and B for the opposite. If students ask about this, remind them that conductance (G) and susceptance (B) are simple two different types of admittances (Y), just as resistance (R) and reactance (X) are simply two different types of impedances (Z).





Question 16. (Click on arrow for answer)

Calculate the total impedances (complete with phase angles) for each of the following capacitor-resistor circuits:

Should be image 02109x01

File Num: 02109

Answer

Should be image 02109x02

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 17. (Click on arrow for answer)

If the source voltage in this circuit is assumed to be the phase reference (that is, the voltage is defined to be at an angle of 0 degrees), determine the relative phase angles of each current in this parallel circuit:

Should be image 02112x01
  • \Theta_{I(R)} =
  • \Theta_{I(C)} =
  • \Theta_{I(total)} =

File Num: 02112

Answer


  • \Theta_{I(R)} = 0^{o}
  • \Theta_{I(C)} = 90^{o}
  • \Theta_{I(total)} = some positive angle between 0^{o} and 90^{o}, exclusive


Notes

Some students will be confused about the positive phase angles, since this is a capacitive circuit and they have learned to associate negative angles with capacitors. It is important for these students to realize, though, that the negative angles they immediately associate with capacitors are in reference to impedance and not necessarily to other variables in the circuit!





Question 18. (Click on arrow for answer)

If the dielectric substance between a capacitor’s plates is not a perfect insulator, there will be a path for direct current (DC) from one plate to the other. This is typically called leakage resistance, and it is modeled as a shunt resistance to an ideal capacitance:

Should be image 01850x01

Calculate the magnitude and phase shift of the current drawn by this real capacitor, if powered by a sinusoidal voltage source of 30 volts RMS at 400 Hz:

Should be image 01850x02

Compare this against the magnitude and phase shift of the current for an ideal capacitor (no leakage).

File Num: 01850

Answer

I = 56.548671 mA \angle 89.98^{o} for the real capacitor with leakage resistance.I = 56.548668 mA \angle 90.00^{o} for the ideal capacitor.

Notes

Discuss with your students the fact that electrolytic capacitors typically have more leakage (less R_{leakage}) than most other capacitor types, due to the thinness of the dielectric oxide layer.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 19. (Click on arrow for answer)

The input impedance of an electrical test instrument is a very important parameter in some applications, because of how the instrument may load the circuit being tested. Oscilloscopes are no different from voltmeters in this regard:

Should be image 02111x01

Typical input impedance for an oscilloscope is 1 M\Omega of resistance, in parallel with a small amount of capacitance. At low frequencies, the reactance of this capacitance is so high that it may be safely ignored. At high frequencies, though, it may become a substantial load to the circuit under test:

Should be image 02111x02

Calculate how many ohms of impedance this oscilloscope input (equivalent circuit shown in the above schematic) will impose on a circuit with a signal frequency of 150 kHz.

File Num: 02111

Answer

Z_{input} = 52.98 k\Omega at 150 kHz

Follow-up question: what are the respective input impedances for ideal voltmeters and ideal ammeters? Explain why each ideal instrument needs to exhibit these impedances in order to accurately measure voltage and current (respectively) with the least “impact” to the circuit under test.


Notes

Mention to your students that this capacitive loading effect only gets worse when a cable is attached to the oscilloscope input. The calculation performed for this question is only for the input of the oscilloscope itself, not including whatever capacitance may be included in the test probe cable!

This is one of the reasons why \times10 probes are used with oscilloscopes: to minimize the loading effect on the tested circuit.





Question 20. (Click on arrow for answer)

Determine the size of capacitor (in Farads) necessary to create a total current of 11.3 mA in this parallel RC circuit:

Should be image 02110x01

File Num: 02110

Answer

C = 562.2 nF

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 21. (Click on arrow for answer)

Explain all the steps necessary to calculate the amount of current in this capacitive AC circuit:

Should be image 01551x01

File Num: 01551

Answer

I = 22.6 mA

Notes

The current is not difficult to calculate, so obviously the most important aspect of this question is not the math. Rather, it is the procedure of calculation: what to do first, second, third, etc., in obtaining the final answer.





Question 22. (Click on arrow for answer)

Calculate the total impedance offered by these two capacitors to a sinusoidal signal with a frequency of 900 Hz:

Should be image 01835x01

Show your work using three different problem-solving strategies:


  • Calculating total capacitance (C_{total}) first, then total impedance (Z_{total}).
  • Calculating individual admittances first (Y_{C1} and Y_{C2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
  • Using complex numbers: calculating individual impedances first (Z_{C1} and Z_{C2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01835

Answer

First strategy:C_{total} = 0.43 \> \mu \hbox{F}X_{total} = 411.3 \> \OmegaZ_{total} = 411.3 \> \Omega \> \angle -90^o or Z_{total} = 0 - j411.3 \> \Omega
Second strategy:Z_{C1} = X_{C1} = 535.9 \> \Omega
Y_{C1} = {1 \over Z_{C1}} = 1.866 \> \hbox{mS}
Z_{C1} = X_{C2} = 1.768 \hbox{ k}\Omega
Y_{C2} = {1 \over Z_{C2}} = 565.5 \> \mu \hbox{S}
Y_{total} = 2.432 \> \hbox{mS}
Z_{total} = {1 \over Y_{total}} = 411.3 \> \Omega
Third strategy: (using complex numbers)X_{C1} = 535.9 \> \Omega Z_{C1} = 535.9 \> \Omega \> \angle -90^oX_{C2} = 1.768 \hbox{ k}\Omega Z_{C1} = 1.768 \hbox{ k}\Omega \> \angle -90^oZ_{total} = 411.3 \> \Omega \> \angle -90^o or Z_{total} = 0 - j411.3 \> \Omega

Notes

A common misconception many students have about capacitive reactances and impedances is that they must interact “oppositely” to how one would normally consider electrical opposition. That is, many students believe capacitive reactances and impedances should add in parallel and diminish in series, because that’s what capacitance (in Farads) does! This is not true, however. Impedances always add in series and diminish in parallel, at least from the perspective of complex numbers. This is one of the reasons I favor AC circuit calculations using complex numbers: because then students may conceptually treat impedance just like they treat DC resistance.

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total capacitance and then calculating impedance from that, or by calculating the impedance of each capacitor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 23. (Click on arrow for answer)

Examine the following circuits, then label the sides of their respective triangles with all the variables that are trigonometrically related in those circuits:

Should be image 03288x01

File Num: 03288

Answer

Should be image 03288x02

Notes

This question asks students to identify those variables in each circuit that vectorially add, discriminating them from those variables which do not add. This is extremely important for students to be able to do if they are to successfully apply “the triangle” to the solution of AC circuit problems.

Note that some of these triangles should be drawn upside-down instead of all the same as they are shown in the question, if we are to properly represent the vertical (imaginary) phasor for capacitive impedance and for inductor admittance. However, the point here is simply to get students to recognize what quantities add and what do not. Attention to the direction (up or down) of the triangle’s opposite side can come later.





Question 24. (Click on arrow for answer)

Draw a phasor diagram showing the trigonometric relationship between resistance, reactance, and impedance in this series circuit:

Should be image 01828x01

Show mathematically how the resistance and reactance combine in series to produce a total impedance (scalar quantities, all). Then, show how to analyze this same circuit using complex numbers: regarding each of the component as having its own impedance, demonstrating mathematically how these impedances add up to comprise the total impedance (in both polar and rectangular forms).

File Num: 01828

Answer

Should be image 01828x02Scalar calculationsR = 2.2 \hbox{ k}\Omega X_C = 2.067 \hbox{ k}\OmegaZ_{series} = \sqrt{R^2 + {X_C}^2}Z_{series} = \sqrt{2200^2 + 2067^2} = 3019 \> \Omega
Complex number calculationsZ_R = 2.2 \hbox{ k}\Omega \> \angle \> 0^o Z_C = 2.067 \hbox{ k}\Omega \> \angle -90^o (Polar form)Z_R = 2.2 \hbox{ k}\Omega + j0 \> \Omega Z_C = 0 \> \Omega - j2.067 \hbox{ k}\Omega (Rectangular form)
Z_{series} = Z_1 + Z_2 + \cdots Z_n (General rule of series impedances)Z_{series} = Z_R + Z_C (Specific application to this circuit)
Z_{series} = 2.2 \hbox{ k}\Omega \> \angle \> 0^o + 2.067 \hbox{ k}\Omega \> \angle -90^o = 3.019 \hbox{ k}\Omega \> \angle -43.2^oZ_{series} = (2.2 \hbox{ k}\Omega + j0 \> \Omega) + (0 \> \Omega - j2.067 \hbox{ k}\Omega) = 2.2 \hbox{ k}\Omega - j2.067 \hbox{ k}\Omega

Notes

I want students to see that there are two different ways of approaching a problem such as this: with scalar math and with complex number math. If students have access to calculators that can do complex-number arithmetic, the “complex” approach is actually simpler for series-parallel combination circuits, and it yields richer (more informative) results.

Ask your students to determine which of the approaches most resembles DC circuit calculations. Incidentally, this is why I tend to prefer complex-number AC circuit calculations over scalar calculations: because of the conceptual continuity between AC and DC. When you use complex numbers to represent AC voltages, currents, and impedances, almost all the rules of DC circuits still apply. The big exception, of course, is calculations involving power.





Question 25. (Click on arrow for answer)

Calculate the total impedance of this RC circuit, once using nothing but scalar numbers, and again using complex numbers:

Should be image 01838x01

File Num: 01838

Answer

Scalar calculationsR_1 = 7.9 \hbox{ k}\Omega G_{R1} = 126.6 \> \mu \hbox{S}X_{C1} = 8.466 \hbox{ k}\Omega B_{C1} = 118.1 \> \mu \hbox{S}Y_{total} = \sqrt{G^2 + B^2} = 173.1 \> \mu \hbox{S}Z_{total} = {1 \over Y_{total}} = 5.776 \hbox{ k}\Omega
Complex number calculationsR_1 = 7.9 \hbox{ k}\Omega Z_{R1} = 7.9 \hbox{ k}\Omega \> \angle \> 0^oX_{C1} = 8.466 \hbox{ k}\Omega Z_{C1} = 8.466 \hbox{ k}\Omega \> \angle -90^oZ_{total} = { 1 \over {{1 \over Z_{R1}} + {1 \over Z_{C1}}}} = 5.776 \hbox{ k}\Omega \> \angle -43.02^o

Notes

Some electronics textbooks (and courses) tend to emphasize scalar impedance calculations, while others emphasize complex number calculations. While complex number calculations provide more informative results (a phase shift given in every variable!) and exhibit conceptual continuity with DC circuit analysis (same rules, similar formulae), the scalar approach lends itself better to conditions where students do not have access to calculators capable of performing complex number arithmetic. Yes, of course, you can do complex number arithmetic without a powerful calculator, but it’s a lot more tedious and prone to errors than calculating with admittances, susceptances, and conductances (primarily because the phase shift angle is omitted for each of the variables).





Question 26. (Click on arrow for answer)

A student is asked to calculate the phase shift for the following circuit’s output voltage, relative to the phase of the source voltage:

Should be image 03748x01

He recognizes this as a series circuit, and therefore realizes that a right triangle would be appropriate for representing component impedances and component voltage drops (because both impedance and voltage are quantities that add in series, and the triangle represents phasor addition):

Should be image 03748x02

The problem now is, which angle does the student solve for in order to find the phase shift of V_{out}? The triangle contains two angles besides the 90^{o} angle, \Theta and \Phi. Which one represents the output phase shift, and more importantly, why?

File Num: 03748

Answer

The proper angle in this circuit is \Theta, and it will be a positive (leading) quantity.


Notes

Too many students blindly use impedance and voltage triangles without really understand what they are and why they work. These same students will have no idea how to approach a problem like this. Work with them to help them understand!





Question 27. (Click on arrow for answer)

Calculate the output voltage of this phase-shifting circuit, expressing it in polar form (magnitude and phase angle relative to the source voltage):

Should be image 02621x01

File Num: 02621

Answer

V_{out} = 2.593 V \angle 61.3^{o}

Notes

This is a very practical application of resistor-capacitor (RC) circuits: to introduce a phase shift to an AC signal. Examples of where a circuit such as this may be used include oscillators (to introduce phase shift into a feedback network for a total phase shift of 360^{o}) and thyristor firing control circuits (phase-shifting the triggering voltage in relation to the source voltage).





Question 28. (Click on arrow for answer)

Calculate the output voltage of this phase-shifting circuit, expressing it in polar form (magnitude and phase angle relative to the source voltage):

Should be image 02620x01

File Num: 02620

Answer

V_{out} = 6.7 V \angle -47.9^{o}

Notes

This is a very practical application of resistor-capacitor (RC) circuits: to introduce a phase shift to an AC signal. Examples of where a circuit such as this may be used include oscillators (to introduce phase shift into a feedback network for a total phase shift of 360^{o}) and thyristor firing control circuits (phase-shifting the triggering voltage in relation to the source voltage).





Question 29. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 70^{o}:

Should be image 02623x01

File Num: 02623

Answer

f = 798 Hz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 30. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 40^{o}:

Should be image 02622x01

File Num: 02622

Answer

f = 6.54 kHz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 31. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -38^{o}:

Should be image 02626x01

File Num: 02626

Answer

f = 465 Hz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 32. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -25^{o}:

Should be image 02625x01

File Num: 02625

Answer

f = 929 Hz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 33. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 25^{o}:

Should be image 03284x01

Also, write an equation that solves for frequency (f), given all the other variables (R, C, and phase angle \theta).

File Num: 03284

Answer

f = 2.143 kHzf = {1 \over {2 \pi R C \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 34. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of 58^{o}:

Should be image 03285x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, C, and phase angle \theta).

File Num: 03285

Answer

R = 669.7 \OmegaR = {1 \over {2 \pi f C \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 35. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of -64^{o}:

Should be image 03287x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, C, and phase angle \theta).

File Num: 03287

Answer

R = 16.734 k\OmegaR = -{{\tan \theta} \over {2 \pi f C}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 36. (Click on arrow for answer)

Use algebraic substitution to generate an equation expressing the output voltage of the following circuit given the input voltage, the input frequency, the capacitor value, and the resistor value:

Should be image 03818x01
V_{out} =

File Num: 03818

Answer

V_{out} = {{R \> V_{in}} \over \sqrt{\left({1 \over 2 \pi f C}\right)^2 + R^2}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 37. (Click on arrow for answer)

Use algebraic substitution to generate an equation expressing the output voltage of the following circuit given the input voltage, the input frequency, the capacitor value, and the resistor value:

Should be image 03819x01
V_{out} =

File Num: 03819

Answer

V_{out} = {{R \> V_{in}} \over \sqrt{\left({{C_1 + C_2} \over 2 \pi f C_1 C_2}\right)^2 + R^2}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 38. (Click on arrow for answer)

Complete the table of values for this circuit, representing all quantities in complex-number form (either polar or rectangular, your choice):

Should be image 03611x01

File Num: 03611

Answer

Should be image 03611x02

Notes

Ask your students to share their problem-solving techniques for this question: how they solved for each parameter and in what order they performed the calculations.





Question 39. (Click on arrow for answer)

This phase-shifting bridge circuit is supposed to provide an output voltage with a variable phase shift from -45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Should be image 03464x01

Suppose, though, that the output signal is stuck at +45^{o} leading the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03464

Answer

A broken connection between the left-hand terminal of the potentiometer and the bridge could cause this to happen:

Should be image 03464x02

I’ll let you figure out why!


Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is much greater than the fixed resistance value.”





Question 40. (Click on arrow for answer)

This phase-shifting bridge circuit is supposed to provide an output voltage with a variable phase shift from -45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Should be image 03465x01

Suppose, though, that the output signal is stuck at -45^{o} lagging the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03465

Answer

A broken connection between the right-hand terminal of the potentiometer and the bridge could cause this to happen:

Should be image 03465x02

I’ll let you figure out why!


Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is much greater than the fixed resistance value.”





Question 41. (Click on arrow for answer)

This phase-shifting bridge circuit is supposed to provide an output voltage with a variable phase shift from -45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Should be image 03466x01

Suppose, though, that the output signal registers as it should with the potentiometer wiper fully to the right, but diminishes greatly in amplitude as the wiper is moved to the left, until there is practically zero output voltage at the full-left position. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03466

Answer

An open failure of the fixed resistor in the upper-left arm of the bridge could cause this to happen:

Should be image 03466x02

Follow-up question: identify another possible component failure that would exhibit the same symptoms.


Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is much greater than the fixed resistance value.”





Question 42. (Click on arrow for answer)

Sketch the approximate waveform of this circuit’s output signal (V_{out}) on the screen of the oscilloscope:

Should be image 03503x01

Hint: use the Superposition Theorem!


File Num: 03503

Answer

Should be image 03503x02

Follow-up question: explain why it is acceptable to use a polarized (polarity-sensitive) capacitor in this circuit when it is clearly connected to a source of AC. Why is it not damaged by the AC voltage when used like this?


Notes

Note that the capacitor size has been chosen for negligible capacitive reactance (X_C) at the specific frequency, such that the 10 k\Omega DC bias resistors present negligible loading to the coupled AC signal. This is typical for this type of biasing circuit.

Aside from giving students an excuse to apply the Superposition Theorem, this question previews a circuit topology that is extremely common in transistor amplifiers.





Question 43. (Click on arrow for answer)

Audio headphones make highly sensitive voltage detectors for AC signals in the audio frequency range. However, the small speakers inside headphones are quite easily damaged by the application of DC voltage.

Explain how a capacitor could be used as a “filtering” device to allow AC signals through to a pair of headphones, yet block any applied DC voltage, so as to help prevent accidental damage of the headphones while using them as an electrical instrument.

The key to understanding how to answer this question is to recognize what a capacitor “appears as” to AC signals versus DC signals.

File Num: 01395

Answer

Connect a capacitor in series with the headphone speakers.


Notes

I highly recommend to students that they should build a transformer-isolation circuit if they intend to use a pair of audio headphones as a test device (see question file number 00983 for a complete schematic diagram).





Question 44. (Click on arrow for answer)

Suppose a friend wanted to install filter networks in the “woofer” section of their stereo system, to prevent high-frequency power from being wasted in speakers incapable of reproducing those frequencies. To this end, your friend installs the following resistor-capacitor networks:

Should be image 00614x01

After examining this schematic, you see that your friend has the right idea in mind, but implemented it incorrectly. These filter circuits would indeed block high-frequency signals from getting to the woofers, but they would not actually accomplish the stated goal of minimizing wasted power.

What would you recommend to your friend in lieu of this circuit design?

File Num: 00614

Answer

Rather than use a “shunting” form of low-pass filter (resistor and capacitor), a “blocking” form of low-pass filter (inductor) should be used instead.


Notes

The reason for this choice in filter designs is very practical. Ask your students to describe how a “shunting” form of filter works, where the reactive component is connected in parallel with the load, receiving power through a series resistor. Contrast this against a “blocking” form of filter circuit, in which a reactive component is connected in series with the load. In one form of filter, a resistor is necessary. In the other form of filter, a resistor is not necessary. What difference does this make in terms of power dissipation within the filter circuit?





Question 45. (Click on arrow for answer)

It is common in audio systems to connect a capacitor in series with each “tweeter” (high-frequency) speaker to act as a simple high-pass filter. The choice of capacitors for this task is important in a high-power audio system.

A friend of mine once had such an arrangement for the tweeter speakers in his car. Unfortunately, though, the capacitors kept blowing up when he operated the stereo at full volume! Tired of replacing these non-polarized electrolytic capacitors, he came to me for advice. I suggested he use mylar or polystyrene capacitors instead of electrolytics. These were a bit more expensive than electrolytic capacitors, but they did not blow up. Explain why.

File Num: 03467

Answer

The issue here was not polarity (AC versus DC), because these were non-polarized electrolytic capacitors which were blowing up. What was an issue was ESR (Equivalent Series Resistance), which electrolytic capacitors are known to have high values of.


Notes

Your students may have to do a bit of refreshing (or first-time research!) on the meaning of ESR before they can understand why large ESR values could cause a capacitor to explode under extreme operating conditions.





All files with file num less than 4100 are Copyright 2003, Tony R. Kuphaldt, released under the Creative Commons Attribution License (v 1.0). All other files are Copyright 2022, David Williams, released under the Creative Commons Attribution License (V 4.0) This means you may do almost anything with this work, so long as you give proper credit.

To view a copy of the license, visit https://creativecommons.org/licenses/by/1.0/, or https://creativecommons.org/licenses/by/4.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.

Practice Problems: Inductors in AC Circuits

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about inductors in AC circuits.

Resistance vs. Reactance vs. Impedance

These three terms represent different forms of opposition to electric current. Despite the fact that they are measured in the same unit (ohms: Omega), they are not the same. Resistance is best thought of as electrical friction, whereas reactance is best thought of as electrical inertia. Whereas resistance creates a voltage drop by dissipating energy, reactance creates a voltage drop by storing and releasing energy. Impedance is a term encompassing both resistance and reactance, usually a combination of both.

Phasors, used to represent AC amplitude and phase relations.

A powerful tool used for understanding the operation of AC circuits is the phasor diagram, consisting of arrows pointing in different directions: the length of each arrow representing the amplitude of some AC quantity (voltage, current, or impedance), and the angle of each arrow representing the shift in phase relative to the other arrows. By representing each AC quantity thusly, we may more easily calculate their relationships to one another, with the phasors showing us how to apply trigonometry (Pythagorean Theorem, sine, cosine, and tangent functions) to the various calculations. An analytical parallel to the graphic tool of phasor diagrams is complex numbers, where we represent each phasor (arrow) by a pair of numbers: either a magnitude and angle (polar notation), or by “real” and “imaginary” magnitudes (rectangular notation). Where phasor diagrams are helpful is in applications where their respective AC quantities add: the resultant of two or more phasors stacked tip-to-tail being the mathematical sum of the phasors. Complex numbers, on the other hand, may be added, subtracted, multiplied, and divided; the last two operations being difficult to graphically represent with arrows.

