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013213215X_ism02.pdf-Circuit Elements a Assessment Problems AP
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013213215X_ism02.pdf-Circuit 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
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
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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Circuits_and_Systems_2
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