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:

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.

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

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).

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.

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):

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?

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

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.

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.

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.

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

$$\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

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

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

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

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

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

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

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

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.

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:

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.

Answer

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.

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

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.

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

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.

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

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).

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

File Num: 04043

Answer

Notes

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

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.

Answer

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.

Answer

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!

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:

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.

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^2

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

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.

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).

Answer

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.

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:

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:

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.

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.

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

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.

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.

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:

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!

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

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

File Num: 02571

Answer

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!

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

File Num: 02098

Answer

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.

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

File Num: 00620

Answer

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?

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!

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:

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

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:

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.

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:

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:

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.

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:

File Num: 00851

Answer

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.

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):

File Num: 00852

Answer

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!

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):

File Num: 00854

Answer

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!

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

File Num: 01541

Answer

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:

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.

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?

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.

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:

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.

Answer

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.

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.

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:

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:

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:

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.

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

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

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!

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

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

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!

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

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

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!

Answer

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.

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:

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

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.

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:

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

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.

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?

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.

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

File Num: 02119

Answer

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.

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:

File Num: 01864

Answer

Equivalent series resistance and reactances:

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.

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:

File Num: 01865

Answer

Equivalent parallel resistance and reactances:

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.

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

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

File Num: 02115

Answer

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.

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

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

File Num: 02116

Answer

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.

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:

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:

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?

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:

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):

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:

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.

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

File Num: 03326

Answer

Notes

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

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

File Num: 03327

Answer

Notes

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

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

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.

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

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:

Solving for A:

Solving for B:

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.

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

Notes

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

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

Notes

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

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.

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

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.

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

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):

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!

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

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

File Num: 02081

Answer

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.

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

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

• 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:

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 Z–R–X 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.

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:

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 Z–R–X 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.

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

File Num: 03288

Answer

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.

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

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

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.

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:

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

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.

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:

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

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.

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:

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

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.

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:

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

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.

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:

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

Notes

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

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:

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

File Num: 02087

Answer

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.

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.).

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:

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

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:

Answer

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

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

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:

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?

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.

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:

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

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.

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:

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:

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

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.

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?

File Num: 02100

Answer

Here are two ways of showing the same addition:

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.

Answer

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

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

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.

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.

Answer

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.

Answer

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

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

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.

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.

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

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.

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

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.).

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

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

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?

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

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

File Num: 02089

Answer

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.

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

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

File Num: 02091

Answer

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.

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?

File Num: 02090

Answer

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.

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:

File Num: 00498

Answer

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.

Answer

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!).

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:

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

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

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.

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.

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

• Frequency
• Period
• Hertz
• Amplitude

File Num: 00054

Answer

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.

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).

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.

Answer

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.

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?

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!

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?

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

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.

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.

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

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

File Num: 00531

Answer

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.

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

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.

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:

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.

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:

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.

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:

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.

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:

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.

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:

File Num: 01668

Answer

400 Hz

Notes

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

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.

File Num: 00541

Answer

Notes

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

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:

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.

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

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:

File Num: 01821

Answer

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

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

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.

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.

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.

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:

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!

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

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.

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

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.

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.

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?

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

File Num: 00493

Answer

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.

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

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!

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?

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

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

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.

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

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

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.

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:

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

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.

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:

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:

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

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.

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?

File Num: 02100

Answer

Here are two ways of showing the same addition:

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.

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

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.

Answer

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.

Answer

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

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

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.

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.

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

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.

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:

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:

File Num: 01480

Answer

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.

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:

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

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).

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.

\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:

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

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).

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?

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.

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.

Answer

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!

Answer

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.

Answer

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.

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

File Num: 01552

Answer

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.

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:

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

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?

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:

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.

Answer

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

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.

Answer

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.

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:

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

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}?

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:

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

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.

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

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):

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!

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

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

File Num: 02081

Answer

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.

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:

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

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.

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

Notes

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

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.

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

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.

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:

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

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.

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):

File Num: 00850

Answer

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.

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

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

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.

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):

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

File Num: 02080

Answer

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.

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

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

File Num: 02103

Answer

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.

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

File Num: 00639

Answer

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.

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

File Num: 01830

Answer

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:

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.

Solve for all voltages and currents in this series LR circuit, and also calculate the phase angle of the total impedance:

File Num: 01831

Answer

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:

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.

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

• 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:

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.

