JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Alternating Current

An LCR series circuit of capacitance 62.5 nF and resistance of

50Ω

, 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

π^{2} = 10

)

Answer: 100

Solution

The current amplitude is largest at resonance, where

f = \frac{1}{2 π L C}

, so

L = \frac{1}{4 π ^{2} f ^{2} C}

. With

f = 2000 Hz

C = 62.5 \times 1 0^{- 9} F

and

π^{2} = 10

L = \frac{1}{4 \times 10 \times ( 2000 ) ^{2} \times 62.5 \times 1 0 ^{- 9}} = \frac{1}{10} = 0.1 H

. The inductance required is

100 mH

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer

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