Conductance, Susceptance, and Admittance.

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance (1 \over R). Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B = {1 \over X} and admittance, the reciprocal of impedance Y = {1 \over Z}) are very necessary in order to draw phasor diagrams for parallel networks.

Question 1. (Click on arrow for answer)

As a general rule, inductors oppose change in (choose: voltage or current), and they do so by . . . (complete the sentence).


Based on this rule, determine how an inductor would react to a constant AC current that increases in frequency. Would an inductor drop more or less voltage, given a greater frequency? Explain your answer.

File Num: 00578

Answer

As a general rule, inductors oppose change in current, and they do so by producing a voltage.


An inductor will drop a greater amount of AC voltage, given the same AC current, at a greater frequency.


Notes

This question is an exercise in qualitative thinking: relating rates of change to other variables, without the use of numerical quantities. The general rule stated here is very, very important for students to master, and be able to apply to a variety of circumstances. If they learn nothing about inductors except for this rule, they will be able to grasp the function of a great many inductor circuits.





Question 2. (Click on arrow for answer)

\int f(x) dx Calculus alert!

We know that the formula relating instantaneous voltage and current in an inductor is this:

e = L{di \over dt}

Knowing this, determine at what points on this sine wave plot for inductor current is the inductor voltage equal to zero, and where the voltage is at its positive and negative peaks. Then, connect these points to draw the waveform for inductor voltage:

Should be image 00576x01

How much phase shift (in degrees) is there between the voltage and current waveforms? Which waveform is leading and which waveform is lagging?

File Num: 00576

Answer

Should be image 00576x02

For an inductor, voltage is leading and current is lagging, by a phase shift of 90^{o}.


Notes

This question is an excellent application of the calculus concept of the derivative: relating one function (instantaneous voltage, e) with the instantaneous rate-of-change of another function (current, di \over dt).





Question 3. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 60 Hz:

Should be image 01832x01

Show your work using two different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01832

Answer

First strategy:L_{total} = 1.1 \hbox{ H}X_{total} = 414.7 \> \OmegaZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega
Second strategy:X_{L1} = 282.7 \> \Omega Z_{L1} = 282.7 \> \Omega \> \angle \> 90^oX_{L2} = 131.9 \> \Omega Z_{L2} = 131.9 \> \Omega \> \angle \> 90^oZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ impedance phasors geometrically add to equal the total impedance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 4. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 01844x01

File Num: 01844

Answer

Z_{total} = \sqrt{R^2 + X^2}

Notes

Ask your students if this equation looks similar to any other mathematical equations they’ve seen before. If not, square both sides of the equation so it looks like Z^2 = R^2 + X^2 and ask them again.





Question 5. (Click on arrow for answer)

Calculate the total impedance of this LR circuit, once using nothing but scalar numbers, and again using complex numbers:

Should be image 01837x01

File Num: 01837

Answer

Scalar calculationsR_1 = 1.5 \hbox{ k}\Omega G_{R1} = 666.7 \> \mu \hbox{S}X_{L1} = 2.513 \hbox{ k}\Omega B_{L1} = 397.9 \> \mu \hbox{S}Y_{total} = \sqrt{G^2 + B^2} = 776.4 \> \mu \hbox{S}Z_{total} = {1 \over Y_{total}} = 1.288 \hbox{ k}\Omega
Complex number calculationsR_1 = 1.5 \hbox{ k}\Omega Z_{R1} = 1.5 \hbox{ k}\Omega \> \angle \> 0^oX_{L1} = 2.513 \hbox{ k}\Omega Z_{L1} = 2.513 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = { 1 \over {{1 \over Z_{R1}} + {1 \over Z_{L1}}}} = 1.288 \hbox{ k}\Omega \> \angle \> 30.83^o

Notes

Some electronics textbooks (and courses) tend to emphasize scalar impedance calculations, while others emphasize complex number calculations. While complex number calculations provide more informative results (a phase shift given in every variable!) and exhibit conceptual continuity with DC circuit analysis (same rules, similar formulae), the scalar approach lends itself better to conditions where students do not have access to calculators capable of performing complex number arithmetic. Yes, of course, you can do complex number arithmetic without a powerful calculator, but it’s a lot more tedious and prone to errors than calculating with admittances, susceptances, and conductances (primarily because the phase shift angle is omitted for each of the variables).





Question 6. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 120 Hz:

Should be image 01833x01

Show your work using three different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual admittances first (Y_{L1} and Y_{L2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
  • Using complex numbers: calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01833

Answer

First strategy:L_{total} = 391.3 \hbox{ mH}X_{total} = 295.0 \> \OmegaZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega
Second strategy:Z_{L1} = X_{L1} = 377.0 \> \Omega
Y_{L1} = {1 \over Z_{L1}} = 2.653 \> \hbox{mS}
Z_{L1} = X_{L2} = 1.357 \hbox{ k}\Omega
Y_{L2} = {1 \over Z_{L2}} = 736.8 \> \mu \hbox{S}
Y_{total} = 3.389 \> \hbox{mS}
Z_{total} = {1 \over Y_{total}} = 295 \> \Omega
Third strategy: (using complex numbers)X_{L1} = 377.0 \> \Omega Z_{L1} = 377.0 \> \Omega \> \angle \> 90^oX_{L2} = 1.357 \hbox{ k}\Omega Z_{L2} = 1.357 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ admittance phasors geometrically add to equal the total admittance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 7. (Click on arrow for answer)

Does an inductor’s opposition to alternating current increase or decrease as the frequency of that current increases? Also, explain why we refer to this opposition of AC current in an inductor as reactance instead of resistance.

File Num: 00580

Answer

The opposition to AC current (“reactance”) of an inductor increases as frequency increases. We refer to this opposition as “reactance” rather than “resistance” because it is non-dissipative in nature. In other words, reactance causes no power to leave the circuit.


Notes

Ask your students to define the relationship between inductor reactance and frequency as either “directly proportional” or “inversely proportional”. These are two phrases used often in science and engineering to describe whether one quantity increases or decreases as another quantity increases. Your students definitely need to be familiar with both these phrases, and be able to interpret and use them in their technical discussions.

Also, discuss the meaning of the word “non-dissipative” in this context. How could we prove that the opposition to current expressed by an inductor is non-dissipative? What would be the ultimate test of this?





Question 8. (Click on arrow for answer)

What will happen to the brightness of the light bulb as the iron core is moved away from the wire coil in this circuit? Explain why this happens.

Should be image 00095x01

File Num: 00095

Answer

The light bulb will glow brighter when the iron core is moved away from the wire coil, due to the change in inductive reactance (X_{L}).


Follow-up question: what circuit failure(s) could cause the light bulb to glow brighter than it should?


Notes

One direction you might want to lead your students in with this question is how AC power may be controlled using this principle. Controlling AC power with a variable reactance has a definite advantage over controlling AC power with a variable resistance: less wasted energy in the form of heat.





Question 9. (Click on arrow for answer)

An inductor rated at 4 Henrys is subjected to a sinusoidal AC voltage of 24 volts RMS, at a frequency of 60 hertz. Write the formula for calculating inductive reactance (X_L), and solve for current through the inductor.

File Num: 00582

Answer

X_L = 2 \pi f L

The current through this inductor is 15.92 mA RMS.


Notes

I have consistently found that qualitative (greater than, less than, or equal) analysis is much more difficult for students to perform than quantitative (punch the numbers on a calculator) analysis. Yet, I have consistently found on the job that people lacking qualitative skills make more “silly” quantitative errors because they cannot validate their calculations by estimation.

In light of this, I always challenge my students to qualitatively analyze formulae when they are first introduced to them. Ask your students to identify what will happen to one term of an equation if another term were to either increase, or decrease (you choose the direction of change). Use up and down arrow symbols if necessary to communicate these changes graphically. Your students will greatly benefit in their conceptual understanding of applied mathematics from this kind of practice!





Question 10. (Click on arrow for answer)

At what frequency does a 350 mH inductor have 4.7 k\Omega of reactance? Write the formula for solving this, in addition to calculating the frequency.

File Num: 00586

Answer

f = 2.137 kHz

Notes

Be sure to ask your students to demonstrate the algebraic manipulation of the original formula, in providing the answer to this question. Algebraic manipulation of equations is a very important skill to have, and it comes only by study and practice.





Question 11. (Click on arrow for answer)

How much inductance would an inductor have to possess in order to provide 540 \Omega of reactance at a frequency of 400 Hz? Write the formula for solving this, in addition to calculating the frequency.

File Num: 03277

Answer

L = 214.9 mH

Notes

Be sure to ask your students to demonstrate the algebraic manipulation of the original formula, in providing the answer to this question. Algebraic manipulation of equations is a very important skill to have, and it comes only by study and practice.





Question 12. (Click on arrow for answer)

Explain all the steps necessary to calculate the amount of current in this inductive AC circuit:

Should be image 01552x01

File Num: 01552

Answer

I = 15.6 mA

Notes

The current is not difficult to calculate, so obviously the most important aspect of this question is not the math. Rather, it is the procedure of calculation: what to do first, second, third, etc., in obtaining the final answer.





Question 13. (Click on arrow for answer)

In this AC circuit, the resistor offers 300 \Omega of resistance, and the inductor offers 400 \Omega of reactance. Together, their series opposition to alternating current results in a current of 10 mA from the 5 volt source:

Should be image 00584x01

How many ohms of opposition does the series combination of resistor and inductor offer? What name do we give to this quantity, and how do we symbolize it, being that it is composed of both resistance (R) and reactance (X)?

File Num: 00584

Answer

Z_{total} = 500 \Omega.

Follow-up question: suppose that the inductor suffers a failure in its wire winding, causing it to “open.” Explain what effect this would have on circuit current and voltage drops.


Notes

Students may experience difficulty arriving at the same quantity for impedance shown in the answer. If this is the case, help them problem-solve by suggesting they simplify the problem: short past one of the load components and calculate the new circuit current. Soon they will understand the relationship between total circuit opposition and total circuit current, and be able to apply this concept to the original problem.

Ask your students why the quantities of 300 \Omega and 400 \Omega do not add up to 700 \Omega like they would if they were both resistors. Does this scenario remind them of another mathematical problem where 3 + 4 = 5? Where have we seen this before, especially in the context of electric circuits?

Once your students make the cognitive connection to trigonometry, ask them the significance of these numbers’ addition. Is it enough that we say a component has an opposition to AC of 400 \Omega, or is there more to this quantity than a single, scalar value? What type of number would be suitable for representing such a quantity, and how might it be written?





Question 14. (Click on arrow for answer)

While studying DC circuit theory, you learned that resistance was an expression of a component’s opposition to electric current. Then, when studying AC circuit theory, you learned that reactance was another type of opposition to current. Now, a third term is introduced: impedance. Like resistance and reactance, impedance is also a form of opposition to electric current.

Explain the difference between these three quantities (resistance, reactance, and impedance) using your own words.

File Num: 01567

Answer

The fundamental distinction between these terms is one of abstraction: impedance is the most general term, encompassing both resistance and reactance. Here is an explanation given in terms of logical sets (using a Venn diagram), along with an analogy from animal taxonomy:

Should be image 01567x01

Resistance is a type of impedance, and so is reactance. The difference between the two has to do with energy exchange.


Notes

The given answer is far from complete. I’ve shown the semantic relationship between the terms resistance, reactance, and impedance, but I have only hinted at the conceptual distinctions between them. Be sure to discuss with your students what the fundamental difference is between resistance and reactance, in terms of electrical energy exchange.





Question 15. (Click on arrow for answer)

In DC circuits, we have Ohm’s Law to relate voltage, current, and resistance together:

E = I R

In AC circuits, we similarly need a formula to relate voltage, current, and impedance together. Write three equations, one solving for each of these three variables: a set of Ohm’s Law formulae for AC circuits. Be prepared to show how you may use algebra to manipulate one of these equations into the other two forms.

File Num: 00590

Answer

E = I ZI = {E \over Z}Z = {E \over I}

If using phasor quantities (complex numbers) for voltage, current, and impedance, the proper way to write these equations is as follows:

E = IZI = {E \over Z}Z = {E \over I}

Bold-faced type is a common way of denoting vector quantities in mathematics.


Notes

Although the use of phasor quantities for voltage, current, and impedance in the AC form of Ohm’s Law yields certain distinct advantages over scalar calculations, this does not mean one cannot use scalar quantities. Often it is appropriate to express an AC voltage, current, or impedance as a simple scalar number.





Question 16. (Click on arrow for answer)

It is often necessary to represent AC circuit quantities as complex numbers rather than as scalar numbers, because both magnitude and phase angle are necessary to consider in certain calculations.

When representing AC voltages and currents in polar form, the angle given refers to the phase shift between the given voltage or current, and a “reference” voltage or current at the same frequency somewhere else in the circuit. So, a voltage of 3.5 \hbox{ V} \angle -45^o means a voltage of 3.5 volts magnitude, phase-shifted 45 degrees behind (lagging) the reference voltage (or current), which is defined to be at an angle of 0 degrees.

But what about impedance (Z)? Does impedance have a phase angle, too, or is it a simple scalar number like resistance or reactance?


Calculate the amount of current that would go through a 100 mH inductor with 36 volts RMS applied to it at a frequency of 400 Hz. Then, based on Ohm’s Law for AC circuits and what you know of the phase relationship between voltage and current for an inductor, calculate the impedance of this inductor in polar form. Does a definite angle emerge from this calculation for the inductor’s impedance? Explain why or why not.

File Num: 00588

Answer

Z_L = 251.33 \Omega \angle 90^{o}

Notes

This is a challenging question, because it asks the student to defend the application of phase angles to a type of quantity that does not really possess a wave-shape like AC voltages and currents do. Conceptually, this is difficult to grasp. However, the answer is quite clear through the Ohm’s Law calculation (Z = {E \over I}).

Although it is natural to assign a phase angle of 0^{o} to the 36 volt supply, making it the reference waveform, this is not actually necessary. Work through this calculation with your students, assuming different angles for the voltage in each instance. You should find that the impedance computes to be the same exact quantity every time.





Question 17. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 75^{o}:

Should be image 03282x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03282

Answer

f = 11.342 kHzf = {R \over {2 \pi L \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 18. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of 44^{o}:

Should be image 03283x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03283

Answer

R = 6.826 k\OmegaR = 2 \pi f L \tan \theta

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 19. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -40^{o}:

Should be image 03280x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03280

Answer

f = 2.804 kHzf = - {{R \tan \theta} \over {2 \pi L}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 20. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of -60^{o}:

Should be image 03281x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03281

Answer

R = 2.902 k\OmegaR = - {{2 \pi f L} \over {\tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 21. (Click on arrow for answer)

If a sinusoidal voltage is applied to an impedance with a phase angle of 0^{o}, the resulting voltage and current waveforms will look like this:

Should be image 00631x01

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit.

File Num: 00631

Answer

Should be image 00631x02

Notes

Ask your students to observe the waveform shown in the answer closely, and determine what sign the power values always are. Note how the voltage and current waveforms alternate between positive and negative, but power does not. Of what significance is this to us? What does this indicate about the nature of a load with an impedance phase angle of 0^{o}?





Question 22. (Click on arrow for answer)

If a sinusoidal voltage is applied to an impedance with a phase angle of 90^{o}, the resulting voltage and current waveforms will look like this:

Should be image 00632x01

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit. Also, explain how the mnemonic phrase “ELI the ICE man” applies to these waveforms.

File Num: 00632

Answer

Should be image 00632x02

The mnemonic phrase, “ELI the ICE man” indicates that this phase shift is due to an inductance rather than a capacitance.


Notes

Ask your students to observe the waveform shown in the answer closely, and determine what sign the power values are. Note how the power waveform alternates between positive and negative values, just as the voltage and current waveforms do. Ask your students to explain what negative power could possibly mean.

Of what significance is this to us? What does this indicate about the nature of a load with an impedance phase angle of 90^{o}?


The phrase, “ELI the ICE man” has been used be generations of technicians to remember the phase relationships between voltage and current for inductors and capacitors, respectively. One area of trouble I’ve noted with students, though, is being able to interpret which waveform is leading and which one is lagging, from a time-domain plot such as this.





Question 23. (Click on arrow for answer)

The impedance triangle is often used to graphically relate Z, R, and X in a series circuit:

Should be image 02076x01

Unfortunately, many students do not grasp the significance of this triangle, but rather memorize it as a “trick” used to calculate one of the three variables given the other two. Explain why a right triangle is an appropriate form to relate these variables, and what each side of the triangle actually represents.

File Num: 02076

Answer

Each side of the impedance triangle is actually a phasor (a vector representing impedance with magnitude and direction):

Should be image 02076x02

Since the phasor for resistive impedance (Z_R) has an angle of zero degrees and the phasor for reactive impedance (Z_C or Z_L) either has an angle of +90 or -90 degrees, the phasor sum representing total series impedance will form the hypotenuse of a right triangle when the first to phasors are added (tip-to-tail).


Follow-up question: as a review, explain why resistive impedance phasors always have an angle of zero degrees, and why reactive impedance phasors always have angles of either +90 degrees or -90 degrees.


Notes

The question is sufficiently open-ended that many students may not realize exactly what is being asked until they read the answer. This is okay, as it is difficult to phrase the question in a more specific manner without giving away the answer!





Question 24. (Click on arrow for answer)

Use the “impedance triangle” to calculate the impedance of this series combination of resistance (R) and inductive reactance (X):

Should be image 02081x01

Explain what equation(s) you use to calculate Z.

File Num: 02081

Answer

Z = 625 \Omega, as calculated by the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 25. (Click on arrow for answer)

Use the “impedance triangle” to calculate the necessary reactance of this series combination of resistance (R) and inductive reactance (X) to produce the desired total impedance of 145 \Omega:

Should be image 02083x01

Explain what equation(s) you use to calculate X, and the algebra necessary to achieve this result from a more common formula.

File Num: 02083

Answer

X = 105 \Omega, as calculated by an algebraically manipulated version of the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 26. (Click on arrow for answer)

Should be image 02084x01

Identify which trigonometric functions (sine, cosine, or tangent) are represented by each of the following ratios, with reference to the angle labeled with the Greek letter “Theta” (\Theta):

{X \over R} = {X \over Z} = {R \over Z} =

File Num: 02084

Answer

Should be image 02084x01{X \over R} = \tan \Theta = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \sin \Theta = {\hbox{Opposite} \over \hbox{Hypotenuse}}{R \over Z} = \cos \Theta = {\hbox{Adjacent} \over \hbox{Hypotenuse}}

Notes

Ask your students to explain what the words “hypotenuse”, “opposite”, and “adjacent” refer to in a right triangle.





Question 27. (Click on arrow for answer)

Trigonometric functions such as sine, cosine, and tangent are useful for determining the ratio of right-triangle side lengths given the value of an angle. However, they are not very useful for doing the reverse: calculating an angle given the lengths of two sides.

Should be image 02086x01

Suppose we wished to know the value of angle \Theta, and we happened to know the values of Z and R in this impedance triangle. We could write the following equation, but in its present form we could not solve for \Theta:

\cos \Theta = {R \over Z}

The only way we can algebraically isolate the angle \Theta in this equation is if we have some way to “undo” the cosine function. Once we know what function will “undo” cosine, we can apply it to both sides of the equation and have \Theta by itself on the left-hand side.

There is a class of trigonometric functions known as inverse or “arc” functions which will do just that: “undo” a regular trigonometric function so as to leave the angle by itself. Explain how we could apply an “arc-function” to the equation shown above to isolate \Theta.

File Num: 02086

Answer

\cos \Theta = {R \over Z} \hbox{ Original equation}\hbox<i>. . . applying the "arc-cosine" function to both sides . . .</i>\arccos \left( \cos \Theta \right) = \arccos \left( {R \over Z} \right)\Theta = \arccos \left( {R \over Z} \right)

Notes

I like to show the purpose of trigonometric arcfunctions in this manner, using the cardinal rule of algebraic manipulation (do the same thing to both sides of an equation) that students are familiar with by now. This helps eliminate the mystery of arcfunctions for students new to trigonometry.





Question 28. (Click on arrow for answer)

A series AC circuit contains 1125 ohms of resistance and 1500 ohms of reactance for a total circuit impedance of 1875 ohms. This may be represented graphically in the form of an impedance triangle:

Should be image 02085x01

Since all side lengths on this triangle are known, there is no need to apply the Pythagorean Theorem. However, we may still calculate the two non-perpendicular angles in this triangle using “inverse” trigonometric functions, which are sometimes called arcfunctions.

Identify which arc-function should be used to calculate the angle \Theta given the following pairs of sides:

R \hbox{ and } ZX \hbox{ and } RX \hbox{ and } Z

Show how three different trigonometric arcfunctions may be used to calculate the same angle \Theta.

File Num: 02085

Answer

\arccos {R \over Z} = 53.13^o\arctan {X \over R} = 53.13^o\arcsin {X \over Z} = 53.13^o

Challenge question: identify three more arcfunctions which could be used to calculate the same angle \Theta.


Notes

Some hand calculators identify arc-trig functions by the letter “A” prepending each trigonometric abbreviation (e.g. “ASIN” or “ATAN”). Other hand calculators use the inverse function notation of a -1 exponent, which is not actually an exponent at all (e.g. \sin^{-1} or \tan^{-1}). Be sure to discuss function notation on your students’ calculators, so they know what to invoke when solving problems such as this.





Question 29. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 00850x01

File Num: 00850

Answer

Z_{total} = \sqrt{R^2 + X^2}

Follow-up question: algebraically manipulate this equation to produce two more; one solving for R and the other solving for X.


Notes

Ask your students if this equation looks similar to any other mathematical equations they’ve seen before. If not, square both sides of the equation so it looks like Z^2 = R^2 + X^2 and ask them again.





