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 522: 39

Answer

$2.8\times10^6J$

Work Step by Step

See equation 17-2. The battery’s stored energy equals the charge it delivers multiplied by its voltage. An amp is a coulomb per second. $$PE=QV$$ $$PE=(65 A\cdot hr)\frac{3600s}{1hr}(12\;V)=2.8\times10^6J$$
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