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: 84

Answer

a) $R_{eq}=6.5\Omega$ b) $I_{14}=0.429A$ c) $I_{12}=0.317A$ d) $P_{4.5}=1.06W$

Work Step by Step

a) The $12\Omega$ and $22\Omega$ are in parallel, so $R_{eq1}=(\frac{1}{12\Omega}+\frac{1}{22\Omega})^{-1}=7.76\Omega$ These are in series with $4.5\Omega$ $R_{eq2}=7.76\Omega+4.5\Omega=12.3\Omega$ This is in parallel with $14\Omega$ $R_{eq}=(\frac{1}{12.3\Omega}+\frac{1}{14\Omega})^{-1}=6.5\Omega$ b) $6.0V-I_{14}(14\Omega)=0$ $I_{14}=0.429A$ c) $6.0V-I_{12}(12\Omega)-I_{4.5}(4.5\Omega)=0$ $6.0V-I_{22}(22\Omega)-I_{4.5}(4.5\Omega)=0$ $I_{12}(12\Omega)=I_{22}(22\Omega)$ $I_{12}=\frac{I_{22}(22\Omega)}{12\Omega}$ $I=\frac{V}{R_{eq}}=\frac{6.0V}{6.5\Omega}=0.92A$ $I=I_{14}+I_{12}+I_{22}$ $0.92A=0.429A+\frac{I_{22}(22\Omega)}{12\Omega}+I_{22}$ $I_{22}=0.17A$ $I_{12}=0.32A$ d) $6.0V-(0.32A)(12\Omega)-I_{4.5}(4.5\Omega)=0$ $I_{4.5}=0.48A$ $P_{4.5}=(I_{4.5})^2(4.5\Omega)=1.06W$
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