The frequencies at which the current amplitude in an LCR series circuit becomes 1 2 times its maximum value, are 212rad s ^ - 1 and 232rad s ^ - 1 . The value of resistance in the circuit is R = 5Ω . The self inductance

Correct answer: 0 Bandwidth of RLC circuit is BW = f _ 2 - f _ 1 = (R)/(2 ) Or, BW = ω _ 2 - ω _ 1 = (R)/(L) BW = 232 - 212 = (R)/(L) Thus, self inductance in the circuit is L = (5)/(20) = 250mH

JEE Main 202228 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 July
Shift: Morning Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2022 · 28 July · Morning Shift · Physics · Alternating Current

The frequencies at which the current amplitude in an LCR series...

The frequencies at which the current amplitude in an LCR series circuit becomes

\dfrac{1}{\sqrt{2}}

times its maximum value, are

212rad{s}^{- 1}

and

232rad{s}^{- 1}

. The value of resistance in the circuit is

R = 5Ω

. The self inductance in the circuit is _____

mH

Correct answer

Step-by-step solution

Bandwidth of RLC circuit is $BW = {f}_{2} - {f}_{1}$ $= \dfrac{R}{2\piL}$
Or, $BW = {\omega}_{2} - {\omega}_{1} = \dfrac{R}{L}$
$BW = 232 - 212 = \dfrac{R}{L}$
Thus, self inductance in the circuit is $L = \dfrac{5}{20} = 250mH$

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