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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 1
2
Circuit Elements
a
Assessment Problems
AP 2.1
[a] Note that the current
i
b
is in the same circuit branch as the 8 A current
source; however,
i
b
is deﬁned in the opposite direction of the current
source. Therefore,
i
b
=

8A
Next, note that the dependent voltage source and the independent
voltage source are in parallel with the same polarity. Therefore, their
voltages are equal, and
v
g
=
i
b
4
=

8
4
=

2V
[b] To ﬁnd the power associated with the 8 A source, we need to ﬁnd the
voltage drop across the source,
v
i
. Note that the two independent sources
are in parallel, and that the voltages
v
g
and
v
1
have the same polarities,
so these voltages are equal:
v
i
=
v
g
=

2V
Using the passive sign convention,
p
s
= (8 A)(
v
i
) = (8 A)(

2 V) =

16 W
Thus the current source generated 16 W of power.
2–1
© 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 2
2–2
CHAPTER 2. Circuit Elements
AP 2.2
[a] Note from the circuit that
v
x
=

25 V. To ﬁnd
α
note that the two
current sources are in the same branch of the circuit but their currents
ﬂow in opposite directions. Therefore
αv
x
=

15 A
Solve the above equation for
α
and substitute for
v
x
,
α
=

15 A
v
x
=

15 A

25 V
=0
.
6 A/V
[b] To ﬁnd the power associated with the voltage source we need to know the
current,
i
v
. Note that this current is in the same branch of the circuit as
the dependent current source and these two currents ﬂow in the same
direction. Therefore, the current
i
v
is the same as the current of the
dependent source:
i
v
=
αv
x
= (0
.
6)(

25) =

15 A
Using the passive sign convention,
p
s
=

(
i
v
)(25 V) =

(

15 A)(25 V) = 375 W
.
Thus the voltage source dissipates 375 W.
AP 2.3
[a] The resistor and the voltage source are in parallel and the resistor voltage
and the voltage source have the same polarities. Therefore these two
voltages are the same:
v
R
=
v
g
= 1 kV
© 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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