Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 24 - Alternating-Current Circuits - Problems and Conceptual Exercises - Page 867: 6

Answer

(a) $192\Omega$ (b) $0.88A$ (c) $150W$

Work Step by Step

(a) We can find the resistance of the light bulb as follows: $R=\frac{V_{rms}^2}{P_{avg}}$ We plug in the known values to obtain: $R=\frac{(120)^2}{75}$ $R=192\Omega$ (b) We can find the maximum current in the bulb as follows: $I_{rms}=\frac{V_{rms}}{R}$ We plug in the known values to obtain: $I_{rms}=\frac{120}{192}=0.625A$ Now $I_{max}=\sqrt{2}I_{rms}$ We plug in the known values to obtain: $I_{max}=\sqrt{2}(0.625)=0.88A$ (c) We can find the maximum power as $P_{max}=I_{max}^2R$ We plug in the known values to obtain: $P_{max}=(0.88)^2(192)$ $P_{max}=150W$
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