Unity Gain Buffer

Unity Gain Buffer Circuit

To build a unity gain buffer, you simply connect the input to the non-inverting pin and feedback the output to the inverting pin. It’s a very simple circuit and gives you an amplifier circuit with a gain of 1. This might not seem like a very useful feature, but check out this video to learn more about this circuit:


This video describes the connections and purposes of a unity gain buffer and shows why an infinite open loop voltage gain leads to unity gain when you have negative feedback.

Introduction to Unity Gain Buffers

Unity Gain Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000, and the inverting input connected directly to its output terminal. In other words, write an equation describing the output voltage of this op-amp (V_{out}) for any given input voltage at the noninverting input (V_{in(+)}):

Should be image 00927x01

Then, once you have an equation written, solve for the over-all voltage gain (A_V = {V_{out} \over V_{in(+)}}) of this amplifier circuit, and calculate the output voltage for a noninverting input voltage of +6 volts.

File Num: 00927

Answer

V_{out} = 100,000(V_{in(+)} - V_{out})(I’ve left it up to you to perform the algebraic simplification here!)A_V = {100,000 \over 100,001} = 0.99999

For an input voltage of +6 volts, the output voltage will be +5.99994 volts.


Notes

The significant point of this question is that students see the over-all voltage gain of the opamp radically attenuated from 100,000 to approximately 1. What is not so evident is just how stable this new voltage gain is, which is one of the purposes for employing negative feedback.





Question 2. (Click on arrow for answer)

How much effect will a change in the op-amp’s open-loop voltage gain have on the overall voltage gain of a negative-feedback circuit such as this?

Should be image 02288x01

If the open-loop gain of this operational amplifier were to change from 100,000 to 200,000, for example, how big of an effect would it have on the voltage gain as measured from the noninverting input to the output?

File Num: 02288

Answer

The different in overall voltage gain will be trivial.


Follow-up question: what advantage is there in building voltage amplifier circuits in this manner, applying negative feedback to a “core” amplifier with very high intrinsic gain?


Notes

Work with your students to calculate a few example scenarios, with the old open-loop gain versus the new open-loop gain. Have the students validate their conclusions with numbers!

Negative feedback is an extremely useful engineering principle, and one that allows us to build very precise amplifiers using imprecise components. Credit for this idea goes to Harold Black, an electrical engineer, in 1920’s. Mr. Black was looking for a way to improve the linearity and stability of amplifiers in telephone systems, and (as legend has it) the idea came to him in a flash of insight as he was commuting on a ferry boat.

An interesting historical side-note is that Black’s 1928 patent application was initially rejected on the grounds that he was trying to submit a perpetual motion device! The concept of negative feedback in an amplifier circuit was so contrary to established engineering thought at the time, that Black experienced significant resistance to the idea within the engineering community. The United States patent office, on the other hand, was inundated with fraudulent “perpetual motion” claims, and so dismissed Black’s invention at first sight.





Question 3. (Click on arrow for answer)

Complete the table of voltages for this opamp “voltage follower” circuit:

Should be image 02289x01$$\vbox{\offinterlineskip \halign{\strut \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil \vrule \cr \noalign{\hrule}V_{in} & V_{out} \cr\noalign{\hrule}

0 volts & 0 volts \cr

\noalign{\hrule}+5 volts & \cr\noalign{\hrule}+10 volts & \cr\noalign{\hrule}+15 volts & \cr\noalign{\hrule}+20 volts & \cr\noalign{\hrule}-5 volts & \cr\noalign{\hrule}-10 volts & \cr\noalign{\hrule}-15 volts & \cr\noalign{\hrule}-20 volts & \cr\noalign{\hrule} } % End of \halign }$$ % End of \vbox

File Num: 02289

Answer

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

0 volts & 0 volts \cr

\noalign{\hrule}+5 volts & +5 volts \cr\noalign{\hrule}+10 volts & +10 volts \cr\noalign{\hrule}+15 volts & +15 volts \cr\noalign{\hrule}+20 volts & +15 volts \cr\noalign{\hrule}-5 volts & -5 volts \cr\noalign{\hrule}-10 volts & -10 volts \cr\noalign{\hrule}-15 volts & -15 volts \cr\noalign{\hrule}-20 volts & -15 volts \cr\noalign{\hrule} } % End of \halign }$$ % End of \vbox

Follow-up question: the output voltage values given in this table are ideal. A real opamp would probably not be able to achieve even what is shown here, due to idiosyncrasies of these amplifier circuits. Explain what would probably be different in a real opamp circuit from what is shown here.


Notes

A common mistake I see students new to opamps make is assuming that the output voltage will magically attain whatever value the gain equation predicts, with no regard for power supply rail voltage limits.

Another good follow-up question to ask your students is this: “How much voltage is there between the two input terminals in each of the situations described in the table?” They will find that the “golden rule” of closed-loop opamp circuits can be violated!

If students have difficulty answering the follow-up question, drop these two hints: (1) Rail-to-rail output swing and (2) Latch-up.





Question 4. (Click on arrow for answer)

This operational amplifier circuit is often referred to as a voltage buffer, because it has unity gain (0 dB) and therefore simply reproduces, or “buffers,” the input voltage:

Should be image 03801x01

What possible use is a circuit such as this, which offers no voltage gain or any other form of signal modification? Wouldn’t a straight piece of wire do the same thing? Explain your answers.

Should be image 03801x02

File Num: 03801

Answer

While this circuit offers no voltage gain, it does offer current gain and impedance transformation. Much like the common-collector (or common-drain) single transistor amplifier circuits which also had voltage gains of (near) unity, opamp buffer circuits are useful whenever one must drive a relatively “heavy” (low impedance) load with a signal coming from a “weak” (high impedance) source.


Notes

I have found that some students have difficulty with the terms “heavy” and “light” in reference to load characteristics. That a “heavy” load would have very few ohms of impedance, and a “light” load would have many ohms of impedance seems counter-intuitive to some. It all makes sense, though, once students realize the terms “heavy” and “light” refer to the amount of current drawn by the respective loads.

Ask your students to explain why the straight piece of wire fails to “buffer” the voltage signal in the same way the the opamp follower circuit does.





Question 5. (Click on arrow for answer)

For all practical purposes, how much voltage exists between the inverting and noninverting input terminals of an op-amp in a functioning negative-feedback circuit?

File Num: 00930

Answer

Zero volts


Notes

Ask your students to explain why there will be (practically) no voltage between the input terminals of an operational amplifier when it is used in a negative feedback circuit.





Question 6. (Click on arrow for answer)

Just as certain assumptions are often made for bipolar transistors in order to simplify their analysis in circuits (an ideal BJT has negligible base current, I_C = I_E, constant \beta, etc.), we often make assumptions about operational amplifiers so we may more easily analyze their behavior in closed-loop circuits. Identify some of these ideal opamp assumptions as they relate to the following parameters:


  • Magnitude of input terminal currents:
  • Input impedance:
  • Output impedance:
  • Input voltage range:
  • Output voltage range:
  • Differential voltage (between input terminals) with negative feedback:

File Num: 02749

Answer


  • Magnitude of input terminal currents: infinitesimal
  • Input impedance: infinite
  • Output impedance: infinitesimal
  • Input voltage range: never exceeding +V/-V
  • Output voltage range: never exceeding +V/-V
  • Differential voltage (between input terminals) with negative feedback: infinitesimal


Notes

Just in case your students are unfamiliar with the words infinite and infinitesimal, tell them they simply mean “bigger than big” and “smaller than small”, respectively.





Question 7. (Click on arrow for answer)

The purpose of this circuit is to provide a pushbutton-adjustable voltage. Pressing one button causes the output voltage to increase, while pressing the other button causes the output voltage to decrease. When neither button is pressed, the voltage remains stable:

Should be image 03773x01

After working just fine for quite a long while, the circuit suddenly fails: now it only outputs zero volts DC all the time.

An experienced technician first checks the power supply voltage to see if it is within normal limits, and it is. Then, the technician checks the voltage across the capacitor. Explain why this is a good test point to check, and what the results of that check would tell the technician about the nature of the fault.

File Num: 03773

Answer

Checking for voltage across the capacitor will tell the technician what voltage the op-amp follower is being “told” to reproduce at the output.


Challenge question: why do you suppose I specify a CA3130 operational amplifier for this particular circuit? What is special about this opamp that qualifies it for the task?


Notes

Knowing where to check for critical signals in a circuit is an important skill, because it usually means the difference between efficiently locating a fault and wasting time. Ask your students to explain in detail the rationale behind checking for voltage across the capacitor, and (again, in detail) what certain voltage measurements at that point would prove about the nature of the fault.


Inverting Amplifier Circuit

Inverting Amplifier Circuit

An inverting amplifier is one where the output is inverted, or phase shifted by 180o compared to the input. This is another simple circuit, but a little more complicated than the unity gain buffer. See this video to learn more:


This video shows why the amplifier is called an “inverting” one and shows how to figure out the voltage gain based on the input and feedback resistor. It also includes an LTSpice simulation to demonstrate how the amplifier works.

Introduction to Inverting Op Amp Circuits

Non-Inverting Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Trace the directions for all currents in this circuit, and calculate the values for voltage at the output (V_{out}) and at test point 1 (V_{TP1}) for several values of input voltage (V_{in}):

Should be image 02467x01
VinVtp1Vout
0.0V
0.4V
1.2V
3.4V
7.1V
10.8V

Then, from the table of calculated values, determine the voltage gain (A_V) for this amplifier circuit.

File Num: 02467

Answer

Should be image 02467x02
VinVtp1Vout
0.0V0.0V0.0V
0.4V0.0V-0.4V
1.2V0.0V-1.2V
3.4V0.0V-3.4V
7.1V0.0V-7.1V
10.8V0.0V-10.8V
A_V = 1 (ratio) = 0 dB

Follow-up question: the point marked “TP1” in this circuit is often referred to as a virtual ground. Explain why this is, based on the voltage figures shown in the above table.


Notes

Some texts describe the voltage gain of an inverting voltage amplifier as being a negative quantity. I tend not to look at things that way, treating all gains as positive quantities and relying on my knowledge of circuit behavior to tell whether the signal is inverted or not. In my teaching experience, I have found that students have a tendency to blindly follow equations rather than think about what it is they are calculating, and that strict adherence to the mathematical signs of gain values only encourages this undesirable behavior (“If the sign of the gain tells me whether the circuit is inverting or not, I can just multiply input voltage by gain and the answer will always be right!”).

This strategy is analogous to problem-solving in electromagnetics, where a common approach is to use math to solve for the absolute values of quantities (potential, induced voltage, magnetic flux), and then to use knowledge of physical principles (Lenz’ Law, right-hand rule) to solve for polarities and directions. The alternative — to try to maintain proper sign convention throughout all calculations — not only complicates the math but it also encourages students to over-focus on calculations and neglect fundamental principles.





Question 2. (Click on arrow for answer)

Calculate the overall voltage gain of this amplifier circuit (A_V), both as a ratio and as a figure in units of decibels (dB). Also, write a general equation for calculating the voltage gain of such an amplifier, given the resistor values of R_1 and R_2:

Should be image 02458x01

File Num: 02458

Answer

A_V = 1 = 0 dBA_V = {R_1 \over R_2} \hbox{ (expressed as a ratio, not dB)}

Follow-up question \#1: sometimes the voltage gain equation for an amplifier of this type is given in the following form:

A_V = -{R_1 \over R_2}

What is the significance of the negative sign in this equation? Is it really necessary, or does it communicate an important concept?


Follow-up question \#2: manipulate the gain equation for this amplifier circuit to solve for the value of resistor R_1.


Notes

Whether inverting amplifier gains are expressed as negative or positive quantities seems to be a matter of taste, from surveying introductory textbooks on the subject. I prefer to stick with absolute (positive) gain values and consider signal inversion separately.





Question 3. (Click on arrow for answer)

Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit (A_V), both as a ratio and as a figure in units of decibels (dB):

Should be image 02468x01

File Num: 02468

Answer

Should be image 02468x02A_V = 0.468 = -6.594 dB

Notes

Operational amplifier circuits provide a great opportunity to review basic concepts of DC circuits: voltage drops, polarity, current directions, Ohm’s Law, Kirchhoff’s Voltage Law, Kirchhoff’s Current Law, etc. This circuit is no exception. Emphasize the fact that a great many opamp circuits may be comprehensively analyzed merely with knowledge of these fundamental principles and the characteristics of an ideal opamp (zero input current, infinite open-loop gain, unlimited output voltage swing, zero voltage between input terminals when negative feedback is in effect).

Some students may arrive at the wrong gain figure because they blindly followed a formula with R_1 and R_2 shown as variables, plugging in this circuit’s values for R_1 and R_2 without considering which resistor is which (is R_1 the feedback resistor or is R_2?). This is by design, as I want students to learn to think about what they are doing rather than thoughtlessly follow instructions.





Question 4. (Click on arrow for answer)

Determine both the input and output voltage in this circuit:

Should be image 02732x01

File Num: 02732

Answer

V_{in} = -10 V. V_{out} = 24 V

Follow-up question: how do we know that the input voltage in this circuit is negative and the output voltage is positive?


Notes

Ask your students how they solved this problem, sharing techniques and strategies to help other students know where to begin and where to proceed from there.





Question 5. (Click on arrow for answer)

The equation for voltage gain (A_V) in a typical inverting, single-ended opamp circuit is as follows:

A_V = {R_{1} \over R_{2}}

Where,

R_1 is the feedback resistor (connecting the output to the inverting input)R_2 is the other resistor (connecting the inverting input to voltage signal input terminal)

Suppose we wished to change the voltage gain in the following circuit from 3.5 to 4.9, but only had the freedom to alter the resistance of R_{2}:

Should be image 02708x01

Algebraically manipulate the gain equation to solve for R_2, then determine the necessary value of R_2 in this circuit to give it a voltage gain of 4.9.

File Num: 02708

Answer

R_2 = {R_1 \over A_V}

For the circuit shown, R_2 would have to be set equal to 1.571 k\Omega.


Notes

Nothing more than a little algebra to obtain the answers for this question!





Question 6. (Click on arrow for answer)

Calculate the necessary resistor value (R_1) in this circuit to give it a voltage gain of 15:

Should be image 02729x01

File Num: 02729

Answer

R_1 = 1.467 k\Omega

Notes

Ask your students how they solved this problem, especially since it is fairly safe to say that they didn’t find the equation directly solving for R_1 in any book. Algebraic manipulation is necessary to take the standard voltage gain equation and put it into a form suitable for use answering this question.





Question 7. (Click on arrow for answer)

Calculate the necessary resistor value (R_1) in this circuit to give it a voltage gain of 7.5:

Should be image 02730x01

File Num: 02730

Answer

R_1 = 62.25 k\Omega

Notes

Ask your students how they solved this problem, especially since it is fairly safe to say that they didn’t find the equation directly solving for R_1 in any book. Algebraic manipulation is necessary to take the standard voltage gain equation and put it into a form suitable for use answering this question.





Question 8. (Click on arrow for answer)

Calculate the output voltage of this op-amp circuit (using negative feedback):

Should be image 00932x01

Also, calculate the DC voltage gain of this circuit.

File Num: 00932

Answer

V_{out} = -8.1 volts
A_V = 5.4

Follow-up question: the midpoint of the voltage divider (connecting to the inverting input of the op-amp) is often called a virtual ground in a circuit like this. Explain why.


Notes

It is important that students learn to analyze the op-amp circuit in terms of voltage drops and currents for each resistor, rather than just calculate the output using a gain formula. Detailed, Ohm’s Law analysis of op-amp circuits is essential for analyzing more complex circuitry.

The “virtual ground” question is an important one for the sake of rapid analysis. Once students understand how and why there is such a thing as a “virtual ground” in an op-amp circuit like this, their analysis of op-amp circuits will be much more efficient.





Question 9. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

Should be image 02470x01

File Num: 02470

Answer


  • Stage 1: A_V = 1.5 = 3.522 dB
  • Stage 2: A_V = 1.545 = 3.781 dB
  • Total: A_V = 2.318 = 7.303 dB

Notes

Not only does this question review calculation of voltage gain for inverting amplifier circuits, but it also reviews decibel calculations (for both single and multi-stage amplifiers). Discuss how the decibel figures for each stage add to equal the total decibel gain, whereas the ratios multiply.





Question 10. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

Should be image 02471x01

File Num: 02471

Answer

  • Stage 1: A_V = 1.766 = 4.940 dB
  • Stage 2: A_V = 0.455 = -6.848 dB
  • Total: A_V = 0.803 = -1.909 dB

Notes

Not only does this question review calculation of voltage gain for inverting amplifier circuits, but it also reviews decibel calculations (for both single and multi-stage amplifiers). Discuss how the decibel figures for each stage add to equal the total decibel gain, whereas the ratios multiply.





Question 11. (Click on arrow for answer)

Operational amplifier circuits employing negative feedback are sometimes referred to as “electronic levers,” because their voltage gains may be understood through the mechanical analogy of a lever. Explain this analogy in your own words, identifying how the lengths and fulcrum location of a lever relate to the component values of an op-amp circuit:

Should be image 00933x01

File Num: 00933

Answer

The analogy of a lever works well to explain how the output voltage of an op-amp circuit relates to the input voltage, in terms of both magnitude and polarity. Resistor values correspond to moment arm lengths, while direction of lever motion (up versus down) corresponds to polarity. The position of the fulcrum represents the location of ground potential in the feedback network.


Notes

I found this analogy in one of the best books I’ve ever read on op-amp circuits: John I. Smith’s Modern Operational Circuit Design. Unfortunately, this book is out of print, but if you can possibly obtain a copy for your library, I highly recommend it!





Question 12. (Click on arrow for answer)

Compare and contrast inverting versus noninverting amplifier circuits constructed using operational amplifiers:

Should be image 02469x01

How do these two general forms of opamp circuit compare, especially in regard to input impedance and the range of voltage gain adjustment?

File Num: 02469

Answer

The noninverting configuration exhibits a far greater input impedance than the inverting amplifier, but has a more limited range of voltage gain: always greater than or equal to unity.


Notes

Just a simple comparison between amplifier configurations, nothing more. Ask your students to elaborate on the inverting amplifier’s range of gain adjustment: how does it differ from the noninverting configuration?





Question 13. (Click on arrow for answer)

What possible benefit is there to adding a voltage buffer to the front end of an inverting amplifier, as shown in the following schematic?

Should be image 02472x01

File Num: 02472

Answer

The voltage buffer raises the amplifier’s input impedance without altering voltage gain.


Notes

Discuss with your students how this is very common: using a voltage buffer as an impedance transformation (or isolation) device so that a weak (high-impedance) source is able to drive an amplifier.





Question 14. (Click on arrow for answer)

The junction between the two resistors and the inverting input of the operational amplifier is often referred to as a virtual ground, the voltage between it and ground being (almost) zero over a wide range of circuit conditions:

Should be image 02473x01

If the operational amplifier is driven into saturation, though, the “virtual ground” will no longer be at ground potential. Explain why this is, and what condition(s) may cause this to happen.

Hint: analyze all currents and voltage drops in the following circuit, assuming an opamp with the ability to swing its output voltage rail-to-rail.

Should be image 02473x02

File Num: 02473

Answer

Any input signal causing the operational amplifier to try to output a voltage beyond either of its supply rails will cause the “virtual ground” node to deviate substantially from ground potential.


Notes

Before students can answer this question, they must understand what saturation means with regard to an operational amplifier. This is where the “hint” scenario comes into play. Students failing to grasp this concept will calculate the voltage drops and currents in the “hint” circuit according to standard procedures and assumptions, and arrive at an output voltage well in excess of +15 volts. Resolving this paradox will lead to insight, and hopefully to a more realistic set of calculations.





Question 15. (Click on arrow for answer)

There is something wrong with this amplifier circuit. Despite an audio signal of normal amplitude detected at test point 1 (TP1), there is no output measured at the “Audio signal out” jack:

Should be image 02474x01

Next, you decide to check for the presence of a good signal at test point 3 (TP3). There, you find 0 volts AC and DC no matter where the volume control is set.

From this information, formulate a plan for troubleshooting this circuit, answering the following questions:


  • What type of signal would you expect to measure at TP3?
  • What would be your next step in troubleshooting this circuit?
  • Are there any elements of this circuit you know to be working properly?
  • What do you suppose would be the most likely failure, assuming this circuit once worked just fine and suddenly stopped working all on it’s own?

File Num: 02474

Answer

The correct voltage signal at TP3 should be an audio waveform with significant crossover distortion (specifically, a vertical “jump” at each point where the waveform crosses zero volts, about 1.4 volts peak to peak). I’ll let you figure out answers to the other questions on your own, or with classmates.


Notes

I have found that troubleshooting scenarios are always good for stimulating class discussions, with students posing strategies for isolating the fault(s) and correcting one another on logical errors. There is not enough information given in this question to ensure a single, correct answer. Discuss this with your students, helping them to use their knowledge of circuit theory and opamps to formulate good diagnostic strategies.


Non-Inverting Amplifier Circuit

Non-Inverting Amplifier

A non-inverting amplifier does not invert that output (as you probably suspected). It does use negative feedback like the inverting amplifier to give a gain that is controllable by the external resistors. Check out this video for the details.


This video shows the configuration of a non-inverting op amp circuit and shows how to derive the equation for the gain of the amplifier. It even includes some simulations to show the op amp in action

Non inverting Op Amp Circuits

Non-Inverting Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000, and the inverting input connected to a voltage divider on its output terminal (so the inverting input receives exactly one-half the output voltage). In other words, write an equation describing the output voltage of this op-amp (V_{out}) for any given input voltage at the noninverting input (V_{in(+)}):

Should be image 00928x01

Then, once you have an equation written, solve for the output voltage if the noninverting input voltage is -2.4 volts.

File Num: 00928

Answer

V_{out} = 100,000(V_{in(+)} - {1 \over 2}V_{out})
(I’ve left it up to you to perform the algebraic simplification here!)\vskip 20pt

For an input voltage of -2.4 volts, the output voltage will be -4.7999 volts.


Follow-up question: what do you notice about the output voltage in this circuit? What value is it very close to being, in relation to the input voltage? Does this pattern hold true for other input voltages as well?


Notes

Your students should see a definite pattern here as they calculate the output voltage for several different input voltage levels. Discuss this phenomenon with your students, asking them to explain it as best they can.





Question 2. (Click on arrow for answer)

Calculate the overall voltage gain of this amplifier circuit (A_V), both as a ratio and as a figure in units of decibels (dB). Also, write a general equation for calculating the voltage gain of such an amplifier, given the resistor values of R_1 and R_2:

Should be image 02457x01

File Num: 02457

Answer

A_V = 2 = 6.02 dBA_V = {R_1 \over R_2} + 1 \hbox{ (expressed as a ratio, not dB)}

Follow-up question: explain how you could modify this particular circuit to have a voltage gain (ratio) of 3 instead of 2.


Notes

Nothing special here — just some practice with voltage gain calculations.





Question 3. (Click on arrow for answer)

What would have to be altered in this circuit to increase its overall voltage gain?

Should be image 00931x01

File Num: 00931

Answer

The voltage divider would have to altered so as to send a smaller proportion of the output voltage to the inverting input.


Notes

Ask your students to explain how they would modify the voltage divider in this circuit to achieve the goal of a smaller voltage division ratio. This should be trivial, but it is always good to review basic principles of electricity even when “deep” into a more advanced topic.





Question 4. (Click on arrow for answer)

Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit (A_V), both as a ratio and as a figure in units of decibels (dB):

Should be image 02459x01

File Num: 02459

Answer

Should be image 02459x02A_V = 1.468 = 3.335 dB

Notes

Operational amplifier circuits provide a great opportunity to review basic concepts of DC circuits: voltage drops, polarity, current directions, Ohm’s Law, Kirchhoff’s Voltage Law, Kirchhoff’s Current Law, etc. This circuit is no exception. Emphasize the fact that a great many opamp circuits may be comprehensively analyzed merely with knowledge of these fundamental principles and the characteristics of an ideal opamp (zero input current, infinite open-loop gain, unlimited output voltage swing, zero voltage between input terminals when negative feedback is in effect).

Some students may arrive at the wrong gain figure because they blindly followed a formula with R_1 and R_2 shown as variables, plugging in this circuit’s values for R_1 and R_2 without considering which resistor is which (is R_1 the feedback resistor or is R_2?). This is by design, as I want students to learn to think about what they are doing rather than thoughtlessly follow instructions.





Question 5. (Click on arrow for answer)

Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit (A_V), both as a ratio and as a figure in units of decibels (dB):

Should be image 02460x01

File Num: 02460

Answer

Should be image 02460x02A_V = 4.704 = 13.449 dB

Follow-up question: how much input impedance does the -2.35 volt source “see” as it drives this amplifier circuit?


Notes

Operational amplifier circuits provide a great opportunity to review basic concepts of DC circuits: voltage drops, polarity, current directions, Ohm’s Law, Kirchhoff’s Voltage Law, Kirchhoff’s Current Law, etc. This circuit is no exception. Emphasize the fact that a great many opamp circuits may be comprehensively analyzed merely with knowledge of these fundamental principles and the characteristics of an ideal opamp (zero input current, infinite open-loop gain, unlimited output voltage swing, zero voltage between input terminals when negative feedback is in effect).

The follow-up question is important because it showcases one of the great advantages of using noninverting opamp amplifier circuits as voltage signal amplifiers: extremely high input impedance. This would be a good opportunity to review typical input impedance values for operational amplifiers, by showing datasheets for some typical opamps and for some non-typical (i.e. MOSFET input) opamps.





Question 6. (Click on arrow for answer)

\int f(x) dx Calculus alert!

You are part of a team building a rocket to carry research instruments into the high atmosphere. One of the variables needed by the on-board flight-control computer is velocity, so it can throttle engine power and achieve maximum fuel efficiency. The problem is, none of the electronic sensors on board the rocket has the ability to directly measure velocity. What is available is an altimeter, which infers the rocket’s altitude (it position away from ground) by measuring ambient air pressure; and also an accelerometer, which infers acceleration (rate-of-change of velocity) by measuring the inertial force exerted by a small mass.

The lack of a “speedometer” for the rocket may have been an engineering design oversight, but it is still your responsibility as a development technician to figure out a workable solution to the dilemma. How do you propose we obtain the electronic velocity measurement the rocket’s flight-control computer needs?

File Num: 02702

Answer

One possible solution is to use an electronic integrator circuit to derive a velocity measurement from the accelerometer’s signal. However, this is not the only possible solution!


Notes

This question simply puts students’ comprehension of basic calculus concepts (and their implementation in electronic circuitry) to a practical test.





Question 7. (Click on arrow for answer)

Calculate the necessary resistor value (R_1) in this circuit to give it a voltage gain of 30:

Should be image 02725x01

File Num: 02725

Answer

R_1 = 1.345 k\Omega

Notes

Ask your students how they solved this problem, especially since it is fairly safe to say that they didn’t find the equation directly solving for R_1 in any book. Algebraic manipulation is necessary to take the standard voltage gain equation and put it into a form suitable for use answering this question.





Question 8. (Click on arrow for answer)

Calculate the necessary resistor value (R_1) in this circuit to give it a voltage gain of 10.5:

Should be image 02724x01

File Num: 02724

Answer

R_1 = 76.95 k\Omega

Notes

Ask your students how they solved this problem, especially since it is fairly safe to say that they didn’t find the equation directly solving for R_1 in any book. Algebraic manipulation is necessary to take the standard voltage gain equation and put it into a form suitable for use answering this question.





Question 9. (Click on arrow for answer)

Determine both the input and output voltage in this circuit:

Should be image 02726x01

File Num: 02726

Answer

V_{in} = 10 V. V_{out} = 46 V

Notes

Ask your students how they solved this problem, sharing techniques and strategies to help other students know where to begin and where to proceed from there.





Question 10. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

Should be image 02727x01

File Num: 02727

Answer


  • A_V = 4.3 = 12.669 dB


  • A_V = 6.745 = 16.579 dB


  • A_V = 29.002 = 29.249 dB


Notes

Not only does this question review calculation of voltage gain for inverting amplifier circuits, but it also reviews decibel calculations (for both single and multi-stage amplifiers). Discuss how the decibel figures for each stage add to equal the total decibel gain, whereas the ratios multiply.





Question 11. (Click on arrow for answer)

How much effect will a change in the op-amp’s open-loop voltage gain have on the overall voltage gain of a negative-feedback circuit such as this?

Should be image 00929x01

If the open-loop gain of this operational amplifier were to change from 100,000 to 200,000, for example, how big of an effect would it have on the voltage gain as measured from the noninverting input to the output?

File Num: 00929

Answer

The different in overall voltage gain will be trivial.


Follow-up question: what advantage is there in building voltage amplifier circuits in this manner, applying negative feedback to a “core” amplifier with very high intrinsic gain?


Notes

Work with your students to calculate a few example scenarios, with the old open-loop gain versus the new open-loop gain. Have the students validate their conclusions with numbers!

Negative feedback is an extremely useful engineering principle, and one that allows us to build very precise amplifiers using imprecise components. Credit for this idea goes to Harold Black, an electrical engineer, in 1920’s. Mr. Black was looking for a way to improve the linearity and stability of amplifiers in telephone systems, and (as legend has it) the idea came to him in a flash of insight as he was commuting on a ferry boat.

An interesting historical side-note is that Black’s 1928 patent application was initially rejected on the grounds that he was trying to submit a perpetual motion device! The concept of negative feedback in an amplifier circuit was so contrary to established engineering thought at the time, that Black experienced significant resistance to the idea within the engineering community. The United States patent office, on the other hand, was inundated with fraudulent “perpetual motion” claims, and so dismissed Black’s invention at first sight.





Question 12. (Click on arrow for answer)

A simple “follower” circuit that boosts the current-output ability of this noninverting amplifier circuit is a set of bipolar junction transistors, connected together in a “push-pull” fashion like this:

Should be image 00935x01

However, if connected exactly as shown, there will be a significant voltage error introduced to the opamp’s output. No longer will the final output voltage (measured across the load) be an exact 3:1 multiple of the input voltage, due to the 0.7 volts dropped by the transistor in active mode:

Should be image 00935x02

There is a very simple way to completely eliminate this error, without adding any additional components. Modify the circuit accordingly.

File Num: 00935

Answer

Should be image 00935x03

If you understand why this circuit works, pat yourself on the back: you truly understand the self-correcting nature of negative feedback. If not, you have a bit more studying to do!


Notes

The answer is not meant to be discouraging for those students of yours who do not understand how the solution works. It is simply a “litmus test” of whether or not your students really comprehend the concept of negative feedback. Although the change made in the circuit is simple, the principle is a bit of a conceptual leap for some people.

It might help your students understand if you label the new wire with the word sense, to indicate its purpose of providing feedback from the very output of the circuit, back to the opamp so it can sense how much voltage the load is receiving.





Question 13. (Click on arrow for answer)

Suppose a technician is checking the operation of the following electronic circuit:

Should be image 02733x01

She decides to measure the voltage on either side of resistor R1 with reference to ground, and obtains these readings:

Should be image 02733x02

On the top side of R1, the voltage with reference to ground is -5.04 volts. On the bottom side of R1, the voltage with reference to ground is -1.87 volts. The color code of resistor R1 is Yellow, Violet, Orange, Gold. From this information, determine the following:


  • Voltage across R1 (between top to bottom):
  • Polarity (+ and -) of voltage across R1:
  • Current (magnitude) through R1:
  • Direction of current through R1:

Additionally, explain how this technician would make each one of these determinations. What rules or laws of electric circuits would she apply?

File Num: 02733

Answer


  • Voltage across R1 (between top to bottom): 3.17 volts
  • Polarity (+ and -) of voltage across R1: (-) on top, (+) on bottom
  • Current (magnitude) through R1: 67.45 \muA
  • Direction of current through R1: upward, following conventional flow


Follow-up question: calculate the range of possible currents, given the specified tolerance of resistor R1 (67.45 \muA assumes 0\% error).


Challenge question: if you recognize the type of circuit this is (by the part number of the IC “chip”: TL082), identify the voltage between pin 3 and ground.


Notes

This is a good example of how Kirchhoff’s Voltage Law is more than just an abstract tool for mathematical analysis — it is also a powerful technique for practical circuit diagnosis. Students must apply KVL to determine the voltage drop across R1, and then use Ohm’s Law to calculate its current.

If students experience difficulty visualizing how KVL plays a part in the solution of this problem, show them this illustration:

Should be image 02733x03

By the way, the answer to the challenge question may only be realized if students recognize this circuit as a noninverting opamp voltage amplifier. The voltage at pin 3 (noninverting input) will be the same as the voltage at pin 2 (inverting input): -1.87 volts.





Question 14. (Click on arrow for answer)

There is something wrong with this amplifier circuit. Note the relative amplitudes of the input and output signals as measured by an oscilloscope:

Should be image 02465x01

This circuit used to function perfectly, but then began to malfunction in this manner: producing a “clipped” output waveform of excessive amplitude. Determine the approximate amplitude that the output voltage waveform should be for the component values given in this circuit, and then identify possible causes of the problem and also elements of the circuit that you know cannot be at fault.

File Num: 02465

Answer

V_{out} (ideal) = 1.01 volts RMS

I’ll let you determine possible faults in the circuit! From what we see here, we know the power supply is functioning (both +V and -V rails) and that there is good signal getting to the noninverting input of the opamp.


Notes

There is definitely more than one possible cause for the observed problem. Discuss alternatives with your students, involving them in the diagnosis process. Ask them why we know that certain elements of the circuit are functioning as they should? Of the possible causes, which are more likely, and why?

Question 15. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

Should be image 02728x01

File Num: 02728

Answer


  • A_V = 1.702 = 4.62 dB


  • A_V = 5.136 = 14.213 dB


  • A_V = 8.743 = 18.833 dB


Notes

Not only does this question review calculation of voltage gain for inverting amplifier circuits, but it also reviews decibel calculations (for both single and multi-stage amplifiers). Discuss how the decibel figures for each stage add to equal the total decibel gain, whereas the ratios multiply.


Summing Amplifier Circuit

Summing (Non-Inverting) Amplifier

A summing amplifier has multiple inputs and sums them (adds them) all together to create the output. Sometimes, the inputs have equal weightings and sometimes they don’t. Sometimes the output is inverted and sometimes it is non-inverted. See this video to find out about the different configurations


This video describes summing amplifiers and shows the two main configurations. Simulations show that the input voltages do not need to be in phase or even the same frequency, the summer will add them together even if they aren’t sinusoidal.

Summing Amplifiers

Summing Amplifier Practice Problems

Question 1. (Click on arrow for answer)

The simple resistor network shown here is known as a passive averager. Describe what the word “passive” means in this context, and write an equation describing the output voltage (V_d) in terms of the input voltages (V_a, V_b, and V_c):

Should be image 01001x01

Hint: there is a network theorem that directly applies to this form of circuit, and it is known as Millman’s Theorem. Research this theorem and use it to generate your equation!

File Num: 01001

Answer

“Passive” means that the circuit contains no amplifying components.V_d = {{V_a + V_b + V_c} \over 3}

Notes

Students need to realize that even passive circuits are able to model (some) mathematical functions! Ask your students if they can think of any network analysis methods to easily calculate the output voltage (V_d) of this circuit, given the input voltages. There is one theorem in particular that works very well for this particular circuit.





Question 2. (Click on arrow for answer)

Add an op-amp circuit to the output of this passive averager network to produce a summer circuit: an operational circuit generating an output voltage equal to the sum of the four input voltages. Then, write an equation describing the whole circuit’s function.

\vskip 30ptShould be image 01002x01

File Num: 01002

Answer

Should be image 01002x02V_{sum} = V_a + V_b + V_c + V_d

Notes

The equation for this circuit is simple enough as to require no explanation. How your students derived this equation, from the base equation of a passive averager network, on the other hand, is worth discussion. Discuss with them the necessary gain of the op-amp circuit, and how this gain figure converts an averaging function into a summing function.





Question 3. (Click on arrow for answer)

Determine all current magnitudes and directions, as well as voltage drops, in this circuit:

Should be image 02515x01

File Num: 02515

Answer

Should be image 02515x02

Follow-up question: what would be required to get this circuit to output the exact sum of the four input voltages?


Notes

This question not only provides practice analyzing the behavior of a summer circuit, but also analyzing the behavior of a passive averager circuit. If your students need some refreshing on how to analyze the passive averager, you might want to review Millman’s Theorem with them.





Question 4. (Click on arrow for answer)

Determine all current magnitudes and directions, as well as voltage drops, in this circuit:

Should be image 02523x01

File Num: 02523

Answer

Should be image 02523x02

Follow-up question: what would be required to get this circuit to output the exact sum of the four input voltages?


Notes

This question not only provides practice analyzing the behavior of a summer circuit, but also analyzing the behavior of a passive averager circuit. If your students need some refreshing on how to analyze the passive averager, you might want to review Millman’s Theorem with them.





Question 5. (Click on arrow for answer)

Predict how the operation of this summer circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Should be image 03780x01
  • Resistor R_1 fails open:
  • Solder bridge (short) across resistor R_3:
  • Resistor R_4 fails open:
  • Resistor R_5 fails open:
  • Solder bridge (short) across resistor R_5:
  • Resistor R_6 fails open:

For each of these conditions, explain why the resulting effects will occur.

File Num: 03780

Answer


  • Resistor R_1 fails open: V_{out becomes equal to {4 \over 3} the sum of voltages V_2, V_3, and V_4.}

  • Solder bridge (short) across resistor R_3: V_{out becomes equal to 4 times V_3.}

  • Resistor R_4 fails open: V_{out becomes equal to {4 \over 3} the sum of voltages V_1, V_2, and V_3.}

  • Resistor R_5 fails open:>Circuit operates as an averager, not a summer.

  • Solder bridge (short) across resistor R_5: V_{out saturates in a positive direction.}

  • Resistor R_6 fails open: V_{out saturates in a positive direction.}


Notes

The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.





Question 6. (Click on arrow for answer)

Determine the amount of current from point A to point B in this circuit:

Should be image 02516x01

File Num: 02516

Answer

I = 6.5 mA

Notes

This question, while being an application of Kirchhoff’s Current Law, is also a prelude to an inverting summer circuit, where an opamp takes that 6.5 mA (total) current and converts it into an output voltage.





Question 7. (Click on arrow for answer)

Determine the amount of current from point A to point B in this circuit, and also the output voltage of the operational amplifier:

Should be image 02517x01

File Num: 02517

Answer

I = 6.5 mA V_{out} = -6.5 V

Notes

This question is best preceded by \#02516, which asks for students to solve for the current between A and B with no opamp in the circuit (simply grounded at point B). When students realize that point B is now a virtual ground instead of a real ground, they see that the same conclusion derived by Kirchhoff’s Current Law in the passive circuit is still valid in this active circuit, and that the result is an output voltage corresponding to that current.





Question 8. (Click on arrow for answer)

Write a mathematical equation for this op-amp circuit, assuming all resistor values are equal:

Should be image 01003x01

What is this circuit typically called?

File Num: 01003

Answer

c = -(a + b)

This type of circuit is typically called an inverting summer.


Follow-up question: explain why the addition of another resistor in this circuit is recommended for optimum accuracy, as shown in the following schematic.

Should be image 01003x02

Challenge question: write an equation describing the proper value of this extra resistor.


Notes

Ask your students about the proper resistor values for an inverting summer circuit. The choices of resistor values are definitely not the same for inverting summer and noninverting summer circuits alike! Discuss why the values are what they are in an inverting summer circuit (using Ohm’s Law to analyze the circuit’s function), emphasizing comprehension over rote memorization.





Question 9. (Click on arrow for answer)

Predict how the operation of this summer circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Should be image 03781x01
  • Resistor R_1 fails open:
  • Resistor R_2 fails open:
  • Solder bridge (short) across resistor R_3:
  • Resistor R_4 fails open:
  • Solder bridge (short) across resistor R_4:

For each of these conditions, explain why the resulting effects will occur.

File Num: 03781

Answer


  • Resistor R_1 fails open: V_{out becomes (inverted) sum of V_2 and V_3 only.}
  • Resistor R_2 fails open: V_{out becomes (inverted) sum of V_1 and V_3 only.}
  • Solder bridge (short) across resistor R_3: V_{out saturates in a negative direction.}
  • Resistor R_4 fails open: V_{out saturates in a negative direction.}
  • Solder bridge (short) across resistor R_4: V_{out goes to 0 volts.}


Notes

The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.





Question 10. (Click on arrow for answer)

Identify some of the distinguishing characteristics of inverting and noninverting summer circuits. How may you identify which is which, and how may you determine the proper resistor values to make each one work as it should?

File Num: 02520

Answer

I won’t directly answer the questions here, but I will give some hints. A noninverting summer circuit is composed of a passive voltage averager circuit coupled to a noninverting voltage amplifier with a voltage gain equal to the number of inputs on the averager. An inverting summer circuit is composed of a passive current summer node coupled to a current-to-voltage converter.


Notes

This question is designed to spur discussion amongst your students, exchanging ideas about each circuit’s defining characteristics. Having students explore each circuit type on their own, reaching their own conclusions about how to differentiate the two, is a far more effective way of making them understand the differences than simply telling them outright.


Difference Amplifier Circuit

A difference amplifier outputs a voltage that is the difference between two input voltages. The inputs might be equally weighted or they might not be. It depends on the values of the resistors around the op amp. Watch this video to learn more:


This video derives the equation for a difference amplifier given arbitrary values for the resistor around it. As you will see if the resistors are all the same or if R1/R3 are equal and R2/R4 are equal, you get a much easier to analyze circuit.

Difference Amplifier Using an Op Amp

Difference Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Complete the table of values for this opamp circuit, calculating the output voltage for each combination of input voltages shown:

Should be image 02518x01
V1V2Vout
0V 0V
+1V 0V
0V +1V
+2V +1.5V
+3.4V +1.2V
-2V +4V
+5V +5V
-3V -3V

What pattern do you notice in the data? What mathematical relationship is there between the two input voltages and the output voltage?

File Num: 02518

Answer

V1V2Vout
0V 0V 0V
+1V 0V -1V
0V +1V +1V
+2V +1.5V -0.5V
+3.4V +1.2V -2.2V
-2V +4V +6V
+5V +5V 0V
-3V -3V 0V

Notes

Though it may be tedious to calculate the output voltage for each set of input voltages, working through all the voltage drops and currents in the opamp circuit one at a time, it shows students how they may be able to discern the function of an opamp circuit merely by applying basic laws of electricity (Ohm’s Law, KVL, and KCL) and the “golden assumptions” of negative feedback opamp circuits (no input currents, zero differential input voltage).





Question 2. (Click on arrow for answer)

This opamp circuit is known as a difference amplifier, sometimes called a subtractor. Assuming that all resistor values are equal in the circuit, write an equation expressing the output (y) as a function of the two input voltages (a and b):

Should be image 01010x01

File Num: 01010

Answer

y = b - a

Notes

Work through some example conditions of input voltages and resistor values to calculate the output voltage using Ohm’s Law and the general principle of negative feedback in an opamp circuit (namely, an assumption of zero voltage differential at the opamp inputs). The goal here is to have students comprehend why this circuit subtracts one voltage from another, rather than just encourage rote memorization.





Question 3. (Click on arrow for answer)

How does the operation of this difference amplifier circuit compare with the resistor values given (2R = twice the resistance of R), versus its operation with all resistor values equal?

Should be image 02525x01

Describe what approach or technique you used to derive your answer, and also explain how your conclusion for this circuit might be generalized for all difference amplifier circuits.

File Num: 02525

Answer

It is very important that you develop the skill of “exploring” a circuit configuration to see what it will do, rather than having to be told what it does (either by your instructor or by a book). All you need to have is a solid knowledge of basic electrical principles (Ohm’s Law, Kirchhoff’s Voltage and Current Laws) and know how opamps behave when configured for negative feedback.


As for a generalized conclusion:

Should be image 02525x02

Notes

It is easy for you (the instructor) to show how and why this circuit acts as it does. The point of this question, however, is to get students to take the initiative to explore the circuit on their own. It is simple enough for any student to set up some hypothetical test conditions (a thought experiment) to analyze what this circuit will do, that the only thing holding them back from doing so is attitude, not aptitude.

This is something I have noticed over years of teaching: so many students who are more than capable of doing the math and applying well-understood electrical rules refuse to do so on their own, because years of educational tradition has indoctrinated them to wait for the instructor’s lead rather than explore a concept on their own.





Question 4. (Click on arrow for answer)

Predict how the operation of this difference amplifier circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Should be image 03782x01
  • Resistor R_1 fails open:
  • Resistor R_2 fails open:
  • Solder bridge (short) across resistor R_3:
  • Resistor R_4 fails open:
  • Solder bridge (short) across resistor R_4:

For each of these conditions, explain why the resulting effects will occur.

File Num: 03782

Answer


  • Resistor R_1 fails open: V_{out} becomes equal to {1 \over 2} V_2.}
  • Resistor R_2 fails open: V_{out} saturates.}
  • Solder bridge (short) across resistor R_3: V_{out} becomes equal to 2 V_2 - V_1 instead of V_2 - V_1.}
  • Resistor R_4 fails open: V_{out} becomes equal to 2 V_2 - V_1 instead of V_2 - V_1.}
  • Solder bridge (short) across resistor R_4: V_{out} becomes equal to -V_1.}


Notes

The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.


Integrator Op Amp Circuit

Integrator Circuit

The integrator circuit does exactly what you would think that it does…it takes the input voltage and integrates it to give the output voltage. Yes, circuits can do calculus. To find out how, see the video below

The integrator op amp circuit is an inverting op amp circuit with a resistor at the input and a capacitor for the feedback. This video goes through the analysis to see why it is called an integrator. The video also includes some simulations that show the integrator working and also not working

Integrator Op Amp Circuit

Differentiator Op Amp Circuit

Differentiator Circuit

Once again, the name of the circuit is the thing that it does. So this circuit differentiates, or takes the derivative of the input voltage to create the output voltage. Well, it doesn’t actually sit there and do the calculation, but the output voltage is proportional to the derivative of the input voltage and in this video, I show you why.

The differentiator op amp circuit does output the derivative of the input, but as you will see in this video, it is a little sensitive and unstable.

Differentiator Op Amp Circuit

Contributors

Contributors to this chapter are listed in chronological order of their contributions, from most recent to first.

David Williams (2022): First edits including videos




Practice Problem Copyright

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

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

Op Amp Basics

Operational amplifiers, also known as op amps, are semiconductor devices characterized with two high impedance inputs, a very high differential voltage gain and a low impedance output. By using external feedback components, a wide variety of different types of circuits can be built. Op amps may very well be the most useful device in analog circuitry because of the wide range of signal processing tasks it can accomplish. They are affordable, durable, and come in a wide variety of packages.

The diagram below shows the circuit symbol for a basic op amp

Together, the non-inverting (Vin+)pin and the inverting (Vin-)pin make up a differential input. VS+ and VS- are the positive and negative power supply respectively. And surprise, surprise Vout is the output. The general equation for the op amp relating inputs to outputs is:

V_{out}=A_{vol}[V_{(in+)}-V_{(in-)}]

Avol is the open loop voltage gain and is typically a very big value and for an ideal op amp, it is infinite.


Video: Intro to Op Amps

To find out more, check out this video:

Introduction to Operational Amplifiers

Op Amp Basics Practice Problems

Question 1. (Click on arrow for answer)

An operational amplifier is a particular type of differential amplifier. Most op-amps receive two input voltage signals and output one voltage signal:

Should be image 00848x01

Here is a single op-amp, shown under two different conditions (different input voltages). Determine the voltage gain of this op-amp, given the conditions shown:

Should be image 00848x02Should be image 00848x03

Also, write a mathematical formula solving for differential voltage gain (A_V) in terms of an op-amp’s input and output voltages.

File Num: 00848

Answer

A_V = 530,000A_V = {\Delta V_{out} \over \Delta (V_{in2} - V_{in1})}

Follow-up question: convert this voltage gain figure (as a ratio) into a voltage gain figure in decibels.


Notes

The calculations for voltage gain here are not that different from the voltage gain calculations for any other amplifier, except that here we’re dealing with a differential amplifier instead of a single-ended amplifier.

A differential voltage gain of 530,000 is not unreasonable for a modern operational amplifier! A gain so extreme may come as a surprise to many students, but they will discover later the utility of such a high gain.





Question 2. (Click on arrow for answer)

Many op-amp circuits require a dual or split power supply, consisting of three power terminals: +V, -V, and Ground. Draw the necessary connections between the 6-volt batteries in this schematic diagram to provide +12 V, -12 V, and Ground to this op-amp:

Should be image 00880x01

File Num: 00880

Answer

Should be image 00880x02

Notes

I encourage your students to learn how to power op-amp circuits with interconnected batteries, because it really helps to build their understanding of what a “split” power supply is, as well as allow them to build functioning op-amp circuits in the absence of a quality benchtop power supply.





Question 3. (Click on arrow for answer)

The 8-pin Dual-Inline-Package (DIP) is a common format in which single and dual operational amplifiers are housed. Shown here are the case outlines for two 8-pin DIPs. Draw the internal op-amp connections for a single op-amp unit, and for a dual op-amp unit:

Should be image 00874x01

You will need to research some op-amp datasheets to find this information. Examples of single op-amp chips include the LM741, CA3130, and TL081. Examples of dual op-amp chips include the LM1458 and TL082.

File Num: 00874

Answer

Should be image 00874x02

Notes

Ask your students to reveal their information sources, and what specific models of op-amp they researched.





Question 4. (Click on arrow for answer)

In this circuit, an op-amp turns on an LED if the proper input voltage conditions are met:

Should be image 00801x01

Trace the complete path of current powering the LED. Where, exactly, does the LED get its power from?

File Num: 00801

Answer

The arrows shown in this diagram trace “conventional” current flow, not electron flow:

Should be image 00801x02

Notes

The important thing to note here is that the load current does not pass through either of the op-amp’s input terminals. All load current is sourced by the op-amp’s power supply! Discuss the importance of this fact with your students.





Question 5. (Click on arrow for answer)

What does it mean if an operational amplifier has the ability to “swing its output rail to rail”? Why is this an important feature to us?

File Num: 00844

Answer

Being able to “swing” the output voltage “rail to rail” means that the full range of an op-amp’s output voltage extends to within millivolts of either power supply “rail” (+V and -V).


Challenge question: identify at least one op-amp model that has this ability, and at least one that does not. Bring the datasheets for these op-amp models with you for reference during discussion time.


Notes

Discuss what this feature means to us as circuit builders in a practical sense. Ask those students who tackled the challenge question to look up the output voltage ranges of their op-amp models. Exactly how close to +V and -V can the output voltage of an op-amp lacking “rail-to-rail” output capability “swing”?





Question 6. (Click on arrow for answer)

Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000. In other words, write an equation describing the output voltage of this op-amp (V_{out}) for any combination of input voltages (V_{in(+)} and V_{in(-)}):

Should be image 00925x01

File Num: 00925

Answer

V_{out} = 100,000(V_{in(+)} - V_{in(-)})

Notes

The concept of a “transfer function” is very useful, and this may be your students’ first exposure to the idea. It is a phrase used quite often in engineering applications, and may denote an equation, a table of numbers, or a graph.

In this particular question, it is important that students know how to derive and use the basic transfer function for a differential amplifier. Challenge your students to express this function in a more general form, so that calculations may be made with different open-loop voltage gains.





Question 7. (Click on arrow for answer)

How much voltage would have to be “dialed up” at the potentiometer in order to stabilize the output at exactly 0 volts, assuming the opamp has no input offset voltage?

Should be image 00924x01

File Num: 00924

Answer

5 volts


Notes

This question is a basic review of an ideal differential amplifier’s function. Ask your students what voltage must be “dialed up” at the potentiometer to produce 0 volts at the output of the op-amp for several different voltages at the other input. If they don’t understand at first, they soon will after discussing these alternate scenarios.





Question 8. (Click on arrow for answer)

An op-amp has +3 volts applied to the inverting input and +3.002 volts applied to the noninverting input. Its open-loop voltage gain is 220,000. Calculate the output voltage as predicted by the following formula:

V_{out} = A_V \left( V_{in(+)} - V_{in(-)} \right)

How much differential voltage (input) is necessary to drive the output of the op-amp to a voltage of -4.5 volts?

File Num: 00926

Answer

V_{out} = 440 volts

Follow-up question: is this voltage figure realistic? Is it possible for an op-amp such as the model 741 to output 440 volts? Why or why not?

The differential input voltage necessary to drive the output of this op-amp to -4.5 volts is -20.455 \muV.


Follow-up question: what does it mean for the input voltage differential to be negative 20.455 microvolts? Provide an example of two input voltages (V_{in(+)} and V_{in(-)}) that would generate this much differential voltage.


Notes

Obviously, there are limitations to the op-amp formula for calculating output voltage, given input voltages and open-loop voltage gain. Students need to realize the practical limits of an op-amp’s output voltage range, and what sets those limits.


Ideal Op Amp

An ideal op amp has the following characteristics:

  1. Infinite open loop voltage gain (Avol)
  2. Infinite input impedance. No current flows into inputs
  3. Zero output impedance. No voltage drop at output due to loads
  4. Infinite bandwidth. Works the same at all frequencies
  5. No offset voltage. When the inputs are zero, the output is zero

Real op amps are designed to be as close to the ideal op amps as possible, and in practice, you generally (but not always) have to look pretty hard to see the non-ideal characteristics of a real op amp.

Comparator

Non-Inverting Comparator Circuit

A comparator is the simplest circuit that you can build with an op amp. It uses the enormous open loop voltage gain of an op amp to indicate which of the two inputs is at a higher voltage.


Video: Op Amp Comparators

This video describes how a comparator works and provides a few examples of the different comparator configurations:

Op Amp Comparator Circuits

Op Amp Comparators Practice Problems

Question 1. (Click on arrow for answer)

Determine the output voltage polarity of this op-amp (with reference to ground), given the following input conditions:

Should be image 00803x01

File Num: 00803

Answer

In these illustrations, I have likened the op-amp’s action to that of a single-pole, double-throw switch, showing the “connection” made between power supply terminals and the output terminal.

Should be image 00803x02

Notes

Determining which “way” the output of an op-amp drives under different input voltage conditions is confusing to many students. Discuss this with them, and ask them to present any principles or analogies they use to remember “which way is which.”





Question 2. (Click on arrow for answer)

Determine the output voltage polarity of this op-amp (with reference to ground), given the following input conditions:

Should be image 03762x01

File Num: 03762

Answer

In these illustrations, I have likened the op-amp’s action to that of a single-pole, double-throw switch, showing the “connection” made between power supply terminals and the output terminal.

Should be image 03762x02

Notes

Determining which “way” the output of an op-amp drives under different input voltage conditions is confusing to many students. Discuss this with them, and ask them to present any principles or analogies they use to remember “which way is which.”





Question 3. (Click on arrow for answer)

In this circuit, a solar cell converts light into voltage for the opamp to “read” on its noninverting input. The opamp’s inverting input connects to the wiper of a potentiometer. Under what conditions does the LED energize?

Should be image 00872x01

File Num: 00872

Answer

The LED energizes under bright-light conditions, de-energizing when the light decreases below the threshold set by the potentiometer.


Follow-up question: determine what would have to be changed in this circuit to make the LED turn on when the solar cell becomes dark.


Notes

There is more than one way to accomplish the task posed by the follow-up question. Be sure to ask your students for their ideas on how to reverse the LED’s operation!





Question 4. (Click on arrow for answer)

A student is operating a simple comparator circuit and documenting the results in a table:

Should be image 00876x01
\settabs \+ \quad MMMM \quad & \quad MMMM \quad & \quad MMMM \quad & \quad MMMM \quad & \cr \+ \hfill & V_{in(+)} & V_{in(-)} & V_{out} \cr \+ \hfill & 3.00 V & 1.45 V & 10.5 V \cr \+ \hfill & 3.00 V & 2.85 V & 10.4 V \cr \+ \hfill & 3.00 V & 3.10 V & 1.19 V \cr \+ \hfill & 3.00 V & 6.75 V & 1.20 V \cr
\+ \hfill & V_{in(+)} & V_{in(-)} & V_{out} \cr \+ \hfill & 2.36 V & 6.50 V & 1.20 V \cr \+ \hfill & 4.97 V & 6.50 V & 1.21 V \cr \+ \hfill & 7.05 V & 6.50 V & 10.5 V \cr \+ \hfill & 9.28 V & 6.50 V & 10.4 V \cr
\+ \hfill & V_{in(+)} & V_{in(-)} & V_{out} \cr \+ \hfill & 10.4 V & 9.87 V & 10.6 V \cr \+ \hfill & 1.75 V & 1.03 V & 10.5 V \cr \+ \hfill & 0.31 V & 1.03 V & 10.5 V \cr \+ \hfill & 5.50 & 5.65 V & 1.19 V \cr

One of these output voltage readings is anomalous. In other words, it does not appear to be “correct”. This is very strange, because these figures are real measurements and not predictions! Perplexed, the student approaches the instructor and asks for help. The instructor sees the anomalous voltage reading and says two words: latch-up. With that, the student goes back to research what this phrase means, and what it has to do with the weird output voltage reading.

Identify which of these output voltage measurements is anomalous, and explain what “latch-up” has to do with it.

File Num: 00876

Answer

Latch-up occurs when one of the input voltage signals approaches too close to one of the power supply rail voltages. The result is the op-amp output saturating “high” even if it isn’t supposed to.


Challenge question: suppose we expected both input voltages to range between 0 and 10 volts during normal operation of this comparator circuit. What could we change in the circuit to allow this range of operation and avoid latch-up?


Notes

Ask your students what they found in their research on “latch-up,” and if this is an idiosyncrasy of all op-amp models, or just some.

Incidentally, the curved op-amp symbol has no special meaning. This symbol was quite popular for representing op-amps during their early years, but has since fallen out of favor. I show it here just to inform your students, in case they ever happen to encounter one of these symbols in an old electronic schematic.





Question 5. (Click on arrow for answer)

In this circuit, an op-amp turns on an LED if the proper input voltage conditions are met:

Should be image 00801x01

Trace the complete path of current powering the LED. Where, exactly, does the LED get its power from?

File Num: 00801

Answer

The arrows shown in this diagram trace “conventional” current flow, not electron flow:

Should be image 00801x02

Notes

The important thing to note here is that the load current does not pass through either of the op-amp’s input terminals. All load current is sourced by the op-amp’s power supply! Discuss the importance of this fact with your students.





Question 6. (Click on arrow for answer)

Trace the output waveform of this comparator circuit:

Should be image 00878x01

File Num: 00878

Answer

Should be image 00878x02

Follow-up question: explain what the phrase duty cycle means with reference to a “square” or “pulse” waveform.


Notes

During discussion, ask your students to explain how the output waveform of this comparator circuit comes to be, step by step. Ask them how they arrived at their solution, and if there is a way this AC/DC problem can be simplified to one that is DC only for easier analysis (determining what the output voltage will do for a certain set of input conditions).





Question 7. (Click on arrow for answer)

Calculate the amount of resistance that the thermistor much reach in order to turn the cooling fan on:

Should be image 04021x01

File Num: 04021

Answer

Thermistor resistance = 5.547 k\Omega


Notes

Ask your students how they arrived at their solution for this question. There is definitely more than one way to do it!





Question 8. (Click on arrow for answer)

Predict how the operation of this thermostat circuit (where the cooling fan motor is supposed to turn on when the temperature gets too high) will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Should be image 03768x01
  • Cable fails open:

  • Comparator U_1 fails with output saturated positive:

  • Resistor R_1 fails open:

  • Capacitor C_1 fails shorted:

  • Transistor Q_1 fails shorted (drain-to-source):

For each of these conditions, explain why the resulting effects will occur.

File Num: 03768

Answer


  • Cable fails open: Fan turns on and never turns off.

  • Comparator U_1 fails with output saturated positive: Fan turns on and never turns off.

  • Resistor R_1 fails open: Fan refuses to turn on.

  • Capacitor C_1 fails shorted: Fan refuses to turn on, transistor Q_1 likely fails due to overheating when it tries to energize fan.

  • Transistor Q_1 fails shorted (drain-to-source): Fan turns on and never turns off.


Notes

The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.





Question 9. (Click on arrow for answer)

Predict how the operation of this thermostat circuit (where the cooling fan motor is supposed to turn on when the temperature gets too high) will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Should be image 03769x01
  • Cable fails open:

  • Comparator U_1 fails with output saturated positive:

  • Resistor R_1 fails open:

  • Cable fails shorted:

  • Transistor Q_1 fails shorted (drain-to-source):

For each of these conditions, explain why the resulting effects will occur.

File Num: 03769

Answer


  • Cable fails open: Fan turns on and never turns off.

  • Comparator U_1 fails with output saturated positive: Fan refuses to turn on.

  • Resistor R_1 fails open: Fan refuses to turn on.

  • Cable fails shorted: Fan refuses to turn on.

  • Transistor Q_1 fails shorted (drain-to-source): Fan turns on and never turns off.


Notes

The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.





Question 10. (Click on arrow for answer)

Explain what a window comparator circuit is (sometimes called a window discriminator), and identify at least one practical application for one.

File Num: 03838

Answer

A “window comparator” circuit detects when a voltage falls between two different reference voltages. I’ll let you figure out some practical applications for such a circuit!


Notes

Ask your students where they found the answer for this question, and further explore some of the practical applications they offer.






Schmitt Trigger

Schmitt Trigger Circuit

Schmitt triggers are comparators with some built in hysteresis so that small changes around the input threshold are ignored.


Video: Schmitt Triggers

To find out more, check out this video:

Intro to Schmitt Triggers

Schmitt Trigger Practice Problems

Question 1. (Click on arrow for answer)

Determine the “trip” voltage of this comparator circuit: the value of input voltage at which the opamp’s output changes state from fully positive to fully negative or visa-versa:

Should be image 02294x01

Now, what do you suppose would happen if the output were fed back to the noninverting input through a resistor? You answer merely has to be qualitative, not quantitative:

Should be image 02294x02

For your information, this circuit configuration is often referred to as a Schmitt trigger.

File Num: 02294

Answer

With no feedback resistor, the “trip” voltage would be 9.21 volts. With the feedback resistor in place, the “trip” voltage would change depending on the state of the opamp’s output!


Follow-up question: describe what effect this changing “trip” voltage value will have on the operation of this comparator circuit.


Notes

Schmitt trigger circuits are very popular for their ability to “cleanly” change states given a noisy input signal. I have intentionally avoided numerical calculations in this question, so that students may concentrate on the concept of positive feedback and how it affects this circuit.





Question 2. (Click on arrow for answer)

A comparator is used as a high wind speed alarm in this circuit, triggering an audio tone to sound whenever the wind speed exceeds a pre-set alarm point:

Should be image 01168x01

The circuit works well to warn of high wind speed, but when the wind speed is just near the threshold level, every little gust causes the alarm to briefly sound, then turn off again. What would be better is for the alarm to sound at a set wind speed, then stay on until the wind speed falls below a substantially lower threshold value (example: alarm at 60 km/h, reset at 50 km/h).

An experienced electronics technician decides to add this functionality to the circuit by adding two resistors:

Should be image 01168x02

Explain why this circuit alteration works to solve the problem.

File Num: 01168

Answer

The added resistors provide positive feedback to the opamp circuit, causing it to exhibit hysteresis.


Challenge question: suppose you wished to increase the gap between the upper and lower alarm thresholds. What resistor value(s) would you have to alter to accomplish this adjustment?


Notes

A practical illustration for positive feedback in an opamp circuit. There is much to discuss here, even beyond the immediate context of positive feedback. Take for instance the oscillator circuit and on/off control transistor. For review, ask your students to explain how both these circuit sections function.





Question 3. (Click on arrow for answer)

Assume that the comparator in this circuit is capable of “swinging” its output fully from rail to rail. Calculate the upper and lower threshold voltages, given the resistor values shown:

Should be image 01169x01V_{UT} = \hskip 80pt V_{LT} =

File Num: 01169

Answer

V_{UT} = +8 volts
V_{LT} = -8 volts

Challenge question: how would you recommend we change the circuit to give threshold voltages of +6 volts and -6 volts, respectively?


Notes

Ask your students to explain what the terms “upper threshold” and “lower threshold” mean with regard to input voltage in a circuit such as this.





Question 4. (Click on arrow for answer)

Assume that the comparator in this circuit is only capable of “swinging” its output to within 1 volt of its power supply rail voltages. Calculate the upper and lower threshold voltages, given the resistor values shown:

Should be image 02662x01V_{UT} = \hskip 80pt V_{LT} =

File Num: 02662

Answer

V_{UT} = +2.093 volts
V_{LT} = -2.093 volts

Notes

As many opamps and comparators are incapable of rail-to-rail output swings, this question is quite realistic.





Question 5. (Click on arrow for answer)

Assume that the comparator in this circuit is only capable of “swinging” its output to within 1 volt of its power supply rail voltages. Calculate the upper and lower threshold voltages, given the resistor values shown:

Should be image 02663x01V_{UT} = \hskip 80pt V_{LT} =

File Num: 02663

Answer

V_{UT} = +4.087 volts
V_{LT} = -4.087 volts

Notes

As many opamps and comparators are incapable of rail-to-rail output swings, this question is quite realistic.





Question 6. (Click on arrow for answer)

A student intends to connect a TL082 opamp as a voltage follower, to “follow” the voltage generated by a potentiometer, but makes a mistake in the breadboard wiring:

Should be image 01148x01

Draw a schematic diagram of this faulty circuit, and determine what the voltmeter’s indication will be, explaining why it is such.

File Num: 01148

Answer

Circuit schematic, as wired:

Should be image 01148x02

The output voltage will saturate at approximately +11 volts, or -11 volts, with the potentiometer having little or no effect.


Notes

Ask your students to characterize the type of feedback exhibited in this circuit. How does this type of feedback affect the opamp’s behavior? Is it possible for the opamp to function as a voltage follower, connected like this?





Question 7. (Click on arrow for answer)

Positive or regenerative feedback is an essential characteristic of all oscillator circuits. Why, then, do comparator circuits utilizing positive feedback not oscillate? Instead of oscillating, the output of a comparator circuit with positive feedback simply saturates to one of its two rail voltage values. Explain this.

File Num: 01172

Answer

The positive feedback used in oscillator circuits is always phase-shifted 360^{o}, while the positive feedback used in comparator circuits has no phase shift at all, being direct-coupled.


Notes

This is a challenging question, and may not be suitable for all students. Basically, what I’m trying to get students to do here is think carefully about the nature of positive feedback as used in comparator circuits, versus as it’s used in oscillator circuits. Students who have simply memorized the concept of “positive feedback causing oscillation” will fail to understand what is being asked in this question, much less understand the given answer.





Op Amp Oscillator

An square wave oscillator can easily be made by using a positive feedback circuit like the one used for the Schmitt trigger along with a negative feedback circuit using an RC circuit like this:

Op Amp Relaxation Oscillator

The voltage at the non-inverting pin is controlled by the output and the positive feedback network. The voltage at the inverting pin is controlled by a charging or discharging RC circuit. The capacitor will charge and discharge towards the voltage set at the non-inverting pin and when it reaches the voltage, the output will switch and the capacitor will then discharge or charge in the opposite direction. This automatic switching back and forth creates an oscillator


Video: Op Amp Oscillator

This video describes how the op amp oscillator works and derives an equation for how to calculate the frequency of the oscillator based on the R and C values

Operational Amplifiers - Relaxation Oscillators

Op Amp Oscillator Practice Problems

Question 1. (Click on arrow for answer)

This is a very common opamp oscillator circuit, technically of the relaxation type:

Should be image 01171x01

Explain how this circuit works, and what waveforms will be measured at points A and B. Be sure to make reference to RC time constants in your explanation.

File Num: 01171

Answer

You will measure a sawtooth-like waveform at point A, and a square wave at point B.


Challenge question: explain how you might go about calculating the frequency of such a circuit, based on what you know about RC time constant circuits. Assume that the opamp can swing its output rail-to-rail, for simplicity.


Notes

This circuit is best understood by building and testing. If you use large capacitor values and/or a large-value resistor in the capacitor’s current path, the oscillation will be slow enough to analyze with a voltmeter rather than an oscilloscope.





Question 2. (Click on arrow for answer)

A variation on the common opamp relaxation oscillator design is this, which gives it variable duty cycle capability:

Should be image 02673x01

Explain how this circuit works, and which direction the potentiometer wiper must be moved to increase the duty cycle (more time spent with the opamp output saturated at +V and less time spent saturated at -V).

File Num: 02673

Answer

Move the wiper up to increase the duty cycle.


Notes

This circuit is best understood by building and testing. If you use large capacitor values and/or a large-value resistor in the capacitor’s current path, the oscillation will be slow enough to analyze with a voltmeter rather than an oscilloscope.

Incidentally, the Schottky diodes are not essential to this circuit’s operation, unless the expected frequency is very high. Really, the purpose of the Schottky diodes, with their low forward voltage drops (0.4 volts typical) and minimal charge storage, is to make the opamp’s job easier at every reversal of output polarity. Remember that this circuit is not exploiting negative feedback! Essentially, it is a positive feedback circuit, and every voltage drop and nonlinearity in the capacitor’s current path will have an effect on capacitor charging/discharging.





Question 3. (Click on arrow for answer)

Most operational amplifiers do not have the ability to swing their output voltages rail-to-rail. Most of those do not swing their output voltages symmetrically. That is, a typical non-rail-to-rail opamp may be able to approach one power supply rail voltage closer than the other; e.g. when powered by a +15/-15 volt split supply, the output saturates positive at +14 volts and saturates negative at -13.5 volts.

What effect do you suppose this non-symmetrical output range will have on a typical relaxation oscillator circuit such as the following, and how might you suggest we fix the problem?

Should be image 02675x01

File Num: 02675

Answer

The duty cycle will not be 50\%. One way to fix the problem is to do something like this:

Should be image 02675x02

Follow-up question: explain how and why this solution works. Now you just knew I was going to ask this question the moment you saw the diagram, didn’t you?


Notes

Note that I added an additional resistor to the circuit, in series with the opamp output terminal. In some cases this is not necessary because the opamp is self-limiting in output current, but it is a good design practice nonetheless. In the event anyone ever swaps out the original opamp for a different model lacking overcurrent protection, the new opamp will not become damaged.





Question 4. (Click on arrow for answer)

Dual, or split, power supplies are very useful in opamp circuits because they allow the output voltage to rise above as well as sink below ground potential, for true AC operation. In some applications, though, it may not be practical or affordable to have a dual power supply to power your opamp circuit. In this event, you need to be able to figure out how to adapt your dual-supply circuit to single-supply operation.

A good example of such a challenge is the familiar opamp relaxation oscillator, shown here:

Should be image 02676x01

First, determine what would happen if we were to simply eliminate the negative portion of the dual power supply and try to run the circuit on a single supply (+V and Ground only):

Should be image 02676x02

Then, modify the schematic so that the circuit will run as well as it did before with the dual supply.

File Num: 02676

Answer

Here is one solution:

Should be image 02676x03

Here is another solution:

Should be image 02676x04

Follow-up question: now you just know what I’m going to ask next, don’t you? How do these modified circuits function?


Notes

Dual power supplies are a luxury in many real-life circumstances, and so your students will need to be able to figure out how to make opamps work in single-supply applications! Work with your students to analyze the function of the suggested solution circuit, to see how it is at once similar and different from its simpler, dual-supply forbear.





Question 5. (Click on arrow for answer)

Predict how the operation of this relaxation oscillator circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Should be image 03798x01
  • Resistor R_1 fails open:

  • Solder bridge (short) across resistor R_1:

  • Capacitor C_1 fails shorted:

  • Solder bridge (short) across resistor R_2:

  • Resistor R_3 fails open:

For each of these conditions, explain why the resulting effects will occur.

File Num: 03798

Answer


  • Resistor R_1 fails open: Opamp output saturates either positive or negative.

  • Solder bridge (short) across resistor R_1: Output voltage settles to 0 volts.

  • Capacitor C_1 fails shorted: Opamp output saturates either positive or negative.

  • Solder bridge (short) across resistor R_2: Output voltage settles to 0 volts.

  • Resistor R_3 fails open: Output voltage settles to 0 volts.


Notes

The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.





Question 6. (Click on arrow for answer)

Identify at least two different component faults that would result in a change in duty cycle for this oscillator circuit:

Should be image 03799x01

File Num: 03799

Answer

A short in either of the two diodes would cause the duty cycle to change.


Follow-up question: what would happen if either of these two diodes failed open?


Notes

Ask your students to explain why the duty cycle would change as a result of either diode failing shorted. This is a good opportunity to further explore the operation of this oscillator circuit.





Contributors

Contributors to this chapter are listed in chronological order of their contributions, from most recent to first.

David Williams (2022): First edits including videos




Practice Problem Copyright

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

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

Practice Problems: Resonance in AC Circuits

Difficult Concepts

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

Conductance, Susceptance, and Admittance

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

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

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

Question 1. (Click on arrow for answer)

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

Should be image 00600x01

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

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

File Num: 00600

Answer

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


Notes

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





Question 2. (Click on arrow for answer)

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

Should be image 00601x01

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

File Num: 00601

Answer

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


Notes

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

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





Question 3. (Click on arrow for answer)

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

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

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

File Num: 01140

Answer

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


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


Notes

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





Question 4. (Click on arrow for answer)

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

Should be image 03290x01

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

File Num: 03290

Answer

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


Notes

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

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





Question 5. (Click on arrow for answer)

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

Should be image 00602x01

File Num: 00602

Answer

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


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


Notes

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





Question 6. (Click on arrow for answer)

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

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

File Num: 00603

Answer

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


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


Notes

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





Question 7. (Click on arrow for answer)

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

File Num: 00606

Answer

45 \Omega \angle 90^{o}


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


Notes

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





Question 8. (Click on arrow for answer)

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

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

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

\noalign{\hrule}

50 Hz & & & \cr

\noalign{\hrule}

60 Hz & & & \cr

\noalign{\hrule}

70 Hz & & & \cr

\noalign{\hrule}

80 Hz & & & \cr

\noalign{\hrule}

90 Hz & & & \cr

\noalign{\hrule}

100 Hz & & & \cr

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

Also, calculate the resonant frequency of this circuit.

File Num: 01873

Answer

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

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

\noalign{\hrule}

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

\noalign{\hrule}

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

\noalign{\hrule}

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

\noalign{\hrule}

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

\noalign{\hrule}

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

\noalign{\hrule}

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

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

Notes

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





Question 9. (Click on arrow for answer)

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

Should be image 00604x01

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


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

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

File Num: 00604

Answer

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


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


Notes

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





Question 10. (Click on arrow for answer)

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

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

File Num: 00607

Answer

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

Notes

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





Question 11. (Click on arrow for answer)

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

Should be image 00608x01

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

File Num: 00608

Answer

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


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


Notes

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

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





Question 12. (Click on arrow for answer)

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

Should be image 00609x01

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

File Num: 00609

Answer

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


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


Notes

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

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





Question 13. (Click on arrow for answer)

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

Should be image 00611x01

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

File Num: 00611

Answer

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


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


Notes

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

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





Question 14. (Click on arrow for answer)

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

Should be image 04043x01

File Num: 04043

Answer

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

Notes

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





Question 15. (Click on arrow for answer)

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

File Num: 01563

Answer

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


Notes

Ask your students to explain their answers mathematically.





Question 16. (Click on arrow for answer)

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

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

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

File Num: 01390

Answer

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

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


Notes

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





Question 17. (Click on arrow for answer)

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

Q = {X_L \over R_{series}}

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

X_L = 2 \pi f L

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

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

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

File Num: 01683

Answer

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

Notes

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





Question 18. (Click on arrow for answer)

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

Should be image 01391x01

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

File Num: 01391

Answer

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


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


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


Notes

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





Question 19. (Click on arrow for answer)

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

Q = {X_s \over R_{s}}

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

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

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

File Num: 02096

Answer

Q = {R_p \over X_p}

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


Notes

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





Question 20. (Click on arrow for answer)

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

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

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

File Num: 01870

Answer

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


Notes

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

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





Question 21. (Click on arrow for answer)

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

File Num: 01872

Answer

Should be image 01872x01

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


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


Notes

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

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





Question 22. (Click on arrow for answer)

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

Should be image 01682x01

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

File Num: 01682

Answer

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


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


Notes

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

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

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




Question 23. (Click on arrow for answer)

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

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

File Num: 00593

Answer

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


Notes

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





Question 24. (Click on arrow for answer)

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

File Num: 00594

Answer

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


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


Notes

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

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





Question 25. (Click on arrow for answer)

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

Should be image 00610x01

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

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

File Num: 00610

Answer

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


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


Notes

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





Question 26. (Click on arrow for answer)

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

File Num: 01869

Answer

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


Notes

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





Question 27. (Click on arrow for answer)

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

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

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


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

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

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

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

File Num: 00605

Answer

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


Notes

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

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

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

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

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

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


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


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

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

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

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





Question 28. (Click on arrow for answer)

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

Should be image 02571x01

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

File Num: 02571

Answer

Should be image 02571x02

Notes

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





Question 29. (Click on arrow for answer)

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

Should be image 02098x01

File Num: 02098

Answer

Should be image 02098x02

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


Notes

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

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

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





Question 30. (Click on arrow for answer)

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

Should be image 00620x01

File Num: 00620

Answer

Should be image 00620x02

Notes

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

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





Question 31. (Click on arrow for answer)

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

File Num: 01871

Answer

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


Notes

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





Question 32. (Click on arrow for answer)

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

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

Should be image 01393x01

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

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

File Num: 01393

Answer

Should be image 01393x02

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


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

Should be image 01393x03

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


Notes

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

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

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





Question 33. (Click on arrow for answer)

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

Should be image 01394x01

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

Should be image 01394x02

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

File Num: 01394

Answer

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


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


Notes

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

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





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

Practice Problems: RLC in AC Circuits

Difficult Concepts

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

Resistance vs. Reactance vs. Impedance

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

Phasors, used to represent AC amplitude and phase relations.

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

Conductance, Susceptance, and Admittance.

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

Question 1. (Click on arrow for answer)

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

Should be image 00851x01

File Num: 00851

Answer

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

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


Notes

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





Question 2. (Click on arrow for answer)

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

Should be image 00852x01

File Num: 00852

Answer

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

Notes

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





Question 3. (Click on arrow for answer)

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

Should be image 00854x01

File Num: 00854

Answer

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

Notes

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





Question 4. (Click on arrow for answer)

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

Should be image 01541x01

File Num: 01541

Answer

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

Notes

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


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


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

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





Question 5. (Click on arrow for answer)

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

Should be image 01554x01

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

Should be image 01554x02

File Num: 01554

Answer

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


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


Notes

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


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


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

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





Question 6. (Click on arrow for answer)

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

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

File Num: 01542

Answer

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

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


Notes

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





Question 7. (Click on arrow for answer)

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

File Num: 00644

Answer

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


Notes

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





Question 8. (Click on arrow for answer)

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

Should be image 00856x01

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

Should be image 00856x02

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

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

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

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

File Num: 00856

Answer

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


As for the reactance relation equation, here it is:

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

Notes

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

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





Question 9. (Click on arrow for answer)

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

Should be image 01540x01

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

File Num: 01540

Answer

R = 150 \Omega
X_C = 200 \Omega

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


Notes

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





Question 10. (Click on arrow for answer)

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

Should be image 00855x01

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

File Num: 00855

Answer

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

Notes

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





Question 11. (Click on arrow for answer)

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

Should be image 00858x01

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

File Num: 00858

Answer

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

Notes

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





Question 12. (Click on arrow for answer)

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

File Num: 00645

Answer

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

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


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


Notes

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





Question 13. (Click on arrow for answer)

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

Should be image 00859x01

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

File Num: 00859

Answer

Should be image 00859x02

Notes

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





Question 14. (Click on arrow for answer)

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

Should be image 02118x01

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

File Num: 02118

Answer

Should be image 02118x02

Notes

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





Question 15. (Click on arrow for answer)

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

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

Should be image 01072x01

File Num: 01072

Answer

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


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


Notes

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





Question 16. (Click on arrow for answer)

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

Should be image 02119x01

File Num: 02119

Answer

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

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


Notes

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

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





Question 17. (Click on arrow for answer)

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

Should be image 01864x01

File Num: 01864

Answer

Equivalent series resistance and reactances:

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

Notes

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





Question 18. (Click on arrow for answer)

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

Should be image 01865x01

File Num: 01865

Answer

Equivalent parallel resistance and reactances:

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

Notes

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





Question 19. (Click on arrow for answer)

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

Should be image 02115x01

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

File Num: 02115

Answer

V_{AB} = 10.491 volts

Notes

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

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





Question 20. (Click on arrow for answer)

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

Should be image 02116x01

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

File Num: 02116

Answer

I_{LR} = 3.290 mA

Notes

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

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

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





Question 21. (Click on arrow for answer)

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

Should be image 00641x01

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

Should be image 00641x02

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

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


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

File Num: 00641

Answer


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


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


Notes

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





Question 22. (Click on arrow for answer)

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

Should be image 00642x01

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

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

Should be image 00642x02

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


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

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

Should be image 00642x03

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


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

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

File Num: 00642

Answer

With no compensating capacitor:


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


With the 20 pF compensating capacitor in place:


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


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


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


Notes

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

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





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

Practice Problems: Complex Numbers and Phasors

Difficult Concepts

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

Resistance vs. Reactance vs. Impedance

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

Phasors, used to represent AC amplitude and phase relations.

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

Conductance, Susceptance, and Admittance.

