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

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Page 5

Problems

2–5

A KCL equation at the upper right node gives

; a KCL equation at

the bottom right node gives

; a KCL equation at the upper left

node gives

. Now replace the currents

and

in the Ohm’s law

equations with

;

Now substitute these expressions for the three voltages into the ﬁrst

equation:

24 =

(

) = 12

Therefore

= 24

12 = 2 A

[b]

2(2) =

[c]

= 3(2) = 6 V

[d]

= 7(2) = 14 V

[e]

A KCL equation at the lower left node gives

. Since

2A. We can now compute the power associated with the voltage

source:

= (24)

= (24)(

2) =

48 W

Therefore 24 V source is delivering 48 W.

AP 2.6

Redraw the circuit labeling all voltages and currents:

We can ﬁnd the value of the unknown resistor if we can ﬁnd the value of its

voltage and its current. To start, write a KVL equation clockwise around the

right loop, starting below the 24 Ω resistor:

120 V +

Use Ohm’s law to calculate the voltage across the 8Ω resistor in terms of its

current:

Substitute the expression for

into the ﬁrst equation:

120 V + 8

120

= 15 A

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 6

2–6

CHAPTER 2. Circuit Elements

Also use Ohm’s law to calculate the value of the current through the 24Ω

resistor:

120 V

24 Ω

=5A

Now write a KCL equation at the top middle node, summing the currents

leaving:

= 5 + 15 = 20 A

Write a KVL equation clockwise around the left loop, starting below the

voltage source:

200 V +

+ 120 V = 0

= 200

120 = 80 V

Now that we know the values of both the voltage and the current for the

unknown resistor, we can use Ohm’s law to calculate the resistance:

=4Ω

AP 2.7

[a]

Plotting a graph of

versus

gives

Note that when

= 0,

= 25 V; therefore the voltage source must be 25

V. 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 25 V source in

series with a 100Ω resistor, as shown 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.

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