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: 84

Answer

a) 3960W b) 18.3W c) 11.5W d) 29.4cents

Work Step by Step

a) $P=IV=(18A)(220V)=3960W$ b) $R=(1.68\times10^{-8}\Omega \cdot m)(\frac{7m}{2.08\times10^{-6}m^2})=0.0565\Omega$ $P=I^2R=(18A)^2(0.0565\Omega)=18.3W$ c) $R=(1.68\times10^{-8}\Omega \cdot m)(\frac{7m}{3.31\times10^{-6}m^2})=0.0355\Omega$ $P=I^2R=(18A)^2(0.0355\Omega)=11.5W$ d) $30days\times12h\times(18.3W-11.5W)\times\frac{1kW}{1000W}\times\frac{12c}{1kWh}=29.4c$
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