CHAPTER 5

Exercises

E5.1 (a) We are given . The angular frequency is the coefficient of t so we have . Then

)30200cos(150)( o−= ttv π 200 radian/sπω =

E5.4 (a) cos(10

Hz 1002/ == πωf ms 10/1 == fT

(b) 10=I

V 1.1062/1502/ === mrms VV

cos(10 (c)

Furthermore, v(t) attains a positive peak when the argument of the cosine function is zero. Thus keeping in mind that has units of radians, the positive peak occurs when

(b) (c) A plot of v(t) is shown in Figure 5.4 in the book.

tω

sin(20

ms 8333.0 180

30 maxmax =⇒×= tt πω

W 225/2 == RVP rmsavg

E5.2 We use the trigonometric identity sin( Thus 100 E5.3 T

The period corresponds to 360 therefore 5 ms corresponds to a phase angle of ( . Thus the voltage is

).90cos() o−= zz)30300cos(100)60300sin( oo −=+ tt ππ

radian/s 3772 ≅= fπω ms 67.16/1 ≅= f V 6.1552 ≅= rmsm VVo

o108=o360)67.16/5 ×)108377cos(6.155)( o−= ttv

ooo 4514.14101090100101 −∠≅−=−∠+∠= jV )45cos(14.14)sin(10) o−=+ ttt ωωω

330.45.25660.8605301 jj −++≅−∠+∠ oo

o44.318.11670.016.11 ∠≅+≅ j)44.3cos(18.11)30sin(5)30 ooo +=+++ ttt ωωω

99.125.702060150202 jj −++≅−∠+∠= ooI o28.2541.3099.125.27 −∠≅−≅ j

)28.25cos(41.30)60cos(15)90 ooo −=−++ ttt ωωω

142

E5.5 The phasors are V

v1 lags v by 60 (or we could say v leads v by 60

ooo 4510 and 3010 3010 321 −∠=+∠=−∠= VV

2 2

1

o

o

o

)o

v1 leads v by 15 (or we could say v lags v by 15

3 3

1

)o

v2 leads v3 by 75 (or we could say v lags v by 75

3

2

)o

E5.6 (a) o905050 ∠=== jLjZL ω o0100∠=LV

(b) The phasor diagram is shown in Figure 5.11a in the book.

o90250/100/ −∠=== jZLLL VI

E5.7 (a)

o905050/1 −∠=−== jCjZC ω o0100∠=CVo902)50/(100/ ∠=−== jZCCC VI

(b) The phasor diagram is shown in Figure 5.11b in the book.

E5.8 (a) (b) The phasor diagram is shown in Figure 5.11c in the book.

o05050 ∠=== RZRo0100∠=RV

o02)50/(100/ ∠=== RRR VI

E5.9 (a) The transformed network is:

mA 13528.28

2502509010 o

o

−∠=+−∠

==jZ

sVI

143

mA )135500cos(28.28)( o−= tti

(b) The phasor diagram is shown in Figure 5.17b in the book. (c) i(t) lags vs(t) by

E5.10 The transformed network is:

o13507.7 −∠== IV RRo4507.7 −∠== IV LjL ω

.45o

Ω−∠=++−+

= 31.5647.55)200/(1)50/(1100/1

1 o

jjZ

V 31.564.277 o−∠== IV Z 69.33547.5)50/( o∠=−= jC VI AA 31.146387.1)200/( o−∠== jL VI

A 31.56774.2)100/( o−∠== VIR

E5.11 The transformed network is:

We write KVL equations for each of the meshes:

100)(100100 211 =−+ IIIj

Simplifying, we have 0)(100100200 1222 =−++− IIII jj

100100)100100( 21 =−+ IIj

Solving we find Thus we have

0)100100(100 21 =−+− II jA. 01 and A 45414.1 21

oo ∠=−∠= II1000cos()( and A )451000 2 ).cos(414.1)(1 ttitti =− o=

144

E5.12 (a) For a power factor of 100%, we have which implies that

the current and voltage are in phase and Thus, Also Thus we have

(b) For a power factor of 20% lagging, we have cos( which implies that the current lags the voltage by Thus,

Also, we have Thus we have

,1)cos( =θ.0=θ .0)tan( == θPQ

( ) A.500/[5000]cos/[ 10)]0cos( === θrmsrms VPI.4014.14 and 14.142 o∠== IrmsI=mI

,2.0) =θ.46. o=θ 78)2.0(cos 1 =−

.kVAR 49.24)tan( == θPQ ( )] = A. 0.50cos/[= θrmsrms VPI.46.3871 o−∠.70 and A 71.70 I2 == rmsm II =

(c) The current ratings would need to be five times higher for the load of part (b) than for that of part (a). Wiring costs would be lower for the load of part (a).

E5.13 The first load is a 10 capacitor for which we have

The second load absorbs an apparent power of with a power factor of 80% lagging from which we have Notice that we select a positive angle for because the load has a lagging power factor. Thus we have and .

F µΩ== 90265)/(1 o−∠3.CjZC ω o90−=Cθ A 770.3/ == CrmsCrms ZVI

0)cos( == CCrmsrmsC IVP θ kVAR 770.3)sin( −== CCrmsrmsC IVQ θ

kVA 10=rmsrmsIV.87.36 o=)8.0(cos 1

2 = −θ

2θkW 0.8)cos( 2 =22 = θrmsrmsIVP

kVAR 6)sin(22 == θrmsrms IVQ Now for the source we have:

kW 82 =+= PPP Cs kVAR 23.22 =+= QQQ Cs

kVA 305.822 =+= sssrmsrms QPIV A 305.8/ == rmssrmsrmssrms VIVI

%33.96%100)/(factor power =×= srmsrmss IVP

E5.14 First, we zero the source and combine impedances in series and parallel to determine the Thévenin impedance.

