A series LCR circuit with R = 250/11 Ω and XL = 70/11 Ω is connected… | JEE Main 2022 (30 Jun, Shift 1)

Q73JEE Main 2022 · 30 JUN · SHIFT 1Easy

Resonance, Bandwidth & Q Factor

A series LCR circuit with $R = \frac{250}{11}\ \Omega$ and $X_L = \frac{70}{11}\ \Omega$ is connected across a 220 V, 50 Hz supply. The value of capacitance needed to maximize the average power of the circuit will be _________ $\mu$ F. (Take: $\pi = \frac{22}{7}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 500

Average power is maximum at resonance: $X_C = X_L = \dfrac{70}{11}\ \Omega$
$C = \dfrac{1}{2\pi f X_C} = \dfrac{1}{2 \times \frac{22}{7} \times 50 \times \frac{70}{11}}$
$2 \times \frac{22}{7} \times 50 \times \frac{70}{11} = 2000 \Rightarrow C = \dfrac{1}{2000}\text{ F} = 500\ \mu\text{F}$
So the answer is 500 $\mu$ F.

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