## Practice Problems: Complex Numbers and Phasors

### Difficult Concepts

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

#### Resistance vs. Reactance vs. Impedance

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

#### Phasors, used to represent AC amplitude and phase relations.

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

#### Conductance, Susceptance, and Admittance.

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

#### Question 1. (Click on arrow for answer)

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

File Num: 03326

#### Answer

x = 10#### Notes

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

#### Question 2. (Click on arrow for answer)

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

File Num: 03327

#### Answer

x = 15#### Notes

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

#### Question 3. (Click on arrow for answer)

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

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

File Num: 02102

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

#### Answer

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

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

#### Notes

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

#### Question 4. (Click on arrow for answer)

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

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

File Num: 03114

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

#### Answer

Standard form of the Pythagorean Theorem:

C = \sqrt{A^2 + B^2}Solving for A:

A = \sqrt{C^2 - B^2}Solving for B:

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

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

#### Question 5. (Click on arrow for answer)

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 6. (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 “Phi” (\phi):

{R \over X} = {X \over Z} = {R \over Z} = File Num: 03113

#### Answer

{R \over X} = \tan \phi = {\hbox{Opposite} \over \hbox{Adjacent}}{X \over Z} = \cos \phi = {\hbox{Adjacent} \over \hbox{Hypotenuse}}{R \over Z} = \sin \phi = {\hbox{Opposite} \over \hbox{Hypotenuse}}#### Notes

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

#### Question 7. (Click on arrow for answer)

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

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

*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.

*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.

#### Answer

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

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

#### Question 8. (Click on arrow for answer)

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

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

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

*why*a right triangle is an appropriate form to relate these variables, and what each side of the triangle actually represents.

#### 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 9. (Click on arrow for answer)

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

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

File Num: 02081

#### Answer

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

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

#### Question 10. (Click on arrow for answer)

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

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

*impedance triangle*very soon in their studies:

*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.

#### 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 11. (Click on arrow for answer)

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

File Num: 02078

*not*proper to use for relating total impedance, resistance, and reactance in parallel circuits as it is for series circuits:

#### 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 12. (Click on arrow for answer)

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

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 13. (Click on arrow for answer)

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

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

File Num: 02107

#### Answer

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

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

#### Question 14. (Click on arrow for answer)

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

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

File Num: 02082

#### Answer

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

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

#### Question 15. (Click on arrow for answer)

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

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

#### Question 16. (Click on arrow for answer)

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

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

File Num: 02092

#### Answer

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

#### Question 17. (Click on arrow for answer)

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

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 *arc*functions.

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

R \hbox{ and } ZX \hbox{ and } RX \hbox{ and } Z
Show how three different trigonometric arcfunctions may be used to calculate the same angle \Theta.

File Num: 02085

*arc*functions.

#### Answer

\arccos {R \over Z} = 53.13^o\arctan {X \over R} = 53.13^o\arcsin {X \over Z} = 53.13^o
Challenge question: identify three *more* arcfunctions which could be used to calculate the same angle \Theta.

#### Notes

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

#### Question 18. (Click on arrow for answer)

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

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

\sin 65^o = {X \over Z}
Solve this equation for the value of X, in ohms.

File Num: 02088

*sine*function relates the sides X and Z of this impedance triangle with the 65 degree angle, because the sine of an angle is the ratio of

*opposite*to

*hypotenuse*, with X being opposite the 65 degree angle. Therefore, we know we can set up the following equation relating these quantities together:

#### Answer

X = 9.063 k\Omega#### Notes

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

#### Question 19. (Click on arrow for answer)

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

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

File Num: 02087

#### Answer

R = 2.165 k\OmegaX = 1.25 k\Omega#### Notes

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

#### Question 20. (Click on arrow for answer)

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

File Num: 04034

*phasor*, as the term applies to electrical calculations.

#### Answer

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

#### Notes

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

#### Question 21. (Click on arrow for answer)

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

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

Written out, we might express the equivalence of these two notations as such:

{\sqrt{3} \over 2} + j{1 \over 2} = 1 \angle {\pi \over 6}\vskip 30pt
Expressed in a more general form, the equivalence between rectangular and polar notations would look like this:

a + jb = c \angle \Theta
However, a problem with the “angle” symbol (\angle) is that we have no standardized way to deal with it mathematically. We would have to invent special rules to describe how to add, subtract, multiply, divide, differentiate, integrate, or otherwise manipulate complex quantities expressed using this symbol. A more profitable alternative to using the “angle” symbol is shown here:

a + jb = c e^{j\Theta}
Explain why this equivalence is mathematically sound.

File Num: 04059

*rectangular*and

*polar*. Take for example the complex quantity {\sqrt{3} \over 2} + j{1 \over 2}. The following illustration shows this point located on the complex plane, along with its rectangular dimensions:

#### Answer

The equivalence shown is based on *Euler’s relation*, which is left to you as an exercise to prove.

#### Notes

This question should probably be preceded by \#04058, which asks students to explore the relationship between the infinite series for e^x, \cos x, and \sin x. In any case, your students will need to know Euler’s relation:

e^{jx} = \cos x + j \sin x#### Question 22. (Click on arrow for answer)

Electrical engineers usually express the frequency of an AC circuit in terms of *angular velocity*, measured in units of *radians per second* rather than cycles per second (Hertz, or Hz).

First, explain what a *radian* is. Next, write an equation relating frequency (f) in Hertz to angular velocity (\omega) in radians per second. Hint: the relationship between the two is perhaps most easily understood in terms of a two-pole AC generator, or alternator, where each revolution of the rotor generates one full cycle of AC.

File Num: 04060

*angular velocity*, measured in units of

*radians per second*rather than cycles per second (Hertz, or Hz).

*radian*is. Next, write an equation relating frequency (f) in Hertz to angular velocity (\omega) in radians per second. Hint: the relationship between the two is perhaps most easily understood in terms of a two-pole AC generator, or alternator, where each revolution of the rotor generates one full cycle of AC.

#### Answer

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

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

\omega = 2 \pi f#### Notes

Personally, I find the rotating alternator model the best way to comprehend the relationship between angular velocity and frequency. If each turn of the rotor is one cycle (2 \pi radians), and frequency is cycles per second, then one revolution per second will be 1 Hertz, which will be 2 \pi radians per second.

#### Question 23. (Click on arrow for answer)

Suppose two people work together to slide a large box across the floor, one pushing with a force of 400 newtons and the other pulling with a force of 300 newtons:

The resultant force from these two persons’ efforts on the box will, quite obviously, be the sum of their forces: 700 newtons (to the right).

What if the person pulling decides to change position and push *sideways* on the box in relation to the first person, so the 400 newton force and the 300 newton force will be perpendicular to each other (the 300 newton force facing into the page, away from you)? What will the resultant force on the box be then?

File Num: 03278

*sideways*on the box in relation to the first person, so the 400 newton force and the 300 newton force will be perpendicular to each other (the 300 newton force facing into the page, away from you)? What will the resultant force on the box be then?

#### Answer

The resultant force on the box will be 500 newtons.

#### Notes

This is a non-electrical application of vector summation, to prepare students for the concept of using vectors to add voltages that are out-of-phase. Note how I chose to use multiples of 3, 4, and 5 for the vector magnitudes.

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

Special types of vectors called *phasors* are often used to depict the magnitude and phase-shifts of sinusoidal AC voltages and currents. Suppose that the following phasors represent the series summation of two AC voltages, one with a magnitude of 3 volts and the other with a magnitude of 4 volts:

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

Also explain the significance of these sums: that we may obtain three *different* values of total voltage (7 volts, 1 volt, or 5 volts) from the same series-connected AC voltages. What does this mean for us as we prepare to analyze AC circuits using the rules we learned for DC circuits?

File Num: 01559

*phasors*are often used to depict the magnitude and phase-shifts of sinusoidal AC voltages and currents. Suppose that the following phasors represent the series summation of two AC voltages, one with a magnitude of 3 volts and the other with a magnitude of 4 volts:

*different*values of total voltage (7 volts, 1 volt, or 5 volts) from the same series-connected AC voltages. What does this mean for us as we prepare to analyze AC circuits using the rules we learned for DC circuits?

#### Answer

Each of the phasor diagrams represents two AC voltages being added together. The dotted phasor represents the sum of the 3-volt and 4-volt signals, for different conditions of phase shift between them.

Please note that these three possibilities are not exhaustive! There are a multitude of other possible total voltages that the series-connected 3 volt and 4 volt sources may create.

Follow-up question: in DC circuits, it is permissible to connect multiple voltage sources in parallel, so long as the voltages (magnitudes) and polarities are the same. Is this also true for AC? Why or why not?

#### Notes

Be sure to discuss with your students that these three conditions shown are not the only conditions possible! I simply chose 0^{o}, 180^{o}, and 90^{o} because they all resulted in round sums for the given quantities.

The follow-up question previews an important subject concerning AC phase: the necessary *synchronization* or paralleled AC voltage sources.

#### Question 25. (Click on arrow for answer)

When drawing phasor diagrams, there is a standardized orientation for all angles used to ensure consistency between diagrams. This orientation is usually referenced to a set of perpendicular lines, like the x and y axes commonly seen when graphing algebraic functions:

The intersection of the two axes is called the *origin*, and straight horizontal to the right is the definition of zero degrees (0^{o}). Thus, a phasor with a magnitude of 6 and an angle of 0^{o} would look like this on the diagram:

Draw a phasor with a magnitude of 10 and an angle of 100 degrees on the above diagram, as well as a phasor with a magnitude of 2 and an angle of -45 degrees. Label what directions 90^{o}, 180^{o}, and 270^{o} would indicate on the same diagram.

File Num: 02099

*origin*, and straight horizontal to the right is the definition of zero degrees (0^{o}). Thus, a phasor with a magnitude of 6 and an angle of 0^{o} would look like this on the diagram:

#### Answer

#### Notes

Graph paper, a ruler, and a protractor may be helpful for your students as they begin to draw and interpret phasor diagrams. Even if they have no prior knowledge of trigonometry or phasors, they should still be able to graphically represent simple phasor systems and even solve for resultant phasors.

#### Question 26. (Click on arrow for answer)

What does it mean to *add* two or more phasors together, in a geometric sense? How would one draw a phasor diagram showing the following two phasors added together?

File Num: 02100

*add*two or more phasors together, in a geometric sense? How would one draw a phasor diagram showing the following two phasors added together?

#### Answer

Here are two ways of showing the same addition:

Follow-up question: how would you verbally explain the process of phasor addition? If you were to describe to someone else how to add phasors together, what would you tell them?

#### Notes

Discuss with your students that phasors may also be subtracted, multiplied, and divided. Subtraction is not too difficult to visualize, but addition and multiplication defies geometric understanding for many.

#### Question 27. (Click on arrow for answer)

Phasors may be symbolically described in two different ways: *polar notation* and *rectangular notation*. Explain what each of these notations means, and why either one may adequately describe a phasor.

File Num: 02101

*polar notation*and

*rectangular notation*. Explain what each of these notations means, and why either one may adequately describe a phasor.

#### Answer

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

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

Follow-up question: why do we need the letter j in rectangular notation? What purpose does it serve, and what does it mean?

#### Notes

When discussing the meaning of j, it might be good to explain what *imaginary numbers* are. Whether or not you choose to do this depends on the mathematical aptitude and background of your students.

#### Question 28. (Click on arrow for answer)

These two phasors are written in a form known as *polar notation*. Re-write them in *rectangular notation*:

4 \> \angle \> 0^o = 3 \> \angle \> 90^o = File Num: 00497

*polar notation*. Re-write them in

*rectangular notation*:

#### Answer

These two phasors, written in rectangular notation, would be 4 + j0 and 0 + j3, respectively, although a mathematician would probably write them as 4 + i0 and 0 + i3, respectively.

Challenge question: what does the lower-case j or i represent, in mathematical terms?

#### Notes

Discuss with your students the two notations commonly used with phasors: *polar* and *rectangular* form. They are merely two different ways of “saying” the same thing. A helpful “prop” for this discussion is the complex number *plane* (as opposed to a number *line* — a one-dimensional field), showing the “real” and “imaginary” axes, in addition to standard angles (right = 0^{o}, left = 180^{o}, up = 90^{o}, down = 270^{o}). Your students should be familiar with this from their research, so have one of them draw the number plane on the whiteboard for all to view.

