Chapter 19-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

Download Chapter 19-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity!

Figure 19.

For Prob. 19.1 and 19.28.

1

1 11 I

V

z

o 1 2

I = I , V 2 (^) = 2 I o = I 1

1

2 21 I

V

z

To get z (^) 22 and z (^) 12 , consider the circuit in Fig. (b).

2

2 22 I

V

z

2 2

' o 6

I I = I

= , V 1 (^) = 6 I o'= I 2

2

1 12 I

V

z

Hence, [ z ] = ⎥Ω ⎦

V 1

V 2

I 1 = 0 1 Ω

(b)

I (^) o '

4 Ω I 2 = 0

V 1

I 1

(a)

I (^) o +

V 2

2 1 ||[ 2 1 ||( 2 1 )]

1

1 11 =^ = + + + I

V

z

11 2 1 ||^2 = + =

z = + +

' o

' o o 4

I I = I

1 1

' o 15

I I = I

o 1 1 15

I = ⋅ I = I

2 o 1 15

V = I = I

12 1

2 21 =^ = = z = I

V

z

To get z (^) 22 , consider the circuit in Fig. (b).

2

2 22 =^ = + + = z = I

V

z

Thus,

[ z ] = ⎥Ω ⎦

(b)

V 1

V 2

I 1 = 0

Chapter 19, Problem 3.

Find the z parameters of the circuit in Fig. 19.67.

Figure 19.

For Prob. 19.3.

Chapter 19, Solution 3.

z 12 (^) = j 6 = z 21

z 11 (^) − z 12 (^) = 4 ⎯⎯→ z 11 (^) = z 12 + 4 = 4 + j 6 Ω

z 22 (^) − z 12 (^) = − j 10 ⎯⎯→ z (^) 22 = z 12 − j 10 = − j 4 Ω

[ ]

j j z j j

j 6 j 4

4 j 6 j 6

Chapter 19, Problem 5.

Obtain the z parameters for the network in Fig. 19.69 as functions of s.

Figure 19.

For Prob. 19.5.

Chapter 19, Solution 5.

Consider the circuit in Fig. (a).

s

1 s s 1

s

1 s s 1

s

|| 1 s

s

s

s

|| 1 s s

z = + +

s 2 s 3 s 1

s s 1 3 2

2

11

z =

1 2

o 1 1

s s 1 s 1

s

s 1

s

s

1 s s 1

s 1

s

1 s s

s

I I I I

o 3 2 1 s 2 s 3 s 1

s I I

s s 2 s 3 s 1

3 2

1 2 o

I

V I

s 2 s 3 s 1

3 2 1

2 21

I

V

z

I 2 = 0

V 1

I 1

(a)

V 2

s

1/s

Io

Consider the circuit in Fig. (b).

s 1

|| 1 s s

s

|| 1 s 1 || s

2

2 22 I

V

z

s 1

s 1 s s

s 1

1 s

s 1

1 s s

s 1

1 s s

2

22

z =

s 2 s 3 s 1

s 2 s 2 3 2

2

22

z =

z (^) 12 = z 21

Hence,

[ z ] =

⎥

s 2 s 3 s 1

s 2 s 2

s 2 s 3 s 1

s 2 s 3 s 1

s 2 s 3 s 1

s s 1

3 2

2

3 2

3 2 3 2

2

V 2

V 1

I 1 = 0 1

(b)

s

1/s

1 1 11 1 1

V I

z I I

1 1

V o = V = I

− V (^) o − 4 I (^) 2 + V 2 (^) = 0 ⎯⎯→ V 2 (^) = Vo + 4 I 1 (^) = 20 I 1 (^) + 4 I 1 (^) = 24 I 1

2 21 1

V

z I

To find z 12 and z 22 , consider the circuit below.

I 1 =0 5 Ω 10 Ω 4I 1 I 2

V 1 20 Ω

V 2

V 2 = (10 + 20) I 2 = 30 I 2

2 22 1

V

z I

V 1 = 20 I 2

1 12 2

V

z I

Thus,

25 20 [ ] 24 30

z

_

Figure 19.

For Prob. 19.7 and 19.80.

Chapter 19, Solution 7.

To get z 11 and z 21 , we consider the circuit below.

I 2 =

I 1 20 Ω 100 Ω

vx 50 Ω 60 Ω

V 1

• V 2

12vx -

Chapter 19, Problem 8.

Find the z parameters of the two-port in Fig. 19.72.

Figure 19.

For Prob. 19.8.

Chapter 19, Solution 8.

To get z 11 and z 21 , consider the circuit below.

j4 Ω

I 1 -j2 Ω 5 Ω I 2 =

j6 Ω j8 Ω

V 2

V 1

10 j 4 I

V

V ( 10 j 2 j 6 )I z 1

1 1 = − + 1 ⎯⎯→ 11 = = +

( 10 j 4 ) I

V

V 10 I j 4 I z 1

2 2 =− 1 − 1 ⎯⎯→ 21 = =− +

To get z 22 and z 12 , consider the circuit below.

j4 Ω

I 1 =0 -j2 Ω 5 Ω I (^2)

j6 Ω j8 Ω

V 2

V 1

15 j 8 I

V

V ( 5 10 j 8 )I z

2

2 2 = + + 2 ⎯⎯→ 22 = = +

( 10 j 4 ) I

V

V ( 10 j 4 )I z

2

1 1 =− + 2 ⎯⎯→ 12 = =− +

Thus,

( 10 j 4 ) ( 15 j 8 )

( 10 j 4 ) ( 10 j 4 ) [z]