Question 30. (Click on arrow for answer)

Draw a phasor diagram showing the trigonometric relationship between resistance, reactance, and impedance in this series circuit:

Should be image 01827x01

Show mathematically how the resistance and reactance combine in series to produce a total impedance (scalar quantities, all). Then, show how to analyze this same circuit using complex numbers: regarding component as having its own impedance, demonstrating mathematically how these impedances add up to comprise the total impedance (in both polar and rectangular forms).

File Num: 01827

Answer

Should be image 01827x02Scalar calculationsR = 2.2 \hbox{ k}\Omega X_L = 1.495 \hbox{ k}\OmegaZ_{series} = \sqrt{R^2 + {X_L}^2}Z_{series} = \sqrt{2200^2 + 1495^2} = 2660 \> \Omega
Complex number calculationsZ_R = 2.2 \hbox{ k}\Omega \> \angle \> 0^o Z_L = 1.495 \hbox{ k}\Omega \> \angle \> 90^o (Polar form)Z_R = 2.2 \hbox{ k}\Omega + j0 \> \Omega Z_L = 0 \> \Omega + j1.495 \hbox{ k}\Omega (Rectangular form)
Z_{series} = Z_1 + Z_2 + \cdots Z_n (General rule of series impedances)Z_{series} = Z_R + Z_L (Specific application to this circuit)
Z_{series} = 2.2 \hbox{ k}\Omega \> \angle \> 0^o + 1.495 \hbox{ k}\Omega \> \angle \> 90^o = 2.66 \hbox{ k}\Omega \> \angle \> 34.2^oZ_{series} = (2.2 \hbox{ k}\Omega + j0 \> \Omega) + (0 \> \Omega + j1.495 \hbox{ k}\Omega) = 2.2 \hbox{ k}\Omega + j1.495 \hbox{ k}\Omega

Notes

I want students to see that there are two different ways of approaching a problem such as this: with scalar math and with complex number math. If students have access to calculators that can do complex-number arithmetic, the “complex” approach is actually simpler for series-parallel combination circuits, and it yields richer (more informative) results.

Ask your students to determine which of the approaches most resembles DC circuit calculations. Incidentally, this is why I tend to prefer complex-number AC circuit calculations over scalar calculations: because of the conceptual continuity between AC and DC. When you use complex numbers to represent AC voltages, currents, and impedances, almost all the rules of DC circuits still apply. The big exception, of course, is calculations involving power.





Question 31. (Click on arrow for answer)

Calculate the total impedance for these two 100 mH inductors at 2.3 kHz, and draw a phasor diagram showing circuit impedances (Z_{total}, R, and X):

Should be image 02080x01

Now, re-calculate impedance and re-draw the phasor impedance diagram supposing the second inductor is replaced by a 1.5 k\Omega resistor:

Should be image 02080x02

File Num: 02080

Answer

Should be image 02080x03Should be image 02080x04

Notes

Phasor diagrams are powerful analytical tools, if one knows how to draw and interpret them. With hand calculators being so powerful and readily able to handle complex numbers in either polar or rectangular form, there is temptation to avoid phasor diagrams and let the calculator handle all the angle manipulation. However, students will have a much better understanding of phasors and complex numbers in AC circuits if you hold them accountable to representing quantities in that form.





Question 32. (Click on arrow for answer)

Calculate the total impedance of this series LR circuit and then calculate the total circuit current:

Should be image 02103x01

Also, draw a phasor diagram showing how the individual component impedances relate to the total impedance.

File Num: 02103

Answer

Z_{total} = 6.944 k\OmegaI = 4.896 mA RMS

Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 33. (Click on arrow for answer)

Calculate the magnitude and phase shift of the current through this inductor, taking into consideration its intrinsic winding resistance:

Should be image 00639x01

File Num: 00639

Answer

I = 7.849 mA \angle -87.08^{o}

Notes

Inductors are the least “pure” of any reactive component, due to significant quantities of resistance in the windings. Discuss this fact with your students, and what it means with reference to choosing inductors versus capacitors in circuit designs that could use either.





Question 34. (Click on arrow for answer)

Solve for all voltages and currents in this series LR circuit:

Should be image 01830x01

File Num: 01830

Answer

V_L = 12.60 \hbox{ volts RMS}V_R = 8.137 \hbox{ volts RMS}I = 11.46 \hbox{ milliamps RMS}

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 35. (Click on arrow for answer)

Solve for all voltages and currents in this series LR circuit, and also calculate the phase angle of the total impedance:

Should be image 01831x01

File Num: 01831

Answer

V_L = 13.04 \hbox{ volts RMS}V_R = 20.15 \hbox{ volts RMS}I = 4.030 \hbox{ milliamps RMS}\Theta_Z = 32.91^o

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 36. (Click on arrow for answer)

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

Should be image 01841x01
  • L_1 = 250 \hbox{ mH}
  • L_2 = 60 \hbox{ mH}
  • R_1 = 6.8 \hbox{ k}\Omega
  • R_2 = 1.2 \hbox{ k}\Omega
  • V_{supply} = 13.4 \hbox{ V RMS}
  • f_{supply} = 6.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01841

Answer

  • I_{total} = 0.895 \hbox{ mA}
  • V_{L1} = 9.14 \hbox{ V}
  • V_{L2} = 2.19 \hbox{ V}
  • V_{R1} = 6.08 \hbox{ V}
  • V_{R2} = 1.07 \hbox{ V}
  • \Theta = 57.71^o

I suggest using a dual-trace oscilloscope to measure total voltage (across the supply terminals) and voltage drop across resistor R_2. Theoretically, measuring the voltage dropped by either resistor would be fine, but R_2 works better for practical reasons (oscilloscope input lead grounding). Phase shift then could be measured either in the time domain or by a Lissajous figure analysis.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 37. (Click on arrow for answer)

One way to vary the amount of power delivered to a resistive AC load is by varying another resistance connected in series:

Should be image 01829x01

A problem with this power control strategy is that power is wasted in the series resistance (I^2R_{series}). A different strategy for controlling power is shown here, using a series inductance rather than resistance:

Should be image 01829x02

Explain why the latter circuit is more power-efficient than the former, and draw a phasor diagram showing how changes in L_{series} affect Z_{total}.

File Num: 01829

Answer

Inductors are reactive rather than resistive components, and therefore do not dissipate power (ideally).

Should be image 01829x03

Follow-up question: the inductive circuit is not just more energy-efficient — it is safer as well. Identify a potential safety hazard that the resistive power-control circuit poses due to the energy dissipation of its variable resistor.


Notes

If appropriate, you may want to mention devices called saturable reactors, which are used to control power in AC circuits by the exact same principle: varying a series inductance.





Question 38. (Click on arrow for answer)

A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = {1 \over R}), and it is measured in the unit of siemens.

Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

Should be image 00853x01

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = {1 \over X}), and the reciprocal of impedance is called admittance (Y = {1 \over Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.

Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Should be image 00853x02

File Num: 00853

Answer

Y_{total} = \sqrt{G^2 + B^2}

Follow-up question \#1: draw a phasor diagram showing how Y, G, and B relate.


Follow-up question \#2: re-write this equation using quantities of resistance (R), reactance (X), and impedance (Z), instead of conductance (G), susceptance (B), and admittance (Y).


Notes

Ask your students if this equation looks familiar to them. It should!


The answer to the second follow-up question is a matter of algebraic substitution. Work through this process with your students, and then ask them to compare the resulting equation with other equations they’ve seen before. Does its form look familiar to them in any way?





Question 39. (Click on arrow for answer)

Students studying AC electrical theory become familiar with the impedance triangle very soon in their studies:

Should be image 02077x01

What these students might not ordinarily discover is that this triangle is also useful for calculating electrical quantities other than impedance. The purpose of this question is to get you to discover some of the triangle’s other uses.

Fundamentally, this right triangle represents phasor addition, where two electrical quantities at right angles to each other (resistive versus reactive) are added together. In series AC circuits, it makes sense to use the impedance triangle to represent how resistance (R) and reactance (X) combine to form a total impedance (Z), since resistance and reactance are special forms of impedance themselves, and we know that impedances add in series.

List all of the electrical quantities you can think of that add (in series or in parallel) and then show how similar triangles may be drawn to relate those quantities together in AC circuits.

File Num: 02077

Answer

Electrical quantities that add:
  • Series impedances
  • Series voltages
  • Parallel admittances
  • Parallel currents
  • Power dissipations

I will show you one graphical example of how a triangle may relate to electrical quantities other than series impedances:

Should be image 02077x02

Notes

It is very important for students to understand that the triangle only works as an analysis tool when applied to quantities that add. Many times I have seen students try to apply the ZRX impedance triangle to parallel circuits and fail because parallel impedances do not add. The purpose of this question is to force students to think about where the triangle is applicable to AC circuit analysis, and not just to use it blindly.

The power triangle is an interesting application of trigonometry applied to electric circuits. You may not want to discuss power with your students in great detail if they are just beginning to study voltage and current in AC circuits, because power is a sufficiently confusing subject on its own.





Question 40. (Click on arrow for answer)

Explain why the “impedance triangle” is not proper to use for relating total impedance, resistance, and reactance in parallel circuits as it is for series circuits:

Should be image 02078x01

File Num: 02078

Answer

Impedances do not add in parallel.


Follow-up question: what kind of a triangle could be properly applied to a parallel AC circuit, and why?


Notes

Trying to apply the ZRX triangle directly to parallel AC circuits is a common mistake many new students make. Key to knowing when and how to use triangles to graphically depict AC quantities is understanding why the triangle works as an analysis tool and what its sides represent.





Question 41. (Click on arrow for answer)

Calculate the total impedance for these two 100 mH inductors at 2.3 kHz, and draw a phasor diagram showing circuit admittances (Y_{total}, G, and B):

Should be image 02079x01

Now, re-calculate impedance and re-draw the phasor admittance diagram supposing the second inductor is replaced by a 1.5 k\Omega resistor:

Should be image 02079x02

File Num: 02079

Answer

Should be image 02079x03Should be image 02079x04

Challenge question: why are the susceptance vectors (B_{L1} and B_{L2}) pointed down instead of up as impedance vectors for inductances typically are?


Notes

Phasor diagrams are powerful analytical tools, if one knows how to draw and interpret them. With hand calculators being so powerful and readily able to handle complex numbers in either polar or rectangular form, there is temptation to avoid phasor diagrams and let the calculator handle all the angle manipulation. However, students will have a much better understanding of phasors and complex numbers in AC circuits if you hold them accountable to representing quantities in that form.





Question 42. (Click on arrow for answer)

Calculate the individual currents through the inductor and through the resistor, the total current, and the total circuit impedance:

Should be image 02104x01

Also, draw a phasor diagram showing how the individual component currents relate to the total current.

File Num: 02104

Answer

I_L = 530.5 \muA RMSI_R = 490.2 \muA RMSI_{total} = 722.3 \muA RMSZ_{total} = 3.461 k\Omega

Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 43. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01839x01

Calculate the current necessary to power this motor if the equivalent resistance and inductance is 20 \Omega and 238 mH, respectively.

File Num: 01839

Answer

I_{supply} = 12.29 \hbox{ A}

Notes

This is a practical example of a parallel LR circuit, as well as an example of how complex electrical devices may be “modeled” by collections of ideal components. To be honest, a loaded AC motor’s characteristics are quite a bit more complex than what the parallel LR model would suggest, but at least it’s a start!


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 44. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01840x01

Calculate the equivalent inductance (L_{eq}) if the measured source current is 27.5 amps and the motor’s equivalent resistance (R_{eq}) is 11.2 \Omega.

File Num: 01840

Answer

L_{eq} = 61.11 \hbox{ mH}

Notes

Here is a case where scalar calculations (R, G, X, B, Y) are much easier than complex number calculations (all Z) would be.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 45. (Click on arrow for answer)

Determine the total current and all component currents in this circuit, stating your answers the way a multimeter would register them:

Should be image 01842x01
  • L_1 = 1.2 \hbox{ H}
  • L_2 = 650 \hbox{ mH}
  • R_1 = 33 \hbox{ k}\Omega
  • R_2 = 27 \hbox{ k}\Omega
  • V_{supply} = 19.7 \hbox{ V RMS}
  • f_{supply} = 4.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01842

Answer

  • I_{total} = 2.12 \hbox{ mA}
  • I_{L1} = 581 \> \mu \hbox{A}
  • I_{L2} = 1.07 \hbox{ mA}
  • I_{R1} = 597 \> \mu \hbox{A}
  • I_{R2} = 730 \> \mu \hbox{A}
  • \Theta = 51.24^o

Measuring \Theta with an oscilloscope requires the addition of a shunt resistor into this circuit, because oscilloscopes are (normally) only able to measure voltage, and there is no phase shift between any voltages in this circuit because all components are in parallel. I leave it to you to suggest where to insert the shunt resistor, what resistance value to select for the task, and how to connect the oscilloscope to the modified circuit.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 46. (Click on arrow for answer)

Calculate the total impedances (complete with phase angles) for each of the following inductor-resistor circuits:

Should be image 02106x01

File Num: 02106

Answer

Should be image 02106x02

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 47. (Click on arrow for answer)

A doorbell ringer has a solenoid with an inductance of 63 mH connected in parallel with a lamp (for visual indication) having a resistance of 150 ohms:

Should be image 02105x01

Calculate the phase shift of the total current (in units of degrees) in relation to the total supply voltage, when the doorbell switch is actuated.

File Num: 02105

Answer

\Theta = 81 degrees

Suppose the lamp turned on whenever the pushbutton switch was actuated, but the doorbell refused to ring. Identify what you think to be the most likely fault which could account for this problem.


Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 48. (Click on arrow for answer)

An AC electric motor operating under loaded conditions draws a current of 11 amps (RMS) from the 120 volt (RMS) 60 Hz power lines. The measured phase shift between voltage and current for this motor is 34^{o}, with voltage leading current.

Determine the equivalent parallel combination of resistance (R) and inductance (L) that is electrically equivalent to this operating motor.

File Num: 01542

Answer

R_{parallel} = 13.16 \Omega
L_{parallel} = 51.75 mH

Challenge question: in the parallel LR circuit, the resistor will dissipate a lot of energy in the form of heat. Does this mean that the electric motor, which is electrically equivalent to the LR network, will dissipate the same amount of heat? Explain why or why not.


Notes

If students get stuck on the challenge question, remind them that an electric motor does mechanical work, which requires energy.





Question 49. (Click on arrow for answer)

Calculate the impedance of a 145 mH inductor connected in series with 750 \Omega resistor at a frequency of 1 kHz, then determine the necessary resistor and inductor values to create the exact same total impedance in a parallel configuration.

File Num: 00645

Answer

Z_{total} = 1.18 k\Omega \angle 50.54^{o}

If connected in parallel: R = 1.857 k\Omega ; L = 243.3 mH.


Hint: if you are having difficulty figuring out where to start in answering this question, consider the fact that these two circuits, if equivalent in total impedance, will draw the exact same amount of current from a common AC source at 1 kHz.


Notes

This is an interesting question, requiring the student to think creatively about how to convert one configuration of circuit into another, while maintaining the same total effect. As usual, the real purpose of a question like this is to develop problem-solving strategies, rather than to simply obtain an answer.





Question 50. (Click on arrow for answer)

It is often useful in AC circuit analysis to be able to convert a series combination of resistance and reactance into an equivalent parallel combination of conductance and susceptance, or visa-versa:

Should be image 00856x01

We know that resistance (R), reactance (X), and impedance (Z), as scalar quantities, relate to one another trigonometrically in a series circuit. We also know that conductance (G), susceptance (B), and admittance (Y), as scalar quantities, relate to one another trigonometrically in a parallel circuit:

Should be image 00856x02

If these two circuits are truly equivalent to one another, having the same total impedance, then their representative triangles should be geometrically similar (identical angles, same proportions of side lengths). With equal proportions, {R \over Z} in the series circuit triangle should be the same ratio as {G \over Y} in the parallel circuit triangle, that is {R \over Z} = {G \over Y}.

Building on this proportionality, prove the following equation to be true:

R_{series} R_{parallel} = {Z_{total}}^2

After this, derive a similar equation relating the series and parallel reactances (X_{series} and X_{parallel}) with total impedance (Z_{total}).

File Num: 00856

Answer

I’ll let you figure out how to turn {R \over Z} = {G \over Y} into R_{series} R_{parallel} = {Z_{total}}^2 on your own!


As for the reactance relation equation, here it is:

X_{series} X_{parallel} = {Z_{total}}^2

Notes

Being able to convert between series and parallel AC networks is a valuable skill for analyzing complex series-parallel combination circuits, because it means any series-parallel combination circuit may then be converted into an equivalent simple-series or simple-parallel, which is mush easier to analyze.

Some students might ask why the conductance/susceptance triangle is “upside-down” compared to the resistance/reactance triangle. The reason has to do with the sign reversal of imaginary quantities when inverted: {1 \over j} = -j. The phase angle of a pure inductance’s impedance is +90 degrees, while the phase angle of the same (pure) inductance’s admittance is -90 degrees, due to reciprocation. Thus, while the X leg of the resistance/reactance triangle points up, the B leg of the conductance/susceptance triangle must point down.





Question 51. (Click on arrow for answer)

Determine an equivalent parallel RC network for the series RC network shown on the left:

Should be image 01540x01

Note that I have already provided a value for the capacitor’s reactance (X_C), which of course will be valid only for a particular frequency. Determine what values of resistance (R) and reactance (X_C) in the parallel network will yield the exact same total impedance (Z_T) at the same signal frequency.

File Num: 01540

Answer

R = 150 \Omega
X_C = 200 \Omega

Follow-up question: explain how you could check your conversion calculations, to ensure both networks are truly equivalent to each other.


Notes

This problem just happens to work out with whole numbers. Believe it or not, I chose these numbers entirely by accident one day, when setting up an example problem to show a student how to convert between series and parallel equivalent networks!





Question 52. (Click on arrow for answer)

Determine the equivalent parallel-connected resistor and inductor values for this series circuit:

Should be image 00855x01

Also, express the total impedance of either circuit (since they are electrically equivalent to one another, they should have the same total impedance) in complex form. That is, express Z as a quantity with both a magnitude and an angle.

File Num: 00855

Answer

R_{parallel} = 2092 \Omega
L_{parallel} = 1.325 H
Z_{total} = 1772 \Omega \angle 32.14^{o}

Notes

There are different methods of solving this problem. Use the discussion time to let students expound on how they approached the problem, pooling together their ideas. Their creativity may surprise you!





Question 53. (Click on arrow for answer)

Convert this series-parallel combination circuit into an equivalent simple-parallel circuit (all components connected in parallel with each other, with nothing in series), and also calculate the circuit’s total impedance:

Should be image 00857x01

File Num: 00857

Answer

Should be image 00857x02
Z_{total} = 963.0 \Omega

Challenge question: from the simple-parallel equivalent circuit shown here, can you generate an equivalent circuit that is simple-series? In other words, can you calculate the proper values of R and L, that when connected in series, will have the same total impedance as this circuit?


Notes

Fundamentally, this question asks students to generate an equivalent parallel R-X circuit from a given series R-X circuit. In this particular circuit, there are two series-connected R-X branches, resulting in an equivalent parallel circuit with four branches.

Calculating the circuit’s total impedance as a scalar figure involves simplifying the circuit once more into two components: a resistance and a reactance.





Question 54. (Click on arrow for answer)

It is not uncommon to see impedances represented in AC circuits as boxes, rather than as combinations of R, L, and/or C. This is simply a convenient way to represent what may be complex sub-networks of components in a larger AC circuit:

Should be image 00859x01

We know that any given impedance may be represented by a simple, two-component circuit: either a resistor and a reactive component connected in series, or a resistor and a reactive component connected in parallel. Assuming a circuit frequency of 250 Hz, determine what combination of series-connected components will be equivalent to this “box” impedance, and also what combination of parallel-connected components will be equivalent to this “box” impedance.

File Num: 00859

Answer

Should be image 00859x02

Notes

Once students learn to convert between complex impedances, equivalent series R-X circuits, and equivalent parallel R-X circuits, it becomes possible for them to analyze the most complex series-parallel impedance combinations imaginable without having to do arithmetic with complex numbers (magnitudes and angles at every step). It does, however, require that students have a good working knowledge of resistance, conductance, reactance, susceptance, impedance, and admittance, and how these quantities relate mathematically to one another in scalar form.





Question 55. (Click on arrow for answer)

It is not uncommon to see impedances represented in AC circuits as boxes, rather than as combinations of R, L, and/or C. This is simply a convenient way to represent what may be complex sub-networks of components in a larger AC circuit:

Should be image 03296x01

We know that any given impedance may be represented by a simple, two-component circuit: either a resistor and a reactive component connected in series, or a resistor and a reactive component connected in parallel. Assuming a circuit frequency of 2 kHz, determine what combination of series-connected components will be equivalent to this “box” impedance, and also what combination of parallel-connected components will be equivalent to this “box” impedance.

File Num: 03296

Answer

Should be image 03296x02

Notes

Once students learn to convert between complex impedances, equivalent series R-X circuits, and equivalent parallel R-X circuits, it becomes possible for them to analyze the most complex series-parallel impedance combinations imaginable without having to do arithmetic with complex numbers (magnitudes and angles at every step). It does, however, require that students have a good working knowledge of resistance, conductance, reactance, susceptance, impedance, and admittance, and how these quantities relate mathematically to one another in scalar form.





All files with file num less than 4100 are Copyright 2003, Tony R. Kuphaldt, released under the Creative Commons Attribution License (v 1.0). All other files are Copyright 2022, David Williams, released under the Creative Commons Attribution License (V 4.0) This means you may do almost anything with this work, so long as you give proper credit.

To view a copy of the license, visit https://creativecommons.org/licenses/by/1.0/, or https://creativecommons.org/licenses/by/4.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.


