Answer
(a) $23.0W$
(b) $0.9W$
(c) $0.9W$
Work Step by Step
(a) We know that
$P_{avg}=\frac{V_{rms}^2}{Z^2}.R$
As $Z=R$
$\implies P_{avg}=\frac{V_{rms}^2}{R^2}.R$
$\implies P_{avg}=\frac{V_{rms}^2}{R}$
We plug in the known values to obtain:
$\implies P_{avg}=\frac{(24)^2}{25}$
$P_{avg}=23.0W$
(b) We know that
$Z=\sqrt{R^2+(X_L-X_C)^2}$
We plug in the known values to obtain:
$Z=\sqrt{(25)^2+(166\pi \times 325\times 10^{-3}-\frac{1}{166\pi \times 45.2\times 10^{-6}})^2}$
$\implies Z=128\Omega$
Now $P{avg}=\frac{V_{rms}^2}{Z^2}\times R$
$\implies P_{avg}=\frac{(24)^2}{(128)^2}\times 25$
$P_{avg}=0.9W$
(c) We know that
$Z=\sqrt{R^2+(X_L-X_C)^2}$
We plug in the known values to obtain:
$Z=\sqrt{(25)^2+(41.5\pi \times 325\times 10^{-3}-\frac{1}{41.5\pi\times 45.2\times 10^{-6}})^2}$
$\implies Z=128\Omega$
Now $P_{avg}=\frac{V_{rms}^2}{Z^2}\times R$
$\implies P_{avg}=\frac{(24)^2}{(128)^2}\times 25$
$P_{avg}=0.9W$
