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013213215X ism02.pdfCircuit Elements a Assessment...
013213215X_ism02.pdfCircuit Elements a Assessment Problems AP
013213215X ism02.pdfCircuit Elemen...
013213215X_ism02.pdfCircuit Elements a Assessment Problems AP
Page 16
2–16
CHAPTER 2. Circuit Elements
Using the values in the table of Fig. P2.13(b)
R
=
(

10)
2
17
.
86
×
10

3
=
(

5)
2
4
.
46
×
10

3
=
(5)
2
4
.
46
×
10

3
=
(10)
2
17
.
86
×
10

3
=
(15)
2
40
.
18
×
10

3
=
(20)
2
71
.
43
×
10

3
≈
5
.
6 kΩ
Note that this value is found in Appendix H.
P 2.14
[a]
Plot the
v
—
i
characteristic:
From the plot:
R
=
Δ
v
Δ
i
=
(180

100)
(16

0)
=5Ω
When
i
t
= 0,
v
t
= 100 V; therefore the ideal current source must have a
current of 100
/
5 = 20 A
[b]
We attach a 20Ω resistor to the device model developed in part (a):
Write a KCL equation at the top node:
20 +
i
t
=
i
1
Write a KVL equation for the right loop, in the direction of the two
currents, using Ohm’s law:
5
i
1
+ 20
i
t
=0
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Page 17
Problems
2–17
Combining the two equations and solving,
5(20 +
i
t
) + 20
i
t
=0
so
25
i
t
=

100;
thus
i
t
=

4A
Now calculate the power dissipated by the resistor:
p
20 Ω
= 20
i
2
t
= 20(

4)
2
= 320 W
P 2.15
[a]
Plot the
v

i
characteristic
From the plot:
R
=
Δ
v
Δ
i
=
(130

50)
(10

0)
=8Ω
When
i
t
= 0,
v
t
= 50 V; therefore the ideal voltage source has a voltage
of 50 V.
[b]
When
v
t
=0
,
i
t
=

50
8
=

6
.
25A
Note that this result can also be obtained by extrapolating the
v

i
characteristic to
v
t
= 0.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Page 18
2–18
CHAPTER 2. Circuit Elements
P 2.16
[a]
[b]
Δ
v
= 25V;
Δ
i
=2
.
5 mA;
R
=
Δ
v
Δ
i
= 10 kΩ
[c]
10
,
000
i
1
= 2500
i
s
,
i
1
=0
.
25
i
s
0
.
02 =
i
1
+
i
s
=1
.
25
i
s
,
i
s
= 16 mA
[d]
v
s
(open circuit) = (20
×
10

3
)(10
×
10
3
) = 200 V
[e]
The open circuit voltage can be found in the table of values (or from the
plot) as the value of the voltage
v
s
when the current
i
s
= 0. Thus,
v
s
(open circuit) = 140 V (from the table)
[f]
Linear model cannot predict the nonlinear behavior of the practical
current source.
P 2.17
[a]
Begin by constructing a plot of voltage versus current:
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
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