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JEE Main 202127 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2021 · 27 July · Evening Shift · Physics · Alternating Current

A 100Ω resistance, a 0 . 1 capacitor and an inductor are connected...

A 100Ω100Ω resistance, a 0.1\muF0 . 1\muF capacitor and an inductor are connected in series across a 250V250V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60Hz60Hz.

Options

  1. A0.70H0 . 70H
  2. B70.3mH70 . 3mH
  3. C7.03×10−5H7 . 03 \times {10}^{- 5}H
  4. D70.3H70 . 3H

Correct answer

D. 70.3H70 . 3H

Step-by-step solution

C=0.1\muF=10−7FC = 0 . 1\muF = {10}^{- 7}F

Resonant frequency =60Hz= 60Hz

ωo=1LC{\omega}_{o} = \dfrac{1}{\sqrt{LC}}

2πf0=1LC⇒L=14π2f02C2\pi{f}_{0} = \dfrac{1}{\sqrt{LC}} ⇒ L = \dfrac{1}{4{\pi}^{2}{f}_{0}^{2}C}

by putting values L≃70.3HzL ≃ 70 . 3Hz.

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