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

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Page 7

Problems

2–7

[b]

Draw the circuit model from part (a) and attach a 25Ω resistor:

To ﬁnd the power delivered to the 25Ω resistor we must calculate the

current through the 25 Ω resistor. Do this by ﬁrst using KCL to recognize

that the current in each of the components is

, ﬂowing in a clockwise

direction. Write a KVL equation in the clockwise direction, starting

below the voltage source, and using Ohm’s law to express the voltage

drop across the resistors in the direction of the current

ﬂowing through

the resistors:

25 V + 100

+ 25

125

= 25

125

Thus, the power delivered to the 25Ω resistor is

= (25)

= (25)(0

=1W

AP 2.8

[a]

From the graph in Assessment Problem 2.7(a), we see that when

= 0,

25 A. Therefore the current source must be 0

25 A. Since the plot

is a straight line, its slope can be used to calculate the value of resistance:

= 100 Ω

A circuit model having the same

characteristic is a 0

25 A current

source in parallel with a 100Ω resistor, as shown below:

[b]

Draw the circuit model from part (a) and attach a 25Ω resistor:

Note that by writing a KVL equation around the right loop we see that

the voltage drop across both resistors is

. Write a KCL equation at the

top center node, summing the currents leaving the node. Use Ohm’s law

to specify the currents through the resistors in terms of the voltage drop

across the resistors and the value of the resistors.

25 +

100

= 25

thus

=5V

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 8

2–8

CHAPTER 2. Circuit Elements

=1W

AP 2.9

First note that we know the current through all elements in the circuit except

the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ

resistor is

; the current in the three elements to the right of the 6 kΩ resistor

is 30

). To ﬁnd the current in the 6 kΩ resistor, write a KCL equation at the

top node:

+ 30

= 31

We can then use Ohm’s law to ﬁnd the voltages across each resistor in terms

. The results are shown in the ﬁgure below:

[a]

To ﬁnd

, write a KVL equation around the left-hand loop, summing

voltages in a clockwise direction starting below the 5V source:

5 V + 54

000

1 V + 186

000

Solving for

000

+ 186

000

=6V

240

000

=6V

Thus,

240

000

= 25

[b]

Now that we have the value of

, we can calculate the voltage for each

component except the dependent source. Then we can write a KVL

equation for the right-hand loop to ﬁnd the voltage

of the dependent

source. Sum the voltages in the clockwise direction, starting to the left of

the dependent source:

000

+8V

186

000

Thus,

= 240

000

8 V = 240

000(25

)

8V=6V

8V=

We now know the values of voltage and current for every circuit element.

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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