Answer
(a) $2.99W$
(b) $5.97W$
Work Step by Step
(a) We can find the average power as
$P_{avg}=\frac{1}{2}\frac{V_{rms}^2}{R}$
We plug in the known values to obtain:
$P_{avg}=\frac{1}{2}\times (\frac{(141)^2}{2(3.3\times 10^3)})=2.99W$
(b) $P_{max}=\frac{v_{rms}^2}{R}$
We plug in the known values to obtain:
$P_{max}=\frac{(141)^2}{3.33\times 10^3}=5.97W$
