Solution manual fundamentals of electric circuits 3rd edition chapter19

Chapter 19, Problem 1.
Obtain the z parameters for the network in Fig. 19.65.

Figure 19.65
For Prob. 19.1 and 19.28.

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Chapter 19, Solution 1.
To get z 11 and z 21 , consider the circuit in Fig. (a).

1Ω

4Ω
Io

+
I1

I2 = 0
+

V1

V2

−

−
(a)

z 11 =

V1
= 1 + 6 || (4 + 2) = 4 Ω
I1

1
I ,
2 1
V2
=
= 1Ω
I1

Io =
z 21

V2 = 2 I o = I 1

To get z 22 and z 12 , consider the circuit in Fig. (b).
I1 = 0

1Ω

4Ω

Io '

+

+

V1

V2

−

−
(b)

z 22 =

V2
= 2 || (4 + 6) = 1.667 Ω
I2

2
1
I2 = I2 ,
2 + 10
6
V1
=
= 1Ω
I2

Io' =
z 12

Hence,

V1 = 6 I o ' = I 2

⎡4
1 ⎤
[z ] = ⎢
⎥Ω
⎣ 1 1.667 ⎦

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Chapter 19, Problem 2.

* Find the impedance parameter equivalent of the network in Fig. 19.66.

Figure 19.66

For Prob. 19.2.

* An asterisk indicates a challenging problem.

Chapter 19, Solution 2.

Consider the circuit in Fig. (a) to get z 11 and z 21 .
1Ω

Io '

1Ω

1Ω
Io

+
I1

1Ω

I2 = 0
+

V1

V2

−

−
1Ω

1Ω

1Ω

1Ω

(a)

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z 11 =

V1
= 2 + 1 || [ 2 + 1 || (2 + 1) ]
I1

⎛
z 11 = 2 + 1 || ⎜ 2 +
⎝

(1)(11 4)
11
3⎞
⎟= 2+
= 2 + = 2.733

15
4⎠
1 + 11 4

1
1
Io' = Io'
1+ 3
4
1
4
Io' =
I1 = I1
1 + 11 4
15
1 4
1
I o = ⋅ I1 = I1
4 15
15

Io =

V2 = I o =
z 21 =

1
I
15 1

V2
1
=
= z 12 = 0.06667
I 1 15

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

I1 = 0

1Ω

1Ω

1Ω

1Ω

+

+

V1

V2

−

−
1Ω

1Ω

1Ω

1Ω

(b)
z 22 =

V2
= 2 + 1 || (2 + 1 || 3) = z 11 = 2.733
I2

Thus,

⎡ 2.733 0.06667 ⎤
[z ] = ⎢
⎥Ω
⎣ 0.06667 2.733 ⎦

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Chapter 19, Problem 3.
Find the z parameters of the circuit in Fig. 19.67.

Figure 19.67
For Prob. 19.3.

Chapter 19, Solution 3.
z12 = j 6 = z21
z11 − z12 = 4
⎯⎯
→ z11 = z12 + 4 = 4 + j 6 Ω

z22 − z12 = − j10

⎯⎯
→ z22 = z12 − j10 = − j 4 Ω

⎡4 + j6 j6 ⎤
⎡4 + j6 j6 ⎤
Ω = ⎢
Ω
[ z] = ⎢
⎥
− j 4⎦
− j4⎥⎦
⎣ j6
⎣ j6

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Chapter 19, Problem 4.
Calculate the z parameters for the circuit in Fig. 19.68.

Figure 19.68
For Prob. 19.4.

Chapter 19, Solution 4.
Transform the Π network to a T network.
Z1

Z3

(12)( j10)
j120
=
12 + j10 − j5 12 + j5
- j60
Z2 =
12 + j5
50
Z3 =
12 + j5

Z1 =

The z parameters are

z 12 = z 21 = Z 2 =

(-j60)(12 - j5)
= -1.775 - j4.26
144 + 25

z 11 = Z1 + z 12 =

( j120)(12 − j5)
+ z 12 = 1.775 + j4.26
169

z 22 = Z 3 + z 21 =

(50)(12 − j5)
+ z 21 = 1.7758 − j5.739
169

Thus,

⎡ 1.775 + j4.26 - 1.775 − j4.26 ⎤
[z ] = ⎢
⎥Ω
⎣ - 1.775 − j4.26 1.775 − j5.739 ⎦
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Chapter 19, Problem 5.
Obtain the z parameters for the network in Fig. 19.69 as functions of s.

Figure 19.69
For Prob. 19.5.

