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JEE MAIN 2024
05-04-24 S1
Question
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20μF is ..............V.
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Solution
$$ \begin{aligned} & X_L=\omega L=100 \times 1=100 \Omega \\ & X_C=\frac{1}{\omega C}=\frac{1}{100 \times 20 \times 10^{-6}}=500 \Omega \\ & Z=\sqrt{\left(X_L-X_C\right)^2+R^2} \\ & \sqrt{(100-500)^2+300^2} \\ & Z=500 \Omega \\ & i_{\text {rms }}=\frac{V_{\text {rms }}}{Z}=\frac{50}{500}=0.1 \mathrm{~A} \end{aligned} $$ rms voltage across capacitor $$ \begin{aligned} & V_{\mathrm{rms}}=X_C i_{\mathrm{rms}} \\ & =500 \times 0.1=50 \mathrm{~V} \end{aligned} $$
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