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

Question 1. (Click on arrow for answer)

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

Should be image 03326x01

File Num: 03326

Answer

x = 10

Notes

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





Question 2. (Click on arrow for answer)

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

Should be image 03327x01

File Num: 03327

Answer

x = 15

Notes

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





Question 3. (Click on arrow for answer)

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

Should be image 02102x01

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

File Num: 02102

Answer

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


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


Notes

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





Question 4. (Click on arrow for answer)

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

Should be image 03114x01

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

File Num: 03114

Answer

Standard form of the Pythagorean Theorem:

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

Solving for A:

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

Solving for B:

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

Notes

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





Question 5. (Click on arrow for answer)

Should be image 02084x01

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

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

File Num: 02084

Answer

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

Notes

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





Question 6. (Click on arrow for answer)

Should be image 03113x01

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

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

File Num: 03113

Answer

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

Notes

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





Question 7. (Click on arrow for answer)

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

Should be image 02086x01

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

\cos \Theta = {R \over Z}

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

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

File Num: 02086

Answer

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

Notes

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





Question 8. (Click on arrow for answer)

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

Should be image 02076x01

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

File Num: 02076

Answer

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

Should be image 02076x02

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


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


Notes

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





Question 9. (Click on arrow for answer)

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

Should be image 02081x01

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

File Num: 02081

Answer

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

Notes

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





Question 10. (Click on arrow for answer)

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

Should be image 02077x01

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

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

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

File Num: 02077

Answer

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

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

Should be image 02077x02

Notes

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

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





Question 11. (Click on arrow for answer)

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

Should be image 02078x01

File Num: 02078

Answer

Impedances do not add in parallel.


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


Notes

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





Question 12. (Click on arrow for answer)

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

Should be image 03288x01

File Num: 03288

Answer

Should be image 03288x02

Notes

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

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





Question 13. (Click on arrow for answer)

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

Should be image 02107x01

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

File Num: 02107

Answer

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

Notes

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





Question 14. (Click on arrow for answer)

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

Should be image 02082x01

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

File Num: 02082

Answer

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

Notes

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





Question 15. (Click on arrow for answer)

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

Should be image 02083x01

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

File Num: 02083

Answer

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

Notes

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





Question 16. (Click on arrow for answer)

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

Should be image 02092x01

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

File Num: 02092

Answer

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

Notes

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





Question 17. (Click on arrow for answer)

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

Should be image 02085x01

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

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

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

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

File Num: 02085

Answer

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

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


Notes

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





Question 18. (Click on arrow for answer)

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

Should be image 02088x01

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

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

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

File Num: 02088

Answer

X = 9.063 k\Omega

Notes

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





Question 19. (Click on arrow for answer)

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

Should be image 02087x01

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

File Num: 02087

Answer

R = 2.165 k\OmegaX = 1.25 k\Omega

Notes

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





Question 20. (Click on arrow for answer)

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

File Num: 04034

Answer

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


Notes

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





Question 21. (Click on arrow for answer)

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

Should be image 04059x01

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

Should be image 04059x02

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

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

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

a + jb = c \angle \Theta

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

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

Explain why this equivalence is mathematically sound.

File Num: 04059

Answer

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


Notes

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

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




Question 22. (Click on arrow for answer)

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

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

File Num: 04060

Answer

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

Should be image 04060x01

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

\omega = 2 \pi f

Notes

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





Question 23. (Click on arrow for answer)

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

Should be image 03278x01

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


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

Should be image 03278x02

File Num: 03278

Answer

The resultant force on the box will be 500 newtons.


Notes

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





Question 24. (Click on arrow for answer)

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

Should be image 01559x01

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

Should be image 01559x02

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

File Num: 01559

Answer

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

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


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


Notes

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

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





Question 25. (Click on arrow for answer)

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

Should be image 02099x01

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

Should be image 02099x02

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

File Num: 02099

Answer

Should be image 02099x03

Notes

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





Question 26. (Click on arrow for answer)

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

Should be image 02100x01

File Num: 02100

Answer

Here are two ways of showing the same addition:

Should be image 02100x02

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


Notes

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





Question 27. (Click on arrow for answer)

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

File Num: 02101

Answer

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

Should be image 02101x01

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

Should be image 02101x02

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


Notes

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





Question 28. (Click on arrow for answer)

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

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

File Num: 00497

Answer

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


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


Notes

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

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





Question 29. (Click on arrow for answer)

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


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

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

File Num: 00495

Answer

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

Notes

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

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





Question 30. (Click on arrow for answer)

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

File Num: 02101

Answer

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

Should be image 02101x01

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

Should be image 02101x02

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


Notes

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





Question 31. (Click on arrow for answer)

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

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

File Num: 00497

Answer

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


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


Notes

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

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





Question 32. (Click on arrow for answer)

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

Should be image 00499x01

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


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

File Num: 00499

Answer

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


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


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


Notes

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

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





Question 33. (Click on arrow for answer)

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

Should be image 03286x01

File Num: 03286

Answer

In this diagram, phasor B is leading phasor A.


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


Notes

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





Question 34. (Click on arrow for answer)

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

Should be image 00496x01

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

Should be image 00496x02

File Num: 00496

Answer

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

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


Notes

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

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





Question 35. (Click on arrow for answer)

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

Should be image 02089x01

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

File Num: 02089

Answer

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

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


Notes

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





Question 36. (Click on arrow for answer)

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

Should be image 02091x01

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

File Num: 02091

Answer

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

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


Notes

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





Question 37. (Click on arrow for answer)

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

Should be image 02090x01

File Num: 02090

Answer

G = 24.75 \muS

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


Notes

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





Question 38. (Click on arrow for answer)

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

Should be image 00498x01

File Num: 00498

Answer

Should be image 00498x02

Notes

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





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

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

Practice Problems: Basic AC Theory

Difficult Concepts

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

Resistance vs. Reactance vs. Impedance

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

Phasors, used to represent AC amplitude and phase relations.

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

Conductance, Susceptance, and Admittance.

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

Common ground connections on oscilloscope inputs

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

Question 1. (Click on arrow for answer)

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

File Num: 00028

Answer

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

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


Notes

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

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





Question 2. (Click on arrow for answer)

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

Should be image 00093x01

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

Should be image 00093x02

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

File Num: 00093

Answer

Should be image 00093x03

Notes

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





Question 3. (Click on arrow for answer)

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

File Num: 03289

Answer

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


Notes

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

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

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





Question 4. (Click on arrow for answer)

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


  • Frequency
  • Period
  • Hertz
  • Amplitude

Should be image 00054x01

File Num: 00054

Answer

Should be image 00054x02

Notes

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





Question 5. (Click on arrow for answer)

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

File Num: 00053

Answer

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


Notes

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





Question 6. (Click on arrow for answer)

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

File Num: 00055

Answer

Period = 2.8571 milliseconds


Notes

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





Question 7. (Click on arrow for answer)

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

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

File Num: 01819

Answer

\lambda \approx 316 meters

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


Notes

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





Question 8. (Click on arrow for answer)

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

Should be image 00387x01

File Num: 00387

Answer

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


Notes

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





Question 9. (Click on arrow for answer)

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

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

Should be image 00530x01

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

Should be image 00530x02

File Num: 00530

Answer

EKG vertical = heart muscle contraction ; EKG horizontal = time

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


Notes

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





Question 10. (Click on arrow for answer)

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

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

File Num: 00536

Answer

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


Notes

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





Question 11. (Click on arrow for answer)

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

Should be image 00531x01

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

Should be image 00531x02

File Num: 00531

Answer

Should be image 00531x03

Notes

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





Question 12. (Click on arrow for answer)

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

Should be image 01672x01

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

File Num: 01672

Answer

The battery voltage is slightly greater than 6.5 volts.


Notes

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





Question 13. (Click on arrow for answer)

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

Should be image 00532x01

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

File Num: 00532

Answer

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


Notes

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





Question 14. (Click on arrow for answer)

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

Should be image 00534x01

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

File Num: 00534

Answer

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


Notes

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





Question 15. (Click on arrow for answer)

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

Should be image 00533x01

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

File Num: 00533

Answer

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


Notes

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





Question 16. (Click on arrow for answer)

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

Should be image 00535x01

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

File Num: 00535

Answer

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


Notes

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





Question 17. (Click on arrow for answer)

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

Should be image 01668x01

File Num: 01668

Answer

400 Hz


Notes

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





Question 18. (Click on arrow for answer)

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

Should be image 00541x01

File Num: 00541

Answer


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

Notes

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





Question 19. (Click on arrow for answer)

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

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

Should be image 01820x01

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

File Num: 01820

Answer

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


Notes

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





Question 20. (Click on arrow for answer)

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

Should be image 01821x01

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

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

Should be image 01821x02

File Num: 01821

Answer

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

Should be image 01821x03

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

Should be image 01821x04

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


Notes

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


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




Question 21. (Click on arrow for answer)

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

File Num: 00051

Answer

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


Notes

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





Question 22. (Click on arrow for answer)

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

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

File Num: 00401

Answer

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


Notes

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





Question 23. (Click on arrow for answer)

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

Should be image 00402x01

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

File Num: 00402

Answer

120 volts AC RMS, by definition.


Notes

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

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





Question 24. (Click on arrow for answer)

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

Should be image 01818x01

File Num: 01818

Answer

The RMS amplitude of this waveform is approximately 0.32 volts.


Notes

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





Question 25. (Click on arrow for answer)

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

Should be image 01824x01

File Num: 01824

Answer

The RMS amplitude of this waveform is 0.5 volt.


Notes

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

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





Question 26. (Click on arrow for answer)

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

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

File Num: 02790

Answer

Analog voltmeter now registers: 111.7 volts


True-RMS digital voltmeter now registers: 110 volts


Notes

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





Question 27. (Click on arrow for answer)

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

Should be image 00493x01

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

File Num: 00493

Answer

Should be image 00493x02

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


Notes

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

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





Question 28. (Click on arrow for answer)

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

Should be image 00492x01

File Num: 00492

Answer

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

Should be image 00492x02

Notes

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





Question 29. (Click on arrow for answer)

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

Should be image 01557x01

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


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


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

File Num: 01557

Answer

Should be image 01557x02

Notes

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





Question 30. (Click on arrow for answer)

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

Should be image 01558x01

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


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

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


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

File Num: 01558

Answer

Should be image 01558x02

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


Notes

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

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





Question 31. (Click on arrow for answer)

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

Should be image 01559x01

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

Should be image 01559x02

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

File Num: 01559

Answer

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

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


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


Notes

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

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





Question 32. (Click on arrow for answer)

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

Should be image 02099x01

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

Should be image 02099x02

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

File Num: 02099

Answer

Should be image 02099x03

Notes

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





Question 33. (Click on arrow for answer)

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

Should be image 02100x01

File Num: 02100

Answer

Here are two ways of showing the same addition:

Should be image 02100x02

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


Notes

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





Question 34. (Click on arrow for answer)

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

Should be image 02102x01

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

File Num: 02102

Answer

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


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


Notes

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





Question 35. (Click on arrow for answer)

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


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

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

File Num: 00495

Answer

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

Notes

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

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





Question 36. (Click on arrow for answer)

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

File Num: 02101

Answer

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

Should be image 02101x01

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

Should be image 02101x02

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


Notes

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





Question 37. (Click on arrow for answer)

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

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

File Num: 00497

Answer

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


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


Notes

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

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





Question 38. (Click on arrow for answer)

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

Should be image 00499x01

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


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

File Num: 00499

Answer

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


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


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


Notes

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

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





Question 39. (Click on arrow for answer)

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

Should be image 01480x01

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

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

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

Should be image 01480x02

File Num: 01480

Answer

Should be image 01480x03

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


Notes

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





Question 40. (Click on arrow for answer)

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

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

Should be image 01481x01

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

File Num: 01481

Answer

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

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


Notes

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





Question 41. (Click on arrow for answer)

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


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

File Num: 00578

Answer

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


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


Notes

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





Question 42. (Click on arrow for answer)

\int f(x) dx Calculus alert!

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

e = L{di \over dt}

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

Should be image 00576x01

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

File Num: 00576

Answer

Should be image 00576x02

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


Notes

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





Question 43. (Click on arrow for answer)

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

File Num: 00580

Answer

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


Notes

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

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





Question 44. (Click on arrow for answer)

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

Should be image 00095x01

File Num: 00095

Answer

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


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


Notes

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





Question 45. (Click on arrow for answer)

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

File Num: 00582

Answer

X_L = 2 \pi f L

The current through this inductor is 15.92 mA RMS.


Notes

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

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





Question 46. (Click on arrow for answer)

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

File Num: 00586

Answer

f = 2.137 kHz

Notes

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





Question 47. (Click on arrow for answer)

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

File Num: 03277

Answer

L = 214.9 mH

Notes

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





Question 48. (Click on arrow for answer)

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

Should be image 01552x01

File Num: 01552

Answer

I = 15.6 mA

Notes

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





Question 49. (Click on arrow for answer)

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

Should be image 00584x01

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

File Num: 00584

Answer

Z_{total} = 500 \Omega.

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


Notes

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

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

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





Question 50. (Click on arrow for answer)

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

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

File Num: 01567

Answer

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

Should be image 01567x01

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


Notes

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





Question 51. (Click on arrow for answer)

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

E = I R

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

File Num: 00590

Answer

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

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

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

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


Notes

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





Question 52. (Click on arrow for answer)

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

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

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


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

File Num: 00588

Answer

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

Notes

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

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





Question 53. (Click on arrow for answer)

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

Should be image 00631x01

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

File Num: 00631

Answer

Should be image 00631x02

Notes

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





Question 54. (Click on arrow for answer)

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

Should be image 00632x01

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

File Num: 00632

Answer

Should be image 00632x02

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


Notes

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

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


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





Question 55. (Click on arrow for answer)

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

Should be image 02076x01

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

File Num: 02076

Answer

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

Should be image 02076x02

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


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


Notes

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





Question 56. (Click on arrow for answer)

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

Should be image 02081x01

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

File Num: 02081

Answer

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

Notes

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





Question 57. (Click on arrow for answer)

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

Should be image 02083x01

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

File Num: 02083

Answer

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

Notes

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





Question 58. (Click on arrow for answer)

Should be image 02084x01

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

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

File Num: 02084

Answer

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

Notes

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





Question 59. (Click on arrow for answer)

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

Should be image 02086x01

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

\cos \Theta = {R \over Z}

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

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

File Num: 02086

Answer

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

Notes

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





Question 60. (Click on arrow for answer)

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

Should be image 02085x01

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

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

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

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

File Num: 02085

Answer

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

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


Notes

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





Question 61. (Click on arrow for answer)

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

Should be image 00850x01

File Num: 00850

Answer

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

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


Notes

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





Question 62. (Click on arrow for answer)

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

Should be image 01827x01

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

File Num: 01827

Answer

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

Notes

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

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





Question 63. (Click on arrow for answer)

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

Should be image 02080x01

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

Should be image 02080x02

File Num: 02080

Answer

Should be image 02080x03Should be image 02080x04

Notes

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





Question 64. (Click on arrow for answer)

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

Should be image 02103x01

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

File Num: 02103

Answer

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

Notes

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





Question 65. (Click on arrow for answer)

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

Should be image 00639x01

File Num: 00639

Answer

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

Notes

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





Question 66. (Click on arrow for answer)

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

Should be image 01830x01

File Num: 01830

Answer

V_L = 12.60 \hbox{ volts RMS}V_R = 8.137 \hbox{ volts RMS}I = 11.46 \hbox{ milliamps RMS}

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 67. (Click on arrow for answer)

Solve for all voltages and currents in this series LR circuit, and also calculate the phase angle of the total impedance:

Should be image 01831x01

File Num: 01831

Answer

V_L = 13.04 \hbox{ volts RMS}V_R = 20.15 \hbox{ volts RMS}I = 4.030 \hbox{ milliamps RMS}\Theta_Z = 32.91^o

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 68. (Click on arrow for answer)

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

Should be image 01841x01
  • L_1 = 250 \hbox{ mH}
  • L_2 = 60 \hbox{ mH}
  • R_1 = 6.8 \hbox{ k}\Omega
  • R_2 = 1.2 \hbox{ k}\Omega
  • V_{supply} = 13.4 \hbox{ V RMS}
  • f_{supply} = 6.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01841

Answer

  • I_{total} = 0.895 \hbox{ mA}
  • V_{L1} = 9.14 \hbox{ V}
  • V_{L2} = 2.19 \hbox{ V}
  • V_{R1} = 6.08 \hbox{ V}
  • V_{R2} = 1.07 \hbox{ V}
  • \Theta = 57.71^o

I suggest using a dual-trace oscilloscope to measure total voltage (across the supply terminals) and voltage drop across resistor R_2. Theoretically, measuring the voltage dropped by either resistor would be fine, but R_2 works better for practical reasons (oscilloscope input lead grounding). Phase shift then could be measured either in the time domain or by a Lissajous figure analysis.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 69. (Click on arrow for answer)

One way to vary the amount of power delivered to a resistive AC load is by varying another resistance connected in series:

Should be image 01829x01

A problem with this power control strategy is that power is wasted in the series resistance (I^2R_{series}). A different strategy for controlling power is shown here, using a series inductance rather than resistance:

Should be image 01829x02

Explain why the latter circuit is more power-efficient than the former, and draw a phasor diagram showing how changes in L_{series} affect Z_{total}.

File Num: 01829

Answer

Inductors are reactive rather than resistive components, and therefore do not dissipate power (ideally).

Should be image 01829x03

Follow-up question: the inductive circuit is not just more energy-efficient — it is safer as well. Identify a potential safety hazard that the resistive power-control circuit poses due to the energy dissipation of its variable resistor.


Notes

If appropriate, you may want to mention devices called saturable reactors, which are used to control power in AC circuits by the exact same principle: varying a series inductance.





Question 70. (Click on arrow for answer)

A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = {1 \over R}), and it is measured in the unit of siemens.

Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

Should be image 00853x01

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = {1 \over X}), and the reciprocal of impedance is called admittance (Y = {1 \over Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.

Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Should be image 00853x02

File Num: 00853

Answer

Y_{total} = \sqrt{G^2 + B^2}

Follow-up question \#1: draw a phasor diagram showing how Y, G, and B relate.


Follow-up question \#2: re-write this equation using quantities of resistance (R), reactance (X), and impedance (Z), instead of conductance (G), susceptance (B), and admittance (Y).


Notes

Ask your students if this equation looks familiar to them. It should!


The answer to the second follow-up question is a matter of algebraic substitution. Work through this process with your students, and then ask them to compare the resulting equation with other equations they’ve seen before. Does its form look familiar to them in any way?





Question 71. (Click on arrow for answer)

Students studying AC electrical theory become familiar with the impedance triangle very soon in their studies:

Should be image 02077x01

What these students might not ordinarily discover is that this triangle is also useful for calculating electrical quantities other than impedance. The purpose of this question is to get you to discover some of the triangle’s other uses.

Fundamentally, this right triangle represents phasor addition, where two electrical quantities at right angles to each other (resistive versus reactive) are added together. In series AC circuits, it makes sense to use the impedance triangle to represent how resistance (R) and reactance (X) combine to form a total impedance (Z), since resistance and reactance are special forms of impedance themselves, and we know that impedances add in series.

List all of the electrical quantities you can think of that add (in series or in parallel) and then show how similar triangles may be drawn to relate those quantities together in AC circuits.

File Num: 02077

Answer

Electrical quantities that add:
  • Series impedances
  • Series voltages
  • Parallel admittances
  • Parallel currents
  • Power dissipations

I will show you one graphical example of how a triangle may relate to electrical quantities other than series impedances:

Should be image 02077x02

Notes

It is very important for students to understand that the triangle only works as an analysis tool when applied to quantities that add. Many times I have seen students try to apply the ZRX impedance triangle to parallel circuits and fail because parallel impedances do not add. The purpose of this question is to force students to think about where the triangle is applicable to AC circuit analysis, and not just to use it blindly.

The power triangle is an interesting application of trigonometry applied to electric circuits. You may not want to discuss power with your students in great detail if they are just beginning to study voltage and current in AC circuits, because power is a sufficiently confusing subject on its own.





Question 72. (Click on arrow for answer)

Explain why the “impedance triangle” is not proper to use for relating total impedance, resistance, and reactance in parallel circuits as it is for series circuits:

Should be image 02078x01

File Num: 02078

Answer

Impedances do not add in parallel.


Follow-up question: what kind of a triangle could be properly applied to a parallel AC circuit, and why?


Notes

Trying to apply the ZRX triangle directly to parallel AC circuits is a common mistake many new students make. Key to knowing when and how to use triangles to graphically depict AC quantities is understanding why the triangle works as an analysis tool and what its sides represent.





Question 73. (Click on arrow for answer)

Calculate the total impedance for these two 100 mH inductors at 2.3 kHz, and draw a phasor diagram showing circuit admittances (Y_{total}, G, and B):

Should be image 02079x01

Now, re-calculate impedance and re-draw the phasor admittance diagram supposing the second inductor is replaced by a 1.5 k\Omega resistor:

Should be image 02079x02

File Num: 02079

Answer

Should be image 02079x03Should be image 02079x04

Challenge question: why are the susceptance vectors (B_{L1} and B_{L2}) pointed down instead of up as impedance vectors for inductances typically are?


Notes

Phasor diagrams are powerful analytical tools, if one knows how to draw and interpret them. With hand calculators being so powerful and readily able to handle complex numbers in either polar or rectangular form, there is temptation to avoid phasor diagrams and let the calculator handle all the angle manipulation. However, students will have a much better understanding of phasors and complex numbers in AC circuits if you hold them accountable to representing quantities in that form.





Question 74. (Click on arrow for answer)

Calculate the individual currents through the inductor and through the resistor, the total current, and the total circuit impedance:

Should be image 02104x01

Also, draw a phasor diagram showing how the individual component currents relate to the total current.

File Num: 02104

Answer

I_L = 530.5 \muA RMSI_R = 490.2 \muA RMSI_{total} = 722.3 \muA RMSZ_{total} = 3.461 k\Omega

Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 75. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01839x01

Calculate the current necessary to power this motor if the equivalent resistance and inductance is 20 \Omega and 238 mH, respectively.

File Num: 01839

Answer

I_{supply} = 12.29 \hbox{ A}

Notes

This is a practical example of a parallel LR circuit, as well as an example of how complex electrical devices may be “modeled” by collections of ideal components. To be honest, a loaded AC motor’s characteristics are quite a bit more complex than what the parallel LR model would suggest, but at least it’s a start!


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 76. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01840x01

Calculate the equivalent inductance (L_{eq}) if the measured source current is 27.5 amps and the motor’s equivalent resistance (R_{eq}) is 11.2 \Omega.

File Num: 01840

Answer

L_{eq} = 61.11 \hbox{ mH}

Notes

Here is a case where scalar calculations (R, G, X, B, Y) are much easier than complex number calculations (all Z) would be.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 77. (Click on arrow for answer)

Determine the total current and all component currents in this circuit, stating your answers the way a multimeter would register them:

Should be image 01842x01
  • L_1 = 1.2 \hbox{ H}
  • L_2 = 650 \hbox{ mH}
  • R_1 = 33 \hbox{ k}\Omega
  • R_2 = 27 \hbox{ k}\Omega
  • V_{supply} = 19.7 \hbox{ V RMS}
  • f_{supply} = 4.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01842

Answer

  • I_{total} = 2.12 \hbox{ mA}
  • I_{L1} = 581 \> \mu \hbox{A}
  • I_{L2} = 1.07 \hbox{ mA}
  • I_{R1} = 597 \> \mu \hbox{A}
  • I_{R2} = 730 \> \mu \hbox{A}
  • \Theta = 51.24^o

Measuring \Theta with an oscilloscope requires the addition of a shunt resistor into this circuit, because oscilloscopes are (normally) only able to measure voltage, and there is no phase shift between any voltages in this circuit because all components are in parallel. I leave it to you to suggest where to insert the shunt resistor, what resistance value to select for the task, and how to connect the oscilloscope to the modified circuit.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 78. (Click on arrow for answer)

Calculate the total impedances (complete with phase angles) for each of the following inductor-resistor circuits:

Should be image 02106x01

File Num: 02106

Answer

Should be image 02106x02

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 79. (Click on arrow for answer)

A doorbell ringer has a solenoid with an inductance of 63 mH connected in parallel with a lamp (for visual indication) having a resistance of 150 ohms:

Should be image 02105x01

Calculate the phase shift of the total current (in units of degrees) in relation to the total supply voltage, when the doorbell switch is actuated.

File Num: 02105

Answer

\Theta = 81 degrees

Suppose the lamp turned on whenever the pushbutton switch was actuated, but the doorbell refused to ring. Identify what you think to be the most likely fault which could account for this problem.


Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 80. (Click on arrow for answer)

An AC electric motor operating under loaded conditions draws a current of 11 amps (RMS) from the 120 volt (RMS) 60 Hz power lines. The measured phase shift between voltage and current for this motor is 34^{o}, with voltage leading current.

Determine the equivalent parallel combination of resistance (R) and inductance (L) that is electrically equivalent to this operating motor.

File Num: 01542

Answer

R_{parallel} = 13.16 \Omega
L_{parallel} = 51.75 mH

Challenge question: in the parallel LR circuit, the resistor will dissipate a lot of energy in the form of heat. Does this mean that the electric motor, which is electrically equivalent to the LR network, will dissipate the same amount of heat? Explain why or why not.


Notes

If students get stuck on the challenge question, remind them that an electric motor does mechanical work, which requires energy.





Question 81. (Click on arrow for answer)

Should be image 01674x01

File Num: 01674

\vfil \eject

Answer

You may use circuit simulation software to set up similar oscilloscope display interpretation scenarios, for practice or for verification of what you see in this exercise.


Notes

Use a sine-wave function generator for the AC voltage source, and be sure set the frequency to some reasonable value (well within the capability of both the oscilloscope and counter to measure).

If this is not the first time students have done this, be sure to “mess up” the oscilloscope controls prior to them making adjustments. Students must learn how to quickly configure an oscilloscope’s controls to display any arbitrary waveform, if they are to be proficient in using an oscilloscope as a diagnostic tool.





Question 82. (Click on arrow for answer)

Should be image 01693x01

File Num: 01693

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Answer

You may use circuit simulation software to set up similar oscilloscope display interpretation scenarios, for practice or for verification of what you see in this exercise.


Notes

Use a sine-wave function generator for the AC voltage source, and be sure set the frequency to some reasonable value (well within the capability of a multimeter to measure). It is very important that students learn to convert between peak and RMS measurements for sine waves, but you might want to mix things up a bit by having them do the same with triangle waves and square waves as well! It is vital that students realize the rule of V_{RMS} = {V_{peak} \over \sqrt{2}} only holds for sinusoidal signals.

If you do choose to challenge students with non-sinusoidal waveshapes, be very sure that they do their voltmeter measurements using true-RMS meters! This means no analog voltmeters, which are “miscalibrated” so their inherently average-responding movements register (sinusoidal) RMS accurately. Your students must use true-RMS digital voltmeters in order for their non-sinusoidal RMS measurements to correlate with their calculations.

Incidentally, this lab exercise also works well as a demonstration of the importance of true-RMS indicating meters, comparing the indications of analog, non-true-RMS digital, and true-RMS digital on the same non-sinusoidal waveform!





Question 83. (Click on arrow for answer)

Should be image 01660x01

File Num: 01660

\vfil \eject

Answer

You may use circuit simulation software to set up similar oscilloscope display interpretation scenarios, for practice or for verification of what you see in this exercise.


Notes

Use a sine-wave function generator for the AC voltage source, and be sure set the frequency to some reasonable value (well within the capability of both the oscilloscope and counter to measure).





Question 84. (Click on arrow for answer)

Should be image 01616x01

File Num: 01616

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Answer

Use circuit simulation software to verify your predicted and measured parameter values.


Notes

Use a sine-wave function generator for the AC voltage source. I recommend against using line-power AC because of strong harmonic frequencies which may be present (due to nonlinear loads operating on the same power circuit). Specify a standard inductor value.

If students are to use a multimeter to make their current and voltage measurements, be sure it is capable of accurate measurement at the circuit frequency! Inexpensive digital multimeters often experience difficulty measuring AC voltage and current toward the high end of the audio-frequency range.