145

50502550

100/1100/112550 jj

jjZt ++−=

++−=

o04.141.103j25100 ∠=+= Then we analyze the circuit to determine the open-circuit voltage.

o4571.70100100

100100 −∠=+

×==joct VV

o04.596858.0/ −∠== tt ZVnI E5.15 (a) For a complex load, maximum power is transferred for

. The Thévenin equivalent with the load attached is:

LLtL jXRjZZ +=−== 25100*

The current is given by

oo

453536.02510025100

4571.70−∠=

−++−∠

=jj

I

The load power is W 25.6)2/3536.0(100 22 === rmsLL IRP

146

(b) For a purely resistive load, maximum power is transferred for The Thévenin equivalent with the load attached is:

. 1.10325100 22 Ω=+== tL ZR

The current is given by

oo

98.373456.0251001.103

4571.70−∠=

−+−∠

=j

I

The load power is

E5.16 The line-to-neutral voltage is 1000 No phase angle was

specified in the problem statement, so we will assume that the phase of Van is zero. Then we have

The circuit for the a phase is shown below. (We can consider a neutral connection to exist in a balanced Y-Y connection even if one is not physically present.)

W 157.6)2/3456.0(1.103 22 === rmsLL IRP

V. 4.5773/ =

ooo 1204.577 1204.577 04.577 ∠=−∠=∠= cnbnan VVV

The a-phase line current is

The currents for phases b and c are the same except for phase.

oo

02.37610.440.75100

04.577−∠=

+∠

==jZL

anaA

VI

oo 98.82610.4 02.157610.4 ∠=−∠= cCbB II

kW 188.3)02.37cos(2

610.44.5773)cos(2

3 =×

== oθLY IVP

147

kVAR 404.2)02.37sin(2

610.44.5773)sin(2

3 =×

== oθLY IVQ

E5.17 The a-phase line-to-neutral voltage is anV

The phase impedance of the equivalent Y is

oo 04.57703/1000 ∠=∠=

. 67.163/503/ Ω=== ∆ZZY

Thus the line current is

A 063.3467.16

04.577 oo

∠=∠

==Y

an

ZVIaA

Similarly, and Finally, the power is

A 12063.34 o−∠=bBI A. 12063.34 o∠=cCI

kW 00.30)2/(3 2 == yaA RIP

Problems

P5.1 (a) Increasing the peak amplitude Vm? 1. Stretches the sinusoidal curve vertically.

(b) Increasing the frequency f? 4. Compresses the sinusoidal curve horizontally. (c) Decreasing θ? 5. Translates the sinusoidal curve to the right. (d) Decreasing the angular frequency ω? 3. Stretches the sinusoidal curve horizontally. (e) Increasing the period? 3. Stretches the sinusoidal curve horizontally.

P5.2 The units of angular frequency ω are radians per second. The units of frequency f are hertz, which are equivalent to inverse seconds. The relationship between them is ω .2 fπ=

148

P5.3*

( ) ( ) ( )oo 601000cos10301000sin10 −=+= ππttv

( ) W 1V 071.72102

ms 21radians 360 angle phase

Hz 500rad/s 1000

2 ==

===

==

−=−==

=

=

RVPVVfT

f

rms

mrms

πθ

πω

o

The first positive peak occurs for

ms 3333.0 031000

=

=−

peak

peak

tt ππ

P5.4

( ) ( ) ( )oo 30500cos50120500sin50 +=+= tttv ππ

( ) W 25V 36.352502

ms 41radians 630 angle phase

Hz 250rad/s 500

2 ==

===

==

===

=

=

RVPVVfT

f

rms

mrms

πθ

πω

o

Positive peak occurs for

=t

ms 3333.0

06500

1

1

−

=+t ππ

The first positive peak after 0 is at ms 667.31 =+= Tt

=t tpeak

149

P5.5* Sinusoidal voltages can be expressed in the form

The peak voltage is V. The frequency is 10 kHz and the angular frequency is ω

radians/s. The phase corresponding to a time interval of 20 µs is . Thus we have V.

).cos()( θω += tVtv m

28.282022 rms =×== VVm

== Tf /1 41022 ππ == f=t∆

o72o360) =×/(= Tt∆θ )72102cos(28. 4 o−28)( = ttv π

P5.6 The angular frequency is radians/s. We have ms. The sinusoid reaches a positive peak one quarter of a cycle after crossing zero with a positive slope. Thus, the first positive peak occurs at t = 3 ms. The phase corresponding to a time interval of 3 ms is

. Thus, we have V. P5.7

ππω 2502 == f 8/1 == fT

=t∆oo 135360)/( =×∆= Ttθ )135o−250cos(15)( = ttv π

( ) ( A 1657cos12 )tti π=( ) ( ) ( ) ( )[ ]

( ) ( ) W 9500212132

W 4000cos195002000cos101922

232

===

+=×==

rmsavg IRP

tttRitp ππ

150

P5.8

P5.9 A Hz

rad/s

Pavg

( ) ( )( ) ( ) ( ) ( )[ ]

( ) W 6240W 1192sin16240596sin12480

V 596sin2503

2

22

==

−===

=

RVttRtvtp

ttv

rms

ππ

π

071.7252 === rmsm II 1001==

Tf

3.6282 == fπω o108360 max −=−=T

tθ

P5.10 A sequence of MATLAB instructions to accomplish the desired plot for

part (a) is

108200cos(071.7)( o−= tti π )

Wx = 2*pi; Wy = 2*pi; Theta = 90*pi/180; t = 0:0.01:20; x = cos(Wx*t); y = cos(Wy*t + Theta); plot(x,y)

By changing the parameters, we can obtain the plots for parts b, c, and d. The resulting plots are

151

(a) (b)

(c) (d)

P5.11*

( ) V 536.32521 1

00

2 === ∫∫ dtdttvTiV

T

rms

P5.12*

( ) [ ] [ ]∫∫∫ +===5.0

0

5.0

0

2

0

2 )4sin(5.125.12)2sin(51 dttdttdttvT

VT

rms ππ

V 5.225.6)4cos(4

5.125.125.0

0==

−=

=

=

t

trms ttV π

π

P5.13* ( ) A 581.114411 2

0

4

20

2 =

+== ∫ ∫∫ dtdtdtti

TI

T

rms

152

P5.14

( ) [ ]∫∫ +==1

0

2

0

2 )2sin()2cos(1 dttBtAdttvT

VT

rms ππ

[ ]

