Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 31 - Fundamentals of Circuits - Exercises and Problems - Page 918: 62

Answer

$I_{20} = 2A$

Work Step by Step

$R_{eq}$ of the circuit is $R_{eq} =10\Omega$ so current through the circuit is $ I = \frac{100}{10}A = 10A$ now potential drop across $2\Omega$ and $4\Omega$ is $V_2 = IR = 10 * 2 V =20 V$ $V_4 = IR = 10 * 4 V =40 V$ since $20\Omega$ and $5\Omega$ are in parallel so potential drop across both will be same and will be equal to $V_{20} =V_5 = 100V - 20V -40 V =40V$ so current through $20\Omega$ i $I_{20} = \frac{V_{20}}{R} = \frac{40}{20}A $ $I_{20} = 2A$
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