Question 85. (Click on arrow for answer)

Should be image 01665x01

File Num: 01665

\vfil \eject

Answer

Use circuit simulation software to verify your predicted and measured parameter values.


Notes

Use a sine-wave function generator for the AC voltage source. I recommend against using line-power AC because of strong harmonic frequencies which may be present (due to nonlinear loads operating on the same power circuit). Specify standard resistor and inductor values.

If students are to use a multimeter to make their current and voltage measurements, be sure it is capable of accurate measurement at the circuit frequency! Inexpensive digital multimeters often experience difficulty measuring AC voltage and current toward the high end of the audio-frequency range.

An extension of this exercise is to incorporate troubleshooting questions. Whether using this exercise as a performance assessment or simply as a concept-building lab, you might want to follow up your students’ results by asking them to predict the consequences of certain circuit faults.





Question 86. (Click on arrow for answer)

Should be image 01823x01

File Num: 01823

\vfil \eject

Answer

There really isn’t much you can do to verify your experimental results. That’s okay, though, because the results are qualitative anyway.


Notes

If the oscilloscope does not have its own internal square-wave signal source, use a function generator set up to output square waves at 1 volt peak-to-peak at a frequency of 1 kHz.

If this is not the first time students have done this, be sure to “mess up” the oscilloscope controls prior to them making adjustments. Students must learn how to quickly configure an oscilloscope’s controls to display any arbitrary waveform, if they are to be proficient in using an oscilloscope as a diagnostic tool.





Question 87. (Click on arrow for answer)

Should be image 03933x01

File Num: 03933

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Answer

I do not provide a grading rubric here, but elsewhere.


Notes

The idea of a troubleshooting log is three-fold. First, it gets students in the habit of documenting their troubleshooting procedure and thought process. This is a valuable habit to get into, as it translates to more efficient (and easier-followed) troubleshooting on the job. Second, it provides a way to document student steps for the assessment process, making your job as an instructor easier. Third, it reinforces the notion that each and every measurement or action should be followed by reflection (conclusion), making the troubleshooting process more efficient.





Question 88. (Click on arrow for answer)

\centerline{NAME: \hskip 80pt \hskip 40pt Troubleshooting Grading Criteria }

You will receive the highest score for which all criteria are met.


100 \% (Must meet or exceed all criteria listed) \item{A.} Absolutely flawless procedure \item{B.} No unnecessary actions or measurements taken \bigskip90 \% (Must meet or exceed these criteria in addition to all criteria for 85\% and below) \item{A.} No reversals in procedure (i.e. changing mind without sufficient evidence) \item{B.} Every single action, measurement, and relevant observation properly documented \bigskip80 \% (Must meet or exceed these criteria in addition to all criteria for 75\% and below) \item{A.} No more than one unnecessary action or measurement \item{B.} No false conclusions or conceptual errors \item{C.} No missing conclusions (i.e. at least one documented conclusion for action / measurement / observation) \bigskip70 \% (Must meet or exceed these criteria in addition to all criteria for 65\%) \item{A.} No more than one false conclusion or conceptual error \item{B.} No more than one conclusion missing (i.e. an action, measurement, or relevant observation without a corresponding conclusion) \bigskip65 \% (Must meet or exceed these criteria in addition to all criteria for 60\%) \item{A.} No more than two false conclusions or conceptual errors \item{B.} No more than two unnecessary actions or measurements \item{C.} No more than one undocumented action, measurement, or relevant observation \item{D.} Proper use of all test equipment \bigskip60 \% (Must meet or exceed these criteria) \item{A.} Fault accurately identified \item{B.} Safe procedures used at all times \bigskip50 \% (Only applicable where students performed significant development/design work — i.e. not a proven circuit provided with all component values) \item{A.} Working prototype circuit built and demonstrated \bigskip0 \% (If any of the following conditions are true) \item{A.} Unsafe procedure(s) used at any point \bigskip

File Num: 03932

Answer

Be sure to document all steps taken and conclusions made in your troubleshooting!


Notes

The purpose of this assessment rubric is to act as a sort of “contract” between you (the instructor) and your student. This way, the expectations are all clearly known in advance, which goes a long way toward disarming problems later when it is time to grade.





Question 89. (Click on arrow for answer)

Determine the frequency of a waveform having a period of 1.4 milliseconds (1.4 ms).

File Num: 03276

Answer

f = 714.29 Hz

Notes

It is important for students to realize the reciprocal relationship between frequency and period. One is cycles per second while the other is seconds per cycle.





Question 90. (Click on arrow for answer)

Assuming the vertical sensitivity control is set to 0.5 volts per division, and the timebase control is set to 2.5 ms per division, calculate the amplitude of this sine wave (in volts peak, volts peak-to-peak, and volts RMS) as well as its frequency.

Should be image 00540x01

File Num: 00540

Answer


\item{} E_{peak} = 2.25 V \item{} E_{peak-to-peak} = 4.50 V \item{} E_{RMS} = 1.59 V \item{} f = 40 Hz

Notes

This question is not only good for introducing basic oscilloscope principles, but it is also excellent for review of AC waveform measurements.





Question 91. (Click on arrow for answer)

Something is wrong with this circuit. Based on the oscilloscope’s display, determine whether the battery or the function generator is faulty:

Should be image 03448x01

File Num: 03448

Answer

The battery is faulty.


Follow-up question: discuss how accidently setting the coupling control on the oscilloscope to “AC” instead of “DC” would also cause this waveform to show on the screen (even with a good battery).


Notes

This question challenges students to apply their knowledge of AC+DC mixed signals to oscilloscope display patterns, in order to determine whether it is the battery or the function generator which has failed.





Question 92. (Click on arrow for answer)

Something is wrong with this circuit. Based on the oscilloscope’s display, determine whether the battery or the function generator is faulty:

Should be image 03449x01

File Num: 03449

Answer

The function generator is faulty.


Follow-up question: explain how this problem could be created simply by connecting the function generator to the circuit with the ground on the left-hand clip instead of the right-hand clip where it should be.

Should be image 03449x02

Notes

This question challenges students to apply their knowledge of AC+DC mixed signals to oscilloscope display patterns, in order to determine whether it is the battery or the function generator which has failed.





Question 93. (Click on arrow for answer)

Shunt resistors are low-value, precision resistors used as current-measuring elements in high-current circuits. The idea is to measure the voltage dropped across this precision resistance and use Ohm’s Law (I = {V \over R}) to infer the amount of current in the circuit:Should be image 03504x01

Since the schematic shows a shunt resistor being used to measure current in an AC circuit, it would be equally appropriate to use an oscilloscope instead of a voltmeter to measure the voltage drop produced by the shunt. However, we must be careful in connecting the oscilloscope to the shunt because of the inherent ground reference of the oscilloscope’s metal case and probe assembly.

Explain why connecting an oscilloscope to the shunt as shown in this second diagram would be a bad idea:

Should be image 03504x02

File Num: 03504

Answer

This would be a bad idea because the oscilloscope’s ground clip would attempt to bypass current around the shunt resistor, through the oscilloscope’s safety ground wire, and back to the grounded terminal of the AC source. Not only would this induce measurement errors, but it could damage the oscilloscope as well.


Follow-up question: identify a better way of connecting this oscilloscope to the shunt resistor.


Notes

The ground-referenced clip on an oscilloscope probe is a constant source of potential trouble for those who do not fully understand it! Even in scenarios where there is little or no potential for equipment damage, placing an earth ground reference on a circuit via the probe clip can make for very strange circuit behavior and erroneous measurements. Problems like this frequently occur when new students attempt to connect their oscilloscopes to circuits powered by signal generators whose outputs are also earth-ground referenced.

In response to the follow-up question, the most obvious answer is to reverse the probe connections: ground clip on the left-hand terminal and probe tip on the right-hand terminal. However, even this might not be the best idea, since it creates a “ground loop” between the oscilloscope and the ground connection at the AC source:

Should be image 03504x03

Ground loops are to be avoided in measurement circuits because they may be the source of some very strange effects, including the coupling of noise voltage from entirely unrelated circuits to the one being measured. To avoid this problem, the best solution for measuring the voltage dropped across the shunt resistor is to use two scope probes and set the scope up for differential voltage measurement:

Should be image 03504x04




Question 94. (Click on arrow for answer)

An electromechanical alternator (AC generator) and a DC-DC inverter both output the same RMS voltage, and deliver the same amount of electrical power to two identical loads:

Should be image 00404x01

However, when measured by an analog voltmeter, the inverter’s output voltage is slightly greater than the alternator’s output voltage. Explain this discrepancy in measurements.

File Num: 00404

Answer

Electromechanical alternators naturally output sinusoidal waveforms. Many DC-AC inverters do not.


Notes

Remember, most analog meter movement designs respond to the average value of a waveform, not its RMS value. If the proportionality between a waveform’s average and RMS values ever change, the relative indications of a true-RMS instrument and an average-based (calibrated to read RMS) instrument will change as well.





Question 95. (Click on arrow for answer)

Is the deflection of an analog AC meter movement proportional to the peak, average, or RMS value of the waveform measured? Explain your answer.

File Num: 00403

Answer

Analog meter deflection is proportional to the average value of the AC waveform measured, for most AC meter movement types. There are some meter movement designs, however, that give indications proportional to the RMS value of the waveform: hot-wire and electrodynamometer movements are of this nature.


Follow-up question: does this mean an average-responding meter movement cannot be calibrated to indicate in RMS units?


Challenge question: why do hot-wire and electrodynamometer meter movements provide true RMS indications, while most other movement designs indicate based on the signal’s average value?


Notes

Students often confuse the terms “average” and “RMS”, thinking they are interchangeable. Discuss the difference between these two terms, both mathematically and practically. While the concepts may seem similar at first, the details are actually quite different.

The question of whether an average-responding instrument can be calibrated to register in RMS units is very practical, since the vast majority of multimeters are calibrated this way. Because the proportionality between the average and RMS values of an AC waveform are dependent on the shape of the waveform, a certain wave-shape must be assumed in order to accurately calibrate an average-responding meter movement for RMS measurement. The assumed wave-shape, of course, is sinusoidal.





Question 96. (Click on arrow for answer)

In calculating the size of wire necessary to carry alternating current to a high-power load, which type of measurement is the best to use for current: peak, average, or RMS? Explain why.

File Num: 00162

Answer

RMS current is the most appropriate type of measurement for calculating wire size.


Notes

A clue to answering this question is this: what actually happens when the ampacity rating of a conductor is exceeded? Why, exactly, is overcurrent a bad thing for conductors in general?

It is important for students to recognize the value of RMS measurements: why do we use them, and in what applications are they the most appropriate type of measurement to use in certain calculations? Ask your students what other applications might best use RMS voltage and current measurements as opposed to peak or average.





Question 97. (Click on arrow for answer)

In calculating the thickness of insulators for high-voltage AC power lines, which type of measurement is the best to use for voltage: peak, average, or RMS? Explain why.

File Num: 00163

Answer

Peak voltage is the most appropriate type of measurement for calculating insulator thickness. The reason why has to do with the time required for an insulator to “flash over.”


Notes

A closely related subject is insulator breakdown, or dielectric strength. What actually happens when the dielectric strength rating of an insulator is exceeded?





Question 98. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03578x01

File Num: 03578

Answer

\Theta \approx 25.9^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 99. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03575x01

File Num: 03575

Answer

\Theta \approx 64.2^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 100. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03577x01

File Num: 03577

Answer

\Theta \approx 34.5^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 101. (Click on arrow for answer)

Calculate the amount of phase shift indicated by this Lissajous figure:

Should be image 03576x01

File Num: 03576

Answer

\Theta \approx 44.4^{o}

Notes

This question is nothing more than an exercise in Lissajous figure interpretation.





Question 102. (Click on arrow for answer)

Determine which way the movable iron piece needs to go in order to brighten the light bulb in this circuit:

Should be image 03446x01

File Num: 03446

Answer

The movable piece needs to move to the left, creating a larger air gap between the poles of the U-shaped inductor core.


Notes

This question is an exercise in applying practical electromagnetic theory (namely, reluctance of an iron/air flux path) to inductance and inductive reactance. Ask your students to explain their reasoning step-by-step as they give their answers to this question.





Question 103. (Click on arrow for answer)

A solenoid valve is a mechanical shutoff device actuated by electricity. An electromagnet coil produces an attractive force on an iron “armature” which then either opens or closes a valve mechanism to control the flow of some fluid. Shown here are two different types of illustrations, both showing a solenoid valve:

Should be image 03445x01

Some solenoid valves are constructed in such a way that the coil assembly may be removed from the valve body, separating these two pieces so that maintenance work may be done on one without interfering with the other. Of course, this means the valve mechanism will no longer be actuated by the magnetic field, but at least one piece may be worked upon without having to remove the other piece from whatever it may be connected to:

Should be image 03445x02

This is commonly done when replacement of the valve mechanism is needed. First, the coil is lifted off the valve mechanism, then the maintenance technician is free to remove the valve body from the pipes and replace it with a new valve body. Lastly, the coil is re-installed on the new valve body and the solenoid is once more ready for service, all without having to electrically disconnect the coil from its power source.

However, if this is done while the coil is energized, it will overheat and burn up in just a few minutes. To prevent this from happening, the maintenance technicians have learned to insert a steel screwdriver through the center hole of the coil while it is removed from the valve body, like this:

Should be image 03445x03

With the steel screwdriver shank taking the place of the iron armature inside the valve body, the coil will not overheat and burn up even if continually powered. Explain the nature of the problem (why the coil tends to burn up when separated from the valve body) and also why a screwdriver put in place of the iron armature works to prevent this from happening.

File Num: 03445

Answer

With the iron armature no longer in the center of the solenoid coil, the coil’s inductance — and therefore its inductive reactance to AC — dramatically diminishes unless the armature is replaced by something else ferromagnetic.


Notes

When I first saw this practice in action, I almost fell over laughing. It is both practical and ingenious, as well as being an excellent example of variable inductance (and inductive reactance) arising from varying reluctance.





Question 104. (Click on arrow for answer)

Doorbell circuits connect a small lamp in parallel with the doorbell pushbutton so that there is light at the button when it is not being pressed. The lamp’s filament resistance is such that there is not enough current going through it to energize the solenoid coil when lit, which means the doorbell will ring only when the pushbutton switch shorts past the lamp:

Should be image 03447x01

Suppose that such a doorbell circuit suddenly stops working one day, and the home owner assumes the power source has quit since the bell will not ring when the button is pressed and the lamp never lights. Although a dead power source is certainly possible, it is not the only possibility. Identify another possible failure in this circuit which would result in no doorbell action (no sound) and no light at the lamp.

File Num: 03447

Answer


  • Solenoid coil failed open
  • Wire broken anywhere in circuit


Notes

After discussing alternative possibilities with your students, shift the discussion to one on how likely any of these failures are. For instance, how likely is it that the solenoid coil has developed an “open” fault compared to the likelihood of a regular wire connection going bad in the circuit? How do either of these possibilities compare with the likelihood of the source failing as a result of a tripped circuit breaker or other power outage?





Question 105. (Click on arrow for answer)

Suppose someone were to ask you to differentiate electrical reactance (X) from electrical resistance (R). How would you distinguish these two similar concepts from one another, using your own words?

File Num: 03301

Answer

It is really important for you to frame this concept in your own words, so be sure to check with your instructor on the accuracy of your answer to this question! To give you a place to start, I offer this distinction: resistance is electrical friction, whereas reactance is electrical energy storage. Fundamentally, the difference between X and R is a matter of energy exchange, and it is understood most accurately in those terms.


Notes

This is an excellent point of crossover with your students’ studies in elementary physics, if they are studying physics now or have studied physics in the past. The energy-storing actions of inductors and capacitors are quite analogous to the energy-storing actions of masses and springs (respectively, if you associate velocity with current and force with voltage). In the same vein, resistance is analogous to kinetic friction between a moving object and a stationary surface. The parallels are so accurate, in fact, that the electrical properties of R, L, and C have been exploited to model mechanical systems of friction, mass, and resilience in circuits known as analog computers.





Question 106. (Click on arrow for answer)

The Pythagorean Theorem is used to calculate the length of the hypotenuse of a right triangle given the lengths of the other two sides:

Should be image 03114x01

Manipulate the standard form of the Pythagorean Theorem to produce a version that solves for the length of A given B and C, and also write a version of the equation that solves for the length of B given A and C.

File Num: 03114

Answer

Standard form of the Pythagorean Theorem:

C = \sqrt{A^2 + B^2}

Solving for A:

A = \sqrt{C^2 - B^2}

Solving for B:

B = \sqrt{C^2 - A^2}

Notes

The Pythagorean Theorem is easy enough for students to find on their own that you should not need to show them. A memorable illustration of this theorem are the side lengths of a so-called 3-4-5 triangle. Don’t be surprised if this is the example many students choose to give.





Question 107. (Click on arrow for answer)

Suppose two people work together to slide a large box across the floor, one pushing with a force of 400 newtons and the other pulling with a force of 300 newtons:

Should be image 03278x01

The resultant force from these two persons’ efforts on the box will, quite obviously, be the sum of their forces: 700 newtons (to the right).


What if the person pulling decides to change position and push sideways on the box in relation to the first person, so the 400 newton force and the 300 newton force will be perpendicular to each other (the 300 newton force facing into the page, away from you)? What will the resultant force on the box be then?

Should be image 03278x02

File Num: 03278

Answer

The resultant force on the box will be 500 newtons.


Notes

This is a non-electrical application of vector summation, to prepare students for the concept of using vectors to add voltages that are out-of-phase. Note how I chose to use multiples of 3, 4, and 5 for the vector magnitudes.





Question 108. (Click on arrow for answer)

A rectangular building foundation with an area of 18,500 square feet measures 100 feet along one side. You need to lay in a diagonal run of conduit from one corner of the foundation to the other. Calculate how much conduit you will need to make the run:

Should be image 03275x01

Also, write an equation for calculating this conduit run length (L) given the rectangular area (A) and the length of one side (x).

File Num: 03275

Answer

Conduit run = 210 feet, 3.6 inches from corner to corner.


Note: the following equation is not the only form possible for calculating the diagonal length. Do not be worried if your equation does not look exactly like this!

L = {{\sqrt{x^4 + A^2}} \over x}

Notes

Determining the necessary length of conduit for this question involves both the Pythagorean theorem and simple geometry.

Most students will probably arrive at this form for their diagonal length equation:

L = \sqrt{x^2 + \left({A \over x}\right)^2}

While this is perfectly correct, it is an interesting exercise to have students convert the equation from this (simple) form to that given in the answer. It is also a very practical question, as equations given in reference books do not always follow the most direct form, but rather are often written in such a way as to look more esthetically pleasing. The simple and direct form of the equation shown here (in the Notes section) looks “ugly” due to the fraction inside the radicand.





Question 109. (Click on arrow for answer)

Should be image 03113x01

Identify which trigonometric functions (sine, cosine, or tangent) are represented by each of the following ratios, with reference to the angle labeled with the Greek letter “Phi” (\phi):

{R \over X} = {X \over Z} = {R \over Z} =

File Num: 03113

Answer

Should be image 03113x01{R \over X} = \tan \phi = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \cos \phi = {\hbox{Adjacent} \over \hbox{Hypotenuse}}{R \over Z} = \sin \phi = {\hbox{Opposite} \over \hbox{Hypotenuse}}

Notes

Ask your students to explain what the words “hypotenuse”, “opposite”, and “adjacent” refer to in a right triangle.





Question 110. (Click on arrow for answer)

In this phasor diagram, determine which phasor is leading and which is lagging the other:

Should be image 03286x01

File Num: 03286

Answer

In this diagram, phasor B is leading phasor A.


Follow-up question: using a protractor, estimate the amount of phase shift between these two phasors.


Notes

It may be helpful to your students to remind them of the standard orientation for phase angles in phasor diagrams (0 degrees to the right, 90 degrees up, etc.).





Question 111. (Click on arrow for answer)

Is it appropriate to assign a phasor angle to a single AC voltage, all by itself in a circuit?

Should be image 00496x01

What if there is more than one AC voltage source in a circuit?

Should be image 00496x02

File Num: 00496

Answer

Phasor angles are relative, not absolute. They have meaning only where there is another phasor to compare against.

Angles may be associated with multiple AC voltage sources in the same circuit, but only if those voltages are all at the same frequency.


Notes

Discuss with your students the notion of “phase angle” in relation to AC quantities. What does it mean, exactly, if a voltage is “3 volts at an angle of 90 degrees”? You will find that such a description only makes sense where there is another voltage (i.e., “4 volts at 0 degrees”) to compare to. Without a frame of reference, phasor angles are meaningless.

Also discuss with your students the nature of phase shifts between different AC voltage sources, if the sources are all at different frequencies. Would the phase angles be fixed, or vary over time? Why? In light of this, why do we not assign phase angles when different frequencies are involved?





Question 112. (Click on arrow for answer)

Determine the total voltage in each of these examples, drawing a phasor diagram to show how the total (resultant) voltage geometrically relates to the source voltages in each scenario:

Should be image 00498x01

File Num: 00498

Answer

Should be image 00498x02

Notes

At first it may confuse students to use polarity marks (+ and -) for AC voltages. After all, doesn’t the polarity of AC alternate back and forth, so as to be continuously changing? However, when analyzing AC circuits, polarity marks are essential for giving a frame of reference to phasor voltages, which like all voltages are measured between two points, and thus may be measured two different ways.





Question 113. (Click on arrow for answer)

Before two or more operating alternators (AC generators) may be electrically coupled, they must be brought into synchronization with each other. If two alternators are out of “sync” (or out of phase) with each other, the result will be a large fault current when the disconnect switch is closed.

A simple and effective means of checking for “sync” prior to closing the disconnect switch for an alternator is to have light bulbs connected in parallel with the disconnect switch contacts, like this:

Should be image 00491x01

What should the alternator operator look for before closing the alternator switch? Do bright lights indicate a condition of being “in-phase” with the bus, or do dim lights indicate this? What does the operator have to do in order to bring an alternator into “phase” with the bus voltage?

Also, describe what the light bulbs would do if the two alternators were spinning at slightly different speeds.

File Num: 00491

Answer

Dim lights indicate a condition of being “in-phase” with the bus. If the two alternators are spinning at slightly different speeds, there will be a heterodyne effect to the light bulbs’ brightness: alternately growing brighter, then dimmer, then brighter again.


Notes

Proper synchronization of alternators with bus voltage is a task that used to be performed exclusively by human operators, but may now be accomplished by automatic controls. It is still important, though, for students of electricity to understand the principles involved in alternator synchronization, and the simple light bulb technique of sync-indication is an excellent means of clarifying the concept.

Discuss with your students the means of bringing an alternator into phase with an AC bus. If the light bulbs are glowing brightly, what should the operator do to make them dim?

It might also be a good idea to discuss with your students what happens once two synchronized alternators become electrically coupled: the two machines become “locked” together as though they were mechanically coupled, thus maintaining synchronization from that point onward.





Question 114. (Click on arrow for answer)

Suppose a power plant operator was about to bring this alternator on-line (connect it to the AC bus), and happened to notice that neither one of the synchronizing lights was lit at all. Thinking this to be unusual, the operator calls you to determine if something is wrong with the system. Describe what you would do to troubleshoot this system.

Should be image 00500x01

File Num: 00500

Answer

Before you proceed with troubleshooting steps, first try to determine if there is anything wrong with this system at all. Could it be that the operator is just overly cautious, or is their caution justified?


Notes

There may be some students who suggest there is nothing wrong at all with the system. Indeed, since dim (or dark) lights normally indicate synchronization, would not the presence of two dim lights indicate that perfect synchronization had already been achieved? Discuss the likelihood of this scenario with your students, that two independent alternators could be maintaining perfect synchronization without being coupled together.

In regard to troubleshooting, this scenario has great potential for group discussion. Despite there being a simple, single, probable condition that could cause this problem, there are several possible component failures that could have created the condition. Different students will undoubtedly have different methods of approaching the problem. Let each one share their views, and discuss together what would be the best approach.





Question 115. (Click on arrow for answer)

When AC power is initially applied to an electric motor (before the motor shaft has an opportunity to start moving), the motor “appears” to the AC power source to be a large inductor:

Should be image 01826x01

If the voltage of the 60 Hz AC power source is 480 volts RMS, and the motor initially draws 75 amps RMS when the double-pole single-throw switch closes, how much inductance (L) must the motor windings have? Ignore any wire resistance, and assume the motor’s only opposition to current in a locked-rotor condition is inductive reactance (X_L).

File Num: 01826

Answer

X_L = 16.98 \hbox{ mH}

Notes

In reality, motor winding resistance plays a substantial part in this sort of calculation, but I simplified things a bit just to give students a practical context for their introductory knowledge of inductive reactance.





Question 116. (Click on arrow for answer)

In analyzing circuits with inductors, we often take the luxury of assuming the constituent inductors to be perfect; i.e., purely inductive, with no “stray” properties such as winding resistance or inter-winding capacitance.

Real life is not so generous. With real inductors, we have to consider these factors. One measure often used to express the “purity” of an inductor is its so-called Q rating, or quality factor.

Write the formula for calculating the quality factor (Q) of a coil, and describe some of the operational parameters that may affect this number.

File Num: 01389

Answer

Q_{coil} = {X_L \over R}

Notes

Your students should be able to immediately understand that Q is not a static property of an inductor. Let them explain what makes Q vary, based on their knowledge of inductive reactance.





Question 117. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 60 Hz:

Should be image 01832x01

Show your work using two different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01832

Answer

First strategy:L_{total} = 1.1 \hbox{ H}X_{total} = 414.7 \> \OmegaZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega
Second strategy:X_{L1} = 282.7 \> \Omega Z_{L1} = 282.7 \> \Omega \> \angle \> 90^oX_{L2} = 131.9 \> \Omega Z_{L2} = 131.9 \> \Omega \> \angle \> 90^oZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ impedance phasors geometrically add to equal the total impedance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 118. (Click on arrow for answer)

Calculate the total impedance of this LR circuit, once using nothing but scalar numbers, and again using complex numbers:

Should be image 01837x01

File Num: 01837

Answer

Scalar calculationsR_1 = 1.5 \hbox{ k}\Omega G_{R1} = 666.7 \> \mu \hbox{S}X_{L1} = 2.513 \hbox{ k}\Omega B_{L1} = 397.9 \> \mu \hbox{S}Y_{total} = \sqrt{G^2 + B^2} = 776.4 \> \mu \hbox{S}Z_{total} = {1 \over Y_{total}} = 1.288 \hbox{ k}\Omega
Complex number calculationsR_1 = 1.5 \hbox{ k}\Omega Z_{R1} = 1.5 \hbox{ k}\Omega \> \angle \> 0^oX_{L1} = 2.513 \hbox{ k}\Omega Z_{L1} = 2.513 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = { 1 \over {{1 \over Z_{R1}} + {1 \over Z_{L1}}}} = 1.288 \hbox{ k}\Omega \> \angle \> 30.83^o

Notes

Some electronics textbooks (and courses) tend to emphasize scalar impedance calculations, while others emphasize complex number calculations. While complex number calculations provide more informative results (a phase shift given in every variable!) and exhibit conceptual continuity with DC circuit analysis (same rules, similar formulae), the scalar approach lends itself better to conditions where students do not have access to calculators capable of performing complex number arithmetic. Yes, of course, you can do complex number arithmetic without a powerful calculator, but it’s a lot more tedious and prone to errors than calculating with admittances, susceptances, and conductances (primarily because the phase shift angle is omitted for each of the variables).





