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: 67

Answer

$3200 per hour per meter

Work Step by Step

$R=(1.68\times10^{-8}\Omega \cdot m)\frac{1m}{1.96\times10^{-5}m^2}=8.57\times10^{-4}\Omega$ $I=\frac{P}{V}=\frac{15\times10^6W}{120V}=1.25\times10^{5}A$ $P=I^2R=(1.25\times10^5A)^2(8.57\times10^{-4}\Omega)=1.34\times10^4kWh$ $P_{2wires}=2P=2.68\times10^4kWh/m$ $2.68\times10^4kWh/m\times(\$0.12/kWh)=\$3200$ per hour per meter
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