Electrical Engineering: Principles & Applications (6th Edition)

Published by Prentice Hall
ISBN 10: 0133116646
ISBN 13: 978-0-13311-664-9

Chapter 1 - 1.6 - Problems - Introduction to Circuit Elements - Page 40: P1.57


$R=100\Omega$ 19% reduction in power

Work Step by Step

$P=\frac{V^2}R$ so $R=\frac{V^2}{P}$ $R=\frac{(100V)^2}{100W} = 100\Omega$ $P=\frac{(90V)^2}{100\Omega}=81W, 100W-81W=19W\div100W=19\%$
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