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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 19
Problems
2–19
[b]
Since the plot is linear for 0
≤
i
s
≤
24 mA amd since
R
=Δ
v/
Δ
i
, we can
calculate
R
from the plotted values as follows:
R
=
Δ
v
Δ
i
=
24

18
0
.
024

0
=
6
0
.
024
= 250 Ω
We can determine the value of the ideal voltage source by considering the
value of
v
s
when
i
s
= 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
≤
i
s
≤
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
i
:

24 V + 250
i
+ 1000
i
=0
so
1250
i
= 24 V
Thus,
i
=
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
i
:

24 V + 250
i
=0
so
250
i
= 24 V
Thus,
i
=
24 V
250 Ω
= 96 mA
© 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 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
i
s
when the voltage
v
s
= 0. Thus,
i
sc
= 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
i
s
). 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:

1
.
5+
i
1
+
i
2
=0
so
i
1
+
i
2
=1
.
5
Write a KVL equation around the right loop:

v
1
+
v
2
+
v
3
=0
From Ohm’s law,
v
1
= 100
i
1
,
v
2
= 150
i
2
,
v
3
= 250
i
2
Substituting,

100
i
1
+ 150
i
2
+ 250
i
2
=0
so

100
i
1
+ 400
i
2
=0
Solving the two equations for
i
1
and
i
2
simultaneously,
i
1
=1
.
2A
and
i
2
=0
.
3A
[b]
Write a KVL equation clockwise around the left loop:

v
o
+
v
1
=0
but
v
1
= 100
i
1
= 100(1
.
2) = 120 V
So
v
o
=
v
1
= 120 V
[c]
Calculate power using
p
=
vi
for the source and
p
=
Ri
2
for the resistors:
p
source
=

v
o
(1
.
5) =

(120)(1
.
5) =

180 W
p
100Ω
=1
.
2
2
(100) = 144 W
p
150Ω
=0
.
3
2
(150) = 13
.
5W
p
250Ω
=0
.
3
2
(250) = 22
.
5W
P
dev
= 180 W
P
abs
= 144 + 13
.
5 + 22
.
5 = 180 W
© 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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