Practice Problems – Inductors in AC

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:

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

File Num: 00576

Answer

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

Notes

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

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:

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 \> \Omega

Z_{total} = 414.7 \> \Omega \> \angle \> 90^o

Z_{total} = 0 + j414.7 \> \Omega

Second strategy:

X_{L1} = 282.7 \> \Omega

Z_{L1} = 282.7 \> \Omega \> \angle \> 90^o

X_{L2} = 131.9 \> \Omega

Z_{L2} = 131.9 \> \Omega \> \angle \> 90^o

Z_{total} = 414.7 \> \Omega \> \angle \> 90^o

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):

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:

File Num: 01837

Answer

Scalar calculations

R_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 calculations

R_1 = 1.5 \hbox{ k}\Omega

Z_{R1} = 1.5 \hbox{ k}\Omega \> \angle \> 0^o

X_{L1} = 2.513 \hbox{ k}\Omega

Z_{L1} = 2.513 \hbox{ k}\Omega \> \angle \> 90^o

Z_{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:

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 \> \Omega

Z_{total} = 295.0 \> \Omega \> \angle \> 90^o

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^o

X_{L2} = 1.357 \hbox{ k}\Omega

Z_{L2} = 1.357 \hbox{ k}\Omega \> \angle \> 90^o

Z_{total} = 295.0 \> \Omega \> \angle \> 90^o

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

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.

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

Question 12. (Click on arrow for answer)

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

File Num: 01552

Answer

I = 15.6

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:

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:

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 Z

I = {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 = IZ

I = {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

^{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 $^{o}$ :

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 kHz

f = {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 $^{o}$ :

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

\Omega

R = 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.

Question 19. (Click on arrow for answer)

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

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

File Num: 03280

Answer

f

= 2.804 kHz

f = - {{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.

Question 20. (Click on arrow for answer)

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

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

File Num: 03281

Answer

R

= 2.902 k

\Omega

R = - {{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.

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:

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

File Num: 00631

Answer

Notes

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

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:

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

File Num: 00632

Answer

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

Notes

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

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

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

Question 23. (Click on arrow for answer)

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

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

File Num: 02076

Answer

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

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

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

Notes

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

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 ( $X$ ):

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 $\Omega$ :

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)

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

{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.

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:

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 } Z

X \hbox{ and } R

X \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)

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

Question 30. (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 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

Scalar calculations

R = 2.2 \hbox{ k}\Omega

X_L = 1.495 \hbox{ k}\Omega

Z_{series} = \sqrt{R^2 + {X_L}^2}

Z_{series} = \sqrt{2200^2 + 1495^2} = 2660 \> \Omega

Complex number calculations

Z_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^o

Z_{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 $X$ ):

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

File Num: 02080

Answer

Notes

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

Question 32. (Click on arrow for answer)

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

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

File Num: 02103

Answer

Z_{total}

= 6.944 k

\Omega

I

= 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:

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:

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.

Step 1: Calculate all reactances (

X

).Step 2: Draw an impedance triangle (

Z

;

R

;

X

), solving for

Z

Step 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 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:

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.

Step 1: Calculate all reactances (

X

).Step 2: Draw an impedance triangle (

Z

;

R

;

X

), solving for

Z

Step 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 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:

$L_1 = 250 \hbox{ mH}$
$L_2 = 60 \hbox{ mH}$
$R_1 = 6.8 \hbox{ k}\Omega$
$R_2 = 1.2 \hbox{ k}\Omega$
$V_{supply} = 13.4 \hbox{ V RMS}$
$f_{supply} = 6.5 \hbox{ kHz}$

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.

Step 1: Calculate all reactances (

X

).Step 2: Draw an impedance triangle (

Z

;

R

;

X

), solving for

Z

Step 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 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:

A problem with this power control strategy is that power is wasted in the series resistance ( $I^2R_{series}$ ). A different strategy for controlling power is shown here, using a series inductance rather than resistance:

Explain why the latter circuit is more power-efficient than the former, and draw a phasor diagram showing how changes in $L_{series}$ affect $Z_{total}$ .

