$\begin{aligned}
& \text{Given,} R=300 \Omega, C=25 \mu \mathrm{F}=25 \times 10^{-6} \mathrm{~F} \\
& V=50 \mathrm{~V}
\end{aligned}$
Frequency of AC source, $v=\frac{50}{\pi} \mathrm{Hz}$
$$
\begin{array}{cc}
\therefore & 2 \pi \nu=\omega \\
\Rightarrow & \omega=2 \pi \times \frac{50}{\pi}=100 \mathrm{rad} / \mathrm{s}
\end{array}
$$
$\therefore$ AC voltage, $V=50 \cos \omega \mathrm{t}=50 \cos (100 t)$
$\therefore$ rms voltage, $V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=\frac{50}{\sqrt{2}}=25 \sqrt{2} \mathrm{~V}$
$\therefore$ Impedance of the $R C$-circuit is
$$
\begin{aligned}
Z & =\sqrt{R^2+X_C^2}=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2} \\
\therefore \quad Z & =\sqrt{300^2+\left(\frac{1}{100 \times 25 \times 10^{-6}}\right)^2} \\
& =\sqrt{300^2+400^2} \\
\Rightarrow \quad Z & =500 \Omega
\end{aligned}
$$
$\therefore$ Current through the circuit, $I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}$
$$
=\frac{25 \sqrt{2}}{500}=\frac{\sqrt{2}}{20} \mathrm{~A}
$$
$\therefore$ Average power dissipated through the circuit is
$$
\begin{aligned}
< P> & =V_{\text {rms }} \times I_{\text {rms }} \times \cos \phi \\
& =25 \sqrt{2} \times \frac{\sqrt{2}}{20} \times \frac{R}{Z} \\
& =\frac{50}{20} \times \frac{300}{500}=1.5 \mathrm{~W}
\end{aligned}
$$