ECE | 232 | Circuits and Systems 2 | lecture notes | 2

013213215X ism02.pdf

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Page 16

2–16

CHAPTER 2. Circuit Elements

Using the values in the table of Fig. P2.13(b)

(

10)

(

(5)

(10)

(15)

(20)

≈

6 kΩ

Note that this value is found in Appendix H.

P 2.14

[a]

Plot the

—

characteristic:

From the plot:

(180

100)

(16

=5Ω

When

= 0,

= 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 +

Write a KVL equation for the right loop, in the direction of the two

currents, using Ohm’s law:

+ 20

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 +

) + 20

100;

thus

Now calculate the power dissipated by the resistor:

20 Ω

= 20

= 20(

= 320 W

P 2.15

[a]

Plot the

characteristic

From the plot:

(130

50)

(10

=8Ω

When

= 0,

= 50 V; therefore the ideal voltage source has a voltage

of 50 V.

[b]

When

25A

Note that this result can also be obtained by extrapolating the

characteristic to

= 0.

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]

= 25V;

5 mA;

= 10 kΩ

[c]

000

= 2500

02 =

= 16 mA

[d]

(open circuit) = (20

)(10

) = 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

when the current

= 0. Thus,

(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:

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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