Chapter 19, Solution 5.
Consider the circuit in Fig. (a).
1

s

I2 = 0
Io

+
I1

1/s

V1

+
V2
−

−
(a)
⎛ 1 ⎞⎛

1⎞
⎜
⎟⎜1 + s + ⎟
⎛
1 ⎞ ⎝ s + 1 ⎠⎝
1 ⎛
1⎞
s⎠
|| ⎜1 + s + ⎟ =
z 11 = 1 || || ⎜1 + s + ⎟ =
1 ⎝
1
s ⎝
s⎠
s⎠ ⎛ 1 ⎞
1+
⎜
⎟ +1+ s +
s
⎝ s + 1⎠
s
2
s + s +1
z 11 = 3
s + 2s 2 + 3s + 1
1
s

Io =

1 ||

1
s

1
s +1

s
s +1

I =
I =
I1
s
1 1
1
1 1
1
2
1 || + 1 + s +
+1+ s +
+ s + s +1
s +1
s
s +1
s
s
s
Io = 3

I1
2
s + 2s + 3s + 1
I1
1
V2 = I o = 3
s
s + 2s 2 + 3s + 1
z 21 =

V2
1
= 3
2
I 1 s + 2s + 3s + 1

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Consider the circuit in Fig. (b).

1

I1 = 0

s

+

+
1/s

V1
−

V2
−

(b)
z 22 =

z 22

V2 1 ⎛
1⎞ 1 ⎛
1 ⎞
⎟
= || ⎜1 + s + 1 || ⎟ = || ⎜1 + s +
I2 s ⎝
s⎠ s ⎝
s + 1⎠

⎛ 1 ⎞⎛
1 ⎞
1
⎜ ⎟⎜1 + s +

⎟
1
+
s
+
⎝ s ⎠⎝
s + 1⎠
s +1
=
=
1
1
s
+1+ s +
1+ s + s2 +
s
s +1
s +1

z 22 =

s 2 + 2s + 2
s 3 + 2s 2 + 3s + 1

z 12 = z 21

Hence,
⎤
⎡
s2 + s + 1

1
⎢ s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1 ⎥
⎥
[z ] = ⎢
1
s 2 + 2s + 2
⎥
⎢
⎣ s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1 ⎦

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Chapter 19, Problem 6.
Compute the z parameters of the circuit in Fig. 19.70.

Figure 19.70
For Prob. 19.6 and 19.73.

Chapter 19, Solution 6.
To find z11 and z21 , consider the circuit below.
I1

5Ω

10Ω

Vo

4I1

I2=0

– +
+

V1

+
_

20 Ω
V2
–

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z11 =

Vo =

V1 (20 + 5) I1

=
= 25 Ω
I1
I1

20
V1 = 20 I1
25

−V o −4 I 2 + V2 = 0

⎯⎯
→

V2 = Vo + 4 I1 = 20 I1 + 4 I1 = 24 I1

V2
= 24 Ω
I1
To find z12 and z22, consider the circuit below.
z21 =

I1=0

5Ω

10Ω

4I1

I2

– +
+
V1

20 Ω

+
_

V2

–

V2 = (10 + 20) I 2 = 30 I 2

V2
= 30 Ω
I1
V1 = 20 I 2
V
z12 = 1 = 20 Ω
I2
z22 =

Thus,

⎡ 25 20 ⎤
[ z] = ⎢

⎥Ω
⎣ 24 30 ⎦

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Chapter 19, Problem 7.
Calculate the impedance-parameter equivalent of the circuit in Fig. 19.71.

Figure 19.71
For Prob. 19.7 and 19.80.

Chapter 19, Solution 7.
To get z11 and z21, we consider the circuit below.
I1

20 Ω

I2=0

100 Ω
+

+
V1
-

vx

50 Ω

60 Ω

12vx

+
V2
-

+

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V1 − Vx Vx Vx + 12Vx
=
+
20
50
160

V − Vx

81 V1
I1 = 1
=
( )
20
121 20

⎯
⎯→

⎯⎯→

Vx =

z11 =

40
V1
121

V1
= 29.88
I1

57 40 20x121
57 40
57
13Vx
I1
)

)V1 = − (
) − 12Vx = − Vx = − (
81
8 121
8 121
8
160
V
= −70.37 I1 ⎯
⎯→ z 21 = 2 = −70.37
I1

V2 = 60(

To get z12 and z22, we consider the circuit below.
I1=0

20 Ω

I2

100 Ω
+

+
V1

50 Ω

vx

60 Ω

-

-

12vx

+
V2
-

+
Vx =

1
50
V2 = V2 ,
100 + 50
3

z 22 =

V2
= 1 / 0.09 = 11.11
I2

I2 =

V2 V2 + 12Vx
+
= 0.09V2
150
60

1
11.11
V1 = Vx = V2 =
I 2 = 3.704I 2
3
3

⎯
⎯→

V
z12 = 1 = 3.704
I2

Thus,
⎡ 29.88 3.704⎤
[z] = ⎢
⎥Ω
⎣− 70.37 11.11⎦
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Chapter 19, Problem 8.

