Question:
Two circuits are shown in the figure (a) & (b). At a frequency of ____________ rad/s the average power dissipated in one cycle will be same in both the circuits.
Two circuits are shown in the figure (a) & (b). At a frequency of ____________ rad/s the average power dissipated in one cycle will be same in both the circuits.
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Circuit (a) dissipates V^2/R. Circuit (b) dissipates the same only when its impedance equals R, i.e. at resonance.
Updated On: Nov 4, 2026
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Correct Answer: 500
Solution and Explanation
Circuit (a): \(P_a = \dfrac{V_{rms}^2}{R}\).
Circuit (b): \(P_b = \dfrac{V_{rms}^2 R}{Z^2}\), with \(Z^2 = R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2\).
\(P_a = P_b\) requires \(Z = R\), i.e. \(\omega L = \dfrac{1}{\omega C}\) (resonance).
\(\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{0.1\times40\times10^{-6}}} = \dfrac{1}{\sqrt{4\times10^{-6}}} = \dfrac{1}{2\times10^{-3}} = 500\ \mathrm{rad/s}\).
Answer: 500
Circuit (b): \(P_b = \dfrac{V_{rms}^2 R}{Z^2}\), with \(Z^2 = R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2\).
\(P_a = P_b\) requires \(Z = R\), i.e. \(\omega L = \dfrac{1}{\omega C}\) (resonance).
\(\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{0.1\times40\times10^{-6}}} = \dfrac{1}{\sqrt{4\times10^{-6}}} = \dfrac{1}{2\times10^{-3}} = 500\ \mathrm{rad/s}\).
Answer: 500
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