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 - General Problems - Page 557: 82

Answer

$I_2=7.848\times10^{-5}A$

Work Step by Step

$I_1=I_2+I_3+I_4$ $0=I_1R_1-E_1+I_4R_4$ $21V=(13000\Omega)I_1+(18000\Omega)I_4$ $0=-I_4R_4-E_3+I_3R_3$ $12V=(12000\Omega)I_3-(18000\Omega)I_4$ $16V=(12000\Omega)I_3-(10000\Omega)I_2$ $17V=(13000\Omega)I_1+(10000\Omega)I_2$ $I_4=\frac{21V-(13000\Omega)I_1}{18000\Omega}$ $I_3=\frac{12V+(18000\Omega)I_4}{12000\Omega}= \frac{33V-(13000\Omega)I_1}{12000\Omega}$ $I_2=\frac{17V-(13000\Omega)I_1}{10000\Omega}$ $I_1=\Big(\frac{17V-(13000\Omega)I_1}{10000\Omega}\Big)+\Big(\frac{33V-(13000\Omega)I_1}{12000\Omega}\Big)+\Big(\frac{21V-(13000\Omega)I_1}{18000\Omega}\Big)=0.00137A$ $17=(13000\Omega\times1.368\times10^{-3})+(10000\Omega)I_2$ $I_2=7.848\times10^{-5}A$
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