The challenge question regards the origin of complex numbers, beginning with the distinction of “imaginary” numbers as being a separate set of quantities from “real” numbers. Electrical engineers, of course, avoid using the lower-case letter i to denote “imaginary” because it would be so easily be confused with the standard notation for instantaneous current i.

#### Question 29. (Click on arrow for answer)

Determine the sum of these two phasors, and draw a phasor diagram showing their geometric addition:

(4 \angle 0^o) + (3 \angle 90^o)

How might a phasor arithmetic problem such as this relate to an AC circuit?

File Num: 00495

#### Answer

(4 \angle 0^o) + (3 \angle 90^o) = (5 \angle 36.87^o)#### Notes

It is very helpful in a question such as this to graphically depict the phasors. Have one of your students draw a phasor diagram on the whiteboard for the whole class to observe and discuss.

The relation of this arithmetic problem to an AC circuit is a very important one for students to grasp. It is one thing for students to be able to mathematically manipulate and combine phasors, but quite another for them to smoothly transition between a phasor operation and comprehension of voltages and/or currents in an AC circuit. Ask your students to describe what the *magnitude* of a phasor means (in this example, the number 5), if that phasor represents an AC voltage. Ask your students to describe what the *angle* of an AC voltage phasor means, as well (in this case, 36.87^{o}), for an AC voltage.

#### Question 30. (Click on arrow for answer)

Phasors may be symbolically described in two different ways: *polar notation* and *rectangular notation*. Explain what each of these notations means, and why either one may adequately describe a phasor.

File Num: 02101

*polar notation*and

*rectangular notation*. Explain what each of these notations means, and why either one may adequately describe a phasor.

#### Answer

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

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

Follow-up question: why do we need the letter j in rectangular notation? What purpose does it serve, and what does it mean?

#### Notes

When discussing the meaning of j, it might be good to explain what *imaginary numbers* are. Whether or not you choose to do this depends on the mathematical aptitude and background of your students.

#### Question 31. (Click on arrow for answer)

These two phasors are written in a form known as *polar notation*. Re-write them in *rectangular notation*:

4 \> \angle \> 0^o = 3 \> \angle \> 90^o = File Num: 00497

*polar notation*. Re-write them in

*rectangular notation*:

#### Answer

These two phasors, written in rectangular notation, would be 4 + j0 and 0 + j3, respectively, although a mathematician would probably write them as 4 + i0 and 0 + i3, respectively.

Challenge question: what does the lower-case j or i represent, in mathematical terms?

#### Notes

Discuss with your students the two notations commonly used with phasors: *polar* and *rectangular* form. They are merely two different ways of “saying” the same thing. A helpful “prop” for this discussion is the complex number *plane* (as opposed to a number *line* — a one-dimensional field), showing the “real” and “imaginary” axes, in addition to standard angles (right = 0^{o}, left = 180^{o}, up = 90^{o}, down = 270^{o}). Your students should be familiar with this from their research, so have one of them draw the number plane on the whiteboard for all to view.

The challenge question regards the origin of complex numbers, beginning with the distinction of “imaginary” numbers as being a separate set of quantities from “real” numbers. Electrical engineers, of course, avoid using the lower-case letter i to denote “imaginary” because it would be so easily be confused with the standard notation for instantaneous current i.

#### Question 32. (Click on arrow for answer)

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

If the 4-volt (peak) sine wave is denoted in phasor notation as 4 \hbox{ V} \angle \> 0^o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.

If the 4-volt (peak) sine wave is denoted in phasor notation as 4 \hbox{ V} \angle \> 90^o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.

File Num: 00499

*leading*and which one is

*lagging*?

#### Answer

The 4-volt (peak) waveform *leads* the 3-volt (peak) waveform. Conversely, the 3-volt waveform *lags* behind the 4-volt waveform.

If the 4-volt waveform is denoted as 4 V \angle 0^o, then the 3-volt waveform should be denoted as 3 V \angle -90^o, or 0 - j3 V.

If the 4-volt waveform is denoted as 4 V \angle 90^o (0 + j4 V in rectangular form), then the 3-volt waveform should be denoted as 3 V \angle 0^o, or 3 + j0 V.

#### Notes

In my years of teaching, I have been surprised at how many students struggle with identifying the “leading” and “lagging” waveforms on a time-domain graph. Be sure to discuss this topic well with your students, identifying methods for correctly distinguishing “leading” waves from “lagging” waves.

This question also provides students with good practice expressing leading and lagging waves in phasor notation. One of the characteristics of phasors made evident in the answer is the relative nature of angles. Be sure to point this out to your students.

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

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

File Num: 03286

*leading*and which is

*lagging*the other:

#### Answer

In this diagram, phasor B is leading phasor A.

Follow-up question: using a protractor, estimate the amount of phase shift between these two phasors.

#### Notes

It may be helpful to your students to remind them of the standard orientation for phase angles in phasor diagrams (0 degrees to the right, 90 degrees up, etc.).

#### Question 34. (Click on arrow for answer)

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

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

File Num: 00496

*angle*to a single AC voltage, all by itself in a circuit?

#### Answer

Phasor angles are *relative*, not *absolute*. They have meaning only where there is another phasor to compare against.

Angles may be associated with multiple AC voltage sources in the same circuit, but *only if those voltages are all at the same frequency*.

#### Notes

Discuss with your students the notion of “phase angle” in relation to AC quantities. What does it mean, exactly, if a voltage is “3 volts at an angle of 90 degrees”? You will find that such a description only makes sense where there is another voltage (i.e., “4 volts at 0 degrees”) to compare to. Without a frame of reference, phasor angles are meaningless.

Also discuss with your students the nature of phase shifts between different AC voltage sources, if the sources are all at different frequencies. Would the phase angles be fixed, or vary over time? Why? In light of this, why do we not assign phase angles when different frequencies are involved?

#### Question 35. (Click on arrow for answer)

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

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

File Num: 02089

#### Answer

I_{total} = 16.12 amps\Theta = 60.26^{o} (negative, if you wish to represent the angle according to the standard coordinate system for phasors).Follow-up question: in calculating \Theta, it is recommended to use the arctangent function instead of either the arcsine or arc-cosine functions. The reason for doing this is accuracy: less possibility of compounded error, due to either rounding and/or calculator-related (keystroke) errors. Explain why the use of the arctangent function to calculate \Theta incurs less chance of error than either of the other two arcfunctions.

#### Notes

The follow-up question illustrates an important principle in many different disciplines: avoidance of unnecessary risk by choosing calculation techniques using given quantities instead of derived quantities. This is a good topic to discuss with your students, so make sure you do so.

#### Question 36. (Click on arrow for answer)

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

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

File Num: 02091

#### Answer

I_{total} = 131.2 mA\Theta = 40.36^{o}Follow-up question: in calculating \Theta, it is recommended to use the arctangent function instead of either the arcsine or arc-cosine functions. The reason for doing this is accuracy: less possibility of compounded error, due to either rounding and/or calculator-related (keystroke) errors. Explain why the use of the arctangent function to calculate \Theta incurs less chance of error than either of the other two arcfunctions.

#### Notes

The follow-up question illustrates an important principle in many different disciplines: avoidance of unnecessary risk by choosing calculation techniques using given quantities instead of derived quantities. This is a good topic to discuss with your students, so make sure you do so.

#### Question 37. (Click on arrow for answer)

A parallel RC circuit has 10 \muS of susceptance (B). How much conductance (G) is necessary to give the circuit a (total) phase angle of 22 degrees?

File Num: 02090

#### Answer

G = 24.75 \muSFollow-up question: how much resistance is this, in ohms?

#### Notes

Ask your students to explain their method(s) of solution, including any ways to double-check the correctness of the answer.

#### Question 38. (Click on arrow for answer)

Determine the total voltage in each of these examples, drawing a phasor diagram to show how the total (resultant) voltage geometrically relates to the source voltages in each scenario:

File Num: 00498

#### Answer

#### Notes

At first it may confuse students to use polarity marks (+ and -) for AC voltages. After all, doesn’t the polarity of AC *alternate* back and forth, so as to be continuously changing? However, when analyzing AC circuits, polarity marks are essential for giving a frame of reference to phasor voltages, which like all voltages are measured *between two points*, and thus may be measured two different ways.

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