Question 119. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 120 Hz:

Should be image 01833x01

Show your work using three different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual admittances first (Y_{L1} and Y_{L2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
  • Using complex numbers: calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01833

Answer

First strategy:L_{total} = 391.3 \hbox{ mH}X_{total} = 295.0 \> \OmegaZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega
Second strategy:Z_{L1} = X_{L1} = 377.0 \> \Omega
Y_{L1} = {1 \over Z_{L1}} = 2.653 \> \hbox{mS}
Z_{L1} = X_{L2} = 1.357 \hbox{ k}\Omega
Y_{L2} = {1 \over Z_{L2}} = 736.8 \> \mu \hbox{S}
Y_{total} = 3.389 \> \hbox{mS}
Z_{total} = {1 \over Y_{total}} = 295 \> \Omega
Third strategy: (using complex numbers)X_{L1} = 377.0 \> \Omega Z_{L1} = 377.0 \> \Omega \> \angle \> 90^oX_{L2} = 1.357 \hbox{ k}\Omega Z_{L2} = 1.357 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ admittance phasors geometrically add to equal the total admittance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 120. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 75^{o}:

Should be image 03282x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03282

Answer

f = 11.342 kHzf = {R \over {2 \pi L \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 121. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of 44^{o}:

Should be image 03283x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03283

Answer

R = 6.826 k\OmegaR = 2 \pi f L \tan \theta

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 122. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -40^{o}:

Should be image 03280x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03280

Answer

f = 2.804 kHzf = - {{R \tan \theta} \over {2 \pi L}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 123. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of -60^{o}:

Should be image 03281x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03281

Answer

R = 2.902 k\OmegaR = - {{2 \pi f L} \over {\tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





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To view a copy of the license, visit https://creativecommons.org/licenses/by/1.0/, or https://creativecommons.org/licenses/by/4.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.

Practice Problems: Inductors in AC Circuits

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about inductors in AC circuits.

Resistance vs. Reactance vs. Impedance

These three terms represent different forms of opposition to electric current. Despite the fact that they are measured in the same unit (ohms: Omega), they are not the same. Resistance is best thought of as electrical friction, whereas reactance is best thought of as electrical inertia. Whereas resistance creates a voltage drop by dissipating energy, reactance creates a voltage drop by storing and releasing energy. Impedance is a term encompassing both resistance and reactance, usually a combination of both.

Phasors, used to represent AC amplitude and phase relations.

A powerful tool used for understanding the operation of AC circuits is the phasor diagram, consisting of arrows pointing in different directions: the length of each arrow representing the amplitude of some AC quantity (voltage, current, or impedance), and the angle of each arrow representing the shift in phase relative to the other arrows. By representing each AC quantity thusly, we may more easily calculate their relationships to one another, with the phasors showing us how to apply trigonometry (Pythagorean Theorem, sine, cosine, and tangent functions) to the various calculations. An analytical parallel to the graphic tool of phasor diagrams is complex numbers, where we represent each phasor (arrow) by a pair of numbers: either a magnitude and angle (polar notation), or by “real” and “imaginary” magnitudes (rectangular notation). Where phasor diagrams are helpful is in applications where their respective AC quantities add: the resultant of two or more phasors stacked tip-to-tail being the mathematical sum of the phasors. Complex numbers, on the other hand, may be added, subtracted, multiplied, and divided; the last two operations being difficult to graphically represent with arrows.

Conductance, Susceptance, and Admittance.

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance (1 \over R). Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B = {1 \over X} and admittance, the reciprocal of impedance Y = {1 \over Z}) are very necessary in order to draw phasor diagrams for parallel networks.

Question 1. (Click on arrow for answer)

Voltage divider circuits may be constructed from reactive components just as easily as they may be constructed from resistors. Take this capacitive voltage divider, for instance:

Should be image 00638x01

Calculate the magnitude and phase shift of V_{out}. Also, describe what advantages a capacitive voltage divider might have over a resistive voltage divider.

File Num: 00638

Answer

V_{out} = 1.754 V \angle 0^{o}

Follow-up question \#1: explain why the division ratio of a capacitive voltage divider remains constant with changes in signal frequency, even though we know that the reactance of the capacitors (X_{C1} and X_{C2}) will change.


Follow-up question \#2: one interesting feature of capacitive voltage dividers is that they harbor the possibility of electric shock after being disconnected from the voltage source, if the source voltage is high enough and if the disconnection happens at just the right time. Explain why a capacitive voltage divider poses this threat whereas a resistive voltage divider does not. Also, identify what the time of disconnection from the AC voltage source has to do with shock hazard.


Notes

Capacitive voltage dividers find use in high-voltage AC instrumentation, due to some of the advantages they exhibit over resistive voltage dividers. Your students should take special note of the phase angle for the capacitor’s voltage drop. Why it is 0 degrees, and not some other angle?





Question 2. (Click on arrow for answer)

A technician needs to know the value of a capacitor, but does not have a capacitance meter nearby. In lieu of this, the technician sets up the following circuit to measure capacitance:

Should be image 02114x01

You happen to walk by this technician’s workbench and ask, “How does this measurement setup work?” The technician responds, “You connect a resistor of known value (R) in series with the capacitor of unknown value (C_x), then adjust the generator frequency until the oscilloscope shows the two voltage drops to be equal, and then you calculate C_x.”

Explain how this system works, in your own words. Also, write the formula you would use to calculate the value of C_x given f and R.

File Num: 02114

Answer

I’ll let you figure out how to explain the operation of this test setup. The formula you would use looks like this:

C_x = {1 \over {2 \pi f R}}

Follow-up question: could you use a similar setup to measure the inductance of an unknown inductor L_x? Why or why not?


Challenge question: astute observers will note that this setup might not work in real life because the ground connection of the oscilloscope is not common with one of the function generator’s leads. Explain why this might be a problem, and suggest a practical solution for it.


Notes

This method of measuring capacitance (or inductance for that matter) is fairly old, and works well if the unknown component has a high Q value.





Question 3. (Click on arrow for answer)

A student measures voltage drops in an AC circuit using three voltmeters and arrives at the following measurements:

Should be image 01566x01

Upon viewing these measurements, the student becomes very perplexed. Aren’t voltage drops supposed to add in series, just as in DC circuits? Why, then, is the total voltage in this circuit only 10.8 volts and not 15.74 volts? How is it possible for the total voltage in an AC circuit to be substantially less than the simple sum of the components’ voltage drops?

Another student, trying to be helpful, suggests that the answer to this question might have something to do with RMS versus peak measurements. A third student disagrees, proposing instead that at least one of the meters is badly out of calibration and thus not reading correctly.

When you are asked for your thoughts on this problem, you realize that neither of the answers proposed thus far are correct. Explain the real reason for the “discrepancy” in voltage measurements, and also explain how you could experimentally disprove the other answers (RMS vs. peak, and bad calibration).

File Num: 01566

Answer

AC voltages still add in series, but phase must also be accounted for when doing so. Unfortunately, multimeters provide no indication of phase whatsoever, and thus do not provide us with all the information we need. (Note: just by looking at this circuit’s components, though, you should still be able to calculate the correct result for total voltage and validate the measurements.)

I’ll let you determine how to disprove the two incorrect explanations offered by the other students!


Challenge question: calculate a set of possible values for the capacitor and resistor that would generate these same voltage drops in a real circuit. Hint: you must also decide on a value of frequency for the power source.


Notes

This question has two different layers: first, how to reconcile the “strange” voltage readings with Kirchhoff’s Voltage Law; and second, how to experimentally validate the accuracy of the voltmeters and the fact that they are all registering the same type of voltage (RMS, peak, or otherwise, it doesn’t matter). The first layer of this question regards the basic concepts of AC phase, while the second exercises troubleshooting and critical thinking skills. Be sure to discuss both of these topics in class with your students.





Question 4. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 01844x01

File Num: 01844

Answer

Z_{total} = \sqrt{R^2 + X^2}

Notes

Ask your students if this equation looks similar to any other mathematical equations they’ve seen before. If not, square both sides of the equation so it looks like Z^2 = R^2 + X^2 and ask them again.





Question 5. (Click on arrow for answer)

Use a triangle to calculate the total voltage of the source for this series RC circuit, given the voltage drop across each component:

Should be image 02107x01

Explain what equation(s) you use to calculate V_{total}, as well as why we must geometrically add these voltages together.

File Num: 02107

Answer

V_{total} = 3.672 volts, as calculated by the Pythagorean Theorem

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 6. (Click on arrow for answer)

Determine the phase angle (\Theta) of the current in this circuit, with respect to the supply voltage:

Should be image 01853x01

File Num: 01853

Answer

\Theta = 26.51^o

Challenge question: explain how the following phasor diagram was determined for this problem:

Should be image 01853x02

Notes

This is an interesting question for a couple of reasons. First, students must determine how they will measure phase shift with just the two voltage indications shown by the meters. This may present a significant challenge for some. Discuss problem-solving strategies in class so that students understand how and why it is possible to determine \Theta.

Secondly, this is an interesting question because it shows how something as abstract as phase angle can be measured with just a voltmeter — no oscilloscope required! Not only that, but we don’t even have to know the component values either! Note that this technique works only for simple circuits.

A practical point to mention here is that multimeters have frequency limits which must be considered when taking measurements on electronic circuits. Some high-quality handheld digital meters have frequency limits of hundred of kilohertz, while others fail to register accurately at only a few thousand hertz. Unless we knew these two digital voltmeters were sufficient for measuring at the signal frequency, their indications would be useless to us.





Question 7. (Click on arrow for answer)

Due to the effects of a changing electric field on the dielectric of a capacitor, some energy is dissipated in capacitors subjected to AC. Generally, this is not very much, but it is there. This dissipative behavior is typically modeled as a series-connected resistance:

Should be image 01847x01

Calculate the magnitude and phase shift of the current through this capacitor, taking into consideration its equivalent series resistance (ESR):

Should be image 01847x02

Compare this against the magnitude and phase shift of the current for an ideal 0.22 \muF capacitor.

File Num: 01847

Answer

I = 3.732206 mA \angle 89.89^{o} for the real capacitor with ESR.
I = 3.732212 mA \angle 90.00^{o} for the ideal capacitor.

Follow-up question \#1: can this ESR be detected by a DC meter check of the capacitor? Why or why not?


Follow-up question \#2: explain how the ESR of a capacitor can lead to physical heating of the component, especially under high-voltage, high-frequency conditions. What safety concerns might arise as a result of this?


Notes

Although capacitors do contain their own parasitic effects, ESR being one of them, they still tend to be much “purer” components than inductors for general use. This is another reason why capacitors are generally favored over inductors in applications where either will suffice.





Question 8. (Click on arrow for answer)

Solve for all voltages and currents in this series RC circuit:

Should be image 01848x01

File Num: 01848

Answer

V_C = 14.39 \hbox{ volts RMS}V_R = 4.248 \hbox{ volts RMS}I = 903.9 \> \mu \hbox{A RMS}

Follow-up question: identify the consequences of a shorted capacitor in this circuit, with regard to circuit current and component voltage drops.


Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 9. (Click on arrow for answer)

Solve for all voltages and currents in this series RC circuit, and also calculate the phase angle of the total impedance:

Should be image 01849x01

File Num: 01849

Answer

V_C = 47.56 \hbox{ volts peak}V_R = 6.508 \hbox{ volts peak}I = 1.972 \hbox{ milliamps peak}\Theta_Z = -82.21^o

Follow-up question: what would we have to do to get these answers in units RMS instead of units “peak”?


Notes

Bring to your students’ attention the fact that total voltage in this circuit is given in “peak” units rather than RMS, and what effect this has on our answers.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 10. (Click on arrow for answer)

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

Should be image 01851x01
  • C_1 = 125 \hbox{ pF}
  • C_2 = 71 \hbox{ pF}
  • R_1 = 6.8 \hbox{ k}\Omega
  • R_2 = 1.2 \hbox{ k}\Omega
  • V_{supply} = 20 \hbox{ V RMS}
  • f_{supply} = 950 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01851

Answer

  • I_{total} = 2.269 \hbox{ mA}
  • V_{C1} = 3.041 \hbox{ V}
  • V_{C2} = 5.354 \hbox{ V}
  • V_{R1} = 15.43 \hbox{ V}
  • V_{R2} = 2.723 \hbox{ V}
  • \Theta = -24.82^o (voltage lagging current)

I suggest using a dual-trace oscilloscope to measure total voltage (across the supply terminals) and voltage drop across resistor R_2. Theoretically, measuring the voltage dropped by either resistor would be fine, but R_2 works better for practical reasons (oscilloscope input lead grounding). Phase shift then could be measured either in the time domain or by a Lissajous figure analysis.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.

It is noteworthy that the low capacitances shown here approach parasitic capacitances between circuit board traces. In other words, whoever designs a circuit to operate at 950 kHz cannot simply place components at will on the board, but must consider the traces themselves to be circuit elements (both capacitive and inductive in nature!). The calculations used to obtain the given answers, of course, assume ideal conditions where the PC board is not considered to possess capacitance or inductance.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 11. (Click on arrow for answer)

Calculate the voltage drops across all components in this circuit, expressing them in complex (polar) form with magnitudes and phase angles each:

Should be image 01852x01

File Num: 01852

Answer

V_{C1} = 0.921 \hbox { V} \> \angle -52.11^oV_{C2} = 0.921 \hbox { V} \> \angle -52.11^oV_{R1} = 1.184 \hbox { V} \> \angle \> 37.90^o

Follow-up question: how much phase shift is there between the capacitors’ voltage drop and the resistor’s voltage drop? Explain why this value is what it is.


Notes

The first challenge of this question is for students to figure out how to reduce this series-parallel combination to something simpler. Fortunately, this is very easy to do if one remembers the properties of parallel capacitances.

Students may be surprised to discover the phase shift between V_C and V_R is the value it is. However, this should not remain a mystery. Discuss this with your class, taking time for all of them to understand why the voltage phasors of a resistor and a capacitor in a simple series circuit will always be orthogonal.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 12. (Click on arrow for answer)

In this circuit, a series resistor-capacitor network creates a phase-shifted voltage for the “gate” terminal of a power-control device known as a TRIAC. All portions of the circuit except for the RC network are “shaded” for de-emphasis:

Should be image 00637x01

Calculate how many degrees of phase shift the capacitor’s voltage is, compared to the total voltage across the series RC network, assuming a frequency of 60 Hz, and a 50\% potentiometer setting.

File Num: 00637

Answer

E_C phase shift = -76.7^{o}

Challenge question: what effect will a change in potentiometer setting have on this phase angle? Specifically, will increasing the resistance make the phase shift approach -90^{o} or approach 0^{o}?


Notes

In this question, I purposely omitted any reference to voltage levels, so the students would have to set up part of the problem themselves. The goal here is to build problem-solving skills.





Question 13. (Click on arrow for answer)

A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = {1 \over R}), and it is measured in the unit of siemens.

Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

Should be image 01845x01

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = {1 \over X}), and the reciprocal of impedance is called admittance (Y = {1 \over Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.

Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Should be image 01845x02

File Num: 01845

Answer

Y_{total} = \sqrt{G^2 + B^2}

Follow-up question \#1: draw a phasor diagram showing how Y, G, and B relate.


Follow-up question \#2: re-write this equation using quantities of resistance (R), reactance (X), and impedance (Z), instead of conductance (G), susceptance (B), and admittance (Y).


Notes

Ask your students if this equation looks familiar to them. It should!


The answer to the challenge question is a matter of algebraic substitution. Work through this process with your students, and then ask them to compare the resulting equation with other equations they’ve seen before. Does its form look familiar to them in any way?





Question 14. (Click on arrow for answer)

Calculate the total impedance offered by these three capacitors to a sinusoidal signal with a frequency of 4 kHz:


  • C_1 = 0.1 \> \mu \hbox{F}
  • C_2 = 0.047 \> \mu \hbox{F}
  • C_3 = 0.033 \> \mu \hbox{F}

Should be image 01846x01

State your answer in the form of a scalar number (not complex), but calculate it using two different strategies:


  • Calculate total capacitance (C_{total}) first, then total impedance (Z_{total}).
  • Calculate individual admittances first (Y_{C1}, Y_{C2}, and Y_{C3}), then total impedance (Z_{total}).

File Num: 01846

Answer

First strategy:C_{total} = 0.18 \> \mu \hbox{F}Z_{total} = 221 \> \Omega
Second strategy:Y_{C1} = 2.51 \hbox{ mS}Y_{C2} = 1.18 \> \hbox{ mS}Y_{C3} = 829 \> \mu \hbox{S}Y_{total} = 4.52 \hbox{ mS}Z_{total} = 221 \> \Omega

Notes

This question is another example of how multiple means of calculation will give you the same answer (if done correctly!). Make note to your students that this indicates an answer-checking strategy!





Question 15. (Click on arrow for answer)

Calculate the total impedance of these parallel-connected components, expressing it in polar form (magnitude and phase angle):

Should be image 02108x01

Also, draw an admittance triangle for this circuit.

File Num: 02108

Answer

Z_{total} = 391.4 \Omega \angle -39.9^{o}Should be image 02108x02

Notes

Some students may wonder why every side of the triangle is represented by a Y term, rather than Y for the hypotenuse, G for the adjacent, and B for the opposite. If students ask about this, remind them that conductance (G) and susceptance (B) are simple two different types of admittances (Y), just as resistance (R) and reactance (X) are simply two different types of impedances (Z).





Question 16. (Click on arrow for answer)

Calculate the total impedances (complete with phase angles) for each of the following capacitor-resistor circuits:

Should be image 02109x01

File Num: 02109

Answer

Should be image 02109x02

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 17. (Click on arrow for answer)

If the source voltage in this circuit is assumed to be the phase reference (that is, the voltage is defined to be at an angle of 0 degrees), determine the relative phase angles of each current in this parallel circuit:

Should be image 02112x01
  • \Theta_{I(R)} =
  • \Theta_{I(C)} =
  • \Theta_{I(total)} =

File Num: 02112

Answer


  • \Theta_{I(R)} = 0^{o}
  • \Theta_{I(C)} = 90^{o}
  • \Theta_{I(total)} = some positive angle between 0^{o} and 90^{o}, exclusive


Notes

Some students will be confused about the positive phase angles, since this is a capacitive circuit and they have learned to associate negative angles with capacitors. It is important for these students to realize, though, that the negative angles they immediately associate with capacitors are in reference to impedance and not necessarily to other variables in the circuit!





Question 18. (Click on arrow for answer)

If the dielectric substance between a capacitor’s plates is not a perfect insulator, there will be a path for direct current (DC) from one plate to the other. This is typically called leakage resistance, and it is modeled as a shunt resistance to an ideal capacitance:

Should be image 01850x01

Calculate the magnitude and phase shift of the current drawn by this real capacitor, if powered by a sinusoidal voltage source of 30 volts RMS at 400 Hz:

Should be image 01850x02

Compare this against the magnitude and phase shift of the current for an ideal capacitor (no leakage).

File Num: 01850

Answer

I = 56.548671 mA \angle 89.98^{o} for the real capacitor with leakage resistance.I = 56.548668 mA \angle 90.00^{o} for the ideal capacitor.

Notes

Discuss with your students the fact that electrolytic capacitors typically have more leakage (less R_{leakage}) than most other capacitor types, due to the thinness of the dielectric oxide layer.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 19. (Click on arrow for answer)

The input impedance of an electrical test instrument is a very important parameter in some applications, because of how the instrument may load the circuit being tested. Oscilloscopes are no different from voltmeters in this regard:

Should be image 02111x01

Typical input impedance for an oscilloscope is 1 M\Omega of resistance, in parallel with a small amount of capacitance. At low frequencies, the reactance of this capacitance is so high that it may be safely ignored. At high frequencies, though, it may become a substantial load to the circuit under test:

Should be image 02111x02

Calculate how many ohms of impedance this oscilloscope input (equivalent circuit shown in the above schematic) will impose on a circuit with a signal frequency of 150 kHz.

File Num: 02111

Answer

Z_{input} = 52.98 k\Omega at 150 kHz

Follow-up question: what are the respective input impedances for ideal voltmeters and ideal ammeters? Explain why each ideal instrument needs to exhibit these impedances in order to accurately measure voltage and current (respectively) with the least “impact” to the circuit under test.


Notes

Mention to your students that this capacitive loading effect only gets worse when a cable is attached to the oscilloscope input. The calculation performed for this question is only for the input of the oscilloscope itself, not including whatever capacitance may be included in the test probe cable!

This is one of the reasons why \times10 probes are used with oscilloscopes: to minimize the loading effect on the tested circuit.





Question 20. (Click on arrow for answer)

Determine the size of capacitor (in Farads) necessary to create a total current of 11.3 mA in this parallel RC circuit:

Should be image 02110x01

File Num: 02110

Answer

C = 562.2 nF

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 21. (Click on arrow for answer)

Explain all the steps necessary to calculate the amount of current in this capacitive AC circuit:

Should be image 01551x01

File Num: 01551

Answer

I = 22.6 mA

Notes

The current is not difficult to calculate, so obviously the most important aspect of this question is not the math. Rather, it is the procedure of calculation: what to do first, second, third, etc., in obtaining the final answer.





Question 22. (Click on arrow for answer)

Calculate the total impedance offered by these two capacitors to a sinusoidal signal with a frequency of 900 Hz:

Should be image 01835x01

Show your work using three different problem-solving strategies:


  • Calculating total capacitance (C_{total}) first, then total impedance (Z_{total}).
  • Calculating individual admittances first (Y_{C1} and Y_{C2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
  • Using complex numbers: calculating individual impedances first (Z_{C1} and Z_{C2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01835

Answer

First strategy:C_{total} = 0.43 \> \mu \hbox{F}X_{total} = 411.3 \> \OmegaZ_{total} = 411.3 \> \Omega \> \angle -90^o or Z_{total} = 0 - j411.3 \> \Omega
Second strategy:Z_{C1} = X_{C1} = 535.9 \> \Omega
Y_{C1} = {1 \over Z_{C1}} = 1.866 \> \hbox{mS}
Z_{C1} = X_{C2} = 1.768 \hbox{ k}\Omega
Y_{C2} = {1 \over Z_{C2}} = 565.5 \> \mu \hbox{S}
Y_{total} = 2.432 \> \hbox{mS}
Z_{total} = {1 \over Y_{total}} = 411.3 \> \Omega
Third strategy: (using complex numbers)X_{C1} = 535.9 \> \Omega Z_{C1} = 535.9 \> \Omega \> \angle -90^oX_{C2} = 1.768 \hbox{ k}\Omega Z_{C1} = 1.768 \hbox{ k}\Omega \> \angle -90^oZ_{total} = 411.3 \> \Omega \> \angle -90^o or Z_{total} = 0 - j411.3 \> \Omega

Notes

A common misconception many students have about capacitive reactances and impedances is that they must interact “oppositely” to how one would normally consider electrical opposition. That is, many students believe capacitive reactances and impedances should add in parallel and diminish in series, because that’s what capacitance (in Farads) does! This is not true, however. Impedances always add in series and diminish in parallel, at least from the perspective of complex numbers. This is one of the reasons I favor AC circuit calculations using complex numbers: because then students may conceptually treat impedance just like they treat DC resistance.

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total capacitance and then calculating impedance from that, or by calculating the impedance of each capacitor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 23. (Click on arrow for answer)

Examine the following circuits, then label the sides of their respective triangles with all the variables that are trigonometrically related in those circuits:

Should be image 03288x01

File Num: 03288

Answer

Should be image 03288x02

Notes

This question asks students to identify those variables in each circuit that vectorially add, discriminating them from those variables which do not add. This is extremely important for students to be able to do if they are to successfully apply “the triangle” to the solution of AC circuit problems.

Note that some of these triangles should be drawn upside-down instead of all the same as they are shown in the question, if we are to properly represent the vertical (imaginary) phasor for capacitive impedance and for inductor admittance. However, the point here is simply to get students to recognize what quantities add and what do not. Attention to the direction (up or down) of the triangle’s opposite side can come later.





Question 24. (Click on arrow for answer)

Draw a phasor diagram showing the trigonometric relationship between resistance, reactance, and impedance in this series circuit:

Should be image 01828x01

Show mathematically how the resistance and reactance combine in series to produce a total impedance (scalar quantities, all). Then, show how to analyze this same circuit using complex numbers: regarding each of the component as having its own impedance, demonstrating mathematically how these impedances add up to comprise the total impedance (in both polar and rectangular forms).

File Num: 01828

Answer

Should be image 01828x02Scalar calculationsR = 2.2 \hbox{ k}\Omega X_C = 2.067 \hbox{ k}\OmegaZ_{series} = \sqrt{R^2 + {X_C}^2}Z_{series} = \sqrt{2200^2 + 2067^2} = 3019 \> \Omega
Complex number calculationsZ_R = 2.2 \hbox{ k}\Omega \> \angle \> 0^o Z_C = 2.067 \hbox{ k}\Omega \> \angle -90^o (Polar form)Z_R = 2.2 \hbox{ k}\Omega + j0 \> \Omega Z_C = 0 \> \Omega - j2.067 \hbox{ k}\Omega (Rectangular form)
Z_{series} = Z_1 + Z_2 + \cdots Z_n (General rule of series impedances)Z_{series} = Z_R + Z_C (Specific application to this circuit)
Z_{series} = 2.2 \hbox{ k}\Omega \> \angle \> 0^o + 2.067 \hbox{ k}\Omega \> \angle -90^o = 3.019 \hbox{ k}\Omega \> \angle -43.2^oZ_{series} = (2.2 \hbox{ k}\Omega + j0 \> \Omega) + (0 \> \Omega - j2.067 \hbox{ k}\Omega) = 2.2 \hbox{ k}\Omega - j2.067 \hbox{ k}\Omega

Notes

I want students to see that there are two different ways of approaching a problem such as this: with scalar math and with complex number math. If students have access to calculators that can do complex-number arithmetic, the “complex” approach is actually simpler for series-parallel combination circuits, and it yields richer (more informative) results.

Ask your students to determine which of the approaches most resembles DC circuit calculations. Incidentally, this is why I tend to prefer complex-number AC circuit calculations over scalar calculations: because of the conceptual continuity between AC and DC. When you use complex numbers to represent AC voltages, currents, and impedances, almost all the rules of DC circuits still apply. The big exception, of course, is calculations involving power.





Question 25. (Click on arrow for answer)

Calculate the total impedance of this RC circuit, once using nothing but scalar numbers, and again using complex numbers:

Should be image 01838x01

File Num: 01838

Answer

Scalar calculationsR_1 = 7.9 \hbox{ k}\Omega G_{R1} = 126.6 \> \mu \hbox{S}X_{C1} = 8.466 \hbox{ k}\Omega B_{C1} = 118.1 \> \mu \hbox{S}Y_{total} = \sqrt{G^2 + B^2} = 173.1 \> \mu \hbox{S}Z_{total} = {1 \over Y_{total}} = 5.776 \hbox{ k}\Omega
Complex number calculationsR_1 = 7.9 \hbox{ k}\Omega Z_{R1} = 7.9 \hbox{ k}\Omega \> \angle \> 0^oX_{C1} = 8.466 \hbox{ k}\Omega Z_{C1} = 8.466 \hbox{ k}\Omega \> \angle -90^oZ_{total} = { 1 \over {{1 \over Z_{R1}} + {1 \over Z_{C1}}}} = 5.776 \hbox{ k}\Omega \> \angle -43.02^o

Notes

Some electronics textbooks (and courses) tend to emphasize scalar impedance calculations, while others emphasize complex number calculations. While complex number calculations provide more informative results (a phase shift given in every variable!) and exhibit conceptual continuity with DC circuit analysis (same rules, similar formulae), the scalar approach lends itself better to conditions where students do not have access to calculators capable of performing complex number arithmetic. Yes, of course, you can do complex number arithmetic without a powerful calculator, but it’s a lot more tedious and prone to errors than calculating with admittances, susceptances, and conductances (primarily because the phase shift angle is omitted for each of the variables).





