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 - General Problems - Page 524: 81

Answer

There is a 32 percent increase in power at the higher voltage.

Work Step by Step

Use equation 18–6b, $P=\frac{V^2}{R}$. $$\frac{P_{13.8V}}{P_{12V}}=\frac{(13.8V)^2}{(12V)^2}=1.3225$$ The power increases by about 32 percent.
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