## 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:

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:

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:

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

*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?

#### 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):

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:

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:

File Num: 01853

#### Answer

\Theta = 26.51^oChallenge question: explain how the following phasor diagram was determined for this problem:

#### Notes

This is an interesting question for a couple of reasons. First, students must determine how they will measure phase shift with just the two voltage indications shown by the meters. This may present a significant challenge for some. Discuss problem-solving strategies in class so that students understand how and why it is possible to determine \Theta.

Secondly, this is an interesting question because it shows how something as abstract as phase angle can be measured with just a voltmeter — no oscilloscope required! Not only that, but we don’t even have to know the component values either! Note that this technique works only for simple circuits.

A practical point to mention here is that multimeters have frequency limits which must be considered when taking measurements on electronic circuits. Some high-quality handheld digital meters have frequency limits of hundred of kilohertz, while others fail to register accurately at only a few thousand hertz. Unless we knew these two digital voltmeters were sufficient for measuring at the signal frequency, their indications would be useless to us.

#### 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:

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

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

File Num: 01847

#### Answer

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:

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:

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^oFollow-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:

- 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:

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^oFollow-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.

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}

#### 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:

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

*TRIAC*. All portions of the circuit except for the RC network are “shaded” for de-emphasis:

#### 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:

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called *susceptance* (B = {1 \over X}), and the reciprocal of impedance is called *admittance* (Y = {1 \over Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.

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

File Num: 01845

*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.

*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.

#### 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}

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):

Also, draw an admittance triangle for this circuit.

File Num: 02108

#### Answer

Z_{total} = 391.4 \Omega \angle -39.9^{o}#### 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:

File Num: 02109

#### Answer

#### Notes

Have your students explain how they solved for each impedance, step by step. You may find different approaches to solving the same problem(s), and your students will benefit from seeing the diversity of solution techniques.

#### 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:

- \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:

Calculate the magnitude and phase shift of the current drawn by this real capacitor, if powered by a sinusoidal voltage source of 30 volts RMS at 400 Hz:

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

File Num: 01850

*leakage resistance*, and it is modeled as a shunt resistance to an ideal capacitance:

#### 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.

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}

#### 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:

Typical input impedance for an oscilloscope is 1 M\Omega of resistance, in parallel with a small amount of capacitance. At low frequencies, the reactance of this capacitance is so high that it may be safely ignored. At high frequencies, though, it may become a substantial load to the circuit under test:

Calculate how many ohms of impedance this oscilloscope input (equivalent circuit shown in the above schematic) will impose on a circuit with a signal frequency of 150 kHz.

File Num: 02111

*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:

#### Answer

Z_{input} = 52.98 k\Omega at 150 kHzFollow-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:

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:

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:

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:

File Num: 03288

#### Answer

#### Notes

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

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

#### Question 24. (Click on arrow for answer)

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

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

File Num: 01828

#### Answer

**Scalar calculations**R = 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 calculations**Z_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:

File Num: 01838

#### Answer

**Scalar calculations**R_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 calculations**R_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:

He recognizes this as a series circuit, and therefore realizes that a right triangle would be appropriate for representing component impedances and component voltage drops (because both impedance and voltage are quantities that add in series, and the triangle represents phasor addition):

The problem now is, which angle does the student solve for in order to find the phase shift of V_{out}? The triangle contains two angles besides the 90^{o} angle, \Theta and \Phi. Which one represents the output phase shift, and more importantly, *why*?

File Num: 03748

*why*?

#### 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):

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):

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}:

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}:

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}:

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}:

File Num: 02625

#### Answer

f = 929 Hz#### Notes

#### Question 33. (Click on arrow for answer)

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

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

File Num: 03284

#### Answer

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}:

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}:

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:

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

__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:

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

__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):

File Num: 03611

#### Answer

#### Notes

Ask your students to share their problem-solving techniques for this question: how they solved for each parameter and in what order they performed the calculations.

#### 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:

Suppose, though, that the output signal is stuck at +45^{o} leading the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03464

#### Answer

A broken connection between the left-hand terminal of the potentiometer and the bridge could cause this to happen:

I’ll let you figure out why!

#### Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is *much* greater than the fixed resistance value.”

#### 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:

Suppose, though, that the output signal is stuck at -45^{o} lagging the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03465

#### Answer

A broken connection between the right-hand terminal of the potentiometer and the bridge could cause this to happen:

I’ll let you figure out why!

#### Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is *much* greater than the fixed resistance value.”

#### 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:

Suppose, though, that the output signal registers as it should with the potentiometer wiper fully to the right, but diminishes greatly in amplitude as the wiper is moved to the left, until there is practically zero output voltage at the full-left position. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.

File Num: 03466

#### Answer

An open failure of the fixed resistor in the upper-left arm of the bridge could cause this to happen:

Follow-up question: identify another possible component failure that would exhibit the same symptoms.

#### Notes

It is essential, of course, that students understand the operational principle of this circuit before they may even attempt to diagnose possible faults. You may find it necessary to discuss this circuit in detail with your students before they are ready to troubleshoot it.

In case anyone asks, the symbolism R_{pot} >> R means “potentiometer resistance is *much* greater than the fixed resistance value.”

#### 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:

Hint: use the Superposition Theorem!

File Num: 03503

#### Answer

Follow-up question: explain why it is acceptable to use a polarized (polarity-sensitive) capacitor in this circuit when it is clearly connected to a source of AC. Why is it not damaged by the AC voltage when used like this?

#### Notes

Note that the capacitor size has been chosen for negligible capacitive reactance (X_C) at the specific frequency, such that the 10 k\Omega DC bias resistors present negligible loading to the coupled AC signal. This is typical for this type of biasing circuit.

Aside from giving students an excuse to apply the Superposition Theorem, this question previews a circuit topology that is extremely common in transistor amplifiers.

#### 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:

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.

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