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Resistors and AC Voltage, Current & Power

Video – AC Voltage, Current, and Resistors

Resistors in AC Circuits (Power)

This video describes the way AC voltage and current behave with resistors. You can also read on for more info

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Pure resistive AC circuit: resistor voltage and current are in phase.

If we were to plot the current and voltage for a very simple AC circuit consisting of a source and a resistor, it would look something like this:

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Voltage and current “in phase” for resistive circuit.

Because the resistor follows Ohm’s law and simply and directly resists the flow of electrons at all periods of time, the waveform for the voltage drop across the resistor is exactly in phase with the waveform for the current through it. In other words, for every point in time, e=iR.

We can look at any point in time along the horizontal axis of the plot and compare the values of current and voltage with each other. When the instantaneous value (any “snapshot” look at the values of a wave are referred to as instantaneous values) for current is zero, the instantaneous voltage across the resistor is also zero. Likewise, at the moment in time where the current through the resistor is at its positive peak, the voltage across the resistor is also at its positive peak, and so on. At any given point in time along the waves, Ohm’s Law holds true for the instantaneous values of voltage and current.

We can also calculate the power dissipated by this resistor (p=ie), and plot those values on the same graph:

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Instantaneous AC power in a pure resistive circuit is always positive.

Note that the power is never a negative value. When the current is positive (above the line), the voltage is also positive, resulting in a power (p=ie) of a positive value. Conversely, when the current is negative (below the line), the voltage is also negative, which results in a positive value for power (a negative number multiplied by a negative number equals a positive number). This consistent “polarity” of power tells us that the resistor is always dissipating power, taking it from the source and releasing it in the form of heat energy. Whether the current is positive or negative, a resistor still dissipates energy.

Ohm’s Law Calculations in AC

Ohm’s law still applies to resistors in AC circuits: E=IR

Or, more correctly: e=iR because we use lower case e and i (and p ) to denote that the voltage/current/power is AC instead of DC.

However, as we learned in chapter 1, the voltage or current of an AC signal can be represented in multiple different ways. You can use the peak voltage (current), the peak-to-peak voltage (current), or RMS voltage (current) to describe it. So which one should you use to calculate Ohm’s Law? Well it turns out that it doesn’t matter, as long as you’re consistent.

For example (assume all voltages and currents are sinusoidal):

  • If you have a 100V peak voltage applied across a 10 Ω resistor, you would have a \frac{100V}{10\Omega} = 10A peak current
  • if you have a 10mA peak-to-peak current going through a 1kΩ resistor, you would have a 0.1A\times1000\Omega=10V peak-to-peak current across it
  • If you have apply 120Vrms to a resistor and get 1A peak current, the resistor would not be \frac{120V}{1A}=120\Omega because the voltage and current are two different types of amplitudes, you would need to convert the peak current to RMS (\frac{1A}{\sqrt{2}} = 0.707A) to get \frac{120V}{0.707A}=169.7\Omega

Since voltage across and current through a resistor are in phase with each other, there is no need to include the phase when doing analysis or calculations on AC circuits that only have resistors. However, when we start to include inductors and capacitors, we will have to take into account phase, so as a simple introduction, using phase in AC circuit analysis, let’s look at how to do that with resistors.

Vector Representation of Resistors: Impedance

When plotting out power for a resistor earlier, we saw that the power is always positive – all of the energy put into a resistor when current flows through it is dissipated as heat. When the power is dissipated like this, it is considered real power.

As we will see with inductors and capacitors, current flow only results in a flow of electrical energy into and out of the device, energy is not dissipated as heat. We’ll talk more about this in the next chapters, but for now, understand that this type of power transfer is considered imaginary because it is not used up.

Now, think back to chapter 2 – having a real part and an imaginary part sounds a lot like the complex plane. Resistors only have a real part, but they can still be represented by a complex number with the imaginary part set to zero. For example a 10 Ω resistor in an AC circuit can be represented as 10+j\Omega or 10\Omega\angle0^o

For resistors (and inductors and capacitors), this complex number is called the impedance of the device and is represented by the letter Z (in bold to indicate that it is a vector).

(Side note: multiple devices can be combined together to create a total impedance for a circuit).

What is the impedance of a 2200 Ohm resistor?

2200+j0 \Omega or 2200 \Omega\angle 0

Another way to think about the affect of resistors (and inductors and capacitors) on voltage and current in AC circuits, is to consider the two different ways that the components affect the relationship between voltage and current. It is not obvious at this point, but not only do these components affect the relationship of magnitude between voltage and current, but they also affect the phase difference between voltage and current (resistors “cause” no phase shift, but inductors and capacitors do).

To represent the two different effects, we can use a vector (the impedance) to represent the magnitude of the effect and the direction of the effect. For resistors, the magnitude of the effect is the resistance (which as we know from Ohm’s law is the ratio of voltage across and current through the resistor) and the direction of the effect is 0 since resistors cause no phase shift.

Examples

Resistors in Series and Parallel

Impedances can be combined. When components are in series and parallel, the overall impedance is calculated in the same way as for DC circuits except that all of the arithmetic is vector arithmetic. Resistors are a simple case though since they have no imaginary component. If you have only resistors in an AC circuit, you can combine them to get an overall equivalent impedance in exactly the same way you would in a DC circuit.

Examples

Contributors

Contributors to this chapter are listed in chronological order of their contributions, from most recent to first.

David Williams (2022): Separated resistors into its own chapter. Added videos, added sections on vector representation of resistors


CC-BY 2000-2020 Tony R. Kuphaldt.   
Creative Commons License


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Mutual inductance and basic operation

Suppose we were to wrap a coil of insulated wire around a loop of ferromagnetic material and energize this coil with an AC voltage source:

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Insulated winding on ferromagnetic loop has inductive reactance, limiting AC current.

As an inductor, we would expect this iron-core coil to oppose the applied voltage with its inductive reactance, limiting current through the coil as predicted by the equations XL = 2πfL and I=E/X (or I=E/Z). For the purposes of this example, though, we need to take a more detailed look at the interactions of voltage, current, and magnetic flux in the device.

Kirchhoff’s voltage law describes how the algebraic sum of all voltages in a loop must equal zero. In this example, we could apply this fundamental law of electricity to describe the respective voltages of the source and of the inductor coil. Here, as in any one-source, one-load circuit, the voltage dropped across the load must equal the voltage supplied by the source, assuming zero voltage dropped along the resistance of any connecting wires. In other words, the load (inductor coil) must produce an opposing voltage equal in magnitude to the source, in order that it may balance against the source voltage and produce an algebraic loop voltage sum of zero. From where does this opposing voltage arise? If the load were a resistor, the voltage drop originates from electrical energy loss, the “friction” of electrons flowing through the resistance. With a perfect inductor (no resistance in the coil wire), the opposing voltage comes from another mechanism: the reaction to a changing magnetic flux in the iron core. When AC current changes, flux Φ changes. Changing flux induces a counter EMF.

Michael Faraday discovered the mathematical relationship between magnetic flux (Φ) and induced voltage with this equation:

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The instantaneous voltage (voltage dropped at any instant in time) across a wire coil is equal to the number of turns of that coil around the core (N) multiplied by the instantaneous rate-of-change in magnetic flux (dΦ/dt) linking with the coil. Graphed, this shows itself as a set of sine waves (assuming a sinusoidal voltage source), the flux wave 90o lagging behind the voltage wave:

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Magnetic flux lags applied voltage by 90o because flux is proportional to a rate of change, dΦ/dt.

Magnetic flux through a ferromagnetic material is analogous to current through a conductor: it must be motivated by some force in order to occur. In electric circuits, this motivating force is voltage (a.k.a. electromotive force, or EMF). In magnetic “circuits,” this motivating force is magnetomotive force, or mmf. Magnetomotive force (mmf) and magnetic flux (Φ) are related to each other by a property of magnetic materials known as reluctance (the latter quantity symbolized by a strange-looking letter “R”):

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In our example, the mmf required to produce this changing magnetic flux (Φ) must be supplied by a changing current through the coil. Magnetomotive force generated by an electromagnet coil is equal to the amount of current through that coil (in amps) multiplied by the number of turns of that coil around the core (the SI unit for mmf is the amp-turn). Because the mathematical relationship between magnetic flux and mmf is directly proportional, and because the mathematical relationship between mmf and current is also directly proportional (no rates-of-change present in either equation), the current through the coil will be in-phase with the flux wave as in

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Magnetic flux, like current, lags applied voltage by 90o.

This is why alternating current through an inductor lags the applied voltage waveform by 90o: because that is what is required to produce a changing magnetic flux whose rate-of-change produces an opposing voltage in-phase with the applied voltage. Due to its function in providing magnetizing force (mmf) for the core, this current is sometimes referred to as the magnetizing current.

It should be mentioned that the current through an iron-core inductor is not perfectly sinusoidal (sine-wave shaped), due to the nonlinear B/H magnetization curve of iron. In fact, if the inductor is cheaply built, using as little iron as possible, the magnetic flux density might reach high levels (approaching saturation), resulting in a magnetizing current waveform that looks something like Figure below

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As flux density approaches saturation, the magnetizing current waveform becomes distorted.

When a ferromagnetic material approaches magnetic flux saturation, disproportionately greater levels of magnetic field force (mmf) are required to deliver equal increases in magnetic field flux (Φ). Because mmf is proportional to current through the magnetizing coil (mmf = NI, where “N” is the number of turns of wire in the coil and “I” is the current through it), the large increases of mmf required to supply the needed increases in flux results in large increases in coil current. Thus, coil current increases dramatically at the peaks in order to maintain a flux waveform that isn’t distorted, accounting for the bell-shaped half-cycles of the current waveform in the above plot.

The situation is further complicated by energy losses within the iron core. The effects of hysteresis and eddy currents conspire to further distort and complicate the current waveform, making it even less sinusoidal and altering its phase to be lagging slightly less than 90o behind the applied voltage waveform. This coil current resulting from the sum total of all magnetic effects in the core (dΦ/dt magnetization plus hysteresis losses, eddy current losses, etc.) is called the exciting current. The distortion of an iron-core inductor’s exciting current may be minimized if it is designed for and operated at very low flux densities. Generally speaking, this requires a core with large cross-sectional area, which tends to make the inductor bulky and expensive. For the sake of simplicity, though, we’ll assume that our example core is far from saturation and free from all losses, resulting in a perfectly sinusoidal exciting current.

As we’ve seen already in the inductors chapter, having a current waveform 90o out of phase with the voltage waveform creates a condition where power is alternately absorbed and returned to the circuit by the inductor. If the inductor is perfect (no wire resistance, no magnetic core losses, etc.), it will dissipate zero power.

Let us now consider the same inductor device, except this time with a second coil wrapped around the same iron core. The first coil will be labeled the primary coil, while the second will be labeled the secondary:

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Ferromagnetic core with primary coil (AC driven) and secondary coil.

If this secondary coil experiences the same magnetic flux change as the primary (which it should, assuming perfect containment of the magnetic flux through the common core), and has the same number of turns around the core, a voltage of equal magnitude and phase to the applied voltage will be induced along its length. In the following graph, the induced voltage waveform is drawn slightly smaller than the source voltage waveform simply to distinguish one from the other:

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Open circuited secondary sees the same flux Φ as the primary. Therefore induced secondary voltage es is the same magnitude and phase as the primary voltage ep.

This effect is called mutual inductance: the induction of a voltage in one coil in response to a change in current in the other coil. Like normal (self-) inductance, it is measured in the unit of Henrys, but unlike normal inductance it is symbolized by the capital letter “M” rather than the letter “L”:

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No current will exist in the secondary coil, since it is open-circuited. However, if we connect a load resistor to it, an alternating current will go through the coil, in-phase with the induced voltage (because the voltage across a resistor and the current through it are always in-phase with each other).

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Resistive load on secondary has voltage and current in-phase.

At first, one might expect this secondary coil current to cause additional magnetic flux in the core. In fact, it does not. If more flux were induced in the core, it would cause more voltage to be induced voltage in the primary coil (remember that e = dΦ/dt). This cannot happen, because the primary coil’s induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff’s voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, what does change is the amount of mmf in the magnetic circuit.

Magnetomotive force is produced any time electrons move through a wire. Usually, this mmf is accompanied by magnetic flux, in accordance with the mmf=ΦR “magnetic Ohm’s Law” equation. In this case, though, additional flux is not permitted, so the only way the secondary coil’s mmf may exist is if a counteracting mmf is generated by the primary coil, of equal magnitude and opposite phase. Indeed, this is what happens, an alternating current forming in the primary coil — 180o out of phase with the secondary coil’s current — to generate this counteracting mmf and prevent additional core flux. Polarity marks and current direction arrows have been added to the illustration to clarify phase relations:

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Flux remains constant with application of a load. However, a counteracting mmf is produced by the loaded secondary.

If you find this process a bit confusing, do not worry. Transformer dynamics is a complex subject. What is important to understand is this: when an AC voltage is applied to the primary coil, it creates a magnetic flux in the core, which induces AC voltage in the secondary coil in-phase with the source voltage. Any current drawn through the secondary coil to power a load induces a corresponding current in the primary coil, drawing current from the source.

Notice how the primary coil is behaving as a load with respect to the AC voltage source, and how the secondary coil is behaving as a source with respect to the resistor. Rather than energy merely being alternately absorbed and returned the primary coil circuit, energy is now being coupled to the secondary coil where it is delivered to a dissipative (energy-consuming) load. As far as the source “knows,” its directly powering the resistor. Of course, there is also an additional primary coil current lagging the applied voltage by 90o, just enough to magnetize the core to create the necessary voltage for balancing against the source (the exciting current).

We call this type of device a transformer, because it transforms electrical energy into magnetic energy, then back into electrical energy again. Because its operation depends on electromagnetic induction between two stationary coils and a magnetic flux of changing magnitude and “polarity,” transformers are necessarily AC devices. Its schematic symbol looks like two inductors (coils) sharing the same magnetic core:

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Schematic symbol for transformer consists of two inductor symbols, separated by lines indicating a ferromagnetic core.

The two inductor coils are easily distinguished in the above symbol. The pair of vertical lines represent an iron core common to both inductors. While many transformers have ferromagnetic core materials, there are some that do not, their constituent inductors being magnetically linked together through the air.

The following photograph shows a power transformer of the type used in gas-discharge lighting. Here, the two inductor coils can be clearly seen, wound around an iron core. While most transformer designs enclose the coils and core in a metal frame for protection, this particular transformer is open for viewing and so serves its illustrative purpose well:

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Example of a gas-discharge lighting transformer.

Both coils of wire can be seen here with copper-colored varnish insulation. The top coil is larger than the bottom coil, having a greater number of “turns” around the core. In transformers, the inductor coils are often referred to as windings, in reference to the manufacturing process where wire is wound around the core material. As modeled in our initial example, the powered inductor of a transformer is called the primary winding, while the unpowered coil is called the secondary winding.

In the next photograph, figure below, a transformer is shown cut in half, exposing the cross-section of the iron core as well as both windings. Like the transformer shown previously, this unit also utilizes primary and secondary windings of differing turn counts. The wire gauge can also be seen to differ between primary and secondary windings. The reason for this disparity in wire gauge will be made clear in the next section of this chapter. Additionally, the iron core can be seen in this photograph to be made of many thin sheets (laminations) rather than a solid piece. The reason for this will also be explained in a later section of this chapter.

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Transformer cross-section cut shows core and windings.

It is easy to demonstrate simple transformer action using SPICE, setting up the primary and secondary windings of the simulated transformer as a pair of “mutual” inductors. The coefficient of magnetic field coupling is given at the end of the “k” line in the SPICE circuit description, this example being set very nearly at perfection (1.000). This coefficient describes how closely “linked” the two inductors are, magnetically. The better these two inductors are magnetically coupled, the more efficient the energy transfer between them should be.

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Spice circuit for coupled inductors.

transformerv1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12 l1 2 0 100 l2 3 5 100 ** This line tells SPICE that the two inductors ** l1 and l2 are magnetically “linked” togetherk l1 l2 0.999 vi1 3 4 ac 0 rload 4 5 1k .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end

Note: the Rbogus resistors are required to satisfy certain quirks of SPICE. The first breaks the otherwise continuous loop between the voltage source and L1 which would not be permitted by SPICE. The second provides a path to ground (node 0) from the secondary circuit, necessary because SPICE cannot function with any ungrounded circuits.

freq v(2) i(v1) 6.000E+01 1.000E+01 9.975E-03 Primary winding freq v(3,5) i(vi1) 6.000E+01 9.962E+00 9.962E-03 Secondary winding

Note that with equal inductances for both windings (100 Henrys each), the AC voltages and currents are nearly equal for the two. The difference between primary and secondary currents is the magnetizing current spoken of earlier: the 90o lagging current necessary to magnetize the core. As is seen here, it is usually very small compared to primary current induced by the load, and so the primary and secondary currents are almost equal. What you are seeing here is quite typical of transformer efficiency. Anything less than 95% efficiency is considered poor for modern power transformer designs, and this transfer of power occurs with no moving parts or other components subject to wear.

If we decrease the load resistance so as to draw more current with the same amount of voltage, we see that the current through the primary winding increases in response. Even though the AC power source is not directly connected to the load resistance (rather, it is electromagnetically “coupled”), the amount of current drawn from the source will be almost the same as the amount of current that would be drawn if the load were directly connected to the source. Take a close look at the next two SPICE simulations, showing what happens with different values of load resistors:

transformer v1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12l1 2 0 100 l2 3 5 100 k l1 l2 0.999 vi1 3 4 ac 0 ** Note load resistance value of 200 ohmsrload 4 5 200 .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end freq v(2) i(v1) 6.000E+01 1.000E+01 4.679E-02 freq v(3,5) i(vi1) 6.000E+01 9.348E+00 4.674E-02

Notice how the primary current closely follows the secondary current. In our first simulation, both currents were approximately 10 mA, but now they are both around 47 mA. In this second simulation, the two currents are closer to equality, because the magnetizing current remains the same as before while the load current has increased. Note also how the secondary voltage has decreased some with the heavier (greater current) load. Let’s try another simulation with an even lower value of load resistance (15 Ω):

transformer v1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12l1 2 0 100 l2 3 5 100 k l1 l2 0.999 vi1 3 4 ac 0 rload 4 5 15 .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end freq v(2) i(v1) 6.000E+01 1.000E+01 1.301E-01 freq v(3,5) i(vi1) 6.000E+01 1.950E+00 1.300E-01

Our load current is now 0.13 amps, or 130 mA, which is substantially higher than the last time. The primary current is very close to being the same, but notice how the secondary voltage has fallen well below the primary voltage (1.95 volts versus 10 volts at the primary). The reason for this is an imperfection in our transformer design: because the primary and secondary inductances aren’t perfectly linked (a k factor of 0.999 instead of 1.000) there is “stray” or “leakage” inductance. In other words, some of the magnetic field isn’t linking with the secondary coil, and thus cannot couple energy to it:

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Leakage inductance is due to magnetic flux not cutting both windings.

Consequently, this “leakage” flux merely stores and returns energy to the source circuit via self-inductance, effectively acting as a series impedance in both primary and secondary circuits. Voltage gets dropped across this series impedance, resulting in a reduced load voltage: voltage across the load “sags” as load current increases.

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Equivalent circuit models leakage inductance as series inductors independent of the “ideal transformer”.

If we change the transformer design to have better magnetic coupling between the primary and secondary coils, the figures for voltage between primary and secondary windings will be much closer to equality again:

transformer v1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12l1 2 0 100 l2 3 5 100** Coupling factor = 0.99999 instead of 0.999k l1 l2 0.99999 vi1 3 4 ac 0 rload 4 5 15 .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end freq v(2) i(v1) 6.000E+01 1.000E+01 6.658E-01freq v(3,5) i(vi1) 6.000E+01 9.987E+00 6.658E-01

Here we see that our secondary voltage is back to being equal with the primary, and the secondary current is equal to the primary current as well. Unfortunately, building a real transformer with coupling this complete is very difficult. A compromise solution is to design both primary and secondary coils with less inductance, the strategy being that less inductance overall leads to less “leakage” inductance to cause trouble, for any given degree of magnetic coupling inefficiency. This results in a load voltage that is closer to ideal with the same (high current heavy) load and the same coupling factor:

transformerv1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12** inductance = 1 henry instead of 100 henrys l1 2 0 1l2 3 5 1k l1 l2 0.999 vi1 3 4 ac 0 rload 4 5 15 .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end freq v(2) i(v1) 6.000E+01 1.000E+01 6.664E-01freq v(3,5) i(vi1) 6.000E+01 9.977E+00 6.652E-01

Simply by using primary and secondary coils of less inductance, the load voltage for this heavy load (high current) has been brought back up to nearly ideal levels (9.977 volts). At this point, one might ask, “If less inductance is all that’s needed to achieve near-ideal performance under heavy load, then why worry about coupling efficiency at all? If its impossible to build a transformer with perfect coupling, but easy to design coils with low inductance, then why not just build all transformers with low-inductance coils and have excellent efficiency even with poor magnetic coupling?”

The answer to this question is found in another simulation: the same low-inductance transformer, but this time with a lighter load (less current) of 1 kΩ instead of 15 Ω:

transformer v1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12l1 2 0 1l2 3 5 1k l1 l2 0.999 vi1 3 4 ac 0 rload 4 5 1k .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end freq v(2) i(v1) 6.000E+01 1.000E+01 2.835E-02freq v(3,5) i(vi1) 6.000E+01 9.990E+00 9.990E-03

With lower winding inductances, the primary and secondary voltages are closer to being equal, but the primary and secondary currents are not. In this particular case, the primary current is 28.35 mA while the secondary current is only 9.990 mA: almost three times as much current in the primary as the secondary. Why is this? With less inductance in the primary winding, there is less inductive reactance, and consequently a much larger magnetizing current. A substantial amount of the current through the primary winding merely works to magnetize the core rather than transfer useful energy to the secondary winding and load.

