Fundamental of Electric Circuits 3rd Edition Solutions Manual Chapter 02

06 04 2012

fundamental of electric circuits 3rd edition solutions manual Chapter 02

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Chapter 2, Problem 81.

In a certain application, the circuit in Figure 2.140 must be designed to meet these two criteria: (a) Vo / Vs = 0.05 (b) Req = 40 kΩ If the load resistor 5 kΩ is fixed, find R1 and R2 to meet the criteria.

Chapter 2, Solution 81

Let R1 and R2 be in kΩ.

R eq

= R1 + R 2 5

V0

=

VS

(1)

5 R2 5 R2

1 From (1) and (2), 0.05 = 5 R 40

From (1), 40 = R 1 + 2

(2)

+ R1

2 = 5 R2

= 5R 2 or R2 = 3.333 kΩ 5+ R2

R1 = 38 kΩ

Thus R1 = 38 kΩ, R2 = 3.333 kΩ


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PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prio written permission of the publisher, or used beyond the limited distribution to teachers and educato permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manua you are using it without permission.

Chapter 2, Problem 82. The pin diagram of a resistance array is shown in Fig. 2.141. Find the equivalent resistance between the following: (a) 1 and 2 (b) 1 and 3 (c) 1 and 4 Chapter 2, Solution 82 (a)

10 Ω 40 Ω 10 Ω 80 Ω 1

2 R12

R12 = 80 + 10 (10 + 40) = 80 +

50 6

=

88.33 Ω

(b)

3 10 Ω

20 Ω 40 Ω

10 Ω

R13

80 Ω 1

R13 = 80 + 10 (10 + 40) + 20 = 100 + 10 50 = 108.33 Ω (c)

4


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(c)


4

20 Ω 10 Ω R14 10 Ω

40 Ω

80 Ω 1

R14 = 80 + 0 (10 + 40 + 10) + 20 = 80 + 0 + 20 = 100 Ω PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.


Chapter 2, Problem 83.

Two delicate devices are rated as shown in Fig. 2.142. Find the values of the resistors and R2 needed to power the devices using a 24-V battery. Chapter 2, Solution 83

The voltage across the fuse should be negligible when compared with 24 V (this can be checked later when we check to see if the fuse rating is exceeded in the final circuit). We can calculate the current through the devices. I1 = I2 =

p1 V1 p2 V2

=

45mW 9V

=

480mW 24

= 5mA = 20mA I2 = 20

Ifuse IR1 R1 24 V

+

-

I1 = 5 MA

R2 IR2


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Let R3 represent the resistance of the first device, we can solve for its value from knowing the voltage across it and the current through it. R3 = 9/0.005 = 1,800 Ω

This is an interesting problem in that it essentially has two unknowns, R 1 and R2 but only

one condition that need to be met and that the voltage across R3 must equal 9 volts. the circuit is powered by a battery we could choose the value of R2 which draws the leas current, R2 = ∞. Thus we can calculate the value of R1 that give 9 volts across R 3. 9 = (24/(R1 + 1800))1800 or R1 = (24/9)1800 – 1800 = 3,000Ω This value of R1 means that we only have a total of 25 mA flowing out of the battery through the fuse which means it will not open and produces a voltage drop across it of 0.05V. This is indeed negligible when compared with the 24-volt source. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.


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Fundamental of Electric Circuits 3rd Edition Solutions Manual Chapter 02

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