Answer
a) $ R=5.11\Omega $
b) $ I_{rms}=2.1A $
c) $ P_{avg}$ increases when resistance $ R $ decreases.
Work Step by Step
(a) We can find the required resistance as follows:
$ R=\frac{V_{rms}^2}{P_{avg}}$
We plug in the known values to obtain:
$ R=\frac{(\frac{15}{\sqrt2})^2}{22}=5.11\Omega $
(b) We know that
$ I_{rms}=\frac{V_{rms}}{R}$
We plug in the known values to obtain:
$ I_{rms}=\frac{(\frac{15}{\sqrt2})}{5.11}$
$ I_{rms}=2.1A $
(c) We know that $ P_{avg}=\frac{V_{rms}^2}{R}$. This equation shows that $ P_{avg}$ is inversely proportional to the resistance. Thus, $ P_{avg}$ increases when resistance $ R $ decreases.
