Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 19 - DC Circuits - Problems - Page 553: 33

Answer

a) $I=\frac{\mathcal{E}}{R}$ b) $R_{eq}=R$

Work Step by Step

a) Number the resistors 1 to 5 from left to right. $I=I_1+I_2$ $I=I_4+I_5$ $I_1=I_3+I_4$ $I_2+I_3=I_5$ $I_1R+I_3R-I_2R=0$ $I_2=I_1+I_3$ $-I_4R+I_5R+I_3R=0$ $I_4=I_3+I_5$ $I_1R+I_3R+I_5R=\mathcal{E}$ $I_1+I_3+I_5=\frac{\mathcal{E}}{R}$ $I_1+I_2+0-I_4-I_5=0$ $I_1+0-I_3-I_4+0=0$ $0+I_2+I_3+0-I_5=0$ $I_1-I_2+I_3+0+0=0$ $I_1+0+I_3+0+I_5=\frac{\mathcal{E}}{R}$ We can solve this by using a 5 by 5 matrix. The results are as follows: $I=\frac{\mathcal{E}}{R}$ b) From Ohm's Law: $\mathcal{E}=IR_{eq}=(\frac{\mathcal{E}}{R})R_{eq}$ $R_{eq}=R$
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