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 ________.

At resonance in a series LCR circuit, the inductive reactance equals the capacitive reactance, i.e., $$X_L = X_C$$. This means the impedance of the circuit is purely resistive, so $$Z = R = 8 \; \Omega$$.

The peak value of the voltage is given as $$V_0 = 250$$ V. The RMS voltage is $$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{250}{\sqrt{2}}$$ V.

The power dissipated at resonance is given by $$P = \frac{V_{rms}^2}{R}$$. Substituting the values, we get $$P = \frac{\left(\frac{250}{\sqrt{2}}\right)^2}{8} = \frac{\frac{250^2}{2}}{8} = \frac{62500}{16} = 3906.25$$ W.

Converting to kilowatts, $$P = 3906.25 \div 1000 \approx 3.9$$ kW. Therefore, to the nearest integer, $$x = 4$$.