circuit analysis chp5

Irwin, Basic Engineering Circuit Analysis, 9/E 1

Chapter 5: Additional Techniques for Circuit Analysis Problem 5.1

+–

5.1

Figure P5.1

P5.1

Ch05.indd 1Ch05.indd 1 12/22/08 4:46:13 PM12/22/08 4:46:13 PM



+

.

.

|

_

|

5.3

Figure P5.3

P5.3




6k

T'-0K

5.5

Figure P5.5

P5.5




5.6

Figure P5.6

P5.6




5.7

Figure P5.7

P5.7



-

-


5.9

Figure P5.9

P5.9




5.10

Figure P5.10

P5.10




5.11

Figure P5.11

P5.11




5.12

Figure P5.12

P5.12




5.13

Figure P5.13

P5.13



'





5.14

Figure P5.14

P5.14



A


5.16

Figure P5.16

P5.16.



eq/

eq/





5.17

Figure P5.17

P5.17



0'


5.19

Figure P5.19

P5.19.



'''





5.21

Figure P5.21

P5.21



+-


5.22

Figure P5.22

P5.22



+_


5.23

Figure P5.23

P5.23




5.24

Figure P5.24

P5.24.




5.25

Figure P5.25

P5.25



+-


5.27

Figure P5.27

P5.27




5.29

Figure P5.29

P5.29




5.30

Figure P5.30

P5.30



"





5.31

Figure P5.31

P5.31



+–


5.32

Figure P5.32

P5.32




5.34

Figure P5.34

P5.34.




5.35

Figure P5.35

P5.35




5.36

Figure P5.36

P5.36.




5.39

Figure P5.39

P5.39,




5.40

Figure P5.40

P5.40




5.41

Figure P5.41

P5.41.



+-


5.43

Figure P5.43

P5.43



+–


5.45

Figure P5.45

P5.45.



+-

5.46

Figure P5.46

P5.46




+-


5.48

Figure P5.48

P5.48



+-


5.49

Figure P5.49

P5.49.




5.51

Figure P5.51

P5.51




5.52

Figure P5.52

P5.52




5.53

Figure P5.53

P5.53,



>


5.54

Figure P5.54

P5.54.




5.56

Figure P5.56

P5.56.



+-

+-


5.58

Figure P5.58

P5.58.




5.60

Figure P5.60

P5.60




5.64

Figure P5.64

P5.64




5.65

Figure P5.65

P5.65



+- +-

2 x

x


5.67

Figure P5.67

P5.67.



-





5.68

Figure P5.68

P5.68



x


5.69

Figure P5.69

P5.69




5.72

Figure P5.72

P5.72.



+–


5.73

Figure P5.73

P5.73




5.74

Figure P5.74

P5.74




5.77

Figure P5.77

P5.77




5.79

Figure P5.79

P5.79




5.80

Figure P5.80

P5.80



+–


5.82

Figure P5.82

P5.82




5.84

Figure P5.84

P5.84




5.85

Figure P5.85

P5.85




5.86

Figure P5.86

P5.86.




5.88

Figure P5.88

P5.88.




5.90

Figure P5.90

P5.90.




5.94

Figure P5.94

P5.94




5.95

Figure P5.95

P5.95




5.97

Figure P5.97

P5.97.




5.98

Figure P5.98

P5.98.




5.99

Figure P5.99

P5.99.




5.101

Figure P5.101

P5.101.



=




++

– ––

+


5.103

Figure P5.103

P5.103.



+-0


5.104

Figure P5.104

P5.104.




5.105

Figure P5.105

P5.105.




5.106

Figure P5.106

P5.106.




5.107

Figure P5.107

P5.107.



-+-

+


5.109

Figure P5.109

P5.109,



-+


5.110

Figure P5.110

P5.110



SOLUTION:


5.111

Figure P5.111

P5.111.



SOLUTION:


5.113

Figure P5.113

P5.113.



SOLUTION:

mA4

1R

�

�

'

ocV

4R2

R

'I

3R

The correct answer is d.Use superposition.

)4('321

1 mRRR

RI ��

��

��

mAmkkk

kI 1)4(211

1' �

��

��

��

)2)(1('3

' kmRIVoc

VVoc 2'

Chapter 5: Additional Techniques for Circuit Analysis Problem 5FE-1



V12

1R

�

�

"

ocV

4R2R

3R

)12(321

3"

��

��

��

RRR

RVoc

)12(211

2" ��

��

��

kkkkVoc

VVoc 6"

VVVV ocococ 862"' ��

1R

4R2R

3R THR

� � �4321RRRRRTH ��

2 (2 )1 2

2 2TH

k kR k kk k

� ��




VVoc 8

� kRTH 2

LR

LLL RIP 2

max

THL RR for maximum power.

THTH

ocL R

RVP

2

2max ��

��

mWkR

VP

TH

ocL 8

)2(4

8

4

22

max




SOLUTION:

xV2V12

�

�

ocV

�k1I

+

-

�k2

�� xV

The correct answer is c.

xVkIkI 21212 ��)2( kIVx

)2(21212 kIkIkI ��

mAI7

12

ocVkI � 212

ocVmk ��

��

7

12212

VVoc 57.8




V12

�k1

+

- xV2

�k2

scI1I

2I

�� xV

mAk

I 62

121 �

��

��

0212

� xVkI0)2(21 12 � kIkI

mAI 242 �

scIII �21

mAmmI sc 30)24(6 ��

� 7.28530

57.8

mIVRsc

ocTH

THL RR for maximum power.

VVoc 57.8

� 7.285THR

LR

mWRV

PTH

ocL 3.64

)7.285(4

)57.8(

4

22

max




SOLUTION:

V12

1R

�

�

ocV

2R

textxI2

I

xI

The correct answer is a.xx III 2�

xII 3II x 12312 �

)3(12312 xx II �

AI x13

4

VIIV xoc 08.1113

4)3(12)3(1212 �

��

��




V12

1R

2R

textxI2

2I

xI

scI

2232 IIIIII xxxsc ��

AI 02

xsc II 3

xI312 AI x 4

AI sc 12)4(3

� 92.012

08.11

sc

ocTH I

VR

VVoc 08.11

� 92.0THR

LR

�12

�� 92.121292.0LR� 92.12LR for maximum power transfer.




SOLUTION:

�3A10

�2�4

�2'I

The correct answer is c.

Use superposition.

AI 4)10(32

2' ��

��

�

�3 �2�4

�2"I

V20




AI 45

20"

"' III �44 ��I

AI 0




SOLUTION:

Use superposition.

�3

A12

�4

�2

�

�

'

ocV

a

b

'I

The correct answer is b.

AI 8)12(42

4' ��

��

�

VIVoc 16)2(8)2(''




�3�4

�2

�

�

"

ocV

a

b

V12

VVoc 4)12(42

2" ��

��

�

VVoc 12416 �



circuit analysis chp5

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