Question 26. (Click on arrow for answer)

A student is asked to calculate the phase shift for the following circuit’s output voltage, relative to the phase of the source voltage:

Should be image 03748x01

He recognizes this as a series circuit, and therefore realizes that a right triangle would be appropriate for representing component impedances and component voltage drops (because both impedance and voltage are quantities that add in series, and the triangle represents phasor addition):

Should be image 03748x02

The problem now is, which angle does the student solve for in order to find the phase shift of V_{out}? The triangle contains two angles besides the 90^{o} angle, \Theta and \Phi. Which one represents the output phase shift, and more importantly, why?

File Num: 03748

Answer

The proper angle in this circuit is \Theta, and it will be a positive (leading) quantity.


Notes

Too many students blindly use impedance and voltage triangles without really understand what they are and why they work. These same students will have no idea how to approach a problem like this. Work with them to help them understand!





Question 27. (Click on arrow for answer)

Calculate the output voltage of this phase-shifting circuit, expressing it in polar form (magnitude and phase angle relative to the source voltage):

Should be image 02621x01

File Num: 02621

Answer

V_{out} = 2.593 V \angle 61.3^{o}

Notes

This is a very practical application of resistor-capacitor (RC) circuits: to introduce a phase shift to an AC signal. Examples of where a circuit such as this may be used include oscillators (to introduce phase shift into a feedback network for a total phase shift of 360^{o}) and thyristor firing control circuits (phase-shifting the triggering voltage in relation to the source voltage).





Question 28. (Click on arrow for answer)

Calculate the output voltage of this phase-shifting circuit, expressing it in polar form (magnitude and phase angle relative to the source voltage):

Should be image 02620x01

File Num: 02620

Answer

V_{out} = 6.7 V \angle -47.9^{o}

Notes

This is a very practical application of resistor-capacitor (RC) circuits: to introduce a phase shift to an AC signal. Examples of where a circuit such as this may be used include oscillators (to introduce phase shift into a feedback network for a total phase shift of 360^{o}) and thyristor firing control circuits (phase-shifting the triggering voltage in relation to the source voltage).





Question 29. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 70^{o}:

Should be image 02623x01

File Num: 02623

Answer

f = 798 Hz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 30. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 40^{o}:

Should be image 02622x01

File Num: 02622

Answer

f = 6.54 kHz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 31. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -38^{o}:

Should be image 02626x01

File Num: 02626

Answer

f = 465 Hz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 32. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -25^{o}:

Should be image 02625x01

File Num: 02625

Answer

f = 929 Hz

Notes

Phase-shifting circuits are very useful, and important to understand. They are particularly important in some types of oscillator circuits, which rely on RC networks such as this to provide certain phase shifts to sustain oscillation.





Question 33. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 25^{o}:

Should be image 03284x01

Also, write an equation that solves for frequency (f), given all the other variables (R, C, and phase angle \theta).

File Num: 03284

Answer

f = 2.143 kHzf = {1 \over {2 \pi R C \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 34. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of 58^{o}:

Should be image 03285x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, C, and phase angle \theta).

File Num: 03285

Answer

R = 669.7 \OmegaR = {1 \over {2 \pi f C \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 35. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of -64^{o}:

Should be image 03287x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, C, and phase angle \theta).

File Num: 03287

Answer

R = 16.734 k\OmegaR = -{{\tan \theta} \over {2 \pi f C}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 36. (Click on arrow for answer)

Use algebraic substitution to generate an equation expressing the output voltage of the following circuit given the input voltage, the input frequency, the capacitor value, and the resistor value:

Should be image 03818x01
V_{out} =

File Num: 03818

Answer

V_{out} = {{R \> V_{in}} \over \sqrt{\left({1 \over 2 \pi f C}\right)^2 + R^2}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 37. (Click on arrow for answer)

Use algebraic substitution to generate an equation expressing the output voltage of the following circuit given the input voltage, the input frequency, the capacitor value, and the resistor value:

Should be image 03819x01
V_{out} =

File Num: 03819

Answer

V_{out} = {{R \> V_{in}} \over \sqrt{\left({{C_1 + C_2} \over 2 \pi f C_1 C_2}\right)^2 + R^2}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 38. (Click on arrow for answer)

Complete the table of values for this circuit, representing all quantities in complex-number form (either polar or rectangular, your choice):

Should be image 03611x01

File Num: 03611

Answer

Should be image 03611x02

Notes

Ask your students to share their problem-solving techniques for this question: how they solved for each parameter and in what order they performed the calculations.





Question 39. (Click on arrow for answer)

This phase-shifting bridge circuit is supposed to provide an output voltage with a variable phase shift from -45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Should be image 03464x01

Suppose, though, that the output signal is stuck at +45^{o} leading the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03464

Answer

A broken connection between the left-hand terminal of the potentiometer and the bridge could cause this to happen:

Should be image 03464x02

I’ll let you figure out why!


Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is much greater than the fixed resistance value.”





Question 40. (Click on arrow for answer)

This phase-shifting bridge circuit is supposed to provide an output voltage with a variable phase shift from -45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Should be image 03465x01

Suppose, though, that the output signal is stuck at -45^{o} lagging the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03465

Answer

A broken connection between the right-hand terminal of the potentiometer and the bridge could cause this to happen:

Should be image 03465x02

I’ll let you figure out why!


Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is much greater than the fixed resistance value.”





Question 41. (Click on arrow for answer)

This phase-shifting bridge circuit is supposed to provide an output voltage with a variable phase shift from -45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Should be image 03466x01

Suppose, though, that the output signal registers as it should with the potentiometer wiper fully to the right, but diminishes greatly in amplitude as the wiper is moved to the left, until there is practically zero output voltage at the full-left position. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03466

Answer

An open failure of the fixed resistor in the upper-left arm of the bridge could cause this to happen:

Should be image 03466x02

Follow-up question: identify another possible component failure that would exhibit the same symptoms.


Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is much greater than the fixed resistance value.”





Question 42. (Click on arrow for answer)

Sketch the approximate waveform of this circuit’s output signal (V_{out}) on the screen of the oscilloscope:

Should be image 03503x01

Hint: use the Superposition Theorem!


File Num: 03503

Answer

Should be image 03503x02

Follow-up question: explain why it is acceptable to use a polarized (polarity-sensitive) capacitor in this circuit when it is clearly connected to a source of AC. Why is it not damaged by the AC voltage when used like this?


Notes

Note that the capacitor size has been chosen for negligible capacitive reactance (X_C) at the specific frequency, such that the 10 k\Omega DC bias resistors present negligible loading to the coupled AC signal. This is typical for this type of biasing circuit.

Aside from giving students an excuse to apply the Superposition Theorem, this question previews a circuit topology that is extremely common in transistor amplifiers.





Question 43. (Click on arrow for answer)

Audio headphones make highly sensitive voltage detectors for AC signals in the audio frequency range. However, the small speakers inside headphones are quite easily damaged by the application of DC voltage.

Explain how a capacitor could be used as a “filtering” device to allow AC signals through to a pair of headphones, yet block any applied DC voltage, so as to help prevent accidental damage of the headphones while using them as an electrical instrument.

The key to understanding how to answer this question is to recognize what a capacitor “appears as” to AC signals versus DC signals.

File Num: 01395

Answer

Connect a capacitor in series with the headphone speakers.


Notes

I highly recommend to students that they should build a transformer-isolation circuit if they intend to use a pair of audio headphones as a test device (see question file number 00983 for a complete schematic diagram).





Question 44. (Click on arrow for answer)

Suppose a friend wanted to install filter networks in the “woofer” section of their stereo system, to prevent high-frequency power from being wasted in speakers incapable of reproducing those frequencies. To this end, your friend installs the following resistor-capacitor networks:

Should be image 00614x01

After examining this schematic, you see that your friend has the right idea in mind, but implemented it incorrectly. These filter circuits would indeed block high-frequency signals from getting to the woofers, but they would not actually accomplish the stated goal of minimizing wasted power.

What would you recommend to your friend in lieu of this circuit design?

File Num: 00614

Answer

Rather than use a “shunting” form of low-pass filter (resistor and capacitor), a “blocking” form of low-pass filter (inductor) should be used instead.


Notes

The reason for this choice in filter designs is very practical. Ask your students to describe how a “shunting” form of filter works, where the reactive component is connected in parallel with the load, receiving power through a series resistor. Contrast this against a “blocking” form of filter circuit, in which a reactive component is connected in series with the load. In one form of filter, a resistor is necessary. In the other form of filter, a resistor is not necessary. What difference does this make in terms of power dissipation within the filter circuit?





Question 45. (Click on arrow for answer)

It is common in audio systems to connect a capacitor in series with each “tweeter” (high-frequency) speaker to act as a simple high-pass filter. The choice of capacitors for this task is important in a high-power audio system.

A friend of mine once had such an arrangement for the tweeter speakers in his car. Unfortunately, though, the capacitors kept blowing up when he operated the stereo at full volume! Tired of replacing these non-polarized electrolytic capacitors, he came to me for advice. I suggested he use mylar or polystyrene capacitors instead of electrolytics. These were a bit more expensive than electrolytic capacitors, but they did not blow up. Explain why.

File Num: 03467

Answer

The issue here was not polarity (AC versus DC), because these were non-polarized electrolytic capacitors which were blowing up. What was an issue was ESR (Equivalent Series Resistance), which electrolytic capacitors are known to have high values of.


Notes

Your students may have to do a bit of refreshing (or first-time research!) on the meaning of ESR before they can understand why large ESR values could cause a capacitor to explode under extreme operating conditions.





All files with file num less than 4100 are Copyright 2003, Tony R. Kuphaldt, released under the Creative Commons Attribution License (v 1.0). All other files are Copyright 2022, David Williams, released under the Creative Commons Attribution License (V 4.0) This means you may do almost anything with this work, so long as you give proper credit.

To view a copy of the license, visit https://creativecommons.org/licenses/by/1.0/, or https://creativecommons.org/licenses/by/4.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.

Practice Problems: Inductors in AC Circuits

Difficult Concepts

These are some concepts that new learners often find challenging. It is probably worthwhile to read through these concepts because they may explain challenges you are facing while learning about inductors in AC circuits.

Resistance vs. Reactance vs. Impedance

These three terms represent different forms of opposition to electric current. Despite the fact that they are measured in the same unit (ohms: Omega), they are not the same. Resistance is best thought of as electrical friction, whereas reactance is best thought of as electrical inertia. Whereas resistance creates a voltage drop by dissipating energy, reactance creates a voltage drop by storing and releasing energy. Impedance is a term encompassing both resistance and reactance, usually a combination of both.

Phasors, used to represent AC amplitude and phase relations.

A powerful tool used for understanding the operation of AC circuits is the phasor diagram, consisting of arrows pointing in different directions: the length of each arrow representing the amplitude of some AC quantity (voltage, current, or impedance), and the angle of each arrow representing the shift in phase relative to the other arrows. By representing each AC quantity thusly, we may more easily calculate their relationships to one another, with the phasors showing us how to apply trigonometry (Pythagorean Theorem, sine, cosine, and tangent functions) to the various calculations. An analytical parallel to the graphic tool of phasor diagrams is complex numbers, where we represent each phasor (arrow) by a pair of numbers: either a magnitude and angle (polar notation), or by “real” and “imaginary” magnitudes (rectangular notation). Where phasor diagrams are helpful is in applications where their respective AC quantities add: the resultant of two or more phasors stacked tip-to-tail being the mathematical sum of the phasors. Complex numbers, on the other hand, may be added, subtracted, multiplied, and divided; the last two operations being difficult to graphically represent with arrows.

Conductance, Susceptance, and Admittance.

Conductance, symbolized by the letter G, is the mathematical reciprocal of resistance (1 \over R). Students typically encounter this quantity in their DC studies and quickly ignore it. In AC calculations, however, conductance and its AC counterparts (susceptance, the reciprocal of reactance B = {1 \over X} and admittance, the reciprocal of impedance Y = {1 \over Z}) are very necessary in order to draw phasor diagrams for parallel networks.

Question 1. (Click on arrow for answer)

As a general rule, inductors oppose change in (choose: voltage or current), and they do so by . . . (complete the sentence).


Based on this rule, determine how an inductor would react to a constant AC current that increases in frequency. Would an inductor drop more or less voltage, given a greater frequency? Explain your answer.

File Num: 00578

Answer

As a general rule, inductors oppose change in current, and they do so by producing a voltage.


An inductor will drop a greater amount of AC voltage, given the same AC current, at a greater frequency.


Notes

This question is an exercise in qualitative thinking: relating rates of change to other variables, without the use of numerical quantities. The general rule stated here is very, very important for students to master, and be able to apply to a variety of circumstances. If they learn nothing about inductors except for this rule, they will be able to grasp the function of a great many inductor circuits.





Question 2. (Click on arrow for answer)

\int f(x) dx Calculus alert!

We know that the formula relating instantaneous voltage and current in an inductor is this:

e = L{di \over dt}

Knowing this, determine at what points on this sine wave plot for inductor current is the inductor voltage equal to zero, and where the voltage is at its positive and negative peaks. Then, connect these points to draw the waveform for inductor voltage:

Should be image 00576x01

How much phase shift (in degrees) is there between the voltage and current waveforms? Which waveform is leading and which waveform is lagging?

File Num: 00576

Answer

Should be image 00576x02

For an inductor, voltage is leading and current is lagging, by a phase shift of 90^{o}.


Notes

This question is an excellent application of the calculus concept of the derivative: relating one function (instantaneous voltage, e) with the instantaneous rate-of-change of another function (current, di \over dt).





Question 3. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 60 Hz:

Should be image 01832x01

Show your work using two different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01832

Answer

First strategy:L_{total} = 1.1 \hbox{ H}X_{total} = 414.7 \> \OmegaZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega
Second strategy:X_{L1} = 282.7 \> \Omega Z_{L1} = 282.7 \> \Omega \> \angle \> 90^oX_{L2} = 131.9 \> \Omega Z_{L2} = 131.9 \> \Omega \> \angle \> 90^oZ_{total} = 414.7 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j414.7 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ impedance phasors geometrically add to equal the total impedance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 4. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 01844x01

File Num: 01844

Answer

Z_{total} = \sqrt{R^2 + X^2}

Notes

Ask your students if this equation looks similar to any other mathematical equations they’ve seen before. If not, square both sides of the equation so it looks like Z^2 = R^2 + X^2 and ask them again.





Question 5. (Click on arrow for answer)

Calculate the total impedance of this LR circuit, once using nothing but scalar numbers, and again using complex numbers:

Should be image 01837x01

File Num: 01837

Answer

Scalar calculationsR_1 = 1.5 \hbox{ k}\Omega G_{R1} = 666.7 \> \mu \hbox{S}X_{L1} = 2.513 \hbox{ k}\Omega B_{L1} = 397.9 \> \mu \hbox{S}Y_{total} = \sqrt{G^2 + B^2} = 776.4 \> \mu \hbox{S}Z_{total} = {1 \over Y_{total}} = 1.288 \hbox{ k}\Omega
Complex number calculationsR_1 = 1.5 \hbox{ k}\Omega Z_{R1} = 1.5 \hbox{ k}\Omega \> \angle \> 0^oX_{L1} = 2.513 \hbox{ k}\Omega Z_{L1} = 2.513 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = { 1 \over {{1 \over Z_{R1}} + {1 \over Z_{L1}}}} = 1.288 \hbox{ k}\Omega \> \angle \> 30.83^o

Notes

Some electronics textbooks (and courses) tend to emphasize scalar impedance calculations, while others emphasize complex number calculations. While complex number calculations provide more informative results (a phase shift given in every variable!) and exhibit conceptual continuity with DC circuit analysis (same rules, similar formulae), the scalar approach lends itself better to conditions where students do not have access to calculators capable of performing complex number arithmetic. Yes, of course, you can do complex number arithmetic without a powerful calculator, but it’s a lot more tedious and prone to errors than calculating with admittances, susceptances, and conductances (primarily because the phase shift angle is omitted for each of the variables).





Question 6. (Click on arrow for answer)

Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 120 Hz:

Should be image 01833x01

Show your work using three different problem-solving strategies:


  • Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
  • Calculating individual admittances first (Y_{L1} and Y_{L2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
  • Using complex numbers: calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).

Do these two strategies yield the same total impedance value? Why or why not?

File Num: 01833

Answer

First strategy:L_{total} = 391.3 \hbox{ mH}X_{total} = 295.0 \> \OmegaZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega
Second strategy:Z_{L1} = X_{L1} = 377.0 \> \Omega
Y_{L1} = {1 \over Z_{L1}} = 2.653 \> \hbox{mS}
Z_{L1} = X_{L2} = 1.357 \hbox{ k}\Omega
Y_{L2} = {1 \over Z_{L2}} = 736.8 \> \mu \hbox{S}
Y_{total} = 3.389 \> \hbox{mS}
Z_{total} = {1 \over Y_{total}} = 295 \> \Omega
Third strategy: (using complex numbers)X_{L1} = 377.0 \> \Omega Z_{L1} = 377.0 \> \Omega \> \angle \> 90^oX_{L2} = 1.357 \hbox{ k}\Omega Z_{L2} = 1.357 \hbox{ k}\Omega \> \angle \> 90^oZ_{total} = 295.0 \> \Omega \> \angle \> 90^o or Z_{total} = 0 + j295.0 \> \Omega

Follow-up question: draw a phasor diagram showing how the two inductors’ admittance phasors geometrically add to equal the total admittance.


Notes

The purpose of this question is to get students to realize that any way they can calculate total impedance is correct, whether calculating total inductance and then calculating impedance from that, or by calculating the impedance of each inductor and then combining impedances to find a total impedance. This should be reassuring, because it means students have a way to check their work when analyzing circuits such as this!





Question 7. (Click on arrow for answer)

Does an inductor’s opposition to alternating current increase or decrease as the frequency of that current increases? Also, explain why we refer to this opposition of AC current in an inductor as reactance instead of resistance.

File Num: 00580

Answer

The opposition to AC current (“reactance”) of an inductor increases as frequency increases. We refer to this opposition as “reactance” rather than “resistance” because it is non-dissipative in nature. In other words, reactance causes no power to leave the circuit.


Notes

Ask your students to define the relationship between inductor reactance and frequency as either “directly proportional” or “inversely proportional”. These are two phrases used often in science and engineering to describe whether one quantity increases or decreases as another quantity increases. Your students definitely need to be familiar with both these phrases, and be able to interpret and use them in their technical discussions.

Also, discuss the meaning of the word “non-dissipative” in this context. How could we prove that the opposition to current expressed by an inductor is non-dissipative? What would be the ultimate test of this?





Question 8. (Click on arrow for answer)

What will happen to the brightness of the light bulb as the iron core is moved away from the wire coil in this circuit? Explain why this happens.

Should be image 00095x01

File Num: 00095

Answer

The light bulb will glow brighter when the iron core is moved away from the wire coil, due to the change in inductive reactance (X_{L}).


Follow-up question: what circuit failure(s) could cause the light bulb to glow brighter than it should?


Notes

One direction you might want to lead your students in with this question is how AC power may be controlled using this principle. Controlling AC power with a variable reactance has a definite advantage over controlling AC power with a variable resistance: less wasted energy in the form of heat.





Question 9. (Click on arrow for answer)

An inductor rated at 4 Henrys is subjected to a sinusoidal AC voltage of 24 volts RMS, at a frequency of 60 hertz. Write the formula for calculating inductive reactance (X_L), and solve for current through the inductor.

File Num: 00582

Answer

X_L = 2 \pi f L

The current through this inductor is 15.92 mA RMS.


Notes

I have consistently found that qualitative (greater than, less than, or equal) analysis is much more difficult for students to perform than quantitative (punch the numbers on a calculator) analysis. Yet, I have consistently found on the job that people lacking qualitative skills make more “silly” quantitative errors because they cannot validate their calculations by estimation.

In light of this, I always challenge my students to qualitatively analyze formulae when they are first introduced to them. Ask your students to identify what will happen to one term of an equation if another term were to either increase, or decrease (you choose the direction of change). Use up and down arrow symbols if necessary to communicate these changes graphically. Your students will greatly benefit in their conceptual understanding of applied mathematics from this kind of practice!





Question 10. (Click on arrow for answer)

At what frequency does a 350 mH inductor have 4.7 k\Omega of reactance? Write the formula for solving this, in addition to calculating the frequency.

File Num: 00586

Answer

f = 2.137 kHz

Notes

Be sure to ask your students to demonstrate the algebraic manipulation of the original formula, in providing the answer to this question. Algebraic manipulation of equations is a very important skill to have, and it comes only by study and practice.





Question 11. (Click on arrow for answer)

How much inductance would an inductor have to possess in order to provide 540 \Omega of reactance at a frequency of 400 Hz? Write the formula for solving this, in addition to calculating the frequency.

File Num: 03277

Answer

L = 214.9 mH

Notes

Be sure to ask your students to demonstrate the algebraic manipulation of the original formula, in providing the answer to this question. Algebraic manipulation of equations is a very important skill to have, and it comes only by study and practice.





Question 12. (Click on arrow for answer)

Explain all the steps necessary to calculate the amount of current in this inductive AC circuit:

Should be image 01552x01

File Num: 01552

Answer

I = 15.6 mA

Notes

The current is not difficult to calculate, so obviously the most important aspect of this question is not the math. Rather, it is the procedure of calculation: what to do first, second, third, etc., in obtaining the final answer.





Question 13. (Click on arrow for answer)

In this AC circuit, the resistor offers 300 \Omega of resistance, and the inductor offers 400 \Omega of reactance. Together, their series opposition to alternating current results in a current of 10 mA from the 5 volt source:

Should be image 00584x01

How many ohms of opposition does the series combination of resistor and inductor offer? What name do we give to this quantity, and how do we symbolize it, being that it is composed of both resistance (R) and reactance (X)?

File Num: 00584

Answer

Z_{total} = 500 \Omega.

Follow-up question: suppose that the inductor suffers a failure in its wire winding, causing it to “open.” Explain what effect this would have on circuit current and voltage drops.


Notes

Students may experience difficulty arriving at the same quantity for impedance shown in the answer. If this is the case, help them problem-solve by suggesting they simplify the problem: short past one of the load components and calculate the new circuit current. Soon they will understand the relationship between total circuit opposition and total circuit current, and be able to apply this concept to the original problem.

Ask your students why the quantities of 300 \Omega and 400 \Omega do not add up to 700 \Omega like they would if they were both resistors. Does this scenario remind them of another mathematical problem where 3 + 4 = 5? Where have we seen this before, especially in the context of electric circuits?

Once your students make the cognitive connection to trigonometry, ask them the significance of these numbers’ addition. Is it enough that we say a component has an opposition to AC of 400 \Omega, or is there more to this quantity than a single, scalar value? What type of number would be suitable for representing such a quantity, and how might it be written?





Question 14. (Click on arrow for answer)

While studying DC circuit theory, you learned that resistance was an expression of a component’s opposition to electric current. Then, when studying AC circuit theory, you learned that reactance was another type of opposition to current. Now, a third term is introduced: impedance. Like resistance and reactance, impedance is also a form of opposition to electric current.

Explain the difference between these three quantities (resistance, reactance, and impedance) using your own words.

File Num: 01567

Answer

The fundamental distinction between these terms is one of abstraction: impedance is the most general term, encompassing both resistance and reactance. Here is an explanation given in terms of logical sets (using a Venn diagram), along with an analogy from animal taxonomy:

Should be image 01567x01

Resistance is a type of impedance, and so is reactance. The difference between the two has to do with energy exchange.


Notes

The given answer is far from complete. I’ve shown the semantic relationship between the terms resistance, reactance, and impedance, but I have only hinted at the conceptual distinctions between them. Be sure to discuss with your students what the fundamental difference is between resistance and reactance, in terms of electrical energy exchange.





Question 15. (Click on arrow for answer)

In DC circuits, we have Ohm’s Law to relate voltage, current, and resistance together:

E = I R

In AC circuits, we similarly need a formula to relate voltage, current, and impedance together. Write three equations, one solving for each of these three variables: a set of Ohm’s Law formulae for AC circuits. Be prepared to show how you may use algebra to manipulate one of these equations into the other two forms.

File Num: 00590

Answer

E = I ZI = {E \over Z}Z = {E \over I}

If using phasor quantities (complex numbers) for voltage, current, and impedance, the proper way to write these equations is as follows:

E = IZI = {E \over Z}Z = {E \over I}

Bold-faced type is a common way of denoting vector quantities in mathematics.


Notes

Although the use of phasor quantities for voltage, current, and impedance in the AC form of Ohm’s Law yields certain distinct advantages over scalar calculations, this does not mean one cannot use scalar quantities. Often it is appropriate to express an AC voltage, current, or impedance as a simple scalar number.





Question 16. (Click on arrow for answer)

It is often necessary to represent AC circuit quantities as complex numbers rather than as scalar numbers, because both magnitude and phase angle are necessary to consider in certain calculations.

When representing AC voltages and currents in polar form, the angle given refers to the phase shift between the given voltage or current, and a “reference” voltage or current at the same frequency somewhere else in the circuit. So, a voltage of 3.5 \hbox{ V} \angle -45^o means a voltage of 3.5 volts magnitude, phase-shifted 45 degrees behind (lagging) the reference voltage (or current), which is defined to be at an angle of 0 degrees.

But what about impedance (Z)? Does impedance have a phase angle, too, or is it a simple scalar number like resistance or reactance?


Calculate the amount of current that would go through a 100 mH inductor with 36 volts RMS applied to it at a frequency of 400 Hz. Then, based on Ohm’s Law for AC circuits and what you know of the phase relationship between voltage and current for an inductor, calculate the impedance of this inductor in polar form. Does a definite angle emerge from this calculation for the inductor’s impedance? Explain why or why not.

File Num: 00588

Answer

Z_L = 251.33 \Omega \angle 90^{o}

Notes

This is a challenging question, because it asks the student to defend the application of phase angles to a type of quantity that does not really possess a wave-shape like AC voltages and currents do. Conceptually, this is difficult to grasp. However, the answer is quite clear through the Ohm’s Law calculation (Z = {E \over I}).

Although it is natural to assign a phase angle of 0^{o} to the 36 volt supply, making it the reference waveform, this is not actually necessary. Work through this calculation with your students, assuming different angles for the voltage in each instance. You should find that the impedance computes to be the same exact quantity every time.





Question 17. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of 75^{o}:

Should be image 03282x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03282

Answer

f = 11.342 kHzf = {R \over {2 \pi L \tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 18. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of 44^{o}:

Should be image 03283x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03283

Answer

R = 6.826 k\OmegaR = 2 \pi f L \tan \theta

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 19. (Click on arrow for answer)

Determine the input frequency necessary to give the output voltage a phase shift of -40^{o}:

Should be image 03280x01

Also, write an equation that solves for frequency (f), given all the other variables (R, L, and phase angle \theta).

File Num: 03280

Answer

f = 2.804 kHzf = - {{R \tan \theta} \over {2 \pi L}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 20. (Click on arrow for answer)

Determine the necessary resistor value to give the output voltage a phase shift of -60^{o}:

Should be image 03281x01

Also, write an equation that solves for this resistance value (R), given all the other variables (f, L, and phase angle \theta).

