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JEE Main 2021 (24 Feb, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2021 (24 Feb, Shift 2)

Q100JEE Main 2021 · 24 FEB · SHIFT 2Difficult

Resonance, Bandwidth & Q Factor

A series L-C-R circuit is designed to resonate at an angular frequency $\omega$ 0 = 105 rad/s. The circuit draws 16W power from 120V source at resonance. The value of resistance 'R' in the circuit is _________ $\Omega$ .

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

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Answer: 900

At resonance $Z = R$ , so $P = \frac{V_{rms}^2}{R}$ .
$R = \frac{V^2}{P} = \frac{120^2}{16} = \frac{14400}{16}$
$R = 900\ \Omega$

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