JEE Advanced 2021 Paper 2 · Q11 · RC Circuits

Correct answer:100.00

Solution

Power dissipated by lamp (resistive): $P = V_R I = 500$ W with $V_R = 100$ V, giving $I = 5$ A. Total source rms voltage $V = 200$ V, so total impedance $Z = V/I = 40~\Omega$. Lamp resistance $R = V_R/I = 100/5 = 20~\Omega$. Capacitive reactance $X_C = \sqrt{Z^2 - R^2} = \sqrt{1600 - 400} = \sqrt{1200} = 20\sqrt{3}~\Omega$. With $X_C = 1/(\omega C)$, $\omega = 2\pi\times 50 = 100\pi$ rad/s. Using $\pi\sqrt{3} \approx 5$: $C = 1/(20\sqrt{3}\cdot 100\pi) = 1/(2000\pi\sqrt{3}) = 1/(2000\cdot 5) = 10^{-4}$ F $= 100\,\mu$F. So $C = 100.00$.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →