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 523: 61

Answer

9.8 hours.

Work Step by Step

The energy provided by the battery, $Q\Delta V$, is consumed by the lights, their power multiplied by the time. $$Q\Delta V=Pt$$ Solve for the time. $$t=\frac{Q\Delta V }{P}=\frac{(75A\cdot h)(3600 s/h)(12\;V)}{92W}=3.52\times10^4s\approx 9.8h$$
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