An ideal transformer with identical primary and secondary windings would manifest equal voltage and current in both sets of windings for any load condition. In a perfect world, transformers would transfer electrical power from primary to secondary as smoothly as though the load were directly connected to the primary power source, with no transformer there at all. However, you can see this ideal goal can only be met if there is perfect coupling of magnetic flux between primary and secondary windings. Being that this is impossible to achieve, transformers must be designed to operate within certain expected ranges of voltages and loads in order to perform as close to ideal as possible. For now, the most important thing to keep in mind is a transformer’s basic operating principle: the transfer of power from the primary to the secondary circuit via electromagnetic coupling.

  • REVIEW:
  • Mutual inductance is where the magnetic flux of two or more inductors are “linked” so that voltage is induced in one coil proportional to the rate-of-change of current in another.
  • A transformer is a device made of two or more inductors, one of which is powered by AC, inducing an AC voltage across the second inductor. If the second inductor is connected to a load, power will be electromagnetically coupled from the first inductor’s power source to that load.
  • The powered inductor in a transformer is called the primary winding. The unpowered inductor in a transformer is called the secondary winding.
  • Magnetic flux in the core (Φ) lags 90o behind the source voltage waveform. The current drawn by the primary coil from the source to produce this flux is called the magnetizing current, and it also lags the supply voltage by 90o.
  • Total primary current in an unloaded transformer is called the exciting current, and is comprised of magnetizing current plus any additional current necessary to overcome core losses. It is never perfectly sinusoidal in a real transformer, but may be made more so if the transformer is designed and operated so that magnetic flux density is kept to a minimum.
  • Core flux induces a voltage in any coil wrapped around the core. The induces voltage(s) are ideally in- phase with the primary winding source voltage and share the same waveshape.
  • Any current drawn through the secondary winding by a load will be “reflected” to the primary winding and drawn from the voltage source, as if the source were directly powering a similar load.

Step-up and step-down transformers

So far, we’ve observed simulations of transformers where the primary and secondary windings were of identical inductance, giving approximately equal voltage and current levels in both circuits. Equality of voltage and current between the primary and secondary sides of a transformer, however, is not the norm for all transformers. If the inductances of the two windings are not equal, something interesting happens:

transformer v1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12l1 2 0 10000 l2 3 5 100 k l1 l2 0.999 vi1 3 4 ac 0 rload 4 5 1k .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end freq v(2) i(v1) 6.000E+01 1.000E+01 9.975E-05 Primary windingfreq v(3,5) i(vi1) 6.000E+01 9.962E-01 9.962E-04 Secondary winding

Notice how the secondary voltage is approximately ten times less than the primary voltage (0.9962 volts compared to 10 volts), while the secondary current is approximately ten times greater (0.9962 mA compared to 0.09975 mA). What we have here is a device that steps voltage down by a factor of ten and current up by a factor of ten:

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Turns ratio of 10:1 yields 10:1 primary:secondary voltage ratio and 1:10 primary:secondary current ratio.

This is a very useful device, indeed. With it, we can easily multiply or divide voltage and current in AC circuits. Indeed, the transformer has made long-distance transmission of electric power a practical reality, as AC voltage can be “stepped up” and current “stepped down” for reduced wire resistance power losses along power lines connecting generating stations with loads. At either end (both the generator and at the loads), voltage levels are reduced by transformers for safer operation and less expensive equipment. A transformer that increases voltage from primary to secondary (more secondary winding turns than primary winding turns) is called a step-up transformer. Conversely, a transformer designed to do just the opposite is called a step-down transformer.

Let’s re-examine a photograph shown in the previous section:

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Transformer cross-section showing primary and secondary windings is a few inches tall (approximately 10 cm).

This is a step-down transformer, as evidenced by the high turn count of the primary winding and the low turn count of the secondary. As a step-down unit, this transformer converts high-voltage, low-current power into low-voltage, high-current power. The larger-gauge wire used in the secondary winding is necessary due to the increase in current. The primary winding, which doesn’t have to conduct as much current, may be made of smaller-gauge wire.

In case you were wondering, it is possible to operate either of these transformer types backwards (powering the secondary winding with an AC source and letting the primary winding power a load) to perform the opposite function: a step-up can function as a step-down and visa-versa. However, as we saw in the first section of this chapter, efficient operation of a transformer requires that the individual winding inductances be engineered for specific operating ranges of voltage and current, so if a transformer is to be used “backwards” like this it must be employed within the original design parameters of voltage and current for each winding, lest it prove to be inefficient (or lest it be damaged by excessive voltage or current!).

Transformers are often constructed in such a way that it is not obvious which wires lead to the primary winding and which lead to the secondary. One convention used in the electric power industry to help alleviate confusion is the use of “H” designations for the higher-voltage winding (the primary winding in a step-down unit; the secondary winding in a step-up) and “X” designations for the lower-voltage winding. Therefore, a simple power transformer will have wires labeled “H1”, “H2”, “X1”, and “X2”. There is usually significance to the numbering of the wires (H1 versus H2, etc.), which we’ll explore a little later in this chapter.

The fact that voltage and current get “stepped” in opposite directions (one up, the other down) makes perfect sense when you recall that power is equal to voltage times current, and realize that transformers cannot produce power, only convert it. Any device that could output more power than it took in would violate the Law of Energy Conservation in physics, namely that energy cannot be created or destroyed, only converted. As with the first transformer example we looked at, power transfer efficiency is very good from the primary to the secondary sides of the device.

The practical significance of this is made more apparent when an alternative is considered: before the advent of efficient transformers, voltage/current level conversion could only be achieved through the use of motor/generator sets. A drawing of a motor/generator set reveals the basic principle involved:

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Motor generator illustrates the basic principle of the transformer.

In such a machine, a motor is mechanically coupled to a generator, the generator designed to produce the desired levels of voltage and current at the rotating speed of the motor. While both motors and generators are fairly efficient devices, the use of both in this fashion compounds their inefficiencies so that the overall efficiency is in the range of 90% or less. Furthermore, because motor/generator sets obviously require moving parts, mechanical wear and balance are factors influencing both service life and performance. Transformers, on the other hand, are able to convert levels of AC voltage and current at very high efficiencies with no moving parts, making possible the widespread distribution and use of electric power we take for granted.

In all fairness it should be noted that motor/generator sets have not necessarily been obsoleted by transformers for all applications. While transformers are clearly superior over motor/generator sets for AC voltage and current level conversion, they cannot convert one frequency of AC power to another, or (by themselves) convert DC to AC or visa-versa. Motor/generator sets can do all these things with relative simplicity, albeit with the limitations of efficiency and mechanical factors already described. Motor/generator sets also have the unique property of kinetic energy storage: that is, if the motor’s power supply is momentarily interrupted for any reason, its angular momentum (the inertia of that rotating mass) will maintain rotation of the generator for a short duration, thus isolating any loads powered by the generator from “glitches” in the main power system.

Looking closely at the numbers in the SPICE analysis, we should see a correspondence between the transformer’s ratio and the two inductances. Notice how the primary inductor (l1) has 100 times more inductance than the secondary inductor (10000 H versus 100 H), and that the measured voltage step-down ratio was 10 to 1. The winding with more inductance will have higher voltage and less current than the other. Since the two inductors are wound around the same core material in the transformer (for the most efficient magnetic coupling between the two), the parameters affecting inductance for the two coils are equal except for the number of turns in each coil. If we take another look at our inductance formula, we see that inductance is proportional to the square of the number of coil turns:

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So, it should be apparent that our two inductors in the last SPICE transformer example circuit — with inductance ratios of 100:1 — should have coil turn ratios of 10:1, because 10 squared equals 100. This works out to be the same ratio we found between primary and secondary voltages and currents (10:1), so we can say as a rule that the voltage and current transformation ratio is equal to the ratio of winding turns between primary and secondary.

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Step-down transformer: (many turns :few turns).

The step-up/step-down effect of coil turn ratios in a transformer is analogous to gear tooth ratios in mechanical gear systems, transforming values of speed and torque in much the same way:

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Torque reducing gear train steps torque down, while stepping speed up.

Step-up and step-down transformers for power distribution purposes can be gigantic in proportion to the power transformers previously shown, some units standing as tall as a home. The following photograph shows a substation transformer standing about twelve feet tall:

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Substation transformer.

  • REVIEW:
  • Transformers “step up” or “step down” voltage according to the ratios of primary to secondary wire turns.
  • a picture

  • A transformer designed to increase voltage from primary to secondary is called a step-up transformer. A transformer designed to reduce voltage from primary to secondary is called a step-down transformer.
  • The transformation ratio of a transformer will be equal to the square root of its primary to secondary inductance (L) ratio.
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Electrical isolation

Aside from the ability to easily convert between different levels of voltage and current in AC and DC circuits, transformers also provide an extremely useful feature called isolation, which is the ability to couple one circuit to another without the use of direct wire connections. We can demonstrate an application of this effect with another SPICE simulation: this time showing “ground” connections for the two circuits, imposing a high DC voltage between one circuit and ground through the use of an additional voltage source:

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Transformer isolates 10 Vac at V1 from 250 VDC at V2.

v1 1 0 ac 10 sinrbogus1 1 2 1e-12 v2 5 0 dc 250 l1 2 0 10000 l2 3 5 100 k l1 l2 0.999 vi1 3 4 ac 0 rload 4 5 1k .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end DC voltages referenced to ground (node 0):(1) 0.0000 (2) 0.0000 (3) 250.0000 (4) 250.0000 (5) 250.0000AC voltages:freq v(2) i(v1) 6.000E+01 1.000E+01 9.975E-05 Primary windingfreq v(3,5) i(vi1) 6.000E+01 9.962E-01 9.962E-04 Secondary winding

SPICE shows the 250 volts DC being impressed upon the secondary circuit elements with respect to ground, but as you can see there is no effect on the primary circuit (zero DC voltage) at nodes 1 and 2, and the transformation of AC power from primary to secondary circuits remains the same as before. The impressed voltage in this example is often called a common-mode voltage because it is seen at more than one point in the circuit with reference to the common point of ground. The transformer isolates the common-mode voltage so that it is not impressed upon the primary circuit at all, but rather isolated to the secondary side. For the record, it does not matter that the common-mode voltage is DC, either. It could be AC, even at a different frequency, and the transformer would isolate it from the primary circuit all the same.

There are applications where electrical isolation is needed between two AC circuit without any transformation of voltage or current levels. In these instances, transformers called isolation transformers having 1:1 transformation ratios are used. A benchtop isolation transformer is shown in Figure below

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Isolation transformer isolates power out from the power line.

  • REVIEW:
  • By being able to transfer power from one circuit to another without the use of interconnecting conductors between the two circuits, transformers provide the useful feature of electrical isolation.
  • Transformers designed to provide electrical isolation without stepping voltage and current either up or down are called isolation transformers.

Phasing

Since transformers are essentially AC devices, we need to be aware of the phase relationships between the primary and secondary circuits. Using our SPICE example from before, we can plot the waveshapes for the primary and secondary circuits and see the phase relations for ourselves:

spice transient analysis file for use with nutmeg:transformerv1 1 0 sin(0 15 60 0 0)rbogus1 1 2 1e-12v2 5 0 dc 250l1 2 0 10000l2 3 5 100k l1 l2 0.999vi1 3 4 ac 0rload 4 5 1k.tran 0.5m 17m.end nutmeg commands:setplot tran1plot v(2) v(3,5)

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Secondary voltage V(3,5) is in-phase with primary voltage V(2), and stepped down by factor of ten.

In going from primary, V(2), to secondary, V(3,5), the voltage was stepped down by a factor of ten, , and the current was stepped up by a factor of 10. Both current and voltage waveforms are in-phase in going from primary to secondary.

nutmeg commands:setplot tran1plot I(L1#branch) I(L2#branch)

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Primary and secondary currents are in-phase. Secondary current is stepped up by a factor of ten.

It would appear that both voltage and current for the two transformer windings are in-phase with each other, at least for our resistive load. This is simple enough, but it would be nice to know which way we should connect a transformer in order to ensure the proper phase relationships be kept. After all, a transformer is nothing more than a set of magnetically-linked inductors, and inductors don’t usually come with polarity markings of any kind. If we were to look at an unmarked transformer, we would have no way of knowing which way to hook it up to a circuit to get in-phase (or 180o out-of-phase) voltage and current:

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As a practical matter, the polarity of a transformer can be ambiguous.

Since this is a practical concern, transformer manufacturers have come up with a sort of polarity marking standard to denote phase relationships. It is called the dot convention, and is nothing more than a dot placed next to each corresponding leg of a transformer winding:

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A pair of dots indicates like polarity.

Typically, the transformer will come with some kind of schematic diagram labeling the wire leads for primary and secondary windings. On the diagram will be a pair of dots similar to what is seen above. Sometimes dots will be omitted, but when “H” and “X” labels are used to label transformer winding wires, the subscript numbers are supposed to represent winding polarity. The “1” wires (H1 and X1) represent where the polarity-marking dots would normally be placed.

The similar placement of these dots next to the top ends of the primary and secondary windings tells us that whatever instantaneous voltage polarity seen across the primary winding will be the same as that across the secondary winding. In other words, the phase shift from primary to secondary will be zero degrees.

On the other hand, if the dots on each winding of the transformer do not match up, the phase shift will be 180o between primary and secondary, like this:

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Out of phase: primary red to dot, secondary black to dot.

Of course, the dot convention only tells you which end of each winding is which, relative to the other winding(s). If you want to reverse the phase relationship yourself, all you have to do is swap the winding connections like this:

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In phase: primary red to dot, secondary red to dot.

  • REVIEW:
  • The phase relationships for voltage and current between primary and secondary circuits of a transformer are direct: ideally, zero phase shift.
  • The dot convention is a type of polarity marking for transformer windings showing which end of the winding is which, relative to the other windings.

Winding configurations

Transformers are very versatile devices. The basic concept of energy transfer between mutual inductors is useful enough between a single primary and single secondary coil, but transformers don’t have to be made with just two sets of windings. Consider this transformer circuit:

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Transformer with multiple secondaries, provides multiple output voltages.

Here, three inductor coils share a common magnetic core, magnetically “coupling” or “linking” them together. The relationship of winding turn ratios and voltage ratios seen with a single pair of mutual inductors still holds true here for multiple pairs of coils. It is entirely possible to assemble a transformer such as the one above (one primary winding, two secondary windings) in which one secondary winding is a step-down and the other is a step-up. In fact, this design of transformer was quite common in vacuum tube power supply circuits, which were required to supply low voltage for the tubes’ filaments (typically 6 or 12 volts) and high voltage for the tubes’ plates (several hundred volts) from a nominal primary voltage of 110 volts AC. Not only are voltages and currents of completely different magnitudes possible with such a transformer, but all circuits are electrically isolated from one another.

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Photograph of multiple-winding transformer with six windings, a primary and five secondaries.

The transformer in Figure above is intended to provide both high and low voltages necessary in an electronic system using vacuum tubes. Low voltage is required to power the filaments of vacuum tubes, while high voltage is required to create the potential difference between the plate and cathode elements of each tube. One transformer with multiple windings suffices elegantly to provide all the necessary voltage levels from a single 115 V source. The wires for this transformer (15 of them!) are not shown in the photograph, being hidden from view.

If electrical isolation between secondary circuits is not of great importance, a similar effect can be obtained by “tapping” a single secondary winding at multiple points along its length, like Figure below.

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A single tapped secondary provides multiple voltages.

A tap is nothing more than a wire connection made at some point on a winding between the very ends. Not surprisingly, the winding turn/voltage magnitude relationship of a normal transformer holds true for all tapped segments of windings. This fact can be exploited to produce a transformer capable of multiple ratios:

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A tapped secondary using a switch to select one of many possible voltages.

Carrying the concept of winding taps further, we end up with a “variable transformer,” where a sliding contact is moved along the length of an exposed secondary winding, able to connect with it at any point along its length. The effect is equivalent to having a winding tap at every turn of the winding, and a switch with poles at every tap position:

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A sliding contact on the secondary continuously varies the secondary voltage.

One consumer application of the variable transformer is in speed controls for model train sets, especially the train sets of the 1950’s and 1960’s. These transformers were essentially step-down units, the highest voltage obtainable from the secondary winding being substantially less than the primary voltage of 110 to 120 volts AC. The variable-sweep contact provided a simple means of voltage control with little wasted power, much more efficient than control using a variable resistor!

Moving-slide contacts are too impractical to be used in large industrial power transformer designs, but multi-pole switches and winding taps are common for voltage adjustment. Adjustments need to be made periodically in power systems to accommodate changes in loads over months or years in time, and these switching circuits provide a convenient means. Typically, such “tap switches” are not engineered to handle full-load current, but must be actuated only when the transformer has been de-energized (no power).

Seeing as how we can tap any transformer winding to obtain the equivalent of several windings (albeit with loss of electrical isolation between them), it makes sense that it should be possible to forego electrical isolation altogether and build a transformer from a single winding. Indeed this is possible, and the resulting device is called an autotransformer:

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This autotransformer steps voltage up with a single tapped winding, saving copper, sacrificing isolation.

The autotransformer depicted above performs a voltage step-up function. A step-down autotransformer would look something like figure below

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This auto transformer steps voltage down with a single copper-saving tapped winding.

Autotransformers find popular use in applications requiring a slight boost or reduction in voltage to a load. The alternative with a normal (isolated) transformer would be to either have just the right primary/secondary winding ratio made for the job or use a step-down configuration with the secondary winding connected in series-aiding (“boosting”) or series-opposing (“bucking”) fashion. Primary, secondary, and load voltages are given to illustrate how this would work.

First, the “boosting” configuration. In Figure below the secondary coil’s polarity is oriented so that its voltage directly adds to the primary voltage.

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Ordinary transformer wired as an autotransformer to boost the line voltage.

Next, the “bucking” configuration. In Figure below the secondary coil’s polarity is oriented so that its voltage directly subtracts from the primary voltage:

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Ordinary transformer wired as an autotransformer to buck the line voltage down.

The prime advantage of an autotransformer is that the same boosting or bucking function is obtained with only a single winding, making it cheaper and lighter to manufacture than a regular (isolating) transformer having both primary and secondary windings.

Like regular transformers, autotransformer windings can be tapped to provide variations in ratio. Additionally, they can be made continuously variable with a sliding contact to tap the winding at any point along its length. The latter configuration is popular enough to have earned itself its own name: the Variac.

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A variac is an autotransformer with a sliding tap.

Small variacs for benchtop use are popular pieces of equipment for the electronics experimenter, being able to step household AC voltage down (or sometimes up as well) with a wide, fine range of control by a simple twist of a knob.

  • REVIEW:
  • Transformers can be equipped with more than just a single primary and single secondary winding pair. This allows for multiple step-up and/or step-down ratios in the same device.
  • Transformer windings can also be “tapped:” that is, intersected at many points to segment a single winding into sections.
  • Variable transformers can be made by providing a movable arm that sweeps across the length of a winding, making contact with the winding at any point along its length. The winding, of course, has to be bare (no insulation) in the area where the arm sweeps.
  • An autotransformer is a single, tapped inductor coil used to step up or step down voltage like a transformer, except without providing electrical isolation.
  • A Variac is a variable autotransformer.

Voltage regulation

As we saw in a few SPICE analyses earlier in this chapter, the output voltage of a transformer varies some with varying load resistances, even with a constant voltage input. The degree of variance is affected by the primary and secondary winding inductances, among other factors, not the least of which includes winding resistance and the degree of mutual inductance (magnetic coupling) between the primary and secondary windings. For power transformer applications, where the transformer is seen by the load (ideally) as a constant source of voltage, it is good to have the secondary voltage vary as little as possible for wide variances in load current.

The measure of how well a power transformer maintains constant secondary voltage over a range of load currents is called the transformer’s voltage regulation. It can be calculated from the following formula:

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“Full-load” means the point at which the transformer is operating at maximum permissible secondary current. This operating point will be determined primarily by the winding wire size (ampacity) and the method of transformer cooling. Taking our first SPICE transformer simulation as an example, let’s compare the output voltage with a 1 kΩ load versus a 200 Ω load (assuming that the 200 Ω load will be our “full load” condition). Recall if you will that our constant primary voltage was 10.00 volts AC:

freq v(3,5) i(vi1) 6.000E+01 9.962E+00 9.962E-03 Output with 1k ohm loadfreq v(3,5) i(vi1) 6.000E+01 9.348E+00 4.674E-02 Output with 200 ohm load

Notice how the output voltage decreases as the load gets heavier (more current). Now let’s take that same transformer circuit and place a load resistance of extremely high magnitude across the secondary winding to simulate a “no-load” condition: (See “transformer” spice list”)

transformer v1 1 0 ac 10 sinrbogus1 1 2 1e-12 rbogus2 5 0 9e12l1 2 0 100 l2 3 5 100 k l1 l2 0.999 vi1 3 4 ac 0 rload 4 5 9e12 .ac lin 1 60 60 .print ac v(2,0) i(v1) .print ac v(3,5) i(vi1) .end freq v(2) i(v1) 6.000E+01 1.000E+01 2.653E-04freq v(3,5) i(vi1) 6.000E+01 9.990E+00 1.110E-12 Output with (almost) no load

So, we see that our output (secondary) voltage spans a range of 9.990 volts at (virtually) no load and 9.348 volts at the point we decided to call “full load.” Calculating voltage regulation with these figures, we get:

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Incidentally, this would be considered rather poor (or “loose”) regulation for a power transformer. Powering a simple resistive load like this, a good power transformer should exhibit a regulation percentage of less than 3%. Inductive loads tend to create a condition of worse voltage regulation, so this analysis with purely resistive loads was a “best-case” condition.

