Explanation:
D Given, $\mathrm{R}=300 \Omega, \mathrm{C}=25 \mu \mathrm{F}=25 \times 10^{-6} \mathrm{~F}, \mathrm{~V}_{0}$ $=50 \mathrm{~V}$
$\mathrm{f}=\frac{50}{\pi} \mathrm{Hz}$
$\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \times \pi \times \frac{50}{\pi} \times 25 \times 10^{-6}}$
$\mathrm{X}_{\mathrm{C}}=\frac{10^{4}}{25} \Omega$
Net impedance of RC circuit,
$\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}\right)^{2}}$
$=\sqrt{(300)^{2}+\left(\frac{10^{4}}{25}\right)^{2}}=\sqrt{(300)^{2}+(400)^{2}}=500 \Omega$
Peak value of current,
$\mathrm{I}_{0}=\frac{\mathrm{V}_{0}}{\mathrm{Z}}=\frac{50}{500}=0.1 \mathrm{Amp}$
Average power dissipated in the circuit,
$\mathrm{P}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\text {rms }} \cos \phi$
$=\frac{\mathrm{E}_{0}}{\sqrt{2}} \times \frac{\mathrm{I}_{0}}{\sqrt{2}} \times \frac{\mathrm{R}}{\mathrm{Z}}=\frac{50 \times 0.1 \times 300}{2 \times 500}=\frac{3}{2}=1.5 \mathrm{~W}$
