Electric Circuits, 9th Edition P34 pdf

2 Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor concepts.. 3 Know how to use the following circuit analysis techniques to solve a circ

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C H A P T E R C O N T E N T S

9.1 The Sinusoidal Source p 308

9.2 The Sinusoidal Response p 311

9.3 The Phasor p 312

9.4 The Passive Circuit Elements in the

Frequency Domain p 317

9.5 Kirchhoff's Laws in the Frequency

Domain p 321

9.6 Series, Parallel, and Delta-to-Wye

Simplifications p 322

9.7 Source Transformations and

Thevenin-Norton Equivalent Circuits p 329

9.8 The Node-Voltage Method p 332

9.9 The Mesh-Current Method p 333

9.10 The Transformer p 334

9.11 The Ideal Transformer p 338

9.12 Phasor Diagrams p 344

1 Understand phasor concepts and be able to

perform a phasor transform and an inverse

phasor transform

2 Be able to transform a circuit with a sinusoidal

source into the frequency domain using phasor

concepts

3 Know how to use the following circuit analysis

techniques to solve a circuit in the frequency

domain:

• Kirchhoffs laws;

• Series, parallel, and delta-to-wye

simplifications;

• Voltage and current division;

• Thevenin and Norton equivalents;

• Node-voltage method; and

• Mesh-current method

4 Be able to analyze circuits containing linear

transformers using phasor methods

5 Understand the ideal transformer constraints

and be able to analyze circuits containing ideal

transformers using phasor methods

306

Sinusoidal Steady-State Analysis

Thus far, we have focused on circuits with constant sources; in

this chapter we are now ready to consider circuits energized by time-varying voltage or current sources In particular, we are inter-ested in sources in which the value of the voltage or current varies sinusoidally Sinusoidal sources and their effect on circuit behavior form an important area of study for several reasons First, the gen-eration, transmission, distribution, and consumption of electric energy occur under essentially sinusoidal steady-state conditions Second, an understanding of sinusoidal behavior makes it possible

to predict the behavior of circuits with nonsinusoidal sources Third, steady-state sinusoidal behavior often simplifies the design

of electrical systems Thus a designer can spell out specifications in terms of a desired steady-state sinusoidal response and design the circuit or system to meet those characteristics If the device satis-fies the specifications, the designer knows that the circuit will respond satisfactorily to nonsinusoidal inputs

The subsequent chapters of this book are largely based on a thorough understanding of the techniques needed to analyze cir-cuits driven by sinusoidal sources Fortunately, the circuit analysis and simplification techniques first introduced in Chapters 1-4 work for circuits with sinusoidal as well as dc sources, so some of the material in this chapter will be very familiar to you The chal-lenges in first approaching sinusoidal analysis include developing the appropriate modeling equations and working in the mathe-matical realm of complex numbers

Practical Perspective

A Household Distribution Circuit

Power systems that generate, transmit, and distribute

electri-cal power are designed to operate in the sinusoidal steady

state The standard household distribution circuit used in the

United States is the three-wire, 240/120 V circuit shown in

the accompanying figure

The transformer is used to reduce the utility distribution

voltage from 13.2 kV to 240 V The center tap on the

second-ary winding provides the 120 V service The operating

fre-quency of power systems in the United States is 60 Hz Both

50 and 60 Hz systems are found outside the United States

^ o y1 0 3

The voltage ratings alluded to above are rms values The rea-son for defining an rms value of a time-varying signal is explained in Chapter 10

307

9.1 The Sinusoidal Source

A sinusoidal voltage source (independent or dependent) produces a

volt-age that varies sinusoidally with time A sinusoidal current source

(inde-pendent or de(inde-pendent) produces a current that varies sinusoidally with

time In reviewing the sinusoidal function, we use a voltage source, but our

observations also apply to current sources

We can express a sinusoidally varying function with either the sine

function or the cosine function Although either works equally well, we

cannot use both functional forms simultaneously We will use the cosine

function throughout our discussion Hence, we write a sinusoidally varying

voltage as

v = Vm cos (art + 4>) (9.1)

To aid discussion of the parameters in Eq 9.1, we show the voltage

versus time plot in Fig 9.1

Note that the sinusoidal function repeats at regular intervals Such a

function is called periodic One parameter of interest is the length of time

required for the sinusoidal function to pass through all its possible values

This time is referred to as the period of the function and is denoted T It is

measured in seconds The reciprocal of T gives the number of cycles per

second, or the frequency, of the sine function and is denoted /, or

f = f- (9.2)

