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An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of \(20 \ \mu \mathrm{F}\) is ......... \(\mathrm{V}\).

An ac source is connected

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Best answer

Correct answer is : 50

\(\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=100 \times 1=100 \Omega\)

\(\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 20 \times 10^{-6}}=500 \Omega\)

\(\mathrm{Z}=\sqrt{\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}+\mathrm{R}^{2}}\)

\(\sqrt{(100-500)^{2}+300^{2}}\)

\(\mathrm{Z}=500 \Omega\)

\(\mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}}=\frac{50}{500}=0.1 \mathrm{A}\)

rms voltage across capacitor

\(\mathrm{V}_{\text {rms }}=\mathrm{X}_{\mathrm{C}} \mathrm{i}_{\text {rms }}\)

\(=500 \times 0.1=50 \mathrm{V}\)

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