Q 12-07-130JEE MainJEE Main 2021 (26 Aug, Shift 1)Easy
A series LCR circuit driven by $300$ V at a frequency of $50$ Hz contains a resistance $R = 3$ k$\Omega$, an inductor of inductive reactance $X_L = 250\pi\ \Omega$ and an unknown capacitor. The value of capacitance to maximise the average power should be: (take $\pi^2 = 10$)
Answer: (B) $4\ \mu$F
Average power is maximum at resonance, $X_C = X_L$:
$$\frac{1}{\omega C} = 250\pi \Rightarrow C = \frac{1}{100\pi\times250\pi} = \frac{1}{25000\pi^2} = \frac{1}{2.5\times10^5}\ \text{F} = 4\ \mu\text{F}$$
Solution by Sreeraj P, M.Sc Physics