+=

++=

++=

=

=

∫

2

20

2

)2(sin)2sin()2cos(2)2(cos

22

1

0

22

5.0

0

2222

BA

tBtA

dttBttABtAV

t

t

rms ππππ

P5.15

( )dttvTiV

T

rms ∫=0

2

[ ]

V 44.13

09.2802.72.0

1

)]20cos(5.1445.144)10cos(204362.0

1

)]10(cos289)10cos(204362.0

1

)]10(cos289)10cos(20436[2.0

1

)]10cos(176[2.0

1

2.0

0

2.0

0

2.0

0

2.0

0

2.0

0

2.0

0

22.0

0

2.0

0

2

2.0

0

2

=

+++=

+++=

++=

++=

+=

∫ ∫∫∫

∫ ∫∫

∫

∫

dttdtdttdt

dttdttdt

dttt

dttVrms

ππ

ππ

ππ

π

P5.16

( ) [ ] [ ]∫∫∫ −=−==1

0

1

0

2

0

2 )2exp(9)exp(31 dttdttdttvT

VT

rms

[ ] V 973.1)]2exp(1[5.4)2exp(5.4 10 =−−=−−= =

=ttt

P5.17

rmsV ( ) ( ) V 887.2

325.6

215.2

211

2

0

32

0

2

0

2 =

===

=

=∫∫

t

t

T tdttdttvT

153

P5.18 The rms values of all periodic waveforms are not equal to their peak values divided by the square root of two. However, they are for sinusoids, which are important special cases.

P5.19 To add sinusoids we: 1. Convert each of the sinusoids to be added into a phasor. 2. Add the phasors and convert the sum to polar form using complex arithimetic. 3. Convert the resulting phasor to a sinusoid. All of the sinusoids to be added must have the same frequency.

P5.20 Two methods to determine the phase relationship between two sinusoids

are: 1. Examine the phasor diagram and consider it to rotate counterclockwise. If phasor A points in a given direction before phasor B by an angle θ, we say that A leads B by the angle θ or that B lags A by the angle θ. 2. Examine plots of the sinusoidal waveforms versus time. If A reaches a point (such as a positive peak or a zero crossing with positive slope) by an interval ∆t before B reaches the corresponding point, we say that A leads B by the angle or that B lags A by the angle θ. °×∆= 360)/( Ttθ

P5.21*

( ) ( )( ) ( ) ( )

( ) ( )oo

o

o

o

45cos4.141454.141100100

100901001000100

90cos100sin100cos100

21s

2

1

2

1

−=

−∠=−=+=

−=−∠=

=∠=

−==

=

ttvjj

tttvttv

s ω

ωωω

VVVVV

o

o

o

45 by leads 45 by lags 90 by lags

2s

1s

12

VVVVVV

154

P5.22*

5.23* )

( ) ( )( )( ) ( )

( ) ( )( ) ( )( ) ( ) o

o

o

o

o

o

60 by leads 60 by lags 120 by lags

90400cos10)150400cos(5

30400cos10

4002

32

31

21

3

2

1

tvtvtvtvtvtv

ttvttvttv

f

+π=

+π=

+π=

π=π=ω

P We are given the expression 5 o

Conver 5

sin(4)75cos(3)75cos( ttt ωωω +−−+ o ting to phasors we obtain

=−∠+−∠−∠ ooo 90475375 +2941.1

)09.82cos(763.375cos(5

o+

+

tt

ωω

The magnitudes of the phasors for the two voltages are and 2 The phase angles are not known. If the ph angles are the same, thephasor sum would have its maximum magnitude which is 27 . On the other hand, if the phase angles differ by 180

=−−− 4)8978.27765.0(8296.4 jjj o09.82763.37274.35176.0 ∠=+ j Thus, we have

)sin(4)75cos(3) oo =+−− tt ωω

P5.24 26

o , the phasor sum would

V. ase

ha

(v

ve its minimum magnitude which is 25of the sum is 7 V and the minimum is 5 V.

) ( )( ) ( ) ( )

o

o

oo

o

759.1293015071.7071.7045100

30cos15060sin15045cos100

1

2

1

−=−∠=

+=∠=

−=+=

+=

jj

tttvtt

ωωω

VV

. Thus, the maximum rms value

P5.25

( ) ( )oo

23.1cos6.20023.16.20029.46.20021s

2

−=

−∠=−=+=

tt

vj

s ωVVV

155

P5.26 V s

o

o

o

77.28 by leads 23.46 by lags

75 by lags

2s

1s

12

VVVVVV

3=mV 5.0=T

21==

Tf π=π=ω 42 f Hz rad/s

V

oo 45360 max −=−=θT

t

)454cos(3)( o−π= ttvo453 −∠=V V

V

121.223

==rmsV

)

P5.27

V

( ) ( )

( ) ( )oo

o

o

50cos071.750071.7

3010071.72

30cos10

1

1

1

1rms1

1

+=

∠=

∠=

=×=

+=

tti

IIttv

m

ω

ω

I

P5.28 We are given the expression sin(5 Converting to phasors we obtain

150cos(5)30cos(5) oo ++++ ttt ωωω

5 5− j Thus, we have

=∠+∠+−∠ ooo 1505305905.23301.4500.23301.4 =+−++ jj 0

0)150cos(5)30cos(5)sin(5 =++++ oo ttt ωωω

156

P5.29 A sequence of MATLAB commands to generate the desired plot is: t = 0:0.01:2; v = cos(19*pi*t) + cos(21*pi*t); plot(v,t)

The resulting plot is

Notice that the first term cos(19πt) has a frequency of 9.5 Hz while the second term cos(21πt) has a frequency of 10.5 Hz. At t = 0, the rotating vectors are in phase and add constructively. At t = 0.5, the vector for the second term has rotated one half turn more than the vector for the first term and the vectors cancel. At t = 1, the first vector has rotated 9.5 turns and the second vector has rotated 10.5 turns so they both point in the same direction and add constructively.

P5.30 i1 leads i2 by the angle

Therefore, the angle for i2 is which is equivalent to

ms 049.4247/1/1 === fT.7.355360)049.4/4(360)/( °=°×=°×∆=∆ Ttθ

°−=∆−= 7.30812 θθθ.3.51 °+

P5.31 For an inductance, we have V .

For a capacitance, we have V .

LL Lj Iω=

CjC

C ωI

=

157

P5.32 For a pure resistance, current and voltage are in phase. For a pure inductance, current lags voltage by 90 .o

For a pure capacitance, current leads voltage by

.90o

P5.33*

( ) ( )

( )( ) ( ) ( ) ( ) ( )ttti

Z

jLjZ

ttv

L

LLL

L

L

L

ππ=−ππ=

−∠π==

∠=

∠π=π=ω=

π=ω

π=

2000sin201902000cos20190201

01090200200

20002000cos10

o

o

o

o

VIV

( ) ( ) o90 by lags tvti LL

P5.34*

I ( )iC

( ) ( )

( ) ( ) ( )tttiZ

jCjZ

ttv

C

C

C

C

ππ

ω

πωπ

2000sin6283.0902000cos6283.0906283.0

010

9092.1592.15

20002000cos10

CC

−=+=

∠==

∠

−∠=−=−

=

=

=

o

o

o

o

V

Ω

C =V

( ) o90 by leads tvt C

158

P5.35 (a) Notice that the voltage is a sine rather than a cosine.

1009010030160100 jZ −=−∠==∠=−∠= ooo

IVIV

Because Z is pure imaginary and negative, the element is a capacitance.

200=ω F501µ=

ω=

ZC

(b) 500=V

Because Z is pure real, the element is a resistance of 250 Ω.

0250025050250 jZ +=∠==∠=∠ ooo

IVI

(c) Notice that the current is a sine rather than a cosine.

Because Z is pure imaginary and positive, the element is an inductance.

1009010060130100 jZ =∠==−∠=∠= ooo

IVIV

400=ω H25.0=ω

=ZL

P5.36 A Matlab m-file that produces the desired plots is:

f=0:1:1000; ZL=2*pi*f*0.01; ZC=1./(2*pi*f*10e-6); R=50 plot(f,ZL) axis([0 1000 0 100]) hold plot(f,ZC) plot(f,R)

The resulting plot is:

159

P5.37 (a) Z = 209020 j−=−∠= o

IV

Because Z is pure imaginary and negative, the element is a capacitance.

(b) 4Z ==IV

Because Z is pure imaginary and positive, the element is an inductance.

500=ω F1001 µω

==Z

C

490 j=∠ o

500=ω mH8==ωZL

(c) 10==VZ

Because Z is pure real, the element is a resistance of 10 Ω.

5.38 (a) From the plot, we see that ms, so we have Hz and Also, we see that the current lags the voltage by 1 ms or 90°, so we have an inductance. Finally, from which we find that 3.18 mH.

100 =∠ o

I

4=T 250/1 == Tf.500πω =

, 5 Ω/ == mm IVLω=L

(b) From the plot, we see that ms, so we have Hz and Also, we see that the current leads the voltage by 2 ms or

8=T 125/1 == Tf.250πω =

160

90°, so we have a capacitance. Finally, 1 from which we find that 0.5093 µF.

P5.39 The steps in analysis of steady-state ac circuits are: 1. Replace sources with their phasors. 2. Replace inductances and capacitances with their complex impedances. 3. Use series/parallel, node voltages, or mesh currents to solve for the quantities of interest.

, 2500// Ω== mm IVCω=C

jω

−

= o41.416.5450.88∠

= jZ

o05.8046.312 ∠==Z

o32.8013.321 ∠==Z

o45

V 45V 45.7

mA 71.70100

010

o

o

o

∠==

−==

∠=

+∠

=

+=

IVIV

VI

LjR

j

LjR

L

R

s

ω

ω

071.7071

45100

o

∠

−

All of the sources must have the same frequency.

P5.40* C

RLjZ ω 1+=

=ω

4.1654 5466.92 :565

=Ω+

−+=

jjω

8.30754 :2027 Ω+ j =ω

55.31654 :2077 Ω+ j

P5.41*

I by lags sV

161

P5.42

P5.43

V 44.63472.4V 56.26944.8

mA 56.2644.8950100

010

o

o

o

o

∠==

−∠==

−∠=

+∠

=

+=

IVIV

VI

LjR

j

LjR

L

R

s

ω

ω

o56.26 by lags sVI

CjLjZZZ

cL ωω +=

+=

11

111

=ω

Ω+= 3.134 :529 jZΩ−== 10.40 :1758 jZωΩ−== 41.69 :1255 jZω

P5.44

CjLjRZZR

ZcL ω+ω+=

++=

1/11

11/11

.capacitive is impedance the negative, is part imaginary the Because26.3058.3 :571 Ω−== jZω

.capacitive is impedance the negative, is part imaginary the Because163.40669.0 :1066 Ω−== jZω

.capacitive is impedance the negative, is part imaginary the Because339.10069.0 :2648 Ω−== jZω

P5.45

I

V 45071.7mA 45071.7

10001000010

o

o

o

∠==

∠=

−∠

=

−=

IV

VI

R

j

CjR

R

s

ω

( ) V 45071.7 o−∠=−= IV CjC ω

o45 by leads sV

162

P5.46*

I

( ) V57.26944.8V 43.63472.4

mA 43.63472.420001000

010

o

o

o

o

−∠=−=

∠==

∠=

−∠

=

−=

IVIV

VI

CjR

j

CjR

C

R

s

ω

ω

o43.63 by leads sV

P5.47

V

o44.63 by leads sI

b lags sIV

P5.48*

o

o

44.6372.4410012001

105.0

∠=

+=

∠=

js

s

IV

I

o

o44.632236.0 ∠== RR VI56.264472.0 −∠== LjL ωVI

( )200110011

mA 0100 o

−+=

∠=

js

s

IV

I

( ) mA 44.6372.44mA 56.2644.89

V 56.26944.8

o

o

o

∠==

−∠==

−∠=

CjR

C

R

ωVIVI

o56.26 y

163

P5.49*

V

( ) mA 9050mA 9050

mA 010V 010

005.0005.010001110

111

mA 010

2

o

o

o

o

o

∠==

−∠==

∠==

∠=

+−=

++=

∠=

−

CjLj

R

jj

CjLjR

C

L

R

s

s

ω

ω

ωω

VIVIVI

I

I

The peak value of is five times larger than the source current! This

is possible because current in the capacitance balances the current in the inductance (i.e., ).

( )tiL

0=+ CL II

P5.50

VC

The peak value of is five times larger than the source voltage! This is possible because the impedance of the capacitor cancels the impedance of the inductance.

P5.51

( )tvL

o

o

o

o

0101009050500

A 01.0500100500

10

010

∠==

∠=×=

∠=

−+=

−+=

∠=

IVIV

VI

V

j

jj

CjRLj

L

s

s

ωω

o9050500 −∠=−= IjR

164

( )

o

o

o

oo

6.5645.77506.1469.154100

6.146549.150100

15060.8650100

5060.8610030100 10090100

21

21

−∠==

−∠==

−∠=

+−−

=+

+−−=

+−

=

∠=−=−∠=

IVIV

VVI

VV

j

jj

jjj

LjR

j

L

R

ω

o

o

90 by lags 6.56 by lags 1

LVIVI

P5.52

P5.53* First we write the KVL equation: −V Then we enclose nodes 1 and 2 in a closed surface to form a supernode

and write a KCL equation:

( )

( )o

o

o

o

oo

o

01100100

10045414.1

901100100

10045414.1

45414.101004571.70

50505050100

01.001.01100

11

∠=

−×−∠=

+=

−∠=

−−

×−∠=+

=

−∠=∠

=

∠=

+=

−+=

++=

++=

jZRR

jj

ZRZ

Z

jjj

jj

CjRLjZ

CC

C

CR

total

total

II

II

I

ωω

o01021 ∠=V

165

+V 0

51520102211 =

−++

jjVVV

The solution to these equations is:

=V

P5.54 The KCL equation is Solving, we

find V V.

o58.29402.91 ∠=Vo45.111986.42 ∠

.01625021

1548330 111 =

−°∠−

++°∠−

jjVVV

o98.1614898.0 ∠=1

P5.55 The KCL equation is Solving, we find

V. P5.56 Writing KCL equations at nodes 1 and 2 we obtain

.0417128

214 11 =°∠++

+°∠−

jVV

°∠= 30.10941.31V

o0115510

211 ∠=+−

+jVVV

Solving these equations, we obtain

o30115510

122 ∠=+−

+− jj

VVV

o54.38735.61 −∠=Vo52.5525.162 −∠=V

P5.57* Writing KVL equations around the meshes, we obtain 4 °∠=−+ 8330)(15 211 III j − 16j Solving, we obtain: =I

°∠−=−+ 25021)(15 122 III j

o60.159191.11 ∠

=I P5.58 The current through the current source is

o26.162140.32 ∠

°∠=− 4121 II Writing KVL around the perimeter of the circuit, we have

Solving, we obtain:

°∠=++ 214)712(8 1 2II j

=I °−∠=−∠ 95.192837.0 and 23.1264.1 21 Io

166

P5.59 Writing KVL equations around the meshes, we obtain 10 )(20 211 =−+ III j 0 20j 5− j Solving, we obtain: =I

10)(15)( 3212 −=−+− IIII)(15 233 =−+ III 0

o4.1509402.01 −∠

=I =I

o0.177051.12 −∠o6.1589972.03 −∠

P5.60 A Matlab program that produces the desired plots is: w=0:1:2000;

L=20e-3; C=50e-6; Zmaga=abs(w*L-1./(w*C)); Zmagb=abs(1./((1./(i*w*L))+i*w*C)); plot(w,Zmaga) hold on plot(w,Zmagb) axis([0 2000 0 100]) hold off

The result is:

167

P5.61 A Matlab program that produces the desired plots is: w=0:1:5000;

L=20e-3; R=50; Zmaga=abs(R+i*w*L); Zmagb=abs(1./((1./(i*w*L))+1/R)); plot(w,Zmaga) hold on plot(w,Zmagb) hold off

The result is:

P5.62 The units for real power are watts (W). For reactive power, the units are

volt-amperes reactive (VAR). For apparent power, the units are volt-amperes (VA).

P5.63 Power factor is the cosine of the power angle. It is often expressed as a

percentage. PF

%100)cos( ×= θ

168

P5.64 (a) For a pure resistance, the power is positive and the reactive power is zero.

(b) For a pure inductance, the power is zero and the reactive power is positive.

(c) For a pure capacitance, the power is zero, and the reactive power is negative.

P5.65 A load with a leading power factor is capacitive and has negative reactive

power. A load with a lagging power factor is inductive and has positive reactive power.

P5.66 See Figure 5.23 in the book. P5.67 Real power represents a net flow, over time, of energy from the source

to the load. This energy must be supplied to the system from hydro, fossil-fuel, or nuclear sources.

Reactive power represents energy that flows back and forth from the source to the load. Aside from losses in the transmission system (lines and transformers), no net energy must be supplied to the system to create the reactive power. Reactive power is important mainly because of the increased system losses associated with it.

P5.68 Usually, power factor correction refers to adding capacitances in parallel

with an inductive load to reduce the reactive power flowing from the power plant through the distribution system to the load. To minimize power system losses, we need 100% power factor for the combined load. Ultimately, an economic analysis is needed to balance the costs of power factor correction against the costs of losses and additional distribution capacity needed because of reactive power.

P5.69* This is a capacitive load because the reactance is negative.

kW 5.22100)15( 22 === RIP rms

power factor cos( ==

kVAR 25.11)50()15( 22 −=−== XIQ rms

o57.26)5.0(tantan 11 =−=

= −−

PQθ

%44.89)θ apparent power 2 += QP KVA 16.252 =

169

P5.70 This is an inductive load because the reactance is positive.

°∠=+= 54.6162.463827 jZ

°−∠=°∠°∠

== 61.28292.3161.5462.46

2621488Z

I V

kW 51.2727)92.31( 22 === RIP rms

kVAR 72.38)38()92.31( 22 === XIQ rms

power factor )cos( == θ apparent power = rmsrmsIV KVA

o61.54=θ%91.57

50.4792.311488 =×=

P5.71 power factor cos( == leading

ooo 306030 −=−=−= iv θθθ%60.86)θ

apparent power ==V KVA

kW 99.12)cos( == θrmsrms IVPkVAR 5.7)sin( −== θrmsrms IVQ

15151000 =×rmsrmsI

P5.72 kV A

o

o

o

3067.6660215

3021000−∠=

∠∠

==IVZ

90=rmsV 20=rmsI ooo 38)23(15 =−−=−= iv θθθlagging %80.78%100cos factor power =×= θ

This is an inductive load.

kW 1418)cos( == θrmsrmsIVP kVAR 1108)sin( == θrmsrmsIVQKVA 1800Power Apparent == rmsrmsIV

P5.73 (a) For a resistance in series with an inductance, real power is positive

and reactive power is positive. (b) For a resistance in series with a capacitance, real power is positive and reactive power is negative.

P5.74 If the inductive reactance is greater than the capacitive reactance, the

total impedance is inductive, real power is zero, and reactive power is positive. If the inductive reactance equals the capacitive reactance, the total impedance is zero, and the current is infinite. Real power and reactive

170

power are indeterminate. This case is rarely, if ever, of practical importance. If the inductive reactance is less than the capacitive reactance, the total impedance is capacitive, real power is zero, and reactive power is negative.

P5.75 If the inductive reactance is greater than the capacitive reactance, the total impedance is capacitive, real power is zero, and reactive power is negative. If the inductive reactance equals the capacitive reactance, the total impedance is infinite, and the current is zero. Real power and reactive power are zero. If the inductive reactance is less than the capacitive reactance, the total impedance is inductive, real power is zero, and reactive power is positive.

P5.76

A Delivered by Source A:

kW AQ kVAR

ooo

55.4522025

272186452268∠=

+∠−∠

=j

I

20=rmsI

272.5)45.5545cos(268 =−= rmsA IP972.0)45.5545sin(268 −=−= rmsI

Absorbed by Source B: kW 3.271)45.5527cos(186 =−= rmsB IP

AQ kVAR Absorbed by resistor:

kW

1.772)45.5527sin(186 −=−= rmsI

2 == RIP rmsR 2Absorbed by inductor: kVAR

P5.77

800.02 == XIQ rmsL

ooo 67.8822.206162280161213)1316( −∠=−∠+∠+= jAVrmsV 2.206rms =AV

Delivered by Source A: kW

AQ kVAR 931.0)16167.88cos()13(2.206 −=−−=AP

513.2)16167.88sin()13(3.206 =−−=

Absorbed by Source B: kW 635.3)16116cos()13(280 −=−−=BP

171

BQ kVAR 191.0)16116sin()13(280 −=−−=

Absorbed by resistor: kW Absorbed by inductor:

704.22 == RIP rmsR

=QL kVAR

P5.78 Apparent power ⇒ ⇒ I

704.22 =XIrms

rmsrmsIV= rmsI2202500 = A11.37=rms

36.19==rms

rms

IVZ

25001500)cos( ==

rmsrmsIVPθ °= 53.13θ ⇒ (We know that θ is positive

because the impedance is inductive.)

Ω L H

LjRjZ ω+=+=°∠= 49.1562.1113.5336.19

62.11=R 0411.060249.15

==π

P5.79

( ) lagging %30.828230.063.34cos factor PowerkVA 55.26 power Apparent

kVAR 09.15sinkW 85.21cos

63.22275.16 5.1661221585

1151221585

===

==

==

==

−∠=

∠+

∠=

o

o

oo

rmsrms

rmsrms

rmsrms

IVIVQIVP

j

θθ

I

P5.80*

kVAR 374.8sinVQ

kW 85.21cos974.8276.14

3001221585

1151221585

rms −==

==

−∠=

−∠

+∠

=

o

oo

rms

rmsrms

IIVP

j

θθ

I

( ) leading %4.93934.097.20cos factor Power

kVA 39.23 power Apparent===

==o

rmsrmsIV

172

P5.81* Load A:

Load B:

( )kVAR 843.4tan

84.259.0coskW 10

1A

==

==

=−

AAA

A

PQ

P

θθ o

Source:

P5.82 oad A:

( )( ) kW 12cos

kVAR 9sin=

=

BBrms

BBrms

II

θθ

kVAR 84.13kW 22

=+=

=+=

BAs

BAs

QQQPPP

( ) 87.368.0cos 1

=

=

== −

rmsB

rmsB

B

Brmsrms

VPVQ

θ o

kVA 15=IV

Pow

( )

Apparent factor er

power Apparent 2

=

+=

s

s

PP ( )

lagging %62.848462.0power

kVA 262

==

=sQ

L ( )

kVAR 421.2tan84.259.0cos

kW 51

A

==

==

=−

AAA

A

PQ

P

θθ o

Load B:

( )( )

kVAR 5.7tanfactor. power leading a for negative is

86.368.0coskW 10

1

−==

−==

=−

BBB

B

B

PQ

P

θθ

θ o

Source:

( ) ( )

leading 94.72%0.9472power Apparent

factor Power

kVA 84.15 power Apparent

kVAR 079.5kW 15

22

===

=+=

−=+=

=+=

s

ss

BAs

BAs

PQP

QQQPPP

173

P5

.83

.84

( )

( ) ( )

leading %48.99 power Apparent

factor Power

VA 3651 power Apparent

VAR 5.370)3.192(

447280447

W 363355

447

22

22

22

22

==

=+=

−=−

+=

+=

+=

===

PQP

XV

XV

QQQR

VP

C

rms

L

rms

CL

rms

P5

( )( ) ( )

PowerVA 1930 power Apparent

VAR 1635

W 102591.572318.4

3.192280550244702447

22

2

==

−=

+=+=

==

−∠=

−+∠

=−+∠

=

rmsrms

rmsCrmsLCL

rms

IV

IXIXQQQIRP

jjCjLjRo

oo

ωωI

( ) leading %13.535313.091.57cos factor ===

( )

.85* (a)

(b)

( ) ( )o52.752400

25.0kV 1cos−∠=IθrmsV

( )

1582.2

103.387

0QkVAR 3.387sin

23

total

θ

−=−=

=×−=

+==

==

X

XVQ

QQIVQ

C

rmsC

Cload

rmsrmsload

F1027 µω

=CCC

o

A 400kW 100cos

52.7525.0cos

===

=

=

=

θθθ

rms

rmsrms

PI

IVP

P5

174

The capacitor must be rated for at least 387.3 kVAR. With the apacitor in place, we have:

(c) ine current is smaller by a factor of 4 with the capacitor in

e, reducing

c

o0100A 100

kW 100

∠=

=

==

Irms

rmsrms

IIVP

The l plac

The ac steady state Thévenin equivalent circuit for a two-terminal circconsists of

RI 2 losses in the line by a factor of 16. P5.86 uit

a phasor voltage source Vt in series with a complex impedance Zt.

ady state Norton equivalent circuit for a two-terminal circuit

zeroing the independent sources and finding the impedance looking into he terminals of the original network. Also, we have

The ac ste

consists of a phasor current source In in parallel with a complex impedance Zt.

Vt is the open-circuit voltage of the original circuit. In is the short

circuit current of the original network. The impedance can be found by

t

To attain maximum power, the load must equal (a ) the complex coof the Théve in

ttt Z IV = .

P5.87 njugate impedance if the load can have any complex value; (b) the

of the Thévenin impedance if the load must be a pure

P5.88

nmagnituderesistance.

The load voltage is given by the voltage divider principle. LZ

LttL ZZ +

= VV

The load voltage LV is larger than the Thévenin voltage in magnitude if the magnitude of Lt ZZ + is less than the magnitude of LZ which can happen when the imaginary parts of Z and Z have opposite signsdoes not hap

t L . This pen in purely resistive circuits.

175

P5.89* (a) Zeroing the current source, we have:

in

Thus, the Thévenin impedance is Ω 57.268.11150100 o∠=+= jZt Under open circuit conditions, there is zero voltage across the

ductance, the current flows through the resistance, and the Thévenin voltage is

o

o

57.26789.1 −∠== ttn ZVIThus, the Thévenin and Norton equivalent circuits are:

0200∠== oct VV

(b) For maximum power transfer, the load impedance is

15010050100

20050100

=−++

=+

=

−=

loadt

tload

load

jjZZ

jZVI

( ) W 50)2/1(100 22 === −loadrmsloadload IRP

(c

) In the case for which the load must be pure resistance, the load for

maximum power transfer is

W 21.47

28.139190.02008.111

−∠===

==

t

tload

I

ZZoV

( )8.11150100

2 ==

+++

−loadrmsloadload

loadtload

IRPjZZ

176

P5.90 Zeroing sources, we have:

Thus, the Thévenin impedance is

4243.63472.451101

1 jj

Zt +=∠=+

= o

Writing a current equation for the node at the upper end of the current source under open circuit conditions, we have

Thus, the Thévenin and Norton equivalent circuits are:

o

o

o

36.3099.1380.9356.62

5510

45100

∠==

∠==

=+∠−

ttn

oct

ococ

Z

j

VIVV

VV

For the maximum power transfer, the load impedance is

loadI

( ) W 6.244

80.9364.154242

80.9356.6242

2 ==

∠=−++

∠=

+=

−=

−loadrmsloadload

loadt

t

load

IRPjjZZ

jZo

oV

In the case for which the load must be pure resistance, the load for maximum power transfer is

loadI

( ) W 2.151

08.62223.8472.44280.9356.62

472.4

2 ==

−∠=++∠

=+

=

==

−loadrmsloadload

loadt

t

tload

IRPjZZ

ZZ

ooV

177

P5.91 At the lower left-hand node under open-circuit conditions, KCL yields

from which we have Then, the voltages across the 5-Ω resistance and the j5-Ω inductance are zero, and KVL yields

xx II 5.0= .0=xI

°∠== − 303ocbat VV

With short circuit conditions, we have

°∠=+

°∠−= 1654243.0

55303jxI

°−∠=−== − 152121.05.0sc xxban IIIIThe Thévenin impedance is given by

Finally, the equivalent circuits are:

°∠=°−∠

°∠== 4514.14

152121.0303

t

ttZ

IV

P5.92 Under open-circuit conditions, we have

V °∠=°∠+== − 87.361002)34(oc jabt VVWith the source zeroed, we look back into the terminals and see Ω Next, the Norton current is

433 =++−= jjZt 4

°∠== 87.365.2t

tn Z

VI

178

P5.93 For maximum power transfer, the impedance of the load should be the

complex conjugate of the Thévenin impedance:

loadZ Setting real parts equal:

( )loadload CjRj ω−=−= 822

Setting imaginary parts equal:

Ω= 22loadR

( )( ) F382.6 81

18µω

ω==

−=−

load

load

CC

P5.94* For maximum power transfer, the impedance of the load should be the

complex conjugate of the Thévenin impedance:

Setting real parts equal:

0146.00401.010146.00401.01

822

jCjRYjZY

jZ

loadloadload

loadload

load

+=+=

+==

−=

ω

Ω=

=

93.240401.01

load

load

RR

Setting imaginary parts equal:

F 7.44

0146.0µ

ω=

=

load

load

CC

P5.95 We are given

( ) ( )o66302cos182 += ttvan

179

(a) By inspection, Hz 06.48

3022=

==

ffπω

(b) (c)

( ) ( )( ) ( )o

o

174302cos18254302cos182

−=

−=

ttvttv

cn

bn

( ) ( )( ) ( )o

o

54302cos182174302cos182

−=

−=

ttvttv

cn

bn

P5.96 We are given:

( )van

( ) ( )

( )( ) ( )ttv

ttttv

cn

bn

ωωω

cos10060cos10060cos100

−=

+=

−=o

o

The phasor diagram is:

For counterclockwise rotation, the sequence of phasors is acb. Thus, this is a negative sequence source. From the phasor diagram, we can determine that

( ) ( )( ) ( )( ) ( )o

o

o

o

o

o

150cos310030cos310090cos3100

1503100

303100

903100

+=

+=

−=

+∠=−=

+∠=−=

−∠=−=

ttvttvttv

ca

bc

ab

anca

cnbnbc

bnanab

ω

ω

ω

VVV

VVV

VVV

cn

180

P5.97*

.98*

P5.99

he power flow pulsates. Reduced vibration in generators nd motors is a potential advantage for the three-phase system. In

ositive seque urce. The phasor diagram is shown in Figure book. Thus, ve

( ) ( )kW 36.19

0cos67.144403cos3V

rms A 67.1430440

rmsV 1.76244033

Y

=

×××==

===

=×=×=

θL

YL

YL

IPR

VI

VV

P5

−∠=

−=

×+=

+=

−

05.6243.2370.2098.10

103775011

11

4

o

Y

jj

CjRZ

ω

Total power flow in a balanced system is constant with time. For a single phase system taaddition, less wire is needed for the same power flow in a balanced three-phase system.

Ω−∠=

=∆

05.6229.703

o

YZZ

P5.100 This is a p nce so

5.41 in the we ha

o03

2430∠=anV

The impedance of an equivalent wye-connected load is

Ω−== ∆ 1667.93

jZZY

The equivalent circuit for the a-phase of an equivalent wye-wye circuit is:

181

Thus, the line current is

( )

( )( ) W 192 1.03

kW 58.1829326.3567.20

36.295.60264.09.347

1.01.026.580.351.01.0

2

2

=×=

=×=

∠=

∠=

−∠=

+−=

∠=

++=

aArmsline

ABrmsload

AB

aAanAn

Y

anaA

IPIP

j

Zj

o

o

o

o

V

IVV

VI

=∆

ABAB Z

VI

P5.101* is shown in Figure

.41 in the book. Thus, we have: This is a positive sequence source. The phasor diagram5

3

o02430∠=anV

The impedance of a equivalent wye-connected load is

Ω−== ∆ 6667.0667.13

jZZY

The equivalent circuit for the a-phase of an equivalent wye-wye circuit is:

182

Thus, the line current is:

( )

( )( ) kW 37.51.03

kW 53.895378.473.109

98.254.58802.47.339

1.01.078.172.189

1.01.0

2

2

aA

=×=

=×=

∠=

=

∠=

−∠=

+−=

∠=

++=

∆

aArmsline

ABrmsload

ABAB

AB

aAanAn

Y

an

IPIP

Z

j

Zj

o

o

o

o

VI

V

IVV

VI

183

P5.102

o

o

o

1503

903

303

−∠×=−=

∠×=−=

−∠×=−=

Yancnca

Ybc

Ybnanab

VVV

VVV

VVV

VVV

cnbn

184

P5.103

rmsV 0.2543

4403

=== LY

VV

o

o

o

1202254

120225402254

∠=

−∠=

∠=

cn

bn

an

V

V

V

o

o

o

1502440902440

302440

∠=

−∠=

∠=

ca

bc

ab

V

V

V

( )

o

o

o

85.532054.215.1862054.2

15.662054.23.037750

∠=

−∠=

−∠=+

=

c

b

ana j

I

I

VI

P5.104 The line-to-line voltage is 240

( ) ( )

( ) ( )VAR 1431

15.66sin054.22543sin3W 9.632

15.66cos054.22543cos3

=

×××==

=

×××==

o

o

θ

θ

LrsmYrms

LrmsYrms

IVQ

IVP

rms.V 7.4153 =The impedance of each arm of the delta is

=∆Z

The equivalent wye has impedances of

o63.444.472)5/(110/1

1∠=

+ j

ZY

Then working with one phase of the wye-wye, we have

lineI A rms

o63.44491.13

∠== ∆Z

0.161240==

YZ power =P

71.44%100)cos(63.44 factor =×= o %kW 83.51)4471.0)(0.161)(240(3 =

185

P5.105 As suggested in the hint given in the book, the impedances of the circuits between terminals a and b with c open must be identical.

Equating the impedances, we obtain:

(1) CBA

CBCA

BACba ZZZ

ZZZZZZZ

ZZ++

+=

++=+

)/(1/11

Similarly for the other pairs of terminals, we obtain

(2)

CBA

CBBA

CABca ZZZ

ZZZZZZZ

ZZ++

+=

++=+

)/(1/11

CBA

BACA

BCAcb ZZZ

ZZZZZZZ

ZZ++

+=

++=+

)/(1/11 (3)

Then adding the respective sides of Equations 1 and 2, subtacting the corresponding sides of Equation 3, and dividing boths sides of the result by 2, we have:

Similarly we obtain: CBA

CBa ZZZ

ZZZ++

=

and

CBA

CAb ZZZ

ZZZ++

=CBA

BAc ZZZ

ZZZ++

=

186

P5.106 As suggested in the hint, consider the circuits shown below. The admittances of the circuits between terminals must be identical.

First, we will solve for the admittances of the delta in terms of the impedances of the wye. Then we will invert the results to obtain relationships between the impedances.

(1) cacbba

cb

cba

BC ZZZZZZZZ

ZZZ

YY++

+=

++

=+

/1/11

1

Similarly working with the other terminials, we obtain

(2)

(3)

Then adding the respective sides of Equations 2 and 3, subtacting the corresponding sides of Equation 1, and dividing boths sides of the result by 2, we have:

cacbba

baBA ZZZZZZ

ZZYY++

+=+

cacbba

caCA ZZZZZZ

ZZYY++

+=+

Inverting both sides of this result yields:

cacbba

aA ZZZZZZ

ZY++

=

Similarly, we obtain:

and

a

cacbbaA Z

ZZZZZZZ ++=

b

cacbbaB Z

ZZZZZZZ ++=

c

cacbbaC Z

ZZZZZZZ ++=

187