Download chapter 19 circuit electrical and more Exercises Electrical Circuit Analysis in PDF only on Docsity!
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
Chapter 19, Problem 1.
Obtain the z parameters for the network in Fig. 19.65.
Figure 19.
For Prob. 19.1 and 19.28.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
Chapter 19, Solution 1.
To get z (^) 11 and z (^) 21 , consider the circuit in Fig. (a).
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
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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
I (^) o
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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
_
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
Chapter 19, Problem 7.
Calculate the impedance-parameter equivalent of the circuit in Fig. 19.71.
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
12vx -
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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]
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
Chapter 19, Problem 10.
Construct a two-port that realizes each of the following z parameters.
(a) [ ] ⎥Ω
z
(b) [ ] Ω
s
s s
s s
1 2
z
Chapter 19, Solution 10.
(a) This is a non-reciprocal circuit so that the two-port looks like the one
shown in Figs. (a) and (b).
z 11 z 22
z 12 I (^2)
(a)
V 2
+
I 1 I 2
V 1
+
20 I 2
(b)
V 2
+
I 1 I 2
V 1
+
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
(b) This is a reciprocal network and the two-port look like the one shown in
Figs. (c) and (d).
- 5 s
s
z 11 − z 12 = 1 + = +
z 22 − z 12 = 2 s
s
z 12 =
z 11 – z 12 z 22 – z 12
(c)
V 2
I 1 I 2
V 1
(d)
V 2
I 1 I 2
V 1
1 Ω 0.5 F^ 2 H
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
Chapter 19, Problem 12.
For the circuit shown in Fig. 19.73, let
[ z ] ⎥ ⎦
Find I (^) 1 , I 2 , V 1 , and V (^) 2.
Figure 19.
For Prob. 19.12.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
Chapter 19, Solution 12.
V 1 = 10 I 1 − 6 I 2 (1)
V 2 = − 4 I 2 + 12 I 2 (2)
V 2 = − 10 I 2 (3)
If we convert the current source to a voltage source, that portion of the circuit becomes
what is shown below.
4 Ω 2 Ω I (^1)
V 1
12 V
− 12 + 6 I 1 + V 1 = 0 ⎯⎯→ V 1 = 12 − 6 I 1 (4)
Substituting (3) and (4) into (1) and (2), we get
12 − 6 I 1 = 10 I 1 − 6 I 2 ⎯⎯→ 12 = 16 I 1 − 6 I 2 (5)
− 10 I 2 = − 4 I 1 + 12 I 2 ⎯⎯→ 0 = − 4 I 1 + 22 I 2 ⎯⎯→ I 1 = 5.5 I 2 (6)
From (5) and (6),
12 = 88 I 2 − 6 I 2 = 82 I 2 ⎯⎯→ I 2 =0.1463 A
I 1 = 5.5 I 2 =0.8049 A
V 2 = − 10 I 2 = −1.463 V
V 1 = 12 − 6 I 1 =7.1706 V
_