There are some applications, however, where poor regulation is actually desired. One such case is in discharge lighting, where a step-up transformer is required to initially generate a high voltage (necessary to “ignite” the lamps), then the voltage is expected to drop off once the lamp begins to draw current. This is because discharge lamps’ voltage requirements tend to be much lower after a current has been established through the arc path. In this case, a step-up transformer with poor voltage regulation suffices nicely for the task of conditioning power to the lamp.

Another application is in current control for AC arc welders, which are nothing more than step-down transformers supplying low-voltage, high-current power for the welding process. A high voltage is desired to assist in “striking” the arc (getting it started), but like the discharge lamp, an arc doesn’t require as much voltage to sustain itself once the air has been heated to the point of ionization. Thus, a decrease of secondary voltage under high load current would be a good thing. Some arc welder designs provide arc current adjustment by means of a movable iron core in the transformer, cranked in or out of the winding assembly by the operator. Moving the iron slug away from the windings reduces the strength of magnetic coupling between the windings, which diminishes no-load secondary voltage and makes for poorer voltage regulation.

No exposition on transformer regulation could be called complete without mention of an unusual device called a ferroresonant transformer. “Ferroresonance” is a phenomenon associated with the behavior of iron cores while operating near a point of magnetic saturation (where the core is so strongly magnetized that further increases in winding current results in little or no increase in magnetic flux).

While being somewhat difficult to describe without going deep into electromagnetic theory, the ferroresonant transformer is a power transformer engineered to operate in a condition of persistent core saturation. That is, its iron core is “stuffed full” of magnetic lines of flux for a large portion of the AC cycle so that variations in supply voltage (primary winding current) have little effect on the core’s magnetic flux density, which means the secondary winding outputs a nearly constant voltage despite significant variations in supply (primary winding) voltage. Normally, core saturation in a transformer results in distortion of the sinewave shape, and the ferroresonant transformer is no exception. To combat this side effect, ferroresonant transformers have an auxiliary secondary winding paralleled with one or more capacitors, forming a resonant circuit tuned to the power supply frequency. This “tank circuit” serves as a filter to reject harmonics created by the core saturation, and provides the added benefit of storing energy in the form of AC oscillations, which is available for sustaining output winding voltage for brief periods of input voltage loss (milliseconds’ worth of time, but certainly better than nothing).

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Ferroresonant transformer provides voltage regulation of the output.

In addition to blocking harmonics created by the saturated core, this resonant circuit also “filters out” harmonic frequencies generated by nonlinear (switching) loads in the secondary winding circuit and any harmonics present in the source voltage, providing “clean” power to the load.

Ferroresonant transformers offer several features useful in AC power conditioning: constant output voltage given substantial variations in input voltage, harmonic filtering between the power source and the load, and the ability to “ride through” brief losses in power by keeping a reserve of energy in its resonant tank circuit. These transformers are also highly tolerant of excessive loading and transient (momentary) voltage surges. They are so tolerant, in fact, that some may be briefly paralleled with unsynchronized AC power sources, allowing a load to be switched from one source of power to another in a “make-before-break” fashion with no interruption of power on the secondary side!

Unfortunately, these devices have equally noteworthy disadvantages: they waste a lot of energy (due to hysteresis losses in the saturated core), generating significant heat in the process, and are intolerant of frequency variations, which means they don’t work very well when powered by small engine-driven generators having poor speed regulation. Voltages produced in the resonant winding/capacitor circuit tend to be very high, necessitating expensive capacitors and presenting the service technician with very dangerous working voltages. Some applications, though, may prioritize the ferroresonant transformer’s advantages over its disadvantages. Semiconductor circuits exist to “condition” AC power as an alternative to ferroresonant devices, but none can compete with this transformer in terms of sheer simplicity.

  • REVIEW:
  • Voltage regulation is the measure of how well a power transformer can maintain constant secondary voltage given a constant primary voltage and wide variance in load current. The lower the percentage (closer to zero), the more stable the secondary voltage and the better the regulation it will provide.
  • A ferroresonant transformer is a special transformer designed to regulate voltage at a stable level despite wide variation in input voltage.

Special transformers and applications

Impedance matching

Because transformers can step voltage and current to different levels, and because power is transferred equivalently between primary and secondary windings, they can be used to “convert” the impedance of a load to a different level. That last phrase deserves some explanation, so let’s investigate what it means.

The purpose of a load (usually) is to do something productive with the power it dissipates. In the case of a resistive heating element, the practical purpose for the power dissipated is to heat something up. Loads are engineered to safely dissipate a certain maximum amount of power, but two loads of equal power rating are not necessarily identical. Consider these two 1000 watt resistive heating elements: (

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Heating elements dissipate 1000 watts, at different voltage and current ratings.

Both heaters dissipate exactly 1000 watts of power, but they do so at different voltage and current levels (either 250 volts and 4 amps, or 125 volts and 8 amps). Using Ohm’s Law to determine the necessary resistance of these heating elements (R=E/I), we arrive at figures of 62.5 Ω and 15.625 Ω, respectively. If these are AC loads, we might refer to their opposition to current in terms of impedance rather than plain resistance, although in this case that’s all they’re composed of (no reactance). The 250 volt heater would be said to be a higher impedance load than the 125 volt heater.

If we desired to operate the 250 volt heater element directly on a 125 volt power system, we would end up being disappointed. With 62.5 Ω of impedance (resistance), the current would only be 2 amps (I=E/R; 125/62.5), and the power dissipation would only be 250 watts (P=IE; 125 x 2), or one-quarter of its rated power. The impedance of the heater and the voltage of our source would be mismatched, and we couldn’t obtain the full rated power dissipation from the heater.

All hope is not lost, though. With a step-up transformer, we could operate the 250 volt heater element on the 125 volt power system like Figure below.

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Step-up transformer operates 1000 watt 250 V heater from 125 V power source

The ratio of the transformer’s windings provides the voltage step-up and current step-down we need for the otherwise mismatched load to operate properly on this system. Take a close look at the primary circuit figures: 125 volts at 8 amps. As far as the power supply “knows,” its powering a 15.625 Ω (R=E/I) load at 125 volts, not a 62.5 Ω load! The voltage and current figures for the primary winding are indicative of 15.625 Ω load impedance, not the actual 62.5 Ω of the load itself. In other words, not only has our step-up transformer transformed voltage and current, but it has transformed impedance as well.

The transformation ratio of impedance is the square of the voltage/current transformation ratio, the same as the winding inductance ratio:

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This concurs with our example of the 2:1 step-up transformer and the impedance ratio of 62.5 Ω to 15.625 Ω (a 4:1 ratio, which is 2:1 squared). Impedance transformation is a highly useful ability of transformers, for it allows a load to dissipate its full rated power even if the power system is not at the proper voltage to directly do so.

Recall from our study of network analysis the Maximum Power Transfer Theorem, which states that the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. Substitute the word “impedance” for “resistance” in that definition and you have the AC version of that Theorem. If we’re trying to obtain theoretical maximum power dissipation from a load, we must be able to properly match the load impedance and source (Thevenin/Norton) impedance together. This is generally more of a concern in specialized electric circuits such as radio transmitter/antenna and audio amplifier/speaker systems. Let’s take an audio amplifier system and see how it works:

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Amplifier with impedance of 500 Ω drives 8 Ω at much less than maximum power.

With an internal impedance of 500 Ω, the amplifier can only deliver full power to a load (speaker) also having 500 Ω of impedance. Such a load would drop higher voltage and draw less current than an 8 Ω speaker dissipating the same amount of power. If an 8 Ω speaker were connected directly to the 500 Ω amplifier as shown, the impedance mismatch would result in very poor (low peak power) performance. Additionally, the amplifier would tend to dissipate more than its fair share of power in the form of heat trying to drive the low impedance speaker.

To make this system work better, we can use a transformer to match these mismatched impedances. Since we’re going from a high impedance (high voltage, low current) supply to a low impedance (low voltage, high current) load, we’ll need to use a step-down transformer:

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Impedance matching transformer matches 500 Ω amplifier to 8 Ω speaker for maximum efficiency.

To obtain an impedance transformation ratio of 500:8, we would need a winding ratio equal to the square root of 500:8 (the square root of 62.5:1, or 7.906:1). With such a transformer in place, the speaker will load the amplifier to just the right degree, drawing power at the correct voltage and current levels to satisfy the Maximum Power Transfer Theorem and make for the most efficient power delivery to the load. The use of a transformer in this capacity is called impedance matching.

Anyone who has ridden a multi-speed bicycle can intuitively understand the principle of impedance matching. A human’s legs will produce maximum power when spinning the bicycle crank at a particular speed (about 60 to 90 revolution per minute). Above or below that rotational speed, human leg muscles are less efficient at generating power. The purpose of the bicycle’s “gears” is to impedance-match the rider’s legs to the riding conditions so that they always spin the crank at the optimum speed.

If the rider attempts to start moving while the bicycle is shifted into its “top” gear, he or she will find it very difficult to get moving. Is it because the rider is weak? No, its because the high step-up ratio of the bicycle’s chain and sprockets in that top gear presents a mismatch between the conditions (lots of inertia to overcome) and their legs (needing to spin at 60-90 RPM for maximum power output). On the other hand, selecting a gear that is too low will enable the rider to get moving immediately, but limit the top speed they will be able to attain. Again, is the lack of speed an indication of weakness in the bicyclist’s legs? No, its because the lower speed ratio of the selected gear creates another type of mismatch between the conditions (low load) and the rider’s legs (losing power if spinning faster than 90 RPM). It is much the same with electric power sources and loads: there must be an impedance match for maximum system efficiency. In AC circuits, transformers perform the same matching function as the sprockets and chain (“gears”) on a bicycle to match otherwise mismatched sources and loads.

Impedance matching transformers are not fundamentally different from any other type of transformer in construction or appearance. A small impedance-matching transformer (about two centimeters in width) for audio-frequency applications is shown in the following photograph:

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Printed circuit board mounted audio impedance matching transformer, top right.

Potential transformers

Transformers can also be used in electrical instrumentation systems. Due to transformers’ ability to step up or step down voltage and current, and the electrical isolation they provide, they can serve as a way of connecting electrical instrumentation to high-voltage, high current power systems. Suppose we wanted to accurately measure the voltage of a 13.8 kV power system (a very common power distribution voltage in American industry):

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Direct measurement of high voltage by a voltmeter is a potential safety hazard.

Designing, installing, and maintaining a voltmeter capable of directly measuring 13,800 volts AC would be no easy task. The safety hazard alone of bringing 13.8 kV conductors into an instrument panel would be severe, not to mention the design of the voltmeter itself. However, by using a precision step-down transformer, we can reduce the 13.8 kV down to a safe level of voltage at a constant ratio, and isolate it from the instrument connections, adding an additional level of safety to the metering system:

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Instrumentation application:“Potential transformer” precisely scales dangerous high voltage to a safe value applicable to a conventional voltmeter.

Now the voltmeter reads a precise fraction, or ratio, of the actual system voltage, its scale set to read as though it were measuring the voltage directly. The transformer keeps the instrument voltage at a safe level and electrically isolates it from the power system, so there is no direct connection between the power lines and the instrument or instrument wiring. When used in this capacity, the transformer is called a Potential Transformer, or simply PT.

Potential transformers are designed to provide as accurate a voltage step-down ratio as possible. To aid in precise voltage regulation, loading is kept to a minimum: the voltmeter is made to have high input impedance so as to draw as little current from the PT as possible. As you can see, a fuse has been connected in series with the PTs primary winding, for safety and ease of disconnecting the PT from the circuit.

A standard secondary voltage for a PT is 120 volts AC, for full-rated power line voltage. The standard voltmeter range to accompany a PT is 150 volts, full-scale. PTs with custom winding ratios can be manufactured to suit any application. This lends itself well to industry standardization of the actual voltmeter instruments themselves, since the PT will be sized to step the system voltage down to this standard instrument level.

Current transformers

Following the same line of thinking, we can use a transformer to step down current through a power line so that we are able to safely and easily measure high system currents with inexpensive ammeters. Of course, such a transformer would be connected in series with the power line, like

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Instrumentation application: “Currrent transformer” steps high current down to a value applicable to a conventional ammeter.

Note that while the PT is a step-down device, the Current Transformer (or CT) is a step-up device (with respect to voltage), which is what is needed to step down the power line current. Quite often, CTs are built as donut-shaped devices through which the power line conductor is run, the power line itself acting as a single-turn primary winding:

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Current conductor to be measured is threaded through the opening. Scaled down current is available on wire leads.

Some CTs are made to hinge open, allowing insertion around a power conductor without disturbing the conductor at all. The industry standard secondary current for a CT is a range of 0 to 5 amps AC. Like PTs, CTs can be made with custom winding ratios to fit almost any application. Because their “full load” secondary current is 5 amps, CT ratios are usually described in terms of full-load primary amps to 5 amps, like this:

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The “donut” CT shown in the photograph has a ratio of 50:5. That is, when the conductor through the center of the torus is carrying 50 amps of current (AC), there will be 5 amps of current induced in the CT’s winding.

Because CTs are designed to be powering ammeters, which are low-impedance loads, and they are wound as voltage step-up transformers, they should never, ever be operated with an open-circuited secondary winding. Failure to heed this warning will result in the CT producing extremely high secondary voltages, dangerous to equipment and personnel alike. To facilitate maintenance of ammeter instrumentation, short-circuiting switches are often installed in parallel with the CT’s secondary winding, to be closed whenever the ammeter is removed for service:

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Short-circuit switch allows ammeter to be removed from an active current transformer circuit.

Though it may seem strange to intentionally short-circuit a power system component, it is perfectly proper and quite necessary when working with current transformers.

Air core transformers

Another kind of special transformer, seen often in radio-frequency circuits, is the air core transformer. True to its name, an air core transformer has its windings wrapped around a nonmagnetic form, usually a hollow tube of some material. The degree of coupling (mutual inductance) between windings in such a transformer is many times less than that of an equivalent iron-core transformer, but the undesirable characteristics of a ferromagnetic core (eddy current losses, hysteresis, saturation, etc.) are completely eliminated. It is in high-frequency applications that these effects of iron cores are most problematic.

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Air core transformers may be wound on cylindrical (a) or toroidal (b) forms. Center tapped primary with secondary (a). Bifilar winding on toroidal form (b).

The inside tapped solenoid winding, without the over winding, could match unequal impedances when DC isolation is not required. When isolation is required the over winding is added over one end of the main winding. Air core transformers are used at radio frequencies when iron core losses are too high. Frequently air core transformers are paralleled with a capacitor to tune it to resonance. The over winding is connected between a radio antenna and ground for one such application. The secondary is tuned to resonance with a variable capacitor. The output may be taken from the tap point for amplification or detection. Small millimeter size air core transformers are used in radio receivers. The largest radio transmitters may use meter sized coils. Unshielded air core solenoid transformers are mounted at right angles to each other to prevent stray coupling.

Stray coupling is minimized when the transformer is wound on a toroid form. Toroidal air core transformers also show a higher degree of coupling, particularly for bifilar windings. Bifilar windings are wound from a slightly twisted pair of wires. This implies a 1:1 turns ratio. Three or four wires may be grouped for 1:2 and other integral ratios. Windings do not have to be bifilar. This allows arbitrary turns ratios. However, the degree of coupling suffers. Toroidal air core transformers are rare except for VHF (Very High Frequency) work. Core materials other than air such as powdered iron or ferrite are preferred for lower radio frequencies.

Tesla Coil

One notable example of an air-core transformer is the Tesla Coil, named after the Serbian electrical genius Nikola Tesla, who was also the inventor of the rotating magnetic field AC motor, polyphase AC power systems, and many elements of radio technology. The Tesla Coil is a resonant, high-frequency step-up transformer used to produce extremely high voltages. One of Tesla’s dreams was to employ his coil technology to distribute electric power without the need for wires, simply broadcasting it in the form of radio waves which could be received and conducted to loads by means of antennas. The basic schematic for a Tesla Coil is shown in figure below

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Tesla Coil: A few heavy primary turns, many secondary turns.

The capacitor, in conjunction with the transformer’s primary winding, forms a tank circuit. The secondary winding is wound in close proximity to the primary, usually around the same nonmagnetic form. Several options exist for “exciting” the primary circuit, the simplest being a high-voltage, low-frequency AC source and spark gap:

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System level diagram of Tesla coil with spark gap drive.

The purpose of the high-voltage, low-frequency AC power source is to “charge” the primary tank circuit. When the spark gap fires, its low impedance acts to complete the capacitor/primary coil tank circuit, allowing it to oscillate at its resonant frequency. The “RFC” inductors are “Radio Frequency Chokes,” which act as high impedances to prevent the AC source from interfering with the oscillating tank circuit.

The secondary side of the Tesla coil transformer is also a tank circuit, relying on the parasitic (stray) capacitance existing between the discharge terminal and earth ground to complement the secondary winding’s inductance. For optimum operation, this secondary tank circuit is tuned to the same resonant frequency as the primary circuit, with energy exchanged not only between capacitors and inductors during resonant oscillation, but also back-and-forth between primary and secondary windings. The visual results are spectacular:

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High voltage high frequency discharge from Tesla coil.

Tesla Coils find application primarily as novelty devices, showing up in high school science fairs, basement workshops, and the occasional low budget science-fiction movie.

It should be noted that Tesla coils can be extremely dangerous devices. Burns caused by radio-frequency (“RF”) current, like all electrical burns, can be very deep, unlike skin burns caused by contact with hot objects or flames. Although the high-frequency discharge of a Tesla coil has the curious property of being beyond the “shock perception” frequency of the human nervous system, this does not mean Tesla coils cannot hurt or even kill you! I strongly advise seeking the assistance of an experienced Tesla coil experimenter if you would embark on building one yourself.

Saturable reactors

So far, we’ve explored the transformer as a device for converting different levels of voltage, current, and even impedance from one circuit to another. Now we’ll take a look at it as a completely different kind of device: one that allows a small electrical signal to exert control over a much larger quantity of electrical power. In this mode, a transformer acts as an amplifier.

The device I’m referring to is called a saturable-core reactor, or simply saturable reactor. Actually, it is not really a transformer at all, but rather a special kind of inductor whose inductance can be varied by the application of a DC current through a second winding wound around the same iron core. Like the ferroresonant transformer, the saturable reactor relies on the principle of magnetic saturation. When a material such as iron is completely saturated (that is, all its magnetic domains are lined up with the applied magnetizing force), additional increases in current through the magnetizing winding will not result in further increases of magnetic flux.

Now, inductance is the measure of how well an inductor opposes changes in current by developing a voltage in an opposing direction. The ability of an inductor to generate this opposing voltage is directly connected with the change in magnetic flux inside the inductor resulting from the change in current, and the number of winding turns in the inductor. If an inductor has a saturated core, no further magnetic flux will result from further increases in current, and so there will be no voltage induced in opposition to the change in current. In other words, an inductor loses its inductance (ability to oppose changes in current) when its core becomes magnetically saturated.

If an inductor’s inductance changes, then its reactance (and impedance) to AC current changes as well. In a circuit with a constant voltage source, this will result in a change in current:

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If L changes in inductance, ZL will correspondingly change, thus changing the circuit current.

A saturable reactor capitalizes on this effect by forcing the core into a state of saturation with a strong magnetic field generated by current through another winding. The reactor’s “power” winding is the one carrying the AC load current, and the “control” winding is one carrying a DC current strong enough to drive the core into saturation:

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DC, via the control winding, saturates the core. Thus, modulating the power winding inductance, impedance, and current.

The strange-looking transformer symbol shown in the above schematic represents a saturable-core reactor, the upper winding being the DC control winding and the lower being the “power” winding through which the controlled AC current goes. Increased DC control current produces more magnetic flux in the reactor core, driving it closer to a condition of saturation, thus decreasing the power winding’s inductance, decreasing its impedance, and increasing current to the load. Thus, the DC control current is able to exert control over the AC current delivered to the load.

The circuit shown would work, but it would not work very well. The first problem is the natural transformer action of the saturable reactor: AC current through the power winding will induce a voltage in the control winding, which may cause trouble for the DC power source. Also, saturable reactors tend to regulate AC power only in one direction: in one half of the AC cycle, the mmf’s from both windings add; in the other half, they subtract. Thus, the core will have more flux in it during one half of the AC cycle than the other, and will saturate first in that cycle half, passing load current more easily in one direction than the other. Fortunately, both problems can be overcome with a little ingenuity:

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Out of phase DC control windings allow symmetrical of control AC.

Notice the placement of the phasing dots on the two reactors: the power windings are “in phase” while the control windings are “out of phase.” If both reactors are identical, any voltage induced in the control windings by load current through the power windings will cancel out to zero at the battery terminals, thus eliminating the first problem mentioned. Furthermore, since the DC control current through both reactors produces magnetic fluxes in different directions through the reactor cores, one reactor will saturate more in one cycle of the AC power while the other reactor will saturate more in the other, thus equalizing the control action through each half-cycle so that the AC power is “throttled” symmetrically. This phasing of control windings can be accomplished with two separate reactors as shown, or in a single reactor design with intelligent layout of the windings and core.

Saturable reactor technology has even been miniaturized to the circuit-board level in compact packages more generally known as magnetic amplifiers. I personally find this to be fascinating: the effect of amplification (one electrical signal controlling another), normally requiring the use of physically fragile vacuum tubes or electrically “fragile” semiconductor devices, can be realized in a device both physically and electrically rugged. Magnetic amplifiers do have disadvantages over their more fragile counterparts, namely size, weight, nonlinearity, and bandwidth (frequency response), but their utter simplicity still commands a certain degree of appreciation, if not practical application.

Saturable-core reactors are less commonly known as “saturable-core inductors” or transductors.

