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 _ _ _ _ .
Solution
At resonance condition, the full voltage is applied across the resistor.
So, power dissipated $= \frac {V^2 _{rms}}{R} = \frac {(\frac {250}{\sqrt 2 })^2} {8} = 3906.25 W \approx 3.91 kW$
$\therefore x = 4 $