Alternating Current

JEE Main level, numerical answer, medium

71 of 190

Question 71 of 190

An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20μF20 \mu \mathrm{F} is _____V.

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Answer: 50

XL=ωL=100×1=100ΩXC=1ωC=1100×20×10−6=500ΩZ=(XL−XC)2+R2(100−500)2+3002Z=500Ωirms =Vrms Z=50500=0.1 A rms voltage across capacitor Vrms =XCirms =500×0.1=50 V\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}

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