Electrical Engineering Question

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Answer 1

R1 = 100 Ohm

V1 =?

Therefore, applying the voltage divider rule

V1 = R1R1 + R2 + R3E

V1 = 100100 + 330 + 680(10)   volts

V1 = 10001110  volts

V1 =0.9  volts

Applying Kirchoff's Law

-E + V1 + V4 = 0

V4 = E - V1

      = (10 - 0.9) volts

      = 9.1 V

R2 = 330 ohms

V2 = ?

Therefore, applying the voltage divider rule

V2 = R2R1 + R2 + R3V4

V2 = 330100 + 330 + 6809.1   volts

V2 = 33011109.1   volts

V2 = 2.705   volts

Now,  V2V1 =2.7050.9 = 3

Thus, V2 is three times V1 similar to R2 and R1.

 

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