ECE | 232 | Circuits and Systems 2 | lecture notes | Problems 2

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Page 9

Problems

2–9

Let’s construct a power table:

Element

Current

Voltage

Power

(

(V)

Equation

(

125

54 kΩ

6kΩ

775

3603

Dep. source

750

1500

8 kΩ

750

1012

750

6000

[c]

The total power generated in the circuit is the sum of the negative power

values in the power table:

125

W+

6000

6150

Thus, the total power generated in the circuit is 6150

[d]

The total power absorbed in the circuit is the sum of the positive power

values in the power table:

W + 3603

W + 1500

W + 1012

W = 6150

Thus, the total power absorbed in the circuit is 6150

AP 2.10 Given that

= 2A, we know the current in the dependent source is

= 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+

10Ω

=5A

4A=

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

, write a KVL equation, summing the voltages counter-clockwise

around the lower right loop. Start below the voltage source.

+ (1 A)(10 Ω) + (2 A)(30 Ω) = 0

= 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+

=4A

1A=3A

The current in the voltage source is 3A, ﬂowing top to bottom. The

power associated with this source is

= (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:

+ (2 A)(30 Ω) = 0

= 60 V

The power associated with this source is

(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:

+ (10 Ω)(1 A) = 0

10 V

The power associated with this source is

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

(

30Ω

)

(30 Ω) + (

10Ω

)

(10 Ω) = (2)

(30 Ω) + (1)

(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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