Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - Problems - Page 523: 54

Answer

a. 1500W. b. 3000W. 0 W.

Work Step by Step

a. Use Equation 18–9c to find the average power. $$\overline{P}=\frac{V_{rms}^2}{R}$$ $$\overline{P}=\frac{(240V)^2}{38\Omega}=1516W\approx 1.5\times10^3 W$$ b. See Figure 18-22. The maximum instantaneous power is twice the average power. The maximum is $3.0\times 10^3 W$. See Figure 18-22. The minimum instantaneous power is zero.
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