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JEE Main 2022 (30 Jun, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2022 (30 Jun, Shift 1)

Q81JEE Main 2022 · 30 JUN · SHIFT 1Medium

Resonance, Bandwidth & Q Factor

In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f2) and new quality factor (Q2) will be:

(f1 = original resonant frequency, Q1 = original quality factor)

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Answer: (a)

$f = \frac{1}{2\pi\sqrt{LC}}$ : with $2L$ and $2C$ , $\sqrt{LC}$ doubles, so $f_2 = \frac{f_1}{2}$ .
$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ : $\frac{2L}{2C} = \frac{L}{C}$ , so $Q_2 = Q_1$ .
So the answer is (a) $f_2 = \frac{f_1}{2}$ and $Q_2 = Q_1$ .

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