Answer
a) $0.11KW$
b) $0.22KW$
Work Step by Step
(a) We can find the average power consumed as
$P_{avg}=I_{rms}^2R$
We plug in the known values to obtain:
$P_{avg}=(0.85)^2(150)=110W=0.11KW$
(b) We can find the maximum power as
$P_{max}=I_{rms}^2(\sqrt{2}I_{rms})^2$
$P_{max}=2(I_{rms} R)$
We plug in the known values to obtain:
$P_{max}=2(110)=0.22KW$
