Fundamentals of Electric Circuits, Second Edition...

Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku

Copyright ©2004 The McGraw-Hill Companies Inc.

1

Chapter 11, Problem 7 (6).

Given the circuit of Fig. 11.40, find the average power absorbed by the 10-Ω resistor.

Figure 11.40

Chapter 11, Solution 7 (6). Applying KVL to the left-hand side of the circuit,

oo 1.04208 VI +=°∠ (1) Applying KCL to the right side of the circuit,

05j105j

8 11o =

−++

VVI

But, o11o 105j10

5j1010

VVVV−

=→−

=

Hence, 01050j

5j108 o

oo =+−

+V

VI

oo 025.0j VI = (2)

Substituting (2) into (1),

)j1(1.0208 o +=°∠ V

j12080

o +°∠

=V

°∠== 25-2

1010

o1

VI

=

== )10(2

10021

R21

P2

1I W250



1

Chapter 11, Problem 12(11).

For each of the circuits in Fig. 11.45, determine the value of load Z for maximum power transfer and the maximum average power transferred.

Figure 11.45

Chapter 11, Solution 12(11).

We find ThZ using the circuit in Fig. (a).

882.1j471.0)4j1(178

j28(8)(-j2)

-j2||8Th −=−=−

==Z

== *

ThL ZZ Ω+ 882.1j471.0 We find ThV using the circuit in Fig. (b).

8 Ω -j2 ΩZth

(a)

4∠0° A 8 Ω -j2 Ω

(b)

+

Vth

-

Io



2

)04(2j8

2j-o °∠

−=I

j28j64-

I8 oTh −==V

=

==)471.0)(8(

6864

R8P

2

L

2

Thmax

VW99.15

We obtain ThZ from the circuit in Fig. (c).

167.1j5.23j9

)3j4)(5(2j)3j4(||52jTh +=

−−

+=−+=Z

== *ThL ZZ Ω− 167.1j5.2

From Fig.(d), we obtain ThV using the voltage division principle.

°∠

−−

=°∠

−−

= 303

10j33j4

)3010(3j93j4

ThV

=

⋅==

)5.2)(8(3

10105

R8P

2

L

2

Thmax

VW389.1

4 Ω

j2 Ω

5 Ω -j3 Ω

(c)

Zth

4 Ω

j2 Ω

5 Ω -j3 Ω

(d)

10∠30° V + - +

Vth

-



1

Chapter 11, Problem 20(17).

The load resistance RL in Fig. 11.53 is adjusted until it absorbs the maximum average power. Calculate the value of RL and the maximum average power.

Figure 11.53

Chapter 11, Solution 20(17). Combine j20 W and -j10 W to get

-j20-j10||20j = To find ThZ , insert a 1-A current source at the terminals of LR , as shown in Fig. (a).

At the supernode,

10j-20j-401 211 VVV

++=

21 4j)2j1(40 VV ++= (1)

Also, o21 4IVV += , where 40

- 1o

VI =

1.11.1 2

121

VVVV =→= (2)

1 A

40 W

-j10 W

V1 V2

(a)

+ -

4 Io

-j20 W

Io



2


22 4j1.1

)2j1(40 VV

+

+= 4.6j1

442 +=V

Ω−== 71.6j05.11

2Th

VZ

== ThLR Z Ω792.6

To find ThV , consider the circuit in Fig. (b).

At the supernode,

j10-j20-40120 211 VVV

+=−

21 4j)2j1(120 VV ++= (3)

Also, o21 4IVV += , where 40

120 1o

VI

−=

1.1122

1

+=

VV (4)


2)818.5j9091.0(82.21j09.109 V+=−

°∠=+−

== 92.43-893.18818.5j9091.082.21j09.109

2Th VV

===)792.6)(8(

)893.18(R8

P2

L

2

Thmax

VW569.6

40 W

-j10 W

V1 V2

(b)

+ -

4 Io

-j20 W

Io

+

Vth

-

120Ð0° V + -



1

Chapter 11, Problem 51(38). For the entire circuit in Fig. 11.71, calculate: (a) the power factor (b) the average power delivered by the source (c) the reactive power (d) the apparent power (e) the complex power

