Q 12-07-122JEE MainJEE Main 2021 (20 Jul, Shift 2)Easy
A series LCR circuit of $R = 5\ \Omega$, $L = 20$ mH and $C = 0.5\ \mu\text{F}$ is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______ $\times10^2$ W.
Numerical value type. Enter your answer.
Answer: 125
At resonance $Z = R$: $P = \dfrac{V^2}{R} = \dfrac{250^2}{5} = 12500\ \text{W} = 125\times10^2$ W.
Solution by Sreeraj P, M.Sc Physics