In a series L-C-R circuit,the frequency of a | vedclass

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(A) The current $i$ through the series $L-C-R$ circuit is given by:$i = \frac{V_{rms}}{\sqrt{R^2 + (X_L - X_C)^2}}$Given: $V_{rms} = 10 \, V$,$R = 3 \

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(A) The current $i$ through the series $L-C-R$ circuit is given by:$i = \frac{V_{rms}}{\sqrt{R^2 + (X_L - X_C)^2}}$Given: $V_{rms} = 10 \, V$,$R = 3 \, \Omega$,$X_L = 15 \, \Omega$,$X_C = 11 \, \Omega$.Substituting the values:$i = \frac{10}{\sqrt{3^2 + (15 - 11)^2}} = \frac{10}{\sqrt{9 + 4^2}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2 \, A$.The potential difference across the inductor is $V_L = i X_L = 2 	imes 15 = 30 \, V$.The potential difference across the capacitor is $V_C = i X_C = 2 	imes 11 = 22 \, V$.Since the inductor and capacitor are in series,the potential difference across their combination is $|V_L - V_C| = |30 - 22| = 8 \, V$.

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