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

Figure 19.72

For Prob. 19.8.

Chapter 19, Solution 8.

To get z11 and z21, consider the circuit below.
j4 Ω

I1 -j2 Ω

•

j6 Ω

5Ω
•

j8 Ω

I2 =0

+

+
V2

V1
10 Ω

-

-

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V1 = (10 − j2 + j6)I1

V2 = −10I1 − j4I1

V
z11 = 1 = 10 + j4
I1

⎯
⎯→

⎯
⎯→

z 21 =

V2
= −(10 + j4)
I1

To get z22 and z12, consider the circuit below.
j4 Ω

I1=0 -j2 Ω

•

5Ω
•

j6 Ω

j8 Ω

I2

+

+
V2

V1
10 Ω

-

-

V2 = (5 + 10 + j8)I 2

V1 = −(10 + j4)I 2

⎯
⎯→

⎯
⎯→

z 22 =

V2
= 15 + j8
I2

V
z12 = 1 = −(10 + j4)
I2

Thus,
⎡ (10 + j4) − (10 + j4)⎤
[z] = ⎢
⎥Ω
⎣− (10 + j4) (15 + j8) ⎦

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Chapter 19, Problem 9.
The y parameters of a network are:

[y ] = ⎡⎢

0.5
⎣− 0.2

− 0.2⎤
0.4 ⎥⎦

Determine the z parameters for the network.

Chapter 19, Solution 9.
y22
0.4
=
= 2.5 , ∆y = y11y22 − y21y12 = 05 x0.4 − 0.2 x0.2 = 0.16
∆y 0.16
−y
0.2
= 1.25 = z21

z12 = 12 =
∆y
0.16
z11 =

z22 =

y11
0.5
=
= 3.125
∆y 0.16

Thus,
⎡ 2.5 1.25 ⎤ ⎡ 2.5 1.25 ⎤
[ z] = ⎢
⎥Ω ⎢
⎥Ω
⎣1.25 3.125 ⎦ ⎣1.25 3.125⎦

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Chapter 19, Problem 10.
Construct a two-port that realizes each of the following z parameters.
⎡25 20⎤

(a) [z ] = ⎢
⎥Ω
⎣ 5 10 ⎦
⎡ 3 1 ⎤
⎢1 + s s ⎥
(b) [z ] = ⎢
⎥Ω
1⎥
⎢1
2s +
⎢⎣ s
s ⎥⎦

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).

I1

z11

z22

I2

+

+

z12 I2

V1

+

V2

+

−

−

(a)

I1

25 Ω

10 Ω

+
V1

I2
+

20 I2

+

+

−

V2
−

(b)

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(b)

This is a reciprocal network and the two-port look like the one shown in
Figs. (c) and (d).

z11 – z12

I1

z22 – z12

I2

+

+

V1

V2

−

−

(c)
z 11 − z 12 = 1 +

2
1
= 1+
s
0.5 s

z 22 − z 12 = 2s
1
z 12 =
s

I1

1Ω

0.5 F

2H

I2

+

+

V1

V2

−

−

(d)

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Chapter 19, Problem 11.
Determine a two-port network that is represented by the following z parameters:

6 + j3
⎣5 − j 2

[z ] = ⎡⎢

5 − j 2⎤
Ω
8 − j ⎥⎦

Chapter 19, Solution 11.
This is a reciprocal network, as shown below.
1+j5
3+j

5-j2

1Ω

j5 Ω

3 j1
ΩΩ

5Ω
-j2 Ω

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Chapter 19, Problem 12.
For the circuit shown in Fig. 19.73, let
⎡ 10 − 6 ⎤
[z] = ⎢
⎥
⎣− 4 12⎦
Find I 1 , I 2 , V1 , and V2 .

Figure 19.73
For Prob. 19.12.

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Chapter 19, Solution 12.
V1 = 10 I1 − 6 I 2
V2 = −4 I 2 + 12 I 2
V2 = −10 I 2

(1)
(2)
(3)

If we convert the current source to a voltage source, that portion of the circuit becomes
what is shown below.
4Ω
2Ω
I1
+
12 V

V1

+
_

–

−12 + 6 I1 + V1 = 0

⎯⎯
→ V1 = 12 − 6 I1

(4)

Substituting (3) and (4) into (1) and (2), we get
12 − 6 I 1 = 10 I1 − 6 I 2

⎯⎯
→ 12 = 16 I1 − 6 I 2

−10 I 2 = −4 I1 + 12 I 2

⎯⎯
→ 0 = −4 I1 + 22 I 2

(5)
⎯⎯
→ I1 = 5.5 I 2

(6)

From (5) and (6),

12 = 88I 2 − 6 I 2 = 82 I 2
⎯⎯
→ I 2 = 0.1463 A
I1 = 5.5 I 2 = 0.8049 A
V2 = −10 I 2 = −1.463 V
V1 = 12 − 6 I1 = 7.1706 V

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Chapter 19, Problem 13.

