College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 1 - Section 1.3 - The Use of Mathematics - Practice Problem: 1.2

Answer

The lightbulb draws 81.0 watts of power during the brownout.

Work Step by Step

$P_0 = V_0^2/R_0 = 100.0 ~W$ $V = 0.900~V_0$ $P = V^2/R = (0.900~V_0)^2/R_0 = 0.810 ~V_0^2/R_0$ $P = 0.810 ~P_0 = 81.0 ~W$ The lightbulb draws 81.0 watts of power during the brownout.
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