JEE PYQ: Alternating Current Question 2
Question 2 - 2021 (16 Mar Shift 1)
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 ____.
Show Answer
Answer: 4
Solution
At resonance, $P = \frac{(V_{rms})^2}{R} = \frac{(250/\sqrt{2})^2}{8} = 3906.25$ W $\approx 4$ kW.
BETA