Determine the average power delivered to Z L = 5 + j 4 in the network of Fig. 19.74.
Note: The voltage is rms.

Figure 19.74

For Prob. 19.13.

Chapter 19, Solution 13.

Consider the circuit as shown below.
10 Ω I1

I2

+
50∠0˚ V

+
_

+
V1
_

[z]
V2

ZL

–
V1 = 40I1 + 60I2
(1)
V2 = 80I1 + 100I2

(2)
V2 = −I2 ZL = −I2(5 + j 4)
(3)
50 = V1 + 10I1
⎯⎯
→ V1 = 50 − 10I1
(4)
Substituting (4) in (1)
50 − 10I1 = 40I1 + 60I2
⎯⎯
→ 5 = 5I1 + 6I2 (5)
Substituting (3) into (2),
−I2(5 + j 4) = 80I1 + 100I2
⎯⎯
→ 0 = 80I1 + (105 + j4)I2
Solving (5) and (6) gives
I2 = –7.423 + j3.299 A

(6)

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We can check the answer using MATLAB.
First we need to rewrite equations 1-4 as follows,
⎡1

⎢0
⎢
⎢0
⎢
⎣1

0 − 40 − 60 ⎤ ⎡ V1 ⎤
⎡0⎤
⎢
⎥
⎥
⎢0⎥
1 − 80 − 100 ⎥ ⎢V2 ⎥
= A*X = ⎢ ⎥ = U
⎢0⎥
1
0
5 + j4⎥ ⎢ I1 ⎥
⎥⎢ ⎥
⎢ ⎥
0 10
0 ⎦⎣ I2 ⎦
⎣50⎦

>> A=[1,0,-40,-60;0,1,-80,-100;0,1,0,(5+4i);1,0,10,0]
A=
1.0e+002 *
0.0100
0
-0.4000

-0.6000
0
0.0100
-0.8000
-1.0000
0
0.0100
0
0.0500 + 0.0400i
0.0100
0
0.1000
0
>> U=[0;0;0;50]
U=
0
0
0
50
>> X=inv(A)*U
X=
-49.0722 +39.5876i
50.3093 +13.1959i
9.9072 - 3.9588i
-7.4227 + 3.2990i
P = |I2|25 = 329.9 W.

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,
you are using it without permission.


Chapter 19, Problem 14.

For the two-port network shown in Fig. 19.75, show that at the output terminals,
Z Th = z 22 −

z 12 z 21
z 11 + Z s

and
VTh =

z 21
Vs
z 11 + Z s

Figure 19.75

For Prob. 19.14 and 19.41.

Chapter 19, Solution 14.

To find Z Th , consider the circuit in Fig. (a).

I1

I2

+
ZS

+
−

V1

−
(a)
V1 = z 11 I 1 + z 12 I 2
V2 = z 21 I 1 + z 22 I 2

(1)
(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,
you are using it without permission.


But
V1 = - Z s I 1

V2 = 1 ,

0 = (z 11 + Z s ) I 1 + z 12 I 2

Hence,

⎯
⎯→ I 1 =

- z 12
I
z 11 + Z s 2

⎛ - z 21 z 12
⎞
1=⎜
+ z 22 ⎟ I 2
⎝ z 11 + Z s
⎠
Z Th =

V2
z z
1
=
= z 22 − 21 12
z 11 + Z s
I2 I2

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

VS

I1

+
−

I2 = 0

+

+

V1

V2 = VTh

−

−
(b)
V1 = Vs − I 1 Z s

I2 = 0 ,

Substituting these into (1) and (2),
Vs − I 1 Z s = z 11 I 1
V2 = z 21 I 1 =

VTh = V2 =

⎯
⎯→ I 1 =

Vs
z 11 + Z s

z 21 Vs
z 11 + Z s

z 21 Vs
z 11 + Z 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,
you are using it without permission.


Chapter 19, Problem 15.

For the two-port circuit in Fig. 19.76,
⎡40 60 ⎤
[z] = ⎢
⎥Ω
⎣80 120⎦
(a) Find Z L for maximum power transfer to the load.
(b) Calculate the maximum power delivered to the load.

Figure 19.76

For Prob. 19.15.

Chapter 19, Solution 15.

(a) From Prob. 18.12,
ZTh = z 22 −

80x 60
z12z 21
= 120 −
= 24
z11 + Zs
40 + 10
ZL = ZTh = 24Ω

(b) VTh =

80
z 21
Vs =
(120) = 192
z11 + Zs
40 + 10
⎛ V
Pmax = ⎜⎜ Th
⎝ 2R Th

2

⎞
⎟⎟ R Th = 4 2 x 24 = 384W
⎠

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,
you are using it without permission.