Figure 11.71


)6j8(||)5j10(2T +−+=Z

j1820j110

2j18

)6j8)(5j10(2T +

++=

++−

+=Z

°∠=+= 5.382188.8768.0j152.8TZ

=°= )5.382cos(pf (lagging)9956.0

)5.382-188.8)(2(

)16(22

1 2

*

2

*

°∠===

ZV

IVS

°∠= 5.38263.15S

=θ= cosSP W56.15

=θ= sinSQ VAR466.1

== SS VA63.15

=°∠= 382.563.15S VA466.1j56.15 +



1

Chapter 11, Problem 57(42). For the circuit in Fig. 11.77, find the average, reactive, and complex power delivered by the dependent voltage source.

Figure 11.77


At node o,

j-1424 1ooo VVVV −

+=−

1o 4j)4j5(24 VV −+= (1)

At node 1, 2j

2j-

1o

1o VV

VV=+

−

o1 )4j2( VV −= (2)


o)168j4j5(24 V−−+= j411

24-o +=V ,

j411j4)-(-24)(2

1 +=V

The voltage across the dependent source is o1o12 4)2)(2( VVVVV +=+=

4j11)4j6)(24-(

)44j2(j411

24-2 +

−=+−⋅

+=V

)2(21

21 *

o2*

2 VVIVS == )4j6(137576

j4-1124-

4j11)4j6)(24-(

−

=⋅+−

=S

=S VA82.16j23.25 −

1 Ω 2 Vo

2 Ω -j1 Ω4 Ω

24∠0° V + - j2 Ω

Vo V1

+

V2

-



1

Chapter 11, Problem 60(45). For the circuit in Fig. 11.80, find Vo and the input power factor.

Figure 11.70


15j20))8.0(sin(cos8.0

20j20S 1-1 +=+=

749.7j16))9.0(sin(cos9.0

16j16S 1-

2 +=+=

°∠=+=+= 29.32585.42749.22j36SSS 21

But o

*o V6IVS ==

==6S

Vo °∠ 29.32098.7

=°= )29.32cos(pf (lagging)8454.0



1

Chapter 11, Problem 62(47). For the circuit in Fig. 11.82, find Vs.

Figure 11.72


25.11j15))8.0(sin(cos8.0

15j15 1-

2 −=−=S

But *222 IVS =

12025.11j15

2

2*2

−==

VS

I

09375.0j125.02 +=I )15.0j3.0(221 ++= IVV

)15.0j3.0)(09375.0j125.0(1201 +++=V 0469.0j02.1201 +=V

843.4j10))9.0(sin(cos9.0

10j10 1-

1 +=+=S

But *111 IVS =

°∠°∠

==02.002.12084.25111.11

1

1*1 V

SI

0405.0j0837.025.82-093.01 −=°∠=I 053.0j2087.021 +=+= III

)04.0j2.0(1s ++= IVV )04.0j2.0)(053.0j2087.0()0469.0j02.120(s ++++=V

0658.0j06.120s +=V =sV V03.006.120 °∠

0.2 + j0.04 Ω I

I1

I2

+ - Vs

0.3 + j0.15 Ω

+

V2

-

+

V1

-



1

Chapter 11, Problem 74(57). A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in Fig. 11.90. (a) Find the power factor of the parallel combination. (b) Calculate the value of the capacitance connected in parallel that will raise the

power factor to unity.

Figure 11.90


°==θ 87.36)8.0(cos-11 kVA30

8.024

cosP

S1

11 ==

θ=

kVAR18)6.0)(30(sinSQ 111 ==θ= kVA18j241 +=S

°==θ 19.18)95.0(cos-12 kVA105.42

95.040

cosP

S2

22 ==

θ=

kVAR144.13sinSQ 222 =θ=

kVA144.13j402 +=S

kVA144.31j6421 +=+= SSS

°=

=θ 95.2564144.31

tan 1- =θ= cospf 8992.0

°=θ 95.252 , °=θ 01 kVAR144.31]0)95.25tan([64]tantan[PQ 12c =−°=θ−θ=

=π

=ω

= 22rms

c

)120)(60)(2(144,31

VQ

C mF74.5

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