The frequencies at which the current amplitude in an LCR series circuit becomes 1/√(2) times its maximum value, are 212 rad s⁻¹ and 232 rad s⁻¹. The value of resistance in the circuit is R = 5 Ω. The self inductance in the circuit is __________ mH. Numerical answer type. Work it out, then open the answer.

Answer: 250. Current falls to 1/√(2) of maximum at the half-power frequencies; their difference is Δω = R/L. Δω = 232 - 212 = 20 rad/s L = (R)/(Δω) = 5/20 = 0.25 H = 250 mH

The frequencies at which the current amplitude in an LCR series… | JEE Main 2022 (28 Jul, Shift 1)

Q83JEE Main 2022 · 28 JUL · SHIFT 1Difficult

Resonance, Bandwidth & Q Factor

The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt{2}}$ times its maximum value, are $212\text{ rad s}^{-1}$ and $232\text{ rad s}^{-1}$ . The value of resistance in the circuit is $R = 5\ \Omega$ . The self inductance in the circuit is __________ $\text{mH}$ .

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

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Answer: 250

Current falls to $\frac{1}{\sqrt{2}}$ of maximum at the half-power frequencies; their difference is $\Delta\omega = \frac{R}{L}$ .
$\Delta\omega = 232 - 212 = 20\text{ rad/s}$
$L = \frac{R}{\Delta\omega} = \frac{5}{20} = 0.25\text{ H} = 250\text{ mH}$

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