An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω… | JEE Main 2023 (25 Jan, Shift 1)

Q56JEE Main 2023 · 25 JAN · SHIFT 1Easy

Resonance, Bandwidth & Q Factor

An LCR series circuit of capacitance 62.5 nF and resistance of 50 $\Omega$ , is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is __________ mH.

(Take $\pi^2 = 10$ )

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

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Answer: 100

Maximum current amplitude at resonance: $L = \dfrac{1}{4\pi^2 f^2 C}$
$4\pi^2 f^2 C = 40 \times (2\times10^3)^2 \times 62.5\times10^{-9} = 40 \times 4\times10^6 \times 62.5\times10^{-9} = 10$
$L = \dfrac{1}{10} = 0.1\text{ H} = 100\text{ mH}$
So the answer is 100 mH.

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