File Num: 03281

Answer

R = 2.902 k\OmegaR = - {{2 \pi f L} \over {\tan \theta}}

Notes

Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 21. (Click on arrow for answer)

If a sinusoidal voltage is applied to an impedance with a phase angle of 0^{o}, the resulting voltage and current waveforms will look like this:

Should be image 00631x01

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit.

File Num: 00631

Answer

Should be image 00631x02

Notes

Ask your students to observe the waveform shown in the answer closely, and determine what sign the power values always are. Note how the voltage and current waveforms alternate between positive and negative, but power does not. Of what significance is this to us? What does this indicate about the nature of a load with an impedance phase angle of 0^{o}?





Question 22. (Click on arrow for answer)

If a sinusoidal voltage is applied to an impedance with a phase angle of 90^{o}, the resulting voltage and current waveforms will look like this:

Should be image 00632x01

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit. Also, explain how the mnemonic phrase “ELI the ICE man” applies to these waveforms.

File Num: 00632

Answer

Should be image 00632x02

The mnemonic phrase, “ELI the ICE man” indicates that this phase shift is due to an inductance rather than a capacitance.


Notes

Ask your students to observe the waveform shown in the answer closely, and determine what sign the power values are. Note how the power waveform alternates between positive and negative values, just as the voltage and current waveforms do. Ask your students to explain what negative power could possibly mean.

Of what significance is this to us? What does this indicate about the nature of a load with an impedance phase angle of 90^{o}?


The phrase, “ELI the ICE man” has been used be generations of technicians to remember the phase relationships between voltage and current for inductors and capacitors, respectively. One area of trouble I’ve noted with students, though, is being able to interpret which waveform is leading and which one is lagging, from a time-domain plot such as this.





Question 23. (Click on arrow for answer)

The impedance triangle is often used to graphically relate Z, R, and X in a series circuit:

Should be image 02076x01

Unfortunately, many students do not grasp the significance of this triangle, but rather memorize it as a “trick” used to calculate one of the three variables given the other two. Explain why a right triangle is an appropriate form to relate these variables, and what each side of the triangle actually represents.

File Num: 02076

Answer

Each side of the impedance triangle is actually a phasor (a vector representing impedance with magnitude and direction):

Should be image 02076x02

Since the phasor for resistive impedance (Z_R) has an angle of zero degrees and the phasor for reactive impedance (Z_C or Z_L) either has an angle of +90 or -90 degrees, the phasor sum representing total series impedance will form the hypotenuse of a right triangle when the first to phasors are added (tip-to-tail).


Follow-up question: as a review, explain why resistive impedance phasors always have an angle of zero degrees, and why reactive impedance phasors always have angles of either +90 degrees or -90 degrees.


Notes

The question is sufficiently open-ended that many students may not realize exactly what is being asked until they read the answer. This is okay, as it is difficult to phrase the question in a more specific manner without giving away the answer!





Question 24. (Click on arrow for answer)

Use the “impedance triangle” to calculate the impedance of this series combination of resistance (R) and inductive reactance (X):

Should be image 02081x01

Explain what equation(s) you use to calculate Z.

File Num: 02081

Answer

Z = 625 \Omega, as calculated by the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 25. (Click on arrow for answer)

Use the “impedance triangle” to calculate the necessary reactance of this series combination of resistance (R) and inductive reactance (X) to produce the desired total impedance of 145 \Omega:

Should be image 02083x01

Explain what equation(s) you use to calculate X, and the algebra necessary to achieve this result from a more common formula.

File Num: 02083

Answer

X = 105 \Omega, as calculated by an algebraically manipulated version of the Pythagorean Theorem.

Notes

Be sure to have students show you the form of the Pythagorean Theorem, rather than showing them yourself, since it is so easy for students to research on their own.





Question 26. (Click on arrow for answer)

Should be image 02084x01

Identify which trigonometric functions (sine, cosine, or tangent) are represented by each of the following ratios, with reference to the angle labeled with the Greek letter “Theta” (\Theta):

{X \over R} = {X \over Z} = {R \over Z} =

File Num: 02084

Answer

Should be image 02084x01{X \over R} = \tan \Theta = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \sin \Theta = {\hbox{Opposite} \over \hbox{Hypotenuse}}{R \over Z} = \cos \Theta = {\hbox{Adjacent} \over \hbox{Hypotenuse}}

Notes

Ask your students to explain what the words “hypotenuse”, “opposite”, and “adjacent” refer to in a right triangle.





Question 27. (Click on arrow for answer)

Trigonometric functions such as sine, cosine, and tangent are useful for determining the ratio of right-triangle side lengths given the value of an angle. However, they are not very useful for doing the reverse: calculating an angle given the lengths of two sides.

Should be image 02086x01

Suppose we wished to know the value of angle \Theta, and we happened to know the values of Z and R in this impedance triangle. We could write the following equation, but in its present form we could not solve for \Theta:

\cos \Theta = {R \over Z}

The only way we can algebraically isolate the angle \Theta in this equation is if we have some way to “undo” the cosine function. Once we know what function will “undo” cosine, we can apply it to both sides of the equation and have \Theta by itself on the left-hand side.

There is a class of trigonometric functions known as inverse or “arc” functions which will do just that: “undo” a regular trigonometric function so as to leave the angle by itself. Explain how we could apply an “arc-function” to the equation shown above to isolate \Theta.

File Num: 02086

Answer

\cos \Theta = {R \over Z} \hbox{ Original equation}\hbox<i>. . . applying the "arc-cosine" function to both sides . . .</i>\arccos \left( \cos \Theta \right) = \arccos \left( {R \over Z} \right)\Theta = \arccos \left( {R \over Z} \right)

Notes

I like to show the purpose of trigonometric arcfunctions in this manner, using the cardinal rule of algebraic manipulation (do the same thing to both sides of an equation) that students are familiar with by now. This helps eliminate the mystery of arcfunctions for students new to trigonometry.





Question 28. (Click on arrow for answer)

A series AC circuit contains 1125 ohms of resistance and 1500 ohms of reactance for a total circuit impedance of 1875 ohms. This may be represented graphically in the form of an impedance triangle:

Should be image 02085x01

Since all side lengths on this triangle are known, there is no need to apply the Pythagorean Theorem. However, we may still calculate the two non-perpendicular angles in this triangle using “inverse” trigonometric functions, which are sometimes called arcfunctions.

Identify which arc-function should be used to calculate the angle \Theta given the following pairs of sides:

R \hbox{ and } ZX \hbox{ and } RX \hbox{ and } Z

Show how three different trigonometric arcfunctions may be used to calculate the same angle \Theta.

File Num: 02085

Answer

\arccos {R \over Z} = 53.13^o\arctan {X \over R} = 53.13^o\arcsin {X \over Z} = 53.13^o

Challenge question: identify three more arcfunctions which could be used to calculate the same angle \Theta.


Notes

Some hand calculators identify arc-trig functions by the letter “A” prepending each trigonometric abbreviation (e.g. “ASIN” or “ATAN”). Other hand calculators use the inverse function notation of a -1 exponent, which is not actually an exponent at all (e.g. \sin^{-1} or \tan^{-1}). Be sure to discuss function notation on your students’ calculators, so they know what to invoke when solving problems such as this.





Question 29. (Click on arrow for answer)

Write an equation that solves for the impedance of this series circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for impedance (in ohms):

Should be image 00850x01

File Num: 00850

Answer

Z_{total} = \sqrt{R^2 + X^2}

Follow-up question: algebraically manipulate this equation to produce two more; one solving for R and the other solving for X.


Notes

Ask your students if this equation looks similar to any other mathematical equations they’ve seen before. If not, square both sides of the equation so it looks like Z^2 = R^2 + X^2 and ask them again.





Question 30. (Click on arrow for answer)

Draw a phasor diagram showing the trigonometric relationship between resistance, reactance, and impedance in this series circuit:

Should be image 01827x01

Show mathematically how the resistance and reactance combine in series to produce a total impedance (scalar quantities, all). Then, show how to analyze this same circuit using complex numbers: regarding component as having its own impedance, demonstrating mathematically how these impedances add up to comprise the total impedance (in both polar and rectangular forms).

File Num: 01827

Answer

Should be image 01827x02Scalar calculationsR = 2.2 \hbox{ k}\Omega X_L = 1.495 \hbox{ k}\OmegaZ_{series} = \sqrt{R^2 + {X_L}^2}Z_{series} = \sqrt{2200^2 + 1495^2} = 2660 \> \Omega
Complex number calculationsZ_R = 2.2 \hbox{ k}\Omega \> \angle \> 0^o Z_L = 1.495 \hbox{ k}\Omega \> \angle \> 90^o (Polar form)Z_R = 2.2 \hbox{ k}\Omega + j0 \> \Omega Z_L = 0 \> \Omega + j1.495 \hbox{ k}\Omega (Rectangular form)
Z_{series} = Z_1 + Z_2 + \cdots Z_n (General rule of series impedances)Z_{series} = Z_R + Z_L (Specific application to this circuit)
Z_{series} = 2.2 \hbox{ k}\Omega \> \angle \> 0^o + 1.495 \hbox{ k}\Omega \> \angle \> 90^o = 2.66 \hbox{ k}\Omega \> \angle \> 34.2^oZ_{series} = (2.2 \hbox{ k}\Omega + j0 \> \Omega) + (0 \> \Omega + j1.495 \hbox{ k}\Omega) = 2.2 \hbox{ k}\Omega + j1.495 \hbox{ k}\Omega

Notes

I want students to see that there are two different ways of approaching a problem such as this: with scalar math and with complex number math. If students have access to calculators that can do complex-number arithmetic, the “complex” approach is actually simpler for series-parallel combination circuits, and it yields richer (more informative) results.

Ask your students to determine which of the approaches most resembles DC circuit calculations. Incidentally, this is why I tend to prefer complex-number AC circuit calculations over scalar calculations: because of the conceptual continuity between AC and DC. When you use complex numbers to represent AC voltages, currents, and impedances, almost all the rules of DC circuits still apply. The big exception, of course, is calculations involving power.





Question 31. (Click on arrow for answer)

Calculate the total impedance for these two 100 mH inductors at 2.3 kHz, and draw a phasor diagram showing circuit impedances (Z_{total}, R, and X):

Should be image 02080x01

Now, re-calculate impedance and re-draw the phasor impedance diagram supposing the second inductor is replaced by a 1.5 k\Omega resistor:

Should be image 02080x02

File Num: 02080

Answer

Should be image 02080x03Should be image 02080x04

Notes

Phasor diagrams are powerful analytical tools, if one knows how to draw and interpret them. With hand calculators being so powerful and readily able to handle complex numbers in either polar or rectangular form, there is temptation to avoid phasor diagrams and let the calculator handle all the angle manipulation. However, students will have a much better understanding of phasors and complex numbers in AC circuits if you hold them accountable to representing quantities in that form.





Question 32. (Click on arrow for answer)

Calculate the total impedance of this series LR circuit and then calculate the total circuit current:

Should be image 02103x01

Also, draw a phasor diagram showing how the individual component impedances relate to the total impedance.

File Num: 02103

Answer

Z_{total} = 6.944 k\OmegaI = 4.896 mA RMS

Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 33. (Click on arrow for answer)

Calculate the magnitude and phase shift of the current through this inductor, taking into consideration its intrinsic winding resistance:

Should be image 00639x01

File Num: 00639

Answer

I = 7.849 mA \angle -87.08^{o}

Notes

Inductors are the least “pure” of any reactive component, due to significant quantities of resistance in the windings. Discuss this fact with your students, and what it means with reference to choosing inductors versus capacitors in circuit designs that could use either.





Question 34. (Click on arrow for answer)

Solve for all voltages and currents in this series LR circuit:

Should be image 01830x01

File Num: 01830

Answer

V_L = 12.60 \hbox{ volts RMS}V_R = 8.137 \hbox{ volts RMS}I = 11.46 \hbox{ milliamps RMS}

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 35. (Click on arrow for answer)

Solve for all voltages and currents in this series LR circuit, and also calculate the phase angle of the total impedance:

Should be image 01831x01

File Num: 01831

Answer

V_L = 13.04 \hbox{ volts RMS}V_R = 20.15 \hbox{ volts RMS}I = 4.030 \hbox{ milliamps RMS}\Theta_Z = 32.91^o

Notes

Nothing special here — just a straightforward exercise in series AC circuit calculations.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 36. (Click on arrow for answer)

Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

Should be image 01841x01
  • L_1 = 250 \hbox{ mH}
  • L_2 = 60 \hbox{ mH}
  • R_1 = 6.8 \hbox{ k}\Omega
  • R_2 = 1.2 \hbox{ k}\Omega
  • V_{supply} = 13.4 \hbox{ V RMS}
  • f_{supply} = 6.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01841

Answer

  • I_{total} = 0.895 \hbox{ mA}
  • V_{L1} = 9.14 \hbox{ V}
  • V_{L2} = 2.19 \hbox{ V}
  • V_{R1} = 6.08 \hbox{ V}
  • V_{R2} = 1.07 \hbox{ V}
  • \Theta = 57.71^o

I suggest using a dual-trace oscilloscope to measure total voltage (across the supply terminals) and voltage drop across resistor R_2. Theoretically, measuring the voltage dropped by either resistor would be fine, but R_2 works better for practical reasons (oscilloscope input lead grounding). Phase shift then could be measured either in the time domain or by a Lissajous figure analysis.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 37. (Click on arrow for answer)

One way to vary the amount of power delivered to a resistive AC load is by varying another resistance connected in series:

Should be image 01829x01

A problem with this power control strategy is that power is wasted in the series resistance (I^2R_{series}). A different strategy for controlling power is shown here, using a series inductance rather than resistance:

Should be image 01829x02

Explain why the latter circuit is more power-efficient than the former, and draw a phasor diagram showing how changes in L_{series} affect Z_{total}.

File Num: 01829

Answer

Inductors are reactive rather than resistive components, and therefore do not dissipate power (ideally).

Should be image 01829x03

Follow-up question: the inductive circuit is not just more energy-efficient — it is safer as well. Identify a potential safety hazard that the resistive power-control circuit poses due to the energy dissipation of its variable resistor.


Notes

If appropriate, you may want to mention devices called saturable reactors, which are used to control power in AC circuits by the exact same principle: varying a series inductance.





Question 38. (Click on arrow for answer)

A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = {1 \over R}), and it is measured in the unit of siemens.

Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

Should be image 00853x01

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = {1 \over X}), and the reciprocal of impedance is called admittance (Y = {1 \over Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.

Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Should be image 00853x02

File Num: 00853

Answer

Y_{total} = \sqrt{G^2 + B^2}

Follow-up question \#1: draw a phasor diagram showing how Y, G, and B relate.


Follow-up question \#2: re-write this equation using quantities of resistance (R), reactance (X), and impedance (Z), instead of conductance (G), susceptance (B), and admittance (Y).


Notes

Ask your students if this equation looks familiar to them. It should!


The answer to the second follow-up question is a matter of algebraic substitution. Work through this process with your students, and then ask them to compare the resulting equation with other equations they’ve seen before. Does its form look familiar to them in any way?





Question 39. (Click on arrow for answer)

Students studying AC electrical theory become familiar with the impedance triangle very soon in their studies:

Should be image 02077x01

What these students might not ordinarily discover is that this triangle is also useful for calculating electrical quantities other than impedance. The purpose of this question is to get you to discover some of the triangle’s other uses.

Fundamentally, this right triangle represents phasor addition, where two electrical quantities at right angles to each other (resistive versus reactive) are added together. In series AC circuits, it makes sense to use the impedance triangle to represent how resistance (R) and reactance (X) combine to form a total impedance (Z), since resistance and reactance are special forms of impedance themselves, and we know that impedances add in series.

List all of the electrical quantities you can think of that add (in series or in parallel) and then show how similar triangles may be drawn to relate those quantities together in AC circuits.

File Num: 02077

Answer

Electrical quantities that add:
  • Series impedances
  • Series voltages
  • Parallel admittances
  • Parallel currents
  • Power dissipations

I will show you one graphical example of how a triangle may relate to electrical quantities other than series impedances:

Should be image 02077x02

Notes

It is very important for students to understand that the triangle only works as an analysis tool when applied to quantities that add. Many times I have seen students try to apply the ZRX impedance triangle to parallel circuits and fail because parallel impedances do not add. The purpose of this question is to force students to think about where the triangle is applicable to AC circuit analysis, and not just to use it blindly.

The power triangle is an interesting application of trigonometry applied to electric circuits. You may not want to discuss power with your students in great detail if they are just beginning to study voltage and current in AC circuits, because power is a sufficiently confusing subject on its own.





Question 40. (Click on arrow for answer)

Explain why the “impedance triangle” is not proper to use for relating total impedance, resistance, and reactance in parallel circuits as it is for series circuits:

Should be image 02078x01

File Num: 02078

Answer

Impedances do not add in parallel.


Follow-up question: what kind of a triangle could be properly applied to a parallel AC circuit, and why?


Notes

Trying to apply the ZRX triangle directly to parallel AC circuits is a common mistake many new students make. Key to knowing when and how to use triangles to graphically depict AC quantities is understanding why the triangle works as an analysis tool and what its sides represent.





Question 41. (Click on arrow for answer)

Calculate the total impedance for these two 100 mH inductors at 2.3 kHz, and draw a phasor diagram showing circuit admittances (Y_{total}, G, and B):

Should be image 02079x01

Now, re-calculate impedance and re-draw the phasor admittance diagram supposing the second inductor is replaced by a 1.5 k\Omega resistor:

Should be image 02079x02

File Num: 02079

Answer

Should be image 02079x03Should be image 02079x04

Challenge question: why are the susceptance vectors (B_{L1} and B_{L2}) pointed down instead of up as impedance vectors for inductances typically are?


Notes

Phasor diagrams are powerful analytical tools, if one knows how to draw and interpret them. With hand calculators being so powerful and readily able to handle complex numbers in either polar or rectangular form, there is temptation to avoid phasor diagrams and let the calculator handle all the angle manipulation. However, students will have a much better understanding of phasors and complex numbers in AC circuits if you hold them accountable to representing quantities in that form.





Question 42. (Click on arrow for answer)

Calculate the individual currents through the inductor and through the resistor, the total current, and the total circuit impedance:

Should be image 02104x01

Also, draw a phasor diagram showing how the individual component currents relate to the total current.

File Num: 02104

Answer

I_L = 530.5 \muA RMSI_R = 490.2 \muA RMSI_{total} = 722.3 \muA RMSZ_{total} = 3.461 k\Omega

Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 43. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01839x01

Calculate the current necessary to power this motor if the equivalent resistance and inductance is 20 \Omega and 238 mH, respectively.

File Num: 01839

Answer

I_{supply} = 12.29 \hbox{ A}

Notes

This is a practical example of a parallel LR circuit, as well as an example of how complex electrical devices may be “modeled” by collections of ideal components. To be honest, a loaded AC motor’s characteristics are quite a bit more complex than what the parallel LR model would suggest, but at least it’s a start!


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 44. (Click on arrow for answer)

A large AC electric motor under load can be considered as a parallel combination of resistance and inductance:

Should be image 01840x01

Calculate the equivalent inductance (L_{eq}) if the measured source current is 27.5 amps and the motor’s equivalent resistance (R_{eq}) is 11.2 \Omega.

File Num: 01840

Answer

L_{eq} = 61.11 \hbox{ mH}

Notes

Here is a case where scalar calculations (R, G, X, B, Y) are much easier than complex number calculations (all Z) would be.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 45. (Click on arrow for answer)

Determine the total current and all component currents in this circuit, stating your answers the way a multimeter would register them:

Should be image 01842x01
  • L_1 = 1.2 \hbox{ H}
  • L_2 = 650 \hbox{ mH}
  • R_1 = 33 \hbox{ k}\Omega
  • R_2 = 27 \hbox{ k}\Omega
  • V_{supply} = 19.7 \hbox{ V RMS}
  • f_{supply} = 4.5 \hbox{ kHz}

Also, calculate the phase angle (\Theta) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.

File Num: 01842

Answer

  • I_{total} = 2.12 \hbox{ mA}
  • I_{L1} = 581 \> \mu \hbox{A}
  • I_{L2} = 1.07 \hbox{ mA}
  • I_{R1} = 597 \> \mu \hbox{A}
  • I_{R2} = 730 \> \mu \hbox{A}
  • \Theta = 51.24^o

Measuring \Theta with an oscilloscope requires the addition of a shunt resistor into this circuit, because oscilloscopes are (normally) only able to measure voltage, and there is no phase shift between any voltages in this circuit because all components are in parallel. I leave it to you to suggest where to insert the shunt resistor, what resistance value to select for the task, and how to connect the oscilloscope to the modified circuit.


Notes

Some students many wonder what type of numerical result best corresponds to a multimeter’s readings, if they do their calculations using complex numbers (“do I use polar or rectangular form, and if rectangular do I use the real or the imaginary part?”). The answers given for this question should clarify that point.

It is very important that students know how to apply this knowledge of AC circuit analysis to real-world situations. Asking students to determine how they would connect an oscilloscope to the circuit to measure \Theta is an exercise in developing their abstraction abilities between calculations and actual circuit scenarios.


Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them. The following is a sample of a written problem-solving strategy for analyzing a series resistive-reactive AC circuit:


Step 1: Calculate all reactances (X).Step 2: Draw an impedance triangle (Z ; R ; X), solving for ZStep 3: Calculate circuit current using Ohm’s Law: I = {V \over Z}Step 4: Calculate series voltage drops using Ohm’s Law: V = {I Z}Step 5: Check work by drawing a voltage triangle (V_{total} ; V_1 ; V_2), solving for V_{total}

By having students outline their problem-solving strategies, everyone gets an opportunity to see multiple methods of solution, and you (the instructor) get to see how (and if!) your students are thinking. An especially good point to emphasize in these “open thinking” activities is how to check your work to see if any mistakes were made.





Question 46. (Click on arrow for answer)

Calculate the total impedances (complete with phase angles) for each of the following inductor-resistor circuits:

Should be image 02106x01

File Num: 02106

Answer

Should be image 02106x02

Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.





Question 47. (Click on arrow for answer)

A doorbell ringer has a solenoid with an inductance of 63 mH connected in parallel with a lamp (for visual indication) having a resistance of 150 ohms:

Should be image 02105x01

Calculate the phase shift of the total current (in units of degrees) in relation to the total supply voltage, when the doorbell switch is actuated.

File Num: 02105

Answer

\Theta = 81 degrees

Suppose the lamp turned on whenever the pushbutton switch was actuated, but the doorbell refused to ring. Identify what you think to be the most likely fault which could account for this problem.


Notes

This would be an excellent question to have students present methods of solution for. Sometimes I have students present nothing but their solution steps on the board in front of class (no arithmetic at all), in order to generate a discussion on problem-solving strategies. The important part of their education here is not to arrive at the correct answer or to memorize an algorithm for solving this type of problem, but rather how to think like a problem-solver, and how to methodically apply the math they know to the problem(s) at hand.





Question 48. (Click on arrow for answer)

An AC electric motor operating under loaded conditions draws a current of 11 amps (RMS) from the 120 volt (RMS) 60 Hz power lines. The measured phase shift between voltage and current for this motor is 34^{o}, with voltage leading current.

Determine the equivalent parallel combination of resistance (R) and inductance (L) that is electrically equivalent to this operating motor.

File Num: 01542

Answer

R_{parallel} = 13.16 \Omega
L_{parallel} = 51.75 mH

Challenge question: in the parallel LR circuit, the resistor will dissipate a lot of energy in the form of heat. Does this mean that the electric motor, which is electrically equivalent to the LR network, will dissipate the same amount of heat? Explain why or why not.


Notes

If students get stuck on the challenge question, remind them that an electric motor does mechanical work, which requires energy.





Question 49. (Click on arrow for answer)

Calculate the impedance of a 145 mH inductor connected in series with 750 \Omega resistor at a frequency of 1 kHz, then determine the necessary resistor and inductor values to create the exact same total impedance in a parallel configuration.

File Num: 00645

Answer

Z_{total} = 1.18 k\Omega \angle 50.54^{o}

If connected in parallel: R = 1.857 k\Omega ; L = 243.3 mH.


Hint: if you are having difficulty figuring out where to start in answering this question, consider the fact that these two circuits, if equivalent in total impedance, will draw the exact same amount of current from a common AC source at 1 kHz.


Notes

This is an interesting question, requiring the student to think creatively about how to convert one configuration of circuit into another, while maintaining the same total effect. As usual, the real purpose of a question like this is to develop problem-solving strategies, rather than to simply obtain an answer.





Question 50. (Click on arrow for answer)

It is often useful in AC circuit analysis to be able to convert a series combination of resistance and reactance into an equivalent parallel combination of conductance and susceptance, or visa-versa:

Should be image 00856x01

We know that resistance (R), reactance (X), and impedance (Z), as scalar quantities, relate to one another trigonometrically in a series circuit. We also know that conductance (G), susceptance (B), and admittance (Y), as scalar quantities, relate to one another trigonometrically in a parallel circuit:

Should be image 00856x02

If these two circuits are truly equivalent to one another, having the same total impedance, then their representative triangles should be geometrically similar (identical angles, same proportions of side lengths). With equal proportions, {R \over Z} in the series circuit triangle should be the same ratio as {G \over Y} in the parallel circuit triangle, that is {R \over Z} = {G \over Y}.

Building on this proportionality, prove the following equation to be true:

R_{series} R_{parallel} = {Z_{total}}^2

After this, derive a similar equation relating the series and parallel reactances (X_{series} and X_{parallel}) with total impedance (Z_{total}).

File Num: 00856

Answer

I’ll let you figure out how to turn {R \over Z} = {G \over Y} into R_{series} R_{parallel} = {Z_{total}}^2 on your own!


As for the reactance relation equation, here it is:

X_{series} X_{parallel} = {Z_{total}}^2

Notes

Being able to convert between series and parallel AC networks is a valuable skill for analyzing complex series-parallel combination circuits, because it means any series-parallel combination circuit may then be converted into an equivalent simple-series or simple-parallel, which is mush easier to analyze.

Some students might ask why the conductance/susceptance triangle is “upside-down” compared to the resistance/reactance triangle. The reason has to do with the sign reversal of imaginary quantities when inverted: {1 \over j} = -j. The phase angle of a pure inductance’s impedance is +90 degrees, while the phase angle of the same (pure) inductance’s admittance is -90 degrees, due to reciprocation. Thus, while the X leg of the resistance/reactance triangle points up, the B leg of the conductance/susceptance triangle must point down.





Question 51. (Click on arrow for answer)

Determine an equivalent parallel RC network for the series RC network shown on the left:

Should be image 01540x01

Note that I have already provided a value for the capacitor’s reactance (X_C), which of course will be valid only for a particular frequency. Determine what values of resistance (R) and reactance (X_C) in the parallel network will yield the exact same total impedance (Z_T) at the same signal frequency.

File Num: 01540

Answer

R = 150 \Omega
X_C = 200 \Omega

Follow-up question: explain how you could check your conversion calculations, to ensure both networks are truly equivalent to each other.


Notes

This problem just happens to work out with whole numbers. Believe it or not, I chose these numbers entirely by accident one day, when setting up an example problem to show a student how to convert between series and parallel equivalent networks!





Question 52. (Click on arrow for answer)

Determine the equivalent parallel-connected resistor and inductor values for this series circuit:

Should be image 00855x01

Also, express the total impedance of either circuit (since they are electrically equivalent to one another, they should have the same total impedance) in complex form. That is, express Z as a quantity with both a magnitude and an angle.

File Num: 00855

Answer