Q90 / 90 · JEE Main 2024 (05 Apr Shift 1) · Physics
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of $20 \mu \mathrm{F}$ is _____V.
Correct answer: 50
Solution
$\begin{aligned} & \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=100 \times 1=100 \Omega \\ & \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 20 \times 10^{-6}}=500 \Omega \\ & \mathrm{Z}=\sqrt{\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2+\mathrm{R}^2} \\ & \sqrt{(100-500)^2+300^2} \\ & \mathrm{Z}=500 \Omega \\ & \mathrm{i}_{\text {rms }}=\frac{\mathrm{V}_{\text {rms }}}{\mathrm{Z}}=\frac{50}{500}=0.1 \mathrm{~A} \\ & \text { rms voltage across capacitor } \\ & \mathrm{V}_{\text {rms }}=\mathrm{X}_{\mathrm{C}} \mathrm{i}_{\text {rms }} \\ & =500 \times 0.1=50 \mathrm{~V}\end{aligned}$
Hence, the correct answer is 50.