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: 88

Answer

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Work Step by Step

a. We are told that the D-cell battery provides 25 mA at 1.5 V for 820 h, and costs \$1.70. $$Energy = Pt=IVt=(0.025A)(1.5V)(820h)\frac{1kW}{1000W}=0.03075kWh$$ $$cost/kWh=\frac{\$1.70}{0.03075kWh }\approx \$55/kWh$$ b. We are told that the AA-cell battery provides 25 mA at 1.5 V for 120 h, and costs \$1.25. $$Energy = Pt=IVt=(0.025A)(1.5V)(120h)\frac{1kW}{1000W}=0.0045kWh$$ $$cost/kWh=\frac{\$1.25}{0.0045kWh }\approx \$280/kWh$$ c. Compare the energy costs per kWh with those from house current. For a D-cell, \$55.28/\$0.10 = 550x as much. For a AA-cell, \$277.78/\$0.10 = 2800x as much.
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