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JEE Main 2021 (16 Mar, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2021 (16 Mar, Shift 1)

Q88JEE Main 2021 · 16 MAR · SHIFT 1Easy

Resonance, Bandwidth & Q Factor

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 ____________.

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

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Answer: 4

At resonance $Z = R$ : $P = \dfrac{V_{rms}^2}{R}$ with $V_{rms} = \dfrac{250}{\sqrt{2}}\text{ V}$
$P = \dfrac{250^2}{2 \times 8} = \dfrac{62500}{16} \approx 3906\text{ W} \approx 3.9\text{ kW}$
Nearest integer: $x = 4$ .

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