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JEE Main 202126 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2021 · 26 August · Morning Shift · Physics · Alternating Current

An inductor coil stores 64J of magnetic field energy and dissipates...

An inductor coil stores 64J64J of magnetic field energy and dissipates energy at the rate of 640W640W when a current of 8A8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit (in ss).
An inductor coil stores 64J of magnetic field energy and dissipates energy at the rate of 640W when a current of 8A is

Options

  1. A0.40 . 4
  2. B0.20 . 2
  3. C0.1250 . 125
  4. D0.80 . 8

Correct answer

B. 0.20 . 2

Step-by-step solution

An inductor coil stores 64J of magnetic field energy and dissipates