Scott-T transformer

Nikola Tesla’s original polyphase power system was based on simple to build 2-phase components. However, as transmission distances increased, the more transmission line efficient 3-phase system became more prominent. Both 2-φ and 3-φ components coexisted for a number of years. The Scott-T transformer connection allowed 2-φ and 3-φ components like motors and alternators to be interconnected. Yamamoto and Yamaguchi:

In 1896, General Electric built a 35.5 km (22 mi) three-phase transmission line operated at 11 kV to transmit power to Buffalo, New York, from the Niagara Falls Project. The two-phase generated power was changed tothree-phase by the use of Scott-T transformations.

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Scott-T transformer converts 2-φ to 3-φ, or vice versa.

The Scott-T transformer set, figure above, consists of a center tapped transformer T1 and an 86.6% tapped transformer T2 on the 3-φ side of the circuit. The primaries of both transformers are connected to the 2-φ voltages. One end of the T2 86.6% secondary winding is a 3-φ output, the other end is connected to the T1 secondary center tap. Both ends of the T1 secondary are the other two 3-φ connections.

Application of 2-φ Niagara generator power produced a 3-φ output for the more efficient 3-φ transmission line. More common these days is the application of 3-φ power to produce a 2-φ output for driving an old 2-φ motor.

In figure below, we use vectors in both polar and complex notation to prove that the Scott-T converts a pair of 2-φ voltages to 3-φ. First, one of the 3-φ voltages is identical to a 2-φ voltage due to the 1:1 transformer T1 ratio, VP12= V2P1. The T1 center tapped secondary produces opposite polarities of 0.5V2P1 on the secondary ends. This ∠0o is vectorially subtracted from T2 secondary voltage due to the KVL equations V31, V23. The T2 secondary voltage is 0.866V2P2 due to the 86.6% tap. Keep in mind that this 2nd phase of the 2-φ is ∠90o. This 0.866V2P2 is added at V31, subtracted at V23 in the KVL equations.

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Scott-T transformer 2-φ to 3-φ conversion equations.

We show “DC” polarities all over this AC only circuit, to keep track of the Kirchhoff voltage loop polarities. Subtracting ∠0o is equivalent to adding ∠180o. The bottom line is when we add 86.6% of ∠90o to 50% of ∠180o we get ∠120o. Subtracting 86.6% of ∠90o from 50% of ∠180o yields ∠-120o or ∠240o.

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Graphical explanation of equations in Figure previous.

In Figure above we graphically show the 2-φ vectors at (a). At (b) the vectors are scaled by transformers T1 and T2 to 0.5 and 0.866 respectively. At (c) 1∠120o = -0.5∠0o + 0.866∠90o, and 1∠240o = -0.5∠0o – 0.866∠90o. The three output phases are 1∠120o and 1∠240o from (c), along with input 1∠0o (a).

Linear Variable Differential Transformer

A linear variable differential transformer (LVDT) has an AC driven primary wound between two secondaries on a cylindrical air core form. A movable ferromagnetic slug converts displacement to a variable voltage by changing the coupling between the driven primary and secondary windings. The LVDT is a displacement or distance measuring transducer. Units are available for measuring displacement over a distance of a fraction of a millimeter to a half a meter. LVDT’s are rugged and dirt resistant compared to linear optical encoders.

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LVDT: linear variable differential transformer.

The excitation voltage is in the range of 0.5 to 10 VAC at a frequency of 1 to 200 Khz. A ferrite core is suitable at these frequencies. It is extended outside the body by an non-magnetic rod. As the core is moved toward the top winding, the voltage across this coil increases due to increased coupling, while the voltage on the bottom coil decreases. If the core is moved toward the bottom winding, the voltage on this coil increases as the voltage decreases across the top coil. Theoretically, a centered slug yields equal voltages across both coils. In practice leakage inductance prevents the null from dropping all the way to 0 V.

With a centered slug, the series-opposing wired secondaries cancel yielding V13 = 0. Moving the slug up increases V13. Note that it is in-phase with with V1, the top winding, and 180o out of phase with V3, bottom winding.

Moving the slug down from the center position increases V13. However, it is 180o out of phase with with V1, the top winding, and in-phase with V3, bottom winding. Moving the slug from top to bottom shows a minimum at the center point, with a 180o phase reversal in passing the center.

  • REVIEW:
  • Transformers can be used to transform impedance as well as voltage and current. When this is done to improve power transfer to a load, it is called impedance matching.
  • A Potential Transformer (PT) is a special instrument transformer designed to provide a precise voltage step-down ratio for voltmeters measuring high power system voltages.
  • A Current Transformer (CT) is another special instrument transformer designed to step down the current through a power line to a safe level for an ammeter to measure.
  • An air-core transformer is one lacking a ferromagnetic core.
  • A Tesla Coil is a resonant, air-core, step-up transformer designed to produce very high AC voltages at high frequency.
  • A saturable reactor is a special type of inductor, the inductance of which can be controlled by the DC current through a second winding around the same core. With enough DC current, the magnetic core can be saturated, decreasing the inductance of the power winding in a controlled fashion.
  • A Scott-T transformer converts 3-φ power to 2-φ power and vice versa.
  • A linear variable differential transformer, also known as an LVDT, is a distance measuring device. It has a movable ferromagnetic core to vary the coupling between the excited primary and a pair of secondaries.

Practical considerations

Power capacity

As has already been observed, transformers must be well designed in order to achieve acceptable power coupling, tight voltage regulation, and low exciting current distortion. Also, transformers must be designed to carry the expected values of primary and secondary winding current without any trouble. This means the winding conductors must be made of the proper gauge wire to avoid any heating problems. An ideal transformer would have perfect coupling (no leakage inductance), perfect voltage regulation, perfectly sinusoidal exciting current, no hysteresis or eddy current losses, and wire thick enough to handle any amount of current. Unfortunately, the ideal transformer would have to be infinitely large and heavy to meet these design goals. Thus, in the business of practical transformer design, compromises must be made.

Additionally, winding conductor insulation is a concern where high voltages are encountered, as they often are in step-up and step-down power distribution transformers. Not only do the windings have to be well insulated from the iron core, but each winding has to be sufficiently insulated from the other in order to maintain electrical isolation between windings.

Respecting these limitations, transformers are rated for certain levels of primary and secondary winding voltage and current, though the current rating is usually derived from a volt-amp (VA) rating assigned to the transformer. For example, take a step-down transformer with a primary voltage rating of 120 volts, a secondary voltage rating of 48 volts, and a VA rating of 1 kVA (1000 VA). The maximum winding currents can be determined as such:

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Sometimes windings will bear current ratings in amps, but this is typically seen on small transformers. Large transformers are almost always rated in terms of winding voltage and VA or kVA.

Energy losses

When transformers transfer power, they do so with a minimum of loss. As it was stated earlier, modern power transformer designs typically exceed 95% efficiency. It is good to know where some of this lost power goes, however, and what causes it to be lost.

There is, of course, power lost due to resistance of the wire windings. Unless superconducting wires are used, there will always be power dissipated in the form of heat through the resistance of current-carrying conductors. Because transformers require such long lengths of wire, this loss can be a significant factor. Increasing the gauge of the winding wire is one way to minimize this loss, but only with substantial increases in cost, size, and weight.

Resistive losses aside, the bulk of transformer power loss is due to magnetic effects in the core. Perhaps the most significant of these “core losses” is eddy-current loss, which is resistive power dissipation due to the passage of induced currents through the iron of the core. Because iron is a conductor of electricity as well as being an excellent “conductor” of magnetic flux, there will be currents induced in the iron just as there are currents induced in the secondary windings from the alternating magnetic field. These induced currents — as described by the perpendicularity clause of Faraday’s Law — tend to circulate through the cross-section of the core perpendicularly to the primary winding turns. Their circular motion gives them their unusual name: like eddies in a stream of water that circulate rather than move in straight lines.

Iron is a fair conductor of electricity, but not as good as the copper or aluminum from which wire windings are typically made. Consequently, these “eddy currents” must overcome significant electrical resistance as they circulate through the core. In overcoming the resistance offered by the iron, they dissipate power in the form of heat. Hence, we have a source of inefficiency in the transformer that is difficult to eliminate.

This phenomenon is so pronounced that it is often exploited as a means of heating ferrous (iron-containing) materials. The photograph of below) shows an “induction heating” unit raising the temperature of a large pipe section. Loops of wire covered by high-temperature insulation encircle the pipe’s circumference, inducing eddy currents within the pipe wall by electromagnetic induction. In order to maximize the eddy current effect, high-frequency alternating current is used rather than power line frequency (60 Hz). The box units at the right of the picture produce the high-frequency AC and control the amount of current in the wires to stabilize the pipe temperature at a pre-determined “set-point.”

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Induction heating: Primary insulated winding induces current into lossy iron pipe (secondary).

The main strategy in mitigating these wasteful eddy currents in transformer cores is to form the iron core in sheets, each sheet covered with an insulating varnish so that the core is divided up into thin slices. The result is very little width in the core for eddy currents to circulate in:

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Dividing the iron core into thin insulated laminations minimizes eddy current loss.

Laminated cores like the one shown here are standard in almost all low-frequency transformers. Recall from the photograph of the transformer cut in half that the iron core was composed of many thin sheets rather than one solid piece. Eddy current losses increase with frequency, so transformers designed to run on higher-frequency power (such as 400 Hz, used in many military and aircraft applications) must use thinner laminations to keep the losses down to a respectable minimum. This has the undesirable effect of increasing the manufacturing cost of the transformer.

Another, similar technique for minimizing eddy current losses which works better for high-frequency applications is to make the core out of iron powder instead of thin iron sheets. Like the lamination sheets, these granules of iron are individually coated in an electrically insulating material, which makes the core nonconductive except for within the width of each granule. Powdered iron cores are often found in transformers handling radio-frequency currents.

Another “core loss” is that of magnetic hysteresis. All ferromagnetic materials tend to retain some degree of magnetization after exposure to an external magnetic field. This tendency to stay magnetized is called “hysteresis,” and it takes a certain investment in energy to overcome this opposition to change every time the magnetic field produced by the primary winding changes polarity (twice per AC cycle). This type of loss can be mitigated through good core material selection (choosing a core alloy with low hysteresis, as evidenced by a “thin” B/H hysteresis curve), and designing the core for minimum flux density (large cross-sectional area).

Transformer energy losses tend to worsen with increasing frequency. The skin effect within winding conductors reduces the available cross-sectional area for electron flow, thereby increasing effective resistance as the frequency goes up and creating more power lost through resistive dissipation. Magnetic core losses are also exaggerated with higher frequencies, eddy currents and hysteresis effects becoming more severe. For this reason, transformers of significant size are designed to operate efficiently in a limited range of frequencies. In most power distribution systems where the line frequency is very stable, one would think excessive frequency would never pose a problem. Unfortunately it does, in the form of harmonics created by nonlinear loads.

As we’ve seen in earlier chapters, nonsinusoidal waveforms are equivalent to additive series of multiple sinusoidal waveforms at different amplitudes and frequencies. In power systems, these other frequencies are whole-number multiples of the fundamental (line) frequency, meaning that they will always be higher, not lower, than the design frequency of the transformer. In significant measure, they can cause severe transformer overheating. Power transformers can be engineered to handle certain levels of power system harmonics, and this capability is sometimes denoted with a “K factor” rating.

Stray capacitance and inductance

Aside from power ratings and power losses, transformers often harbor other undesirable limitations which circuit designers must be made aware of. Like their simpler counterparts — inductors — transformers exhibit capacitance due to the insulation dielectric between conductors: from winding to winding, turn to turn (in a single winding), and winding to core. Usually this capacitance is of no concern in a power application, but small signal applications (especially those of high frequency) may not tolerate this quirk well. Also, the effect of having capacitance along with the windings’ designed inductance gives transformers the ability to resonate at a particular frequency, definitely a design concern in signal applications where the applied frequency may reach this point (usually the resonant frequency of a power transformer is well beyond the frequency of the AC power it was designed to operate on).

Flux containment (making sure a transformer’s magnetic flux doesn’t escape so as to interfere with another device, and making sure other devices’ magnetic flux is shielded from the transformer core) is another concern shared both by inductors and transformers.

Closely related to the issue of flux containment is leakage inductance. We’ve already seen the detrimental effects of leakage inductance on voltage regulation with SPICE simulations early in this chapter. Because leakage inductance is equivalent to an inductance connected in series with the transformer’s winding, it manifests itself as a series impedance with the load. Thus, the more current drawn by the load, the less voltage available at the secondary winding terminals. Usually, good voltage regulation is desired in transformer design, but there are exceptional applications. As was stated before, discharge lighting circuits require a step-up transformer with “loose” (poor) voltage regulation to ensure reduced voltage after the establishment of an arc through the lamp. One way to meet this design criterion is to engineer the transformer with flux leakage paths for magnetic flux to bypass the secondary winding(s). The resulting leakage flux will produce leakage inductance, which will in turn produce the poor regulation needed for discharge lighting.

Core saturation

Transformers are also constrained in their performance by the magnetic flux limitations of the core. For ferromagnetic core transformers, we must be mindful of the saturation limits of the core. Remember that ferromagnetic materials cannot support infinite magnetic flux densities: they tend to “saturate” at a certain level (dictated by the material and core dimensions), meaning that further increases in magnetic field force (mmf) do not result in proportional increases in magnetic field flux (Φ).

When a transformer’s primary winding is overloaded from excessive applied voltage, the core flux may reach saturation levels during peak moments of the AC sinewave cycle. If this happens, the voltage induced in the secondary winding will no longer match the wave-shape as the voltage powering the primary coil. In other words, the overloaded transformer will distort the waveshape from primary to secondary windings, creating harmonics in the secondary winding’s output. As we discussed before, harmonic content in AC power systems typically causes problems.

Special transformers known as peaking transformers exploit this principle to produce brief voltage pulses near the peaks of the source voltage waveform. The core is designed to saturate quickly and sharply, at voltage levels well below peak. This results in a severely cropped sine-wave flux waveform, and secondary voltage pulses only when the flux is changing (below saturation levels):

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Voltage and flux waveforms for a peaking transformer.

Another cause of abnormal transformer core saturation is operation at frequencies lower than normal. For example, if a power transformer designed to operate at 60 Hz is forced to operate at 50 Hz instead, the flux must reach greater peak levels than before in order to produce the same opposing voltage needed to balance against the source voltage. This is true even if the source voltage is the same as before.

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Magnetic flux is higher in a transformer core driven by 50 Hz as compared to 60 Hz for the same voltage.

Since instantaneous winding voltage is proportional to the instantaneous magnetic flux’s rate of change in a transformer, a voltage waveform reaching the same peak value, but taking a longer amount of time to complete each half-cycle, demands that the flux maintain the same rate of change as before, but for longer periods of time. Thus, if the flux has to climb at the same rate as before, but for longer periods of time, it will climb to a greater peak value.

Mathematically, this is another example of calculus in action. Because the voltage is proportional to the flux’s rate-of-change, we say that the voltage waveform is the derivative of the flux waveform, “derivative” being that calculus operation defining one mathematical function (waveform) in terms of the rate-of-change of another. If we take the opposite perspective, though, and relate the original waveform to its derivative, we may call the original waveform the integral of the derivative waveform. In this case, the voltage waveform is the derivative of the flux waveform, and the flux waveform is the integral of the voltage waveform.

The integral of any mathematical function is proportional to the area accumulated underneath the curve of that function. Since each half-cycle of the 50 Hz waveform accumulates more area between it and the zero line of the graph than the 60 Hz waveform will — and we know that the magnetic flux is the integral of the voltage — the flux will attain higher values in Figure below.

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Flux changing at the same rate rises to a higher level at 50 Hz than at 60 Hz.

Yet another cause of transformer saturation is the presence of DC current in the primary winding. Any amount of DC voltage dropped across the primary winding of a transformer will cause additional magnetic flux in the core. This additional flux “bias” or “offset” will push the alternating flux waveform closer to saturation in one half-cycle than the other.

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DC in primary, shifts the waveform peaks toward the upper saturation limit.

For most transformers, core saturation is a very undesirable effect, and it is avoided through good design: engineering the windings and core so that magnetic flux densities remain well below the saturation levels. This ensures that the relationship between mmf and Φ is more linear throughout the flux cycle, which is good because it makes for less distortion in the magnetization current waveform. Also, engineering the core for low flux densities provides a safe margin between the normal flux peaks and the core saturation limits to accommodate occasional, abnormal conditions such as frequency variation and DC offset.

Inrush current

When a transformer is initially connected to a source of AC voltage, there may be a substantial surge of current through the primary winding called inrush current. This is analogous to the inrush current exhibited by an electric motor that is started up by sudden connection to a power source, although transformer inrush is caused by a different phenomenon.

We know that the rate of change of instantaneous flux in a transformer core is proportional to the instantaneous voltage drop across the primary winding. Or, as stated before, the voltage waveform is the derivative of the flux waveform, and the flux waveform is the integral of the voltage waveform. In a continuously-operating transformer, these two waveforms are phase-shifted by 90o. Since flux (Φ) is proportional to the magnetomotive force (mmf) in the core, and the mmf is proportional to winding current, the current waveform will be in-phase with the flux waveform, and both will be lagging the voltage waveform by 90o:

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Continuous steady-state operation: Magnetic flux, like current, lags applied voltage by 90o.

Let us suppose that the primary winding of a transformer is suddenly connected to an AC voltage source at the exact moment in time when the instantaneous voltage is at its positive peak value. In order for the transformer to create an opposing voltage drop to balance against this applied source voltage, a magnetic flux of rapidly increasing value must be generated. The result is that winding current increases rapidly, but actually no more rapidly than under normal conditions:

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Connecting transformer to line at AC volt peak: Flux increases rapidly from zero, same as steady-state operation.

Both core flux and coil current start from zero and build up to the same peak values experienced during continuous operation. Thus, there is no “surge” or “inrush” or current in this scenario. above)

Alternatively, let us consider what happens if the transformer’s connection to the AC voltage source occurs at the exact moment in time when the instantaneous voltage is at zero. During continuous operation (when the transformer has been powered for quite some time), this is the point in time where both flux and winding current are at their negative peaks, experiencing zero rate-of-change (dΦ/dt = 0 and di/dt = 0). As the voltage builds to its positive peak, the flux and current waveforms build to their maximum positive rates-of-change, and on upward to their positive peaks as the voltage descends to a level of zero:

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Starting at e=0 V is not the same as running continuously in above.These expected waveforms are incorrect– Φ and i should start at zero.

A significant difference exists, however, between continuous-mode operation and the sudden starting condition assumed in this scenario: during continuous operation, the flux and current levels were at their negative peaks when voltage was at its zero point; in a transformer that has been sitting idle, however, both magnetic flux and winding current should start at zero. When the magnetic flux increases in response to a rising voltage, it will increase from zero upward, not from a previously negative (magnetized) condition as we would normally have in a transformer that’s been powered for awhile. Thus, in a transformer that’s just “starting,” the flux will reach approximately twice its normal peak magnitude as it “integrates” the area under the voltage waveform’s first half-cycle:

a picture

Starting at e=0 V, Φ starts at initial condition Φ=0, increasing to twice the normal value, assuming it doesn’t saturate the core.

In an ideal transformer, the magnetizing current would rise to approximately twice its normal peak value as well, generating the necessary mmf to create this higher-than-normal flux. However, most transformers aren’t designed with enough of a margin between normal flux peaks and the saturation limits to avoid saturating in a condition like this, and so the core will almost certainly saturate during this first half-cycle of voltage. During saturation, disproportionate amounts of mmf are needed to generate magnetic flux. This means that winding current, which creates the mmf to cause flux in the core, will disproportionately rise to a value easily exceeding twice its normal peak:

a picture

Starting at e=0 V, Current also increases to twice the normal value for an unsaturated core, or considerably higher in the (designed for) case of saturation.

This is the mechanism causing inrush current in a transformer’s primary winding when connected to an AC voltage source. As you can see, the magnitude of the inrush current strongly depends on the exact time that electrical connection to the source is made. If the transformer happens to have some residual magnetism in its core at the moment of connection to the source, the inrush could be even more severe. Because of this, transformer overcurrent protection devices are usually of the “slow-acting” variety, so as to tolerate current surges such as this without opening the circuit.

Heat and Noise

In addition to unwanted electrical effects, transformers may also exhibit undesirable physical effects, the most notable being the production of heat and noise. Noise is primarily a nuisance effect, but heat is a potentially serious problem because winding insulation will be damaged if allowed to overheat. Heating may be minimized by good design, ensuring that the core does not approach saturation levels, that eddy currents are minimized, and that the windings are not overloaded or operated too close to maximum ampacity.

Large power transformers have their core and windings submerged in an oil bath to transfer heat and muffle noise, and also to displace moisture which would otherwise compromise the integrity of the winding insulation. Heat-dissipating “radiator” tubes on the outside of the transformer case provide a convective oil flow path to transfer heat from the transformer’s core to ambient air:

a picture

Large power transformers are submerged in heat dissipating insulating oil.

Oil-less, or “dry,” transformers are often rated in terms of maximum operating temperature “rise” (temperature increase beyond ambient) according to a letter-class system: A, B, F, or H. These letter codes are arranged in order of lowest heat tolerance to highest:

  • Class A: No more than 55o Celsius winding temperature rise, at 40o Celsius (maximum) ambient air temperature.
  • Class B: No more than 80o Celsius winding temperature rise, at 40o Celsius (maximum)ambient air temperature.
  • Class F: No more than 115o Celsius winding temperature rise, at 40o Celsius (maximum)ambient air temperature.
  • Class H: No more than 150o Celsius winding temperature rise, at 40o Celsius (maximum)ambient air temperature.

Audible noise is an effect primarily originating from the phenomenon of magnetostriction: the slight change of length exhibited by a ferromagnetic object when magnetized. The familiar “hum” heard around large power transformers is the sound of the iron core expanding and contracting at 120 Hz (twice the system frequency, which is 60 Hz in the United States) — one cycle of core contraction and expansion for every peak of the magnetic flux waveform — plus noise created by mechanical forces between primary and secondary windings. Again, maintaining low magnetic flux levels in the core is the key to minimizing this effect, which explains why ferroresonant transformers — which must operate in saturation for a large portion of the current waveform — operate both hot and noisy.

