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) $30days\times12h\times(18.3W-11.5W)\times\frac{1kW}{1000W}\times\frac{12c}{1kWh}=29.4c$