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: 50

Answer

330 volts

Work Step by Step

First find the rms current by using the relationship between peak and rms values, equation 18–8a. $$I_{rms}=\frac{ I_{peak}}{\sqrt2}$$ $$ =\frac{6.4A}{\sqrt2 }$$ Next, relate the rms values to the power by using equation 19-9a. $$P=I_{rms}V_{rms}$$ $$V_{rms}=\frac{P}{I_{rms}}$$ $$V_{rms}=\frac{1500W\sqrt2}{6.4A}=330V$$
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