A cycle per second is referred to as a hertz, abbreviated Hz (The term

cycles per second rarely is used in contemporary technical literature.) The

coefficient of t in Eq 9.1 contains the numerical value of Torf Omega (co)

represents the angular frequency of the sinusoidal function, or

Equation 9.3 is based on the fact that the cosine (or sine) function passes

through a complete set of values each time its argument, cot, passes

through 2-7T rad (360°) From Eq 9.3, note that, whenever t is an integral

multiple of T, the argument cot increases by an integral multiple of 2TT rad

The coefficient V m gives the maximum amplitude of the sinusoidal

voltage Because ±1 bounds the cosine function, ±V m bounds the

ampli-tude Figure 9.1 shows these characteristics

The angle cp in Eq 9.1 is known as the phase angle of the sinusoidal

voltage It determines the value of the sinusoidal function at t = 0;

there-fore, it fixes the point on the periodic wave at which we start measuring

time Changing the phase angle cp shifts the sinusoidal function along the

time axis but has no effect on either the amplitude (V m ) or the angular

fre-quency (co) Note, for example, that reducing cp to zero shifts the sinusoidal

function shown in Fig 9.1 cp/co time units to the right, as shown in Fig 9.2

Note also that if cp is positive, the sinusoidal function shifts to the left,

whereas if cp is negative, the function shifts to the right (See Problem 9.5.)

A comment with regard to the phase angle is in order: cot and cp must

carry the same units, because they are added together in the argument of

the sinusoidal function With cot expressed in radians, you would expect cp

to be also However, cp normally is given in degrees, and cot is converted

from radians to degrees before the two quantities are added We continue

9.1 The Sinusoidal Source 3 0 9

this bias t o w a r d degrees by expressing t h e phase angle in degrees Recall

from your studies of trigonometry that t h e conversion from radians to

degrees is given by

( n u m b e r of degrees) 18CT

TT ( n u m b e r of radians) (9.4)

Another important characteristic of the sinusoidal voltage (or

cur-rent) is its rms value The rms value of a periodic function is defined as the

square root of the mean value of the squared function Hence, if

v = V m cos (cot + 4>), the rms value of v is

t»+ T

V 2m COS 2 (cot + ¢) dt (9.5)

Note from Eq 9.5 that we obtain the mean value of the squared voltage by

integrating v 2 over one period (that is, from t 0 to t Q + T) and then dividing

by the range of integration, T Note further that the starting point for the

integration t (] is arbitrary

The quantity under the radical sign in Eq 9.5 reduces to V 2 „/2 (See

Problem 9.6.) Hence the rms value of v is

V

-y rms

Vm

vr

(9.6) M rms value of a sinusoidal voltage source

T h e r m s value of t h e sinusoidal voltage d e p e n d s only o n t h e m a x i m u m

amplitude of v, namely, V m T h e r m s value is n o t a function of either t h e

frequency or t h e p h a s e angle We stress t h e i m p o r t a n c e of t h e rms value as

it relates t o p o w e r calculations in C h a p t e r 10 (see Section 10.3)

Thus, we can completely describe a specific sinusoidal signal if we know

its frequency, phase angle, a n d amplitude (either t h e m a x i m u m o r the r m s

value) Examples 9.1, 9.2, and 9.3 illustrate these basic properties of the

sinusoidal function In Example 9.4, we calculate t h e rms value of a periodic

function, and in so doing we clarify the meaning of root mean square

Example 9 1 Finding the Characteristics of a Sinusoidal Current

A sinusoidal current has a maximum amplitude of

20 A The current passes through o n e complete cycle

in 1 ms T h e m a g n i t u d e of the current at zero time

is 10 A

a) What is the frequency of the current in hertz?

b) W h a t is t h e frequency in radians p e r second?

c) Write t h e expression for i(t) using t h e cosine

function Express <£ in degrees

d) What is the rms value of the current?

Solution

a) From the statement of the problem, T = 1 ms;

h e n c e / = 1/T = 1000 Hz

b) to « 277-/ = 2000TT r a d / s c) We h a v e i(t) = I m cos (<ot + ¢) = 20 COS(2000TT/

+ <f>), b u t /(0) = 10 A Therefore 10 = 20 cos 4>

a n d cl> = 60° Thus t h e expression for i(t) becomes

/(f) = 20cos(20007rf + 60°)

d) From the derivation of Eq 9.6, the rms value of a sinusoidal current is / „ , / V 2 Therefore the rms value is 2 0 / V 2 , or 14.14 A

310 Sinusoidal Steady-State Analysis

Finding the Characteristics of a Sinusoidal Voltage

A sinusoidal voltage is given by the expression

v = 300 cos (12()77/ + 30°)

a) What is the period of the voltage in milliseconds?

b) What is the frequency in hertz?

c) What is the magnitude of v at t = 2.778 ms?

d) What is the rms value of V?

Solution

a) From the expression for v, to = 12077 rad/s

Because (0 = 2TT/7\ T = 2TT/<O = ^ s,

or 16.667 ms

b) The frequency is 1/71, or 60 Hz

c) From (a), co = 2 77-/ 16.667; thus, at t = 2.778 ms,

at is nearly 1.047 rad, or 60° Therefore,

y(2.778ms) = 300 cos (60° + 30°) = 0 V

d)V ms = 300/ V2 = 212.13 V

Example 9.3 Translating a Sine Expression to a Cosine Expression

We can translate the sine function to the cosine

function by subtracting 90° (TT/2 rad) from the

argu-ment of the sine function

a) Verify this translation by showing that

sin (tot + 0) = cos (tot + 8 - 90°)

b) Use the result in (a) to express sin (cot + 30°) as

a cosine function

Solution

a) Verification involves direct application of the trigonometric identity

cos(a — /3) = cos a cos /3 + sin a sin /3

We let a = ait + 0 and /3 = 90° As cos 90° = 0 and

sin 90° = 1, we have

cos(a -/3)= sin a = sin(atf + 0) = cos(a>/ + 0 - 90°)

b) From (a) we have sin(wr + 30°) = cos(o>/ + 30° - 90°) = cos(atf - 60°)

Example 9.4 Calculating the rms Value of a Triangular Waveform

Calculate the rms value of the periodic triangular

current shown in Fig 9.3 Express your answer in

terms of the peak current I p

7 7 2 \

-Figure 9.3 A Periodic triangular current

Solution

From Eq 9.5, the rms value of i is

^rms \l

-T-Interpreting the integral under the radical sign as the area under the squared function for an interval

of one period is helpful in finding the rms value The squared function with the area between 0 and

T shaded is shown in Fig 9.4, which also indicates

that for this particular function, the area under the

9.2 The Sinusoidal Response 311

squared current for an interval of one period is

equal to four times the area under the squared

cur-rent for the interval 0 to TfA seconds; that is,

t«+T /.7/4

i 2 dt = 4 / i 2 dt

etc

- 7 / 2 - 7 / 4 0

Figure 9.4 • r versus t

7/4 7/2 37/4 7

The analytical expression for /' in the interval 0 to

774is

4 7

i = -jrt, 0 < / < 774

The area under the squared function for one period is

/ i 2 dt = 4 /

Tlj

r2 3

The mean, or average, value of the function is simply the area for one period divided by the period Thus

1 ^ - i,2

— in'

7 3 3 p '

The rms value of the current is the square root of this mean value Hence

rms V 3 '

NOTE: Assess your understanding of this material by trying Chapter Problems 9.1, 9.4, 9.8

9.2 The Sinusoidal Response

Before focusing on the steady-state response to sinusoidal sources, let's

consider the problem in broader terms, that is, in terms of the total

response Such an overview will help you keep the steady-state solution in

perspective The circuit shown in Fig 9.5 describes the general nature of

the problem There, v s is a sinusoidal voltage, or

v s = V m c o s i^t + </>)• (9.7)

For convenience, we assume the initial current in the circuit to be zero and

measure time from the moment the switch is closed The task is to derive

the expression for /(0 when t > 0 It is similar to finding the step response

of an RL circuit, as in Chapter 7 The only difference is that the voltage

source is now a time-varying sinusoidal voltage rather than a constant, or

dc, voltage Direct application of Kirchhoffs voltage law to the circuit

shown in Fig 9.5 leads to the ordinary differential equation

Figure 9.5 • An RL circuit excited by a sinusoidal

voltage source

L— + Ri = V m cos {a)t + 4>\ (9.8)

the formal solution of which is discussed in an introductory course in

dif-ferential equations We ask those of you who have not yet studied

differ-ential equations to accept that the solution for / is

-v.,

VR 2 + <o 2 L 2 cos (0 -6)e~W L)t + V„

VR 2 + <o2L2

cos (cot + 4> - 0),

(9.9)

where 0 is defined as the angle whose tangent is coL/R Thus we can easily

determine 0 for a circuit driven by a sinusoidal source of known frequency

We can check the validity of Eq 9.9 by determining that it satisfies

Eq 9.8 for all values of t > 0; this exercise is left for your exploration in

Problem 9.10

The first term on the right-hand side of Eq 9.9 is referred to as the

transient component of the current because it becomes infinitesimal as

time elapses The second term on the right-hand side is known as the

steady-state component of the solution It exists as long as the switch

remains closed and the source continues to supply the sinusoidal voltage

In this chapter, we develop a technique for calculating the steady-state

response directly, thus avoiding the problem of solving the differential

equation However, in using this technique we forfeit obtaining either the

transient component or the total response, which is the sum of the

tran-sient and steady-state components

We now focus on the steady-state portion of Eq 9.9 It is important to

remember the following characteristics of the steady-state solution:

1 The steady-state solution is a sinusoidal function

2 The frequency of the response signal is identical to the frequency of

the source signal This condition is always true in a linear circuit

when the circuit parameters, R, L, and C, are constant (If

frequen-cies in the response signals are not present in the source signals,

there is a nonlinear element in the circuit.)

3 The maximum amplitude of the steady-state response, in general,

differs from the maximum amplitude of the source For the circuit

being discussed, the maximum amplitude of the response signal is

VJ\/R 2 + arL 2 , and the maximum amplitude of the signal source

is V m

4 The phase angle of the response signal, in general, differs from the

phase angle of the source For the circuit being discussed, the phase

angle of the current is 4> - 0 and that of the voltage source is <f>

These characteristics are worth remembering because they help you

understand the motivation for the phasor method, which we introduce in

Section 9.3 In particular, note that once the decision has been made to

find only the steady-state response, the task is reduced to finding the

max-imum amplitude and phase angle of the response signal The waveform

and frequency of the response are already known

NOTE: Assess your understanding of this material by trying Chapter

Problem 9.9

9.3 The Phasor

The phasor is a complex number that carries the amplitude and phase

angle information of a sinusoidal function.1 The phasor concept is rooted

in Euler's identity, which relates the exponential function to the

trigono-metric function:

e ±jd = cos6» ± / s i n 0 (9.10)

Equation 9.10 is important here because it gives us another way of

express-ing the cosine and sine functions We can think of the cosine function as the

If you feel a bit uneasy about complex numbers, peruse Appendix B

9.3 The Phasor 313

real part of the exponential function and the sine function as the imaginary

part of the exponential function; that is,

cosfl = Sfc{tf**}, (9.11) and

sin0 = 3{<?'0}, (9.12) where 5ft means "the real part of1 and S means "the imaginary part of."

Because we have already chosen to use the cosine function in

analyz-ing the sinusoidal steady state (see Section 9.1), we can apply Eq 9.11

directly In particular, we write the sinusoidal voltage function given by

Eq 9.1 in the form suggested by Eq 9.11:

v = V m cos (cot 4- (f>)

We can move the coefficient V m inside the argument of the real part of the

function without altering the result We can also reverse the order of the

two exponential functions inside the argument and write Eq 9.13 as

In Eq 9.14, note that the quantity V m e® is a complex number that carries

the amplitude and phase angle of the given sinusoidal function This

complex number is by definition the phasor representation, or phasor

transform, of the given sinusoidal function Thus

V = V m ef* = V{V m cos(cot + ¢)), (9,15) < Phasor transform

where the notation V{V m cos (cot + <f>)} is read "the phasor transform of

Vmcos (cot 4- <£)."Thus the phasor transform transfers the sinusoidal

func-tion from the time domain to the complex-number domain, which is also

called the frequency domain, since the response depends, in general, on to

As in Eq 9.15, throughout this book we represent a phasor quantity by

using a boldface letter

Equation 9.15 is the polar form of a phasor, but we also can express a

phasor in rectangular form Thus we rewrite Eq 9.15 as

V = V m cos $ + jV m sin </> (9.16)

Both polar and rectangular forms are useful in circuit applications of the

phasor concept

One additional comment regarding Eq 9.15 is in order The frequent

occurrence of the exponential function e^ has led to an abbreviation that

lends itself to text material This abbreviation is the angle notation

We use this notation extensively in the material that follows

Inverse Phasor Transform

So far we have emphasized moving from the sinusoidal function to its

pha-sor transform However, we may also reverse the process That is, for a

phasor we may write the expression for the sinusoidal function Thus for

V = l ( ) 0 / - 2 6 ° , the expression for v is 100cos (a)t - 26°) because we

have decided to use the cosine function for all sinusoids Observe that we

cannot deduce the value of co from the phasor The phasor carries only

amplitude and phase information The step of going from the phasor

transform to the time-domain expression is referred to as finding the

inverse phasor transform and is formalized by the equation

V~ x {V m e^} = $t{V m e i4 >e> m }, (9.17)

where the notation V~ l {V m e^} is read as "the inverse phasor transform of

V m e'^." Equation 9.17 indicates that to find the inverse phasor transform, we

multiply the phasor by <?/W and then extract the real part of the product

The phasor transform is useful in circuit analysis because it reduces

the task of finding the maximum amplitude and phase angle of the

steady-state sinusoidal response to the algebra of complex numbers The

follow-ing observations verify this conclusion:

1 The transient component vanishes as time elapses, so the

steady-state component of the solution must also satisfy the differential

equation (See Problem 9.10[b].)

2 In a linear circuit driven by sinusoidal sources, the steady-state

response also is sinusoidal, and the frequency of the sinusoidal

response is the same as the frequency of the sinusoidal source

3 Using the notation introduced in Eq 9.11, we can postulate that the

steady-state solution is of the form lR{Ae' li e JM '}, where A is the

maximum amplitude of the response and /3 is the phase angle of the

response

4 When we substitute the postulated steady-state solution into the

differential equation, the exponential term e ,u>l cancels out, leaving

the solution for A and ft in the domain of

complex numbers

We illustrate these observations with the circuit shown in Fig 9.5 (see

p 311) We know that the steady-state solution for the current i is of the form

W O = K { / / f l , (9.18)

where the subscript uss" emphasizes that we are dealing with the

steady-state solution When we substitute Eq 9.18 into Eq 9.8, we generate the

expression

®.{}<oLI m e®ei«*} + n{RI m e iP e J(M } = ^{V m e^ wt \ (9.19)

In deriving Eq 9.19 we recognized that both differentiation and

multiplica-tion by a constant can be taken inside the real part of an operamultiplica-tion We also

rewrote the right-hand side of Eq 9.8, using the notation of Eq 9.11 From

the algebra of complex numbers, we know that the sum of the real parts is the

same as the real part of the sum Therefore we may reduce the left-hand side

of Eq 9.19 to a single term:

9R{(/wL + K ^ ' V * } = ftlV^V*"} (9.20)

Recall that our decision to use the cosine function in analyzing the

response of a circuit in the sinusoidal steady state results in the use of

the 5ft operator in deriving Eq 9.20 If instead we had chosen to use the

sine function in our sinusoidal steady-state analysis, we would have

applied Eq 9.12 directly, in place of Eq 9.11, and the result would be

Eq.9.21:

Q{(j(ol + R)I m e ifi e^} = ^{V m e j ^ wt } (9.21)

Note that the complex quantities on either side of Eq 9.21 are identical to

those on either side of Eq 9.20 When both the real and imaginary parts of

two complex quantities are equal, then the complex quantities are

them-selves equal Therefore, from Eqs 9.20 and 9.21,

(J(oL + R)I m e® = V m e i<!>

or

Note that e Ja>t has been eliminated from the determination of the

ampli-tude (/,„) and phase angle (/3) of the response Thus, for this circuit, the

task of finding /,„ and /3 involves the algebraic manipulation of the

com-plex quantities V m e® and R + jwL Note that we encountered both polar

and rectangular forms

An important warning is in order: The phasor transform, along with

the inverse phasor transform, allows you to go back and forth between

the time domain and the frequency domain Therefore, when you obtain

a solution, you are either in the time domain or the frequency domain

You cannot be in both domains simultaneously Any solution that

con-tains a mixture of time domain and phasor domain nomenclature is

nonsensical

The phasor transform is also useful in circuit analysis because it applies

directly to the sum of sinusoidal functions Circuit analysis involves

sum-ming currents and voltages, so the importance of this observation is

obvi-ous We can formalize this property as follows: If

where all the voltages on the right-hand side are sinusoidal voltages of the

same frequency, then