## 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, 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: File Num: 00851

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): File Num: 00852

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): File Num: 00854

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: File Num: 01541

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? Now, suppose we take these same components and re-connect them in parallel rather than series. Does this change the circuit’s overall “appearance” to the source? Does the source now “see” an equivalent capacitor or an equivalent inductor? Explain your answer. File Num: 01554

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

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

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: We know that resistance (R), reactance (X), and impedance (Z), as scalar quantities, relate to one another trigonometrically in a series circuit. We also know that conductance (G), susceptance (B), and admittance (Y), as scalar quantities, relate to one another trigonometrically in a parallel circuit: If these two circuits are truly equivalent to one another, having the same total impedance, then their representative triangles should be geometrically similar (identical angles, same proportions of side lengths). With equal proportions, {R \over Z} in the series circuit triangle should be the same ratio as {G \over Y} in the parallel circuit triangle, that is {R \over Z} = {G \over Y}.

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

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

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

File Num: 00856

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: 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

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: 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

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: 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

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

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: 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 #### 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: 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 #### 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? File Num: 01072

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: File Num: 02119

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: File Num: 01864

Equivalent series resistance and reactances: Z_{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: File Num: 01865

Equivalent parallel resistance and reactances: Z_{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: Hint: convert the parallel RC sub-network into a series equivalent first.

File Num: 02115

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

File Num: 02116

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: When measuring high-frequency AC voltages, however, the parasitic capacitance and inductance of the coaxial cable may present problems. We may represent these distributed characteristics of the cable as “lumped” parameters: a single capacitor and a single inductor modeling the cable’s behavior: Typical parasitic values for a 10-foot cable would be 260 pF of capacitance and 650 \muH of inductance. The voltmeter itself, of course, is not without its own inherent impedances, either. For the sake of this example, let’s consider the meter’s “input impedance” to be a simple resistance of 1 M\Omega.

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

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

File Num: 00641

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

While this technique works very well when measuring DC voltage, it does not do so well when measuring AC voltage, due to the parasitic capacitance of the cable connecting the test probes to the instrument (parasitic cable inductance has been omitted from this diagram for simplicity): To see the effects of this capacitance for yourself, calculate the voltage at the instrument input terminals assuming a parasitic capacitance of 180 pF and an AC voltage source of 10 volts, for the following frequencies:

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

The debilitating effect of cable capacitance may be compensated for with the addition of another capacitor, connected in parallel with the 9 M\Omega range resistor. If we are trying to maintain a voltage division ratio of 10:1, this “compensating” capacitor must be {1 \over 9} the value of the capacitance parallel to the instrument input: Re-calculate the voltage at the instrument input terminals with this compensating capacitor in place. You should notice quite a difference in instrument voltages across this frequency range!

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

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

File Num: 00642

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.