013213215X ism02.pdf-Circuit Elements a ...
013213215X_ism02.pdf-Circuit Elements a Assessment Problems AP
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013213215X ism02.pdf-Circuit Elemen...
013213215X_ism02.pdf-Circuit 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.
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 counter-clockwise
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 right-hand 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
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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