Question:

An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20 μF is .............. V.
Circuit

Updated On: Mar 22, 2025
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Correct Answer: 50

Solution and Explanation

The inductive reactance is:

\[ X_L = \omega L = 100 \times 1 = 100 \, \Omega. \]

The capacitive reactance is:

\[ X_C = \frac{1}{\omega C} = \frac{1}{100 \times 20 \times 10^{-6}} = 500 \, \Omega. \]

The total impedance is:

\[ Z = \sqrt{(X_L - X_C)^2 + R^2} = \sqrt{(100 - 500)^2 + 300^2}. \]

Simplifying:

\[ Z = \sqrt{(-400)^2 + 300^2} = \sqrt{160000 + 90000} = 500 \, \Omega. \]

The rms current is:

\[ i_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{50}{500} = 0.1 \, \text{A}. \]

The rms voltage across the capacitor is:
\[ V_{\text{rms, capacitor}} = X_C \cdot i_{\text{rms}} = 500 \times 0.1 = 50 \, \text{V}. \]

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