Basic Engineering Circuit Analysis 9th (From Brad-Pc - 3)

Irwin, Basic Engineering Circuit Analysis, 9/E 1

Chapter 8: AC Circuit Analysis Techniques Problem 8.1

8.1

Ch08.indd 1Ch08.indd 1 12/18/08 10:42:39 AM12/18/08 10:42:39 AM






8.3

Figure P8.3

P8.3







) A

8.5

Figure P8.5

P8.5













8.9

Figure P8.9

P8.9.







8.11

Figure P8.11

P8.11.







8.13

Figure P8.13

P8.13.







8.15

Figure P8.15

P8.15









8.17

Figure P8.17

P8.17







8.19

Figure P8.19

P8.19




-

8.20

Figure P8.20

P8.20













8.22

Figure P8.22

P8.22










8.24

Figure P8.24

P8.24







- t

8.26

Figure P8.26

P8.26.




8.27

Figure P8.27

P8.27







8.28

Figure P8.28

P8.28




-




8.29

Figure P8.29

P8.29.










8.31

Figure P8.31

P8.31.







-j26.53

8.32

Figure P8.32

P8.32.







8.33

Figure P8.33

P8.33.







8.34

Figure P8.34

P8.34.




"




8.35

Figure P8.35

P8.35.







8.36

Figure P8.36

P8.36.










8.37

Figure P8.37

P8.37.







x

x

o

Figure P8.38

8.38 P8.38.







8.39

Figure P8.39

P8.39.




-

--

-

-

-

-







8.41

Figure P8.41

P8.41.




8.42

Figure P8.42

P8.42.




8.43

Figure P8.43

P8.43.




8.44

Figure P8.44

P8.44.







8.46

Figure P8.46

P8.46.







o

o

_

o_

"

8.48

Figure P8.48

P8.48.










8.50

Figure P8.50

P8.50










8.52

Figure P8.52

P8.52.







8.53

Figure P8.53

P8.53.







"

8.54

Figure P8.54

P8.54.










8.56

Figure P8.56

P8.56.







8.57

Figure P8.57

P8.57,













8.59

Figure P8.59

P8.59
















8.61

Figure P8.61

P8.61,










Figure P8.62







8.63

Figure P8.63

P8.63.



















-

8.67

Figure P8.67

P8.67.







8.68

Figure P8.68

P8.68.




8.69

Figure P8.69

P8.69
















8.71

Figure P8.71

P8.71.










x

>o

8.72

Figure P8.72

P8.72.







8.73

Figure P8.73

P8.73




8.74

Figure P8.74

P8.74.













8.76

Figure P8.76

P8.76.






















8.79

Figure P8.79

P8.79.
















8.81

Figure P8.81

P8.81.




-




8.82

Figure P8.82

P8.82.







8.83

Figure P8.83

P8.83.













8.85

Figure P8.85

P8.85.






















8.89

Figure P8.89

P8.89.




o

o










8.90

Figure P8.90

P8.90




''













8.92

Figure P8.92

P8.92.



















8.96

Figure P8.96

P8.96.







8.97

Figure P8.97

P8.97.







8.98

Figure P8.98

P8.98













8.100

Figure P8.100

P8.100







8.101

Figure P8.101

P8.101.




o













8.103

Figure P8.103

P8.103,
















8.105

Figure P8.105

P8.105













8.106

Figure P8.106

P8.106













8.108

Figure P8.108

P8.108






















8.111

Figure P8.111

P8.111.







8.112

Figure P8.112

P8.112







8.113

Figure P8.113

P8.113



















- - -

--

--

---

- -

- - -

- -

8.115

Figure P8.115

P8.115.







+







8.116

Figure P8.116

P8.116






















8.118

Figure P8.118

P8.118.













8.119

Figure P8.119

P8.119.











































8.123

Figure P8.123

P8.123






















8.125

Figure P8.125

P8.125













8.126

Figure P8.126

P8.126.
















8.127

Figure P8.127

P8.127.













8.128

Figure P8.128

P8.128.




8.129

Figure P8.129

P8.129







SOLUTION:

V��04 �1

�1j

�� 1j

�2 V��012+

V0

-

The correct answer is d.

A��08�1

�1j

�� 1j�2

+

V0

-

Chapter 8: AC Circuit Analysis Techniques Problem 8FE-1



��

� 8.04.012)1(2 jjjZ

A��08�1

�1j

Z

I0

� � Ajjj

jI 8.46.108118.04.0

8.04.00 ��

��

��

��

� �� VjV �� 6.7106.518.46.10




SOLUTION:

1I

�� 1j

�1

V��012

xI2

1V 2V

text

0V

2I

xI

A��02

�2

The correct answer is c.




VV �� 0122

11

jVI x �

�

02 1 �� III xx

01

01211

2 111 ��

��

�

��

��

VjV

jV

0)012(112 111 �� jVjVV

Vjj

jV 6.32.113

)012(11 ��

��

�

KCL at node 0:0022 2 �� II x

00221

2 021 ��

�

��

��

VVjV

0022

0121

6.32.12 0 ��

�

��

�� Vjj

VV �� 97.88.300




SOLUTION:

V��012

�� 1j

�4

�2j

02V

�2

2I1I

�

�

0V+

-

I

The correct answer is a.

21 III ��

21 III ��KVL around the left loop:

01 222012 VIjI ��

20 4IV �

�� 012)28()22( 21 IjIjKVL around the right loop:

220 4122 IIjIjV ��

0)14(2 21 �� IjIj




Two equations and two unknowns:�� 012)28()22( 21 IjIj

0)14(2 21 �� IjIjIt follows that:

AI �� 4524.41

AI �� 96.3006.22

)96.3006.2(40 ��VVV �� 96.3024.80




SOLUTION:

The correct answer is b.

cZ is small at midband.

65

155

��

�kk

kVV

s

x

160126

)12(61040 30 ��

� �

��

kkkkx

VV

x

� � 33.1331606500 ��

��

�

��

��

xs

x

s VV

VV

VV




SOLUTION:

V��06Zeq

sI

The correct answer is c.

��

� 34.653.4343.1858.133111)31)(11(jjjjZeq

AI s ��

�� 34.63245.1

34.653.406




V��06�1

�3j

�1

�3 �� 1jIo

KVL around the outer loop:)11()34.63245.1(306 0 jI ��

AI �� 92.3248.10



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Basic Engineering Circuit Analysis 9th (From Brad-Pc - 3)