Question:
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8 \(\Omega\), L = 24 mH and C = 60 \(\mu\)F. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____________.
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8 \(\Omega\), L = 24 mH and C = 60 \(\mu\)F. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____________.
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At resonance the impedance equals R, so \(P = \frac{V_{rms}^2}{R}\) with \(V_{rms} = \frac{250}{\sqrt{2}}\) V.
Updated On: Nov 4, 2026
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Correct Answer: 4
Solution and Explanation
At resonance \(X_L = X_C\), so \(Z = R = 8\ \Omega\).
\(V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{250}{\sqrt{2}}\ \mathrm{V}\).
\(P = \frac{V_{rms}^2}{R} = \frac{250^2}{2\times8} = \frac{62500}{16} = 3906.25\ \mathrm{W}\approx 3.9\ \mathrm{kW}\).
To the nearest integer, \(x = 4\).
Answer: 4
\(V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{250}{\sqrt{2}}\ \mathrm{V}\).
\(P = \frac{V_{rms}^2}{R} = \frac{250^2}{2\times8} = \frac{62500}{16} = 3906.25\ \mathrm{W}\approx 3.9\ \mathrm{kW}\).
To the nearest integer, \(x = 4\).
Answer: 4
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