One way to vary the amount of power delivered to a resistive AC load is by varying another resistance connected in series:

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:

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).

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.

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:

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):

File Num: 00853

Answer

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?

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

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

• 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:

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 Z–R–X 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.

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:

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 Z–R–X 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.

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):

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

File Num: 02079

Answer

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.

Calculate the individual currents through the inductor and through the resistor, the total current, and the total circuit impedance:

Also, draw a phasor diagram showing how the individual component currents relate to the total current.

File Num: 02104

Answer

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.

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

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

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:

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.

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

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

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:

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.

Determine the total current and all component currents in this circuit, stating your answers the way a multimeter would register them:

• 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:

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.

Calculate the total impedances (complete with phase angles) for each of the following inductor-resistor circuits:

File Num: 02106

Answer

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.

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:

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

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.

Answer

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.

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.

File Num: 01693

\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 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!

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).

File Num: 01616

\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 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.

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.

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.

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.

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.

Answer

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.

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.

File Num: 00540

Answer

Notes

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

Something is wrong with this circuit. Based on the oscilloscope’s display, determine whether the battery or the function generator is faulty:

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.

Something is wrong with this circuit. Based on the oscilloscope’s display, determine whether the battery or the function generator is faulty:

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.

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.

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:

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:

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:

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:

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:

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.

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.

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.

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?

Calculate the amount of phase shift indicated by this Lissajous figure:

File Num: 03578

Answer

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.

Calculate the amount of phase shift indicated by this Lissajous figure:

File Num: 03575

Answer

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.

Calculate the amount of phase shift indicated by this Lissajous figure:

File Num: 03577

Answer

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.

Calculate the amount of phase shift indicated by this Lissajous figure:

File Num: 03576

Answer

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.

Determine which way the movable iron piece needs to go in order to brighten the light bulb in this circuit:

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.

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:

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:

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:

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.

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:

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?

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.

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

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:

Solving for A:

Solving for B:

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.

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:

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?

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.

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:

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!

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:

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.

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

Notes

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

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

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.).

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

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

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?

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:

File Num: 00498

Answer

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.

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:

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.

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.

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.

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:

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

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.

Answer

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.

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 60 Hz:

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

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!

Calculate the total impedance of this LR circuit, once using nothing but scalar numbers, and again using complex numbers:

File Num: 01837

Answer

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).

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 120 Hz:

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

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!

Determine the input frequency necessary to give the output voltage a phase shift of 75^{o}:

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03282

Answer

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.

Determine the necessary resistor value to give the output voltage a phase shift of 44^{o}:

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

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.

Determine the input frequency necessary to give the output voltage a phase shift of -40^{o}:

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03280

Answer

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.

Determine the necessary resistor value to give the output voltage a phase shift of -60^{o}:

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

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.

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:

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

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?

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:

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:

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.

A student measures voltage drops in an AC circuit using three voltmeters and arrives at the following measurements:

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.

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):

File Num: 01844

Answer

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.

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

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

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.

Determine the phase angle (\Theta) of the current in this circuit, with respect to the supply voltage:

File Num: 01853

Answer

Challenge question: explain how the following phasor diagram was determined for this problem:

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.

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:

Calculate the magnitude and phase shift of the current through this capacitor, taking into consideration its equivalent series resistance (ESR):

Compare this against the magnitude and phase shift of the current for an ideal 0.22 \muF capacitor.

File Num: 01847

Answer

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.

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

File Num: 01848

Answer

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:

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.

Solve for all voltages and currents in this series RC circuit, and also calculate the phase angle of the total impedance:

File Num: 01849

Answer

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:

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.

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

• 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:

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.

Calculate the voltage drops across all components in this circuit, expressing them in complex (polar) form with magnitudes and phase angles each:

File Num: 01852

Answer

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:

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.

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:

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

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.

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:

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):

File Num: 01845

Answer

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?

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}

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

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!

Calculate the total impedance of these parallel-connected components, expressing it in polar form (magnitude and phase angle):

Also, draw an admittance triangle for this circuit.

File Num: 02108

Answer

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).

Calculate the total impedances (complete with phase angles) for each of the following capacitor-resistor circuits:

File Num: 02109

Answer

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.

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:

• \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!

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:

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:

Compare this against the magnitude and phase shift of the current for an ideal capacitor (no leakage).

File Num: 01850

Answer

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:

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.

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:

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:

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

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.

Determine the size of capacitor (in Farads) necessary to create a total current of 11.3 mA in this parallel RC circuit:

File Num: 02110

Answer

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.

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

File Num: 01551

Answer

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.

Calculate the total impedance offered by these two capacitors to a sinusoidal signal with a frequency of 900 Hz:

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

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!

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

File Num: 03288

Answer

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.

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

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

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.

Calculate the total impedance of this RC circuit, once using nothing but scalar numbers, and again using complex numbers:

File Num: 01838

Answer

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).

A student is asked to calculate the phase shift for the following circuit’s output voltage, relative to the phase of the source voltage:

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):

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!

Calculate the output voltage of this phase-shifting circuit, expressing it in polar form (magnitude and phase angle relative to the source voltage):

File Num: 02621

Answer

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).

Calculate the output voltage of this phase-shifting circuit, expressing it in polar form (magnitude and phase angle relative to the source voltage):

File Num: 02620

Answer

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).

Determine the input frequency necessary to give the output voltage a phase shift of 70^{o}:

File Num: 02623

Answer

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.

Determine the input frequency necessary to give the output voltage a phase shift of 40^{o}:

File Num: 02622

Answer

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.

Determine the input frequency necessary to give the output voltage a phase shift of -38^{o}:

File Num: 02626

Answer

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.

Determine the input frequency necessary to give the output voltage a phase shift of -25^{o}:

File Num: 02625

Answer

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.

Determine the input frequency necessary to give the output voltage a phase shift of 25^{o}:

Also, write an equation that solves for frequency (f), given all the other variables (R, C, and phase angle \theta).

File Num: 03284

Answer

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.

Determine the necessary resistor value to give the output voltage a phase shift of 58^{o}:

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

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.

Determine the necessary resistor value to give the output voltage a phase shift of -64^{o}:

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

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.

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:

V_{out} =

File Num: 03818

Answer

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.

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:

V_{out} =

File Num: 03819

Answer

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.

Complete the table of values for this circuit, representing all quantities in complex-number form (either polar or rectangular, your choice):

File Num: 03611

Answer

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.

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:

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:

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.”

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:

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:

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.”

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:

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:

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.”

Sketch the approximate waveform of this circuit’s output signal (V_{out}) on the screen of the oscilloscope:

Hint: use the Superposition Theorem!

File Num: 03503

Answer

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.

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).

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:

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?

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.

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.

\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:

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

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).

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 60 Hz:

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

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!

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):

File Num: 01844

Answer

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.

Calculate the total impedance of this LR circuit, once using nothing but scalar numbers, and again using complex numbers:

File Num: 01837

Answer

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).

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 120 Hz:

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

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!

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?

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.

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.

Answer

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!

Answer

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.

Answer

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.

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

File Num: 01552

Answer

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.

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:

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

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?

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:

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.

Answer

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

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.

Answer

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.

Determine the input frequency necessary to give the output voltage a phase shift of 75^{o}:

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03282

Answer

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.

Determine the necessary resistor value to give the output voltage a phase shift of 44^{o}:

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

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.

Determine the input frequency necessary to give the output voltage a phase shift of -40^{o}:

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03280

Answer

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.

Determine the necessary resistor value to give the output voltage a phase shift of -60^{o}:

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

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.

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:

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

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}?

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:

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

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.

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

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):

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!

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

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

File Num: 02081

Answer

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.

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:

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

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.

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

Notes

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

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.

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

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.

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:

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

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.

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):

File Num: 00850

Answer

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.

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

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

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.

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):

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

File Num: 02080

Answer

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.

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

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

File Num: 02103

Answer

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.

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

File Num: 00639

Answer

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.

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

File Num: 01830

Answer

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:

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.

Solve for all voltages and currents in this series LR circuit, and also calculate the phase angle of the total impedance:

File Num: 01831

Answer

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:

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.

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

• 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:

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.

One way to vary the amount of power delivered to a resistive AC load is by varying another resistance connected in series:

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:

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).

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.

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:

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):

File Num: 00853

Answer

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?

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

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

• 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:

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 Z–R–X 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.

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:

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 Z–R–X 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.

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):

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

File Num: 02079

Answer

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.

Calculate the individual currents through the inductor and through the resistor, the total current, and the total circuit impedance:

Also, draw a phasor diagram showing how the individual component currents relate to the total current.

File Num: 02104

Answer

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.

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

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

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: