Electrical Engineering Chapter 5 Homework The period corresponds to 360 therefore 5

1

CHAPTER 5

Exercises

E5.1 (a) We are given

)30200cos(150)( 



ttv 

. The angular frequency is

the coefficient of

t

so we have

radian/s 200





. Then

Hz 1002/ 

f

ms 10/1 

fT

V 1.1062/1502/ 

mrms VV

Furthermore,

v

(

t

) attains a positive peak when the argument of the

cosine function is zero. Thus keeping in mind that

t

has units of

radians, the positive peak occurs when

ms 8333.0

180

30 maxmax 

tt 



(b)

W 225/

2

RVP rmsavg

(c) A plot of

v

(

t

) is shown in Figure 5.4 in the book.

)30300cos(100)60300sin(100  

tt 

The period corresponds to



360

therefore 5 ms corresponds to a phase

angle of

 108360)67.16/5( 

. Thus the voltage is

)108377cos(6.155)( 



ttv

V

)45cos(14.14)sin(10)cos(10 



ttt 

(b)

330.45.25660.86053010

1

jj

 

I



44.318.11670.016.11 

j

)44.3cos(18.11)30sin(5)30cos(10  

ttt 

(c)

99.125.70206015020

2

jj

 

I



28.2541.3099.125.27 

j

)28.25cos(41.30)60cos(15)90sin(20  

ttt 

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2

E5.5 The phasors are

 4510 and 3010 3010 321  VVV

v

1 lags

v

2 by



60

(or we could say

v

2

leads

v

1 by

)60

v

1 leads

v

3 by



15

(or we could say

v

3

lags

v

1 by

)15

v

2 leads

v

3 by



75

(or we could say

v

3

lags

v

2 by

)75

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3



13507.7  IV

R



4507.7  IV

Lj

L

(b) The phasor diagram is shown in Figure 5.17b in the book.

(c)

i

(

t

) lags

vs

(

t

) by

.45

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4

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

,1)cos( 



which implies that

the current and voltage are in phase and

.0



Thus,

.0)tan( 

PQ

Also

 

A. 10)]0cos(500/[5000]cos/[ 



rmsrms VPI

Thus we have

.4014.14 and 14.142

 I

rmsm II

implies that the current lags the voltage by

.46.78)2.0(cos 1

 



Thus,

.kVAR 49.24)tan( 

PQ

Also, we have

 

A. 0.50]cos/[ 



rmsrms VPI

Thus we have

.46.3871.70 and A 71.702

 I

rmsm II

of part (b) than for that of part (a). Wiring costs would be lower for the

load of part (a).

 903.265)/(1

CjZC



90

C



A 770.3/ 

C

rms

Crms ZVI

0)cos( 

CCrms

rms

CIVP 

kVAR 770.3)sin( 

CCrms

rms

CIVQ 

power factor of 80% lagging from which we have

.87.36)8.0(cos 1

2



 



Notice that we select a positive angle for

2



because the load has a

lagging power factor. Thus we have

kW 0.8)cos( 222 



rms

rms IVP

and

kVAR 6)sin(

22 



rms

rms IVQ

.

kW 8

2

PPP C

s

kVAR 23.2

2

QQQ C

s

kVA 305.8

22 

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.

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5

50502550

100/1100/1

1

2550

jj

j

jZt









04.141.103j25100 



4571.70

100100

100

100 





j

oct

VV



04.596858.0/ 

tt ZV

n

I



453536.0

2510025100

4571.70 







jj

I

The load power is

W 25.6)2/3536.0(100 22 

rms

LL IRP

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6

(b) For a purely resistive load, maximum power is transferred

for

. 1.10325100 22 

t

LZR

The Thévenin equivalent with the

load attached is:



98.373456.0

251001.103

4571.70 







j

I

The load power is

W 157.6)2/3456.0(1.103 22 

rms

LL IRP

Van is zero. Then we have

 1204.577 1204.577 04.577 

cn

bn

an

VVV

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.)



02.37610.4

40.75100

04.577 







jZ L

an

aA

V

I

The currents for phases

b

and

c

are the same except for phase.

 98.82610.4 02.157610.4 

cC

bB

II

kW 188.3)02.37cos(

2

610.44.577

3)cos(

2

3



 



L

YIV

P

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or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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7

kVAR 404.2)02.37sin(

2

610.44.577

3)sin(

2

3



 



L

YIV

Q

 04.57703/1000 

an

V

The phase impedance of the equivalent Y is

. 67.163/503/  

ZZY

A 063.34

67.16

04.577 









Y

an

Z

V

IaA

Similarly,

A 12063.34 



bB

I

and

A. 12063.34 



cC

I

Finally, the power is

kW 00.30)2/(3 2

y

aA RIP



601

805030100

211 





jj

VVV



302

805050

122 





jj

VVV











































































302

601

8050

1

50

1

8050

1

8050

1

8050

1

30100

1

2

1

V

jjj

Y = [(1/(100+j*30)+1/(50-j*80)) (-1/(50-j*80));…

(-1/(50-j*80)) (1/(j*50)+1/(50-j*80))];

I = [pin(1,60); pin(2,30)];

V = inv(Y)*I;

pout(V(1))

pout(V(2))



21.10698.79

1V

and



30.11613.124

2V

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8

Problems

P5.1 The units of angular frequency



are radians per second. The units of

frequency

f

are hertz, which are equivalent to inverse seconds. The

radian measure of angle is length divided by length. In terms of physical

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9

P5.5

 

   

 30500cos50120500sin50 

tttv 

 

W 25

V 36.352502

ms 41

Hz 250

rad/s 500

2







RVP

VV

fT

f

rms

mrms



ms 3333.0

06500

1





t

ππ

The first positive peak after

0

t

is at

ms 667.3

1

Tttpeak

The peak voltage is

28.282022 rms 

VVm

V. The frequency is



Tf

/1

10 kHz and the angular frequency is

4

1022 ππω 

f

radians/s. The phase corresponding to a time interval of



t

Δ

20



s is

 72360)/( 

Tt

Δθ

. Thus, we have

)72102cos(28.28)( 4



ttv

π

V.

28.282202

T



20002

f

rad/s



108360 max 

T

t



)1082000cos(28.28)( 



tti 

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10

P5.8

 

V 250sin10

ttv 



   

 

W 500sin15.2250sin5 22

ttRtvtp 



2

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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11

values divided by the square root of two. However, they are for

sinusoids, which are important special cases.

P5.12*

 

A 808.3425

4

11 2

0

4

20

2











  

dtdtdtti

T

I

T

rms

P5.13

 

V 61.1015

2

11

0

2

0

2 

dtdttv

T

i

V

rms

P5.14

 

   

 

1

0

1

0

2

0

2)2exp(100)exp(10

1

dttdttdttv

T

V

T

rms

 

V 575.6)]2exp(1[50)2exp(50 1

0 



t

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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12

P5.15

 

 

 

1

0

2

0

2)2sin(2)2cos(

1

dttBtAdttv

T

V

T

rms 

 

































2

4

2

4

0

2

)2(sin4)2sin()2cos(4)2(cos

22

1

0

22

1

0

2222

BA

t

B

t

A

dttBttABtAV

t

rms



syms Vrms t A B

Vrms = sqrt((int((A*cos(2*pi*t)+2*B*sin(2*pi*t))^2,t,0,1)))

Vrms =

(A^2/2 + 2*B^2)^(1/2)

 

   

 



025.0

2

2/

2)40cos(8820)20cos(420

1

dttdttdttv

T

V

T

rms 

V 228)40sin(

40

160

025.0



















t

rms ttV 



P5.17

 

V 774.5

3

25

2

1

5

2

11 2

0

3

2

0

2

0

2



















t

T

rms

t

dttdttv

T

V

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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13

P5.18

 

dttv

T

i

V

T

rms





0

2

 

V 5

06.109.0

1.0

1

)]40cos(1616)20cos(2249

1.0

1

)]20(cos32)20cos(2249

1.0

1

)]20(cos32)20cos(2249[

1.0

1

)]20cos(243[

1.0

1

1.0

0

1.0

0

1.0

0

1.0

0

1.0

0

1.0

0

2

1.0

0

1.0

0

2

1.0

0

2

































 

 



dttdtdttdt

dttdttdt

dttt

dttVrms





complex number whose magnitude equals the peak amplitude of the

sinusoid and whose phase is the phase angle of the sinusoid written as a

cosine function.

2. Add the phasors and convert the sum to polar form using complex

arithimetic.

3. Convert the resulting phasor to a sinusoid.

counterclockwise. If phasor

A

points in a given direction before phasor

B

by an angle



, we say that

A

leads

B

by the angle



or that

B

lags

A

by the

angle

.

(b) 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

 360)/(

Tt

or that

B

lags

A

by the angle

.

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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14

P5.21* We are given the expression

)sin(4)75cos(3)75cos(5

ttt

ωωω  

Converting to phasors we obtain

  904753755

 4)8978.27765.0(8296.42941.1

jjj



09.82763.37274.35176.0 

j

Thus, we have

)09.82cos(763.3

)sin(4)75cos(3)75cos(5









t

ttt

ω

ωωω

P5.23*

   

 



45cos4.141

454.141100100

10090100

1000100

90cos100sin100

21

s

2

1

2











ttv

j

tttv

s

ω

ωω

VVV

V



45 by leads

45 by lags

90 by lags

2

s

1

s

12

VV

 

   



60 by leads

60 by lags

120 by lags

90400cos10

)150400cos(5

30400cos10

4002

32

31

21

3

2

1

tvtv

ttv

f





is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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15

P5.24 The magnitudes of the phasors for the two voltages are

212

and

27

V. The phase angles are not known. If the phase angles are the same, the

phasor sum would have its maximum magnitude which is

219

. On the

other hand, if the phase angles differ by



180

, the phasor sum would

have its minimum magnitude, which is

25

. Thus, the maximum rms value

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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16

 66.5982.22

Thus, we have

)66.59cos(82.22

)120cos(10)30cos(15)45sin(5









t

ttt







3015

1V

V



10071.7

1I

A

 



10cos071.7

1

tti 

A

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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17

P5.33*

   



90 by leads

tvti CC

409040605.230100

jZ

 

I

V

IV

Because

Z

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

200



mH200



Z

L

25902530460100

jZ

 

I

V

IV

Because

Z

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

200



F200

1







Z

C

Because

Z

is pure real, the element is a resistance of 20 .

and

.500





Also, we see that the current lags the voltage by 1 ms or

 

ttti

Z

j

C

j

Z

ttv

C

ππ

ω

πω

π

2000sin6283.0902000cos6283.0

906283.0

010

9092.1592.15

2000

2000cos10

CC

C















VI

V

Ω

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

18

90, so we have an inductance. . Finally,

, 5000/

mm IVL

from

which we find that



L

3.183 H.

(b) From the plot, we see that

16

T

ms, so we have

5.62/1 

Tf

Hz

and

.125





Also, we see that the current leads the voltage by 4 ms or

90, so we have a capacitance. Finally,

, 2500//1 

mm IVC

from

which we find that



C

1.019 F.

Because

Z

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

1000



F50

1







Z

C

Because

Z

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

1000



mH10



Z

L

(c)

505  

I

V

Z

Because

Z

is pure real, the element is a resistance of 5 .

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.

All of the sources must have the same frequency.



45 by lags

s

VI

V 45071.7

mA 4571.70

100100

010















IV

Lj

R

j

L

R

ω

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

19

P5.39

mA 43.6372.44

200100

010 















jLjR

s



V

I

V 43.63472.4 

 IV

R

V 57.26944.8 

 IV

Lj

L



43.63 by lags

s

VI



43.63 by leads s

VI

V 45071.7 

 IV

R

 

V 45071.7 

 IV

Cj

C

I leads Vs by 45

V

s

I

V

R

V

L

63.43

 

V57.26944.8

V 43.63472.4

mA 43.63472.4

20001000

010

















IV

Cj

R

j

CjR

C

R

ω

V

s

I

V

R

V

C

45

is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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20

P5.42*

C

jRLjZ

ω

ω1





57.711.158 01550

2005050 :500





Ω

j

jjZ

ω



050 50

10050100 :1000





Ω

jjZ

ω



57.711.158 15050

5050200 :2000





Ω

j

jjZ

ω

 







31.5647.55153846769230

01.0005.0501

1

:500

.j .

jj

Z

 01000100 :1000

jZ



 31.5647.55153846769230 :2000

.j .Z

:500







57.26944.848

05.0)55/(1

1

j

jj

Z



 57.2636.221020 :1000

jZ



 60.85519.65.65.0 :2000

jZ



is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

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is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,

or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to:

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.