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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 9
Problems
2–9
Let’s construct a power table:
Element
Current
Voltage
Power
Power
(
μ
A)
(V)
Equation
(
μ
W)
5V
25
5
p
=

vi

125
54 kΩ
25
1
.
35
p
=
Ri
2
33
.
75
1V
25
1
p
=

vi

25
6kΩ
775
4
.
65
p
=
Ri
2
3603
.
75
Dep. source
750

2
p
=

vi
1500
1
.
8 kΩ
750
1
.
35
p
=
Ri
2
1012
.
5
8V
750
8
p
=

vi

6000
[c]
The total power generated in the circuit is the sum of the negative power
values in the power table:

125
μ
W+

25
μ
W+

6000
μ
W=

6150
μ
W
Thus, the total power generated in the circuit is 6150
μ
W.
[d]
The total power absorbed in the circuit is the sum of the positive power
values in the power table:
33
.
75
μ
W + 3603
.
75
μ
W + 1500
μ
W + 1012
.
5
μ
W = 6150
μ
W
Thus, the total power absorbed in the circuit is 6150
μ
W.
AP 2.10 Given that
i
φ
= 2A, we know the current in the dependent source is
2
i
φ
= 4A. We can write a KCL equation at the left node to ﬁnd the current in
the 10 Ω resistor. Summing the currents leaving the node,

5A+2A+4A+
i
10Ω
=0
so
i
10Ω
=5A

2A

4A=

1A
Thus, the current in the 10Ω resistor is 1A, ﬂowing right to left, as seen in
the circuit below.
© 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 10
2–10
CHAPTER 2. Circuit Elements
[a]
To ﬁnd
v
s
, write a KVL equation, summing the voltages counterclockwise
around the lower right loop. Start below the voltage source.

v
s
+ (1 A)(10 Ω) + (2 A)(30 Ω) = 0
so
v
s
= 10 V + 60 V = 70 V
[b]
The current in the voltage source can be found by writing a KCL equation
at the righthand node. Sum the currents leaving the node

4A+1A+
i
v
=0
so
i
v
=4A

1A=3A
The current in the voltage source is 3A, ﬂowing top to bottom. The
power associated with this source is
p
=
vi
= (70 V)(3 A) = 210 W
Thus, 210 W are absorbed by the voltage source.
[c]
The voltage drop across the independent current source can be found by
writing a KVL equation around the left loop in a clockwise direction:

v
5
A
+ (2 A)(30 Ω) = 0
so
v
5
A
= 60 V
The power associated with this source is
p
=

v
5
A
i
=

(60 V)(5 A) =

300 W
This source thus delivers 300W of power to the circuit.
[d]
The voltage across the controlled current source can be found by writing a
KVL equation around the upper right loop in a clockwise direction:
+
v
4
A
+ (10 Ω)(1 A) = 0
so
v
4
A
=

10 V
The power associated with this source is
p
=
v
4
A
i
=(

10 V)(4 A) =

40 W
This source thus delivers 40W of power to the circuit.
[e]
The total power dissipated by the resistors is given by
(
i
30Ω
)
2
(30 Ω) + (
i
10Ω
)
2
(10 Ω) = (2)
2
(30 Ω) + (1)
2
(10 Ω) = 120 + 10 = 130 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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