Another noise-producing phenomenon in power transformers is the physical reaction force between primary and secondary windings when heavily loaded. If the secondary winding is open-circuited, there will be no current through it, and consequently no magneto-motive force (mmf) produced by it. However, when the secondary is “loaded” (current supplied to a load), the winding generates an mmf, which becomes counteracted by a “reflected” mmf in the primary winding to prevent core flux levels from changing. These opposing mmf’s generated between primary and secondary windings as a result of secondary (load) current produce a repulsive, physical force between the windings which will tend to make them vibrate. Transformer designers have to consider these physical forces in the construction of the winding coils, to ensure there is adequate mechanical support to handle the stresses. Under heavy load (high current) conditions, though, these stresses may be great enough to cause audible noise to emanate from the transformer.

  • REVIEW:
  • Power transformers are limited in the amount of power they can transfer from primary to secondary winding(s). Large units are typically rated in VA (volt-amps) or kVA (kilo volt-amps).
  • Resistance in transformer windings contributes to inefficiency, as current will dissipate heat, wasting energy.
  • Magnetic effects in a transformer’s iron core also contribute to inefficiency. Among the effects are eddy currents (circulating induction currents in the iron core) and hysteresis (power lost due to overcoming the tendency of iron to magnetize in a particular direction).
  • Increased frequency results in increased power losses within a power transformer. The presence of harmonics in a power system is a source of frequencies significantly higher than normal, which may cause overheating in large transformers.
  • Both transformers and inductors harbor certain unavoidable amounts of capacitance due to wire insulation (dielectric) separating winding turns from the iron core and from each other. This capacitance can be significant enough to give the transformer a natural resonant frequency, which can be problematic in signal applications.
  • Leakage inductance is caused by magnetic flux not being 100% coupled between windings in a transformer. Any flux not involved with transferring energy from one winding to another will store and release energy, which is how (self-) inductance works. Leakage inductance tends to worsen a transformer’s voltage regulation (secondary voltage “sags” more for a given amount of load current).
  • Magnetic saturation of a transformer core may be caused by excessive primary voltage, operation at too low of a frequency, and/or by the presence of a DC current in any of the windings. Saturation may be minimized or avoided by conservative design, which provides an adequate margin of safety between peak magnetic flux density values and the saturation limits of the core.
  • Transformers often experience significant inrush currents when initially connected to an AC voltage source. Inrush current is most severe when connection to the AC source is made at the moment instantaneous source voltage is zero.
  • Noise is a common phenomenon exhibited by transformers — especially power transformers — and is primarily caused by magnetostriction of the core. Physical forces causing winding vibration may also generate noise under conditions of heavy (high current) secondary winding load.

Contributors

Contributors to this chapter are listed in chronological order of their contributions, from most recent to first. See Appendix 2 (Contributor List) for dates and contact information.

David Williams (2022): Minor edits and import to website.

Bart Anderson (January 2004): Corrected conceptual errors regarding Tesla coil operation and safety.

Jason Starck (June 2000): HTML document formatting, which led to a much better-looking second edition.

Bibliography

  1. [MYA]Mitsuyoshi Yamamoto, Mitsugi Yamaguchi, “Electric Power In Japan, Rapid Electrification a Century Ago”, EDN, (4/11/2002). http://www.ieee.org/organizations/pes/public/2005/mar/peshistory.html

CC-BY 2000-2020 Tony R. Kuphaldt.   
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What is a filter?

It is sometimes desirable to have circuits capable of selectively filtering one frequency or range of frequencies out of a mix of different frequencies in a circuit. A circuit designed to perform this frequency selection is called a filter circuit, or simply a filter. A common need for filter circuits is in high-performance stereo systems, where certain ranges of audio frequencies need to be amplified or suppressed for best sound quality and power efficiency. You may be familiar with equalizers, which allow the amplitudes of several frequency ranges to be adjusted to suit the listener’s taste and acoustic properties of the listening area. You may also be familiar with crossover networks, which block certain ranges of frequencies from reaching speakers. A tweeter (high-frequency speaker) is inefficient at reproducing low-frequency signals such as drum beats, so a crossover circuit is connected between the tweeter and the stereo’s output terminals to block low-frequency signals, only passing high-frequency signals to the speaker’s connection terminals. This gives better audio system efficiency and thus better performance. Both equalizers and crossover networks are examples of filters, designed to accomplish filtering of certain frequencies.

Another practical application of filter circuits is in the “conditioning” of non-sinusoidal voltage waveforms in power circuits. Some electronic devices are sensitive to the presence of harmonics in the power supply voltage, and so require power conditioning for proper operation. If a distorted sine-wave voltage behaves like a series of harmonic waveforms added to the fundamental frequency, then it should be possible to construct a filter circuit that only allows the fundamental waveform frequency to pass through, blocking all (higher-frequency) harmonics.

We will be studying the design of several elementary filter circuits in this lesson. To reduce the load of math on the reader, I will make extensive use of SPICE as an analysis tool, displaying Bode plots (amplitude versus frequency) for the various kinds of filters. Bear in mind, though, that these circuits can be analyzed over several points of frequency by repeated series-parallel analysis, much like the previous example with two sources (60 and 90 Hz), if the student is willing to invest a lot of time working and re-working circuit calculations for each frequency.

REVIEW:

  • A filter is an AC circuit that separates some frequencies from others within mixed-frequency signals.
  • Audio equalizers and crossover networks are two well-known applications of filter circuits.
  • A Bode plot is a graph plotting waveform amplitude or phase on one axis and frequency on the other.

Low-pass filters

By definition, a low-pass filter is a circuit offering easy passage to low-frequency signals and difficult passage to high-frequency signals. There are two basic kinds of circuits capable of accomplishing this objective, and many variations of each one: The inductive low-pass filter in figure belowand the capacitive low-pass filter in

a picture

Inductive low-pass filter

The inductor’s impedance increases with increasing frequency. This high impedance in series tends to block high-frequency signals from getting to the load. This can be demonstrated with a SPICE analysis:

inductive lowpass filter v1 1 0 ac 1 sin l1 1 2 3rload 2 0 1k .ac lin 20 1 200.plot ac v(2) .end

a picture

The response of an inductive low-pass filter falls off with increasing frequency.

a picture

Capacitive low-pass filter.

The capacitor’s impedance decreases with increasing frequency. This low impedance in parallel with the load resistance tends to short out high-frequency signals, dropping most of the voltage across series resistor R1.

capacitive lowpass filter v1 1 0 ac 1 sin r1 1 2 500 c1 2 0 7u rload 2 0 1k .ac lin 20 30 150 .plot ac v(2) .end

a picture

The response of a capacitive low-pass filter falls off with increasing frequency.

The inductive low-pass filter is the pinnacle of simplicity, with only one component comprising the filter. The capacitive version of this filter is not that much more complex, with only a resistor and capacitor needed for operation. However, despite their increased complexity, capacitive filter designs are generally preferred over inductive because capacitors tend to be “purer” reactive components than inductors and therefore are more predictable in their behavior. By “pure” I mean that capacitors exhibit little resistive effects than inductors, making them almost 100% reactive. Inductors, on the other hand, typically exhibit significant dissipative (resistor-like) effects, both in the long lengths of wire used to make them, and in the magnetic losses of the core material. Capacitors also tend to participate less in “coupling” effects with other components (generate and/or receive interference from other components via mutual electric or magnetic fields) than inductors, and are less expensive.

However, the inductive low-pass filter is often preferred in AC-DC power supplies to filter out the AC “ripple” waveform created when AC is converted (rectified) into DC, passing only the pure DC component. The primary reason for this is the requirement of low filter resistance for the output of such a power supply. A capacitive low-pass filter requires an extra resistance in series with the source, whereas the inductive low-pass filter does not. In the design of a high-current circuit like a DC power supply where additional series resistance is undesirable, the inductive low-pass filter is the better design choice. On the other hand, if low weight and compact size are higher priorities than low internal supply resistance in a power supply design, the capacitive low-pass filter might make more sense.

All low-pass filters are rated at a certain cutoff frequency. That is, the frequency above which the output voltage falls below 70.7% of the input voltage. This cutoff percentage of 70.7 is not really arbitrary, all though it may seem so at first glance. In a simple capacitive/resistive low-pass filter, it is the frequency at which capacitive reactance in ohms equals resistance in ohms. In a simple capacitive low-pass filter (one resistor, one capacitor), the cutoff frequency is given as:

a picture

Inserting the values of R and C from the last SPICE simulation into this formula, we arrive at a cutoff frequency of 45.473 Hz. However, when we look at the plot generated by the SPICE simulation, we see the load voltage well below 70.7% of the source voltage (1 volt) even at a frequency as low as 30 Hz, below the calculated cutoff point. What’s wrong? The problem here is that the load resistance of 1 kΩ affects the frequency response of the filter, skewing it down from what the formula told us it would be. Without that load resistance in place, SPICE produces a Bode plot whose numbers make more sense:

capacitive lowpass filter v1 1 0 ac 1 sin r1 1 2 500 c1 2 0 7u* note: no load resistor! .ac lin 20 40 50.plot ac v(2) .end

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For the capacitive low-pass filter with R = 500 Ω and C = 7 µF, the Output should be 70.7% at 45.473 Hz.

  fcutoff = 1/(2πRC) = 1/(2π(500 Ω)(7 µF)) = 45.473 Hz

When dealing with filter circuits, it is always important to note that the response of the filter depends on the filter’s component values and the impedance of the load. If a cutoff frequency equation fails to give consideration to load impedance, it assumes no load and will fail to give accurate results for a real-life filter conducting power to a load.

One frequent application of the capacitive low-pass filter principle is in the design of circuits having components or sections sensitive to electrical “noise.” As mentioned at the beginning of the last chapter, sometimes AC signals can “couple” from one circuit to another via capacitance (Cstray) and/or mutual inductance (Mstray) between the two sets of conductors. A prime example of this is unwanted AC signals (“noise”) becoming impressed on DC power lines supplying sensitive circuits:

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Noise is coupled by stray capacitance and mutual inductance into “clean” DC power.

The oscilloscope-meter on the left shows the “clean” power from the DC voltage source. After coupling with the AC noise source via stray mutual inductance and stray capacitance, though, the voltage as measured at the load terminals is now a mix of AC and DC, the AC being unwanted. Normally, one would expect Eload to be precisely identical to Esource, because the uninterrupted conductors connecting them should make the two sets of points electrically common. However, power conductor impedance allows the two voltages to differ, which means the noise magnitude can vary at different points in the DC system.

If we wish to prevent such “noise” from reaching the DC load, all we need to do is connect a low-pass filter near the load to block any coupled signals. In its simplest form, this is nothing more than a capacitor connected directly across the power terminals of the load, the capacitor behaving as a very low impedance to any AC noise, and shorting it out. Such a capacitor is called a decoupling capacitor:

a picture

Decoupling capacitor, applied to load, filters noise from DC power supply.

A cursory glance at a crowded printed-circuit board (PCB) will typically reveal decoupling capacitors scattered throughout, usually located as close as possible to the sensitive DC loads. Capacitor size is usually 0.1 µF or more, a minimum amount of capacitance needed to produce a low enough impedance to short out any noise. Greater capacitance will do a better job at filtering noise, but size and economics limit decoupling capacitors to meager values.

REVIEW:

  • A low-pass filter allows for easy passage of low-frequency signals from source to load, and difficult passage of high-frequency signals.
  • Inductive low-pass filters insert an inductor in series with the load; capacitive low-pass filters insert a resistor in series and a capacitor in parallel with the load. The former filter design tries to “block” the unwanted frequency signal while the latter tries to short it out.
  • The cutoff frequency for a low-pass filter is that frequency at which the output (load) voltage equals 70.7% of the input (source) voltage. Above the cutoff frequency, the output voltage is lower than 70.7% of the input, and vice versa.

High-pass filters

A high-pass filter’s task is just the opposite of a low-pass filter: to offer easy passage of a high-frequency signal and difficult passage to a low-frequency signal. As one might expect, the inductive and capacitive igure below versions of the high-pass filter are just the opposite of their respective low-pass filter designs:

a picture

Capacitive high-pass filter.

The capacitor’s impedanceincreases with decreasing frequency. This high impedance in series tends to block low-frequency signals from getting to load.

capacitive highpass filter v1 1 0 ac 1 sin c1 1 2 0.5u rload 2 0 1k .ac lin 20 1 200.plot ac v(2) .end

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The response of the capacitive high-pass filter increases with frequency.

a picture

Inductive high-pass filter.

The inductor’s impedance decreases with decreasing frequency. This low impedance in parallel tends to short out low-frequency signals from getting to the load resistor. As a consequence, most of the voltage gets dropped across series resistor R1.

inductive highpass filter v1 1 0 ac 1 sin r1 1 2 200 l1 2 0 100m rload 2 0 1k .ac lin 20 1 200.plot ac v(2) .end

a picture

The response of the inductive high-pass filter increases with frequency.

This time, the capacitive design is the simplest, requiring only one component above and beyond the load. And, again, the reactive purity of capacitors over inductors tends to favor their use in filter design, especially with high-pass filters where high frequencies commonly cause inductors to behave strangely due to the skin effect and electromagnetic core losses.

As with low-pass filters, high-pass filters have a rated cutoff frequency, above which the output voltage increases above 70.7% of the input voltage. Just as in the case of the capacitive low-pass filter circuit, the capacitive high-pass filter’s cutoff frequency can be found with the same formula:

a picture

In the example circuit, there is no resistance other than the load resistor, so that is the value for R in the formula.

Using a stereo system as a practical example, a capacitor connected in series with the tweeter (treble) speaker will serve as a high-pass filter, imposing a high impedance to low-frequency bass signals, thereby preventing that power from being wasted on a speaker inefficient for reproducing such sounds. In like fashion, an inductor connected in series with the woofer (bass) speaker will serve as a low-pass filter for the low frequencies that particular speaker is designed to reproduce. In this simple example circuit, the midrange speaker is subjected to the full spectrum of frequencies from the stereo’s output. More elaborate filter networks are sometimes used, but this should give you the general idea. Also bear in mind that I’m only showing you one channel (either left or right) on this stereo system. A real stereo would have six speakers: 2 woofers, 2 midranges, and 2 tweeters.

a picture

High-pass filter routes high frequencies to tweeter, while low-pass filter routes lows to woofer.

For better performance yet, we might like to have some kind of filter circuit capable of passing frequencies that are between low (bass) and high (treble) to the midrange speaker so that none of the low- or high-frequency signal power is wasted on a speaker incapable of efficiently reproducing those sounds. What we would be looking for is called a band-pass filter, which is the topic of the next section.

REVIEW:

  • A high-pass filter allows for easy passage of high-frequency signals from source to load, and difficult passage of low-frequency signals.
  • Capacitive high-pass filters insert a capacitor in series with the load; inductive high-pass filters insert a resistor in series and an inductor in parallel with the load. The former filter design tries to “block” the unwanted frequency signal while the latter tries to short it out.
  • The cutoff frequency for a high-pass filter is that frequency at which the output (load) voltage equals 70.7% of the input (source) voltage. Above the cutoff frequency, the output voltage is greater than 70.7% of the input, and vice versa.

Band-pass filters

There are applications where a particular band, or spread, or frequencies need to be filtered from a wider range of mixed signals. Filter circuits can be designed to accomplish this task by combining the properties of low-pass and high-pass into a single filter. The result is called a band-pass filter. Creating a bandpass filter from a low-pass and high-pass filter can be illustrated using block diagrams:

a picture

System level block diagram of a band-pass filter.

What emerges from the series combination of these two filter circuits is a circuit that will only allow passage of those frequencies that are neither too high nor too low. Using real components, here is what a typical schematic might look like the figure belowThe response of the band-pass filter is shown in

a picture

Capacitive band-pass filter.

capacitive bandpass filter v1 1 0 ac 1 sin r1 1 2 200 c1 2 0 2.5u c2 2 3 1u rload 3 0 1k .ac lin 20 100 500 .plot ac v(3) .end

a picture

The response of a capacitive bandpass filter peaks within a narrow frequency range.

Band-pass filters can also be constructed using inductors, but as mentioned before, the reactive “purity” of capacitors gives them a design advantage. If we were to design a bandpass filter using inductors, it might look something like the figure below

a picture

Inductive band-pass filter.

The fact that the high-pass section comes “first” in this design instead of the low-pass section makes no difference in its overall operation. It will still filter out all frequencies too high or too low.

While the general idea of combining low-pass and high-pass filters together to make a bandpass filter is sound, it is not without certain limitations. Because this type of band-pass filter works by relying on either section to block unwanted frequencies, it can be difficult to design such a filter to allow unhindered passage within the desired frequency range. Both the low-pass and high-pass sections will always be blocking signals to some extent, and their combined effort makes for an attenuated (reduced amplitude) signal at best, even at the peak of the “pass-band” frequency range. Notice the curve peak on the previous SPICE analysis: the load voltage of this filter never rises above 0.59 volts, although the source voltage is a full volt. This signal attenuation becomes more pronounced if the filter is designed to be more selective (steeper curve, narrower band of passable frequencies).

There are other methods to achieve band-pass operation without sacrificing signal strength within the pass-band. We will discuss those methods a little later in this chapter.

REVIEW:

  • A band-pass filter works to screen out frequencies that are too low or too high, giving easy passage only to frequencies within a certain range.
  • Band-pass filters can be made by stacking a low-pass filter on the end of a high-pass filter, or vice versa.
  • “Attenuate” means to reduce or diminish in amplitude. When you turn down the volume control on your stereo, you are “attenuating” the signal being sent to the speakers.

Band-stop filters

Also called band-elimination, band-reject, or notch filters, this kind of filter passes all frequencies above and below a particular range set by the component values. Not surprisingly, it can be made out of a low-pass and a high-pass filter, just like the band-pass design, except that this time we connect the two filter sections in parallel with each other instead of in series.

a picture

System level block diagram of a band-stop filter.

Constructed using two capacitive filter sections, it looks something like

a picture

“Twin-T” band-stop filter.

The low-pass filter section is comprised of R1, R2, and C1 in a “T” configuration. The high-pass filter section is comprised of C2, C3, and R3 in a “T” configuration as well. Together, this arrangement is commonly known as a “Twin-T” filter, giving sharp response when the component values are chosen in the following ratios:

a picture

Given these component ratios, the frequency of maximum rejection (the “notch frequency”) can be calculated as follows:

a picture

The impressive band-stopping ability of this filter is illustrated by the following SPICE analysis:

twin-t bandstop filterv1 1 0 ac 1 sin r1 1 2 200 c1 2 0 2u r2 2 3 200 c2 1 4 1u r3 4 0 100 c3 4 3 1u rload 3 0 1k .ac lin 20 200 1.5k .plot ac v(3) .end

a picture

Response of “twin-T” band-stop filter.

REVIEW:

  • A band-stop filter works to screen out frequencies that are within a certain range, giving easy passage only to frequencies outside of that range. Also known as band-elimination, band-reject, or notch filters.
  • Band-stop filters can be made by placing a low-pass filter in parallel with a high-pass filter. Commonly, both the low-pass and high-pass filter sections are of the “T” configuration, giving the name “Twin-T” to the band-stop combination.
  • The frequency of maximum attenuation is called the notch frequency.

Resonant filters

So far, the filter designs we’ve concentrated on have employed either capacitors or inductors, but never both at the same time. We should know by now that combinations of L and C will tend to resonate, and this property can be exploited in designing band-pass and band-stop filter circuits.

Series LC circuits give minimum impedance at resonance, while parallel LC (“tank”) circuits give maximum impedance at their resonant frequency. Knowing this, we have two basic strategies for designing either band-pass or band-stop filters.

For band-pass filters, the two basic resonant strategies are this: series LC to pass a signalor parallel LC to short a signal. The two schemes will be contrasted and simulated here:

a picture

Series resonant LC band-pass filter.

Series LC components pass signal at resonance, and block signals of any other frequencies from getting to the load.

series resonant bandpass filterv1 1 0 ac 1 sin l1 1 2 1c1 2 3 1u rload 3 0 1k .ac lin 20 50 250 .plot ac v(3) .end

a picture

Series resonant band-pass filter: voltage peaks at resonant frequency of 159.15 Hz.

A couple of points to note: see how there is virtually no signal attenuation within the “pass band” (the range of frequencies near the load voltage peak), unlike the band-pass filters made from capacitors or inductors alone. Also, since this filter works on the principle of series LC resonance, the resonant frequency of which is unaffected by circuit resistance, the value of the load resistor will not skew the peak frequency. However, different values for the load resistor will change the “steepness” of the Bode plot (the “selectivity” of the filter).

The other basic style of resonant band-pass filters employs a tank circuit (parallel LC combination) to short out signals too high or too low in frequency from getting to the load:

a picture

Parallel resonant band-pass filter.

The tank circuit will have a lot of impedance at resonance, allowing the signal to get to the load with minimal attenuation. Under or over resonant frequency, however, the tank circuit will have a low impedance, shorting out the signal and dropping most of it across series resistor R1.

parallel resonant bandpass filter v1 1 0 ac 1 sin r1 1 2 500 l1 2 0 100m c1 2 0 10u rload 2 0 1k .ac lin 20 50 250 .plot ac v(2) .end

a picture

Parallel resonant filter: voltage peaks a resonant frequency of 159.15 Hz.

Just like the low-pass and high-pass filter designs relying on a series resistance and a parallel “shorting” component to attenuate unwanted frequencies, this resonant circuit can never provide full input (source) voltage to the load. That series resistance will always be dropping some amount of voltage so long as there is a load resistance connected to the output of the filter.

It should be noted that this form of band-pass filter circuit is very popular in analog radio tuning circuitry, for selecting a particu