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: 30

Answer

$R_{120\Omega}$: $I_1=0.029A$ down $R_{74\Omega}$: $I_1=0.029A$ right $R_{110\Omega}$: $I_2=0.025A$ up $R_{25\Omega}$: $I_2=0.025A$ left $R_{56\Omega}$: $I_3=0.004A$ up

Work Step by Step

Name the currents: $I_1$ from resistor $R_{120\Omega}$ to $R_{74\Omega}$ $I_2$ from resistor $R_{110\Omega}$ to $R_{25\Omega}$ $I_3$ from resistor $R_{56\Omega}$ $I_2=I_1+I_3$ $5.8V=194\Omega I_1-56\Omega I_3$ $3.0V=135\Omega I_2+56\Omega I_3$ $3.0V=135\Omega I_1+214\Omega I_3$ $I_3=0.014A-0.631I_1$ $5.8V=194\Omega I_1-56\Omega (0.014A-0.631I_1)$ $5.8V=229\Omega I_1-0.784V$ $I_1=0.029A$ $I_3=0.014A-0.631I_1=-0.004A$ $I_2=I_1+I_3=0.029A-0.004A=0.025A$
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