A sinusoidal voltage of peak value 250, V is | vedclass

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(B) At resonance,the impedance of the $LCR$ circuit is purely resistive,i.e.,$Z = R = 8\, \Omega$.The peak voltage is $V_0 = 250\, V$. The root mean s

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$4$

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(B) At resonance,the impedance of the $LCR$ circuit is purely resistive,i.e.,$Z = R = 8\, \Omega$.The peak voltage is $V_0 = 250\, V$. The root mean square $(RMS)$ voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{250}{\sqrt{2}}\, V$.The power dissipated at resonance is given by the formula $P = \frac{(V_{rms})^2}{R}$.Substituting the values: $P = \frac{(250 / \sqrt{2})^2}{8} = \frac{62500 / 2}{8} = \frac{31250}{8} = 3906.25\, W$.Converting to $kW$: $P = 3.90625\, kW$.Rounding to the nearest integer,we get $x = 4$.

$A$ 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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