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

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Page 19

Problems

2–19

[b]

Since the plot is linear for 0

≤

24 mA amd since

=Δ

, we can

calculate

from the plotted values as follows:

024

= 250 Ω

We can determine the value of the ideal voltage source by considering the

value of

when

= 0. When there is no current, there is no voltage

drop across the resistor, so all of the voltage drop at the output is due to

the voltage source. Thus the value of the voltage source must be 24 V.

The model, valid for 0

≤

24 mA, is shown below:

[c]

The circuit is shown below:

Write a KVL equation in the clockwise direction, starting below the

voltage source. Use Ohm’s law to express the voltage drop across the

resistors in terms of the current

24 V + 250

+ 1000

1250

= 24 V

Thus,

24 V

1250 Ω

= 19

2 mA

[d]

The circuit is shown below:

Write a KVL equation in the clockwise direction, starting below the

voltage source. Use Ohm’s law to express the voltage drop across the

resistors in terms of the current

24 V + 250

250

= 24 V

Thus,

24 V

250 Ω

= 96 mA

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 20

2–20

CHAPTER 2. Circuit Elements

[e]

The short circuit current can be found in the table of values (or from the

plot) as the value of the current

when the voltage

= 0. Thus,

= 48 mA

(from table)

[f]

The plot of voltage versus current constructed in part (a) is not linear (it

is piecewise linear, but not linear for all values of

). Since the proposed

circuit model is a linear model, it cannot be used to predict the nonlinear

behavior exhibited by the plotted data.

P 2.18

[a]

Write a KCL equation at the top node:

Write a KVL equation around the right loop:

From Ohm’s law,

= 100

= 150

= 250

Substituting,

100

+ 150

+ 250

100

+ 400

Solving the two equations for

and

simultaneously,

and

[b]

Write a KVL equation clockwise around the left loop:

but

= 100

= 100(1

2) = 120 V

= 120 V

[c]

Calculate power using

for the source and

for the resistors:

source

5) =

(120)(1

5) =

180 W

100Ω

(100) = 144 W

150Ω

(150) = 13

250Ω

(250) = 22



dev

= 180 W



abs

= 144 + 13

5 + 22

5 = 180 W

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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