File Num: 01829

Answer

Inductors are reactive rather than resistive components, and therefore do not dissipate power (ideally).

Follow-up question: the inductive circuit is not just more energy-efficient — it is safer as well. Identify a potential safety hazard that the resistive power-control circuit poses due to the energy dissipation of its variable resistor.

Notes

If appropriate, you may want to mention devices called saturable reactors, which are used to control power in AC circuits by the exact same principle: varying a series inductance.

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:

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

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

File Num: 00853

Answer

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:

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:

Notes

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

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

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:

File Num: 02078

Answer

Impedances do not add in parallel.

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

Notes

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

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 $B$ ):

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

File Num: 02079

Answer

Challenge question: why are the susceptance vectors ( $B_{L1}$ and $B_{L2}$ ) pointed down instead of up as impedance vectors for inductances typically are?

Notes

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:

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

File Num: 02104

Answer

I_L

= 530.5

\mu

A RMS

I_R

= 490.2

\mu

A RMS

I_{total}

= 722.3

\mu

A RMS

Z_{total}

= 3.461 k

\Omega

Notes

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:

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!

Step 1: Calculate all reactances (

X

).Step 2: Draw an impedance triangle (

Z

;

R

;

X

), solving for

Z

Step 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 44. (Click on arrow for answer)

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

Calculate the equivalent inductance ( $L_{eq}$ ) if the measured source current is 27.5 amps and the motor’s equivalent resistance ( $R_{eq}$ ) is 11.2 $\Omega$ .

File Num: 01840

Answer

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.

Step 1: Calculate all reactances (

X

).Step 2: Draw an impedance triangle (

Z

;

R

;

X

), solving for

Z

Step 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 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:

$L_1 = 1.2 \hbox{ H}$
$L_2 = 650 \hbox{ mH}$
$R_1 = 33 \hbox{ k}\Omega$
$R_2 = 27 \hbox{ k}\Omega$
$V_{supply} = 19.7 \hbox{ V RMS}$
$f_{supply} = 4.5 \hbox{ kHz}$

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

Step 1: Calculate all reactances (

X

).Step 2: Draw an impedance triangle (

Z

;

R

;

X

), solving for

Z

Step 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 46. (Click on arrow for answer)

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

File Num: 02106

Answer

Notes

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

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:

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

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:

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

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

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

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

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

File Num: 00856

Answer

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

As for the reactance relation equation, here it is:

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:

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:

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

File Num: 00855

Answer

R_{parallel} =

2092

\Omega

L_{parallel} =

1.325 H

Z_{total} =

1772

\Omega

\angle

32.14

^{o}

Notes

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

Question 53. (Click on arrow for answer)

Convert this series-parallel combination circuit into an equivalent simple-parallel circuit (all components connected in parallel with each other, with nothing in series), and also calculate the circuit’s total impedance:

File Num: 00857

Answer

Z_{total} =

963.0

\Omega

Challenge question: from the simple-parallel equivalent circuit shown here, can you generate an equivalent circuit that is simple-series? In other words, can you calculate the proper values of $R$ and $L$ , that when connected in series, will have the same total impedance as this circuit?

Notes

Fundamentally, this question asks students to generate an equivalent parallel R-X circuit from a given series R-X circuit. In this particular circuit, there are two series-connected R-X branches, resulting in an equivalent parallel circuit with four branches.

Calculating the circuit’s total impedance as a scalar figure involves simplifying the circuit once more into two components: a resistance and a reactance.

Question 54. (Click on arrow for answer)

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

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

File Num: 00859

Answer

Notes

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

Question 55. (Click on arrow for answer)

We know that any given impedance may be represented by a simple, two-component circuit: either a resistor and a reactive component connected in series, or a resistor and a reactive component connected in parallel. Assuming a circuit frequency of 2 kHz, determine what combination of series-connected components will be equivalent to this “box” impedance, and also what combination of parallel-connected components will be equivalent to this “box” impedance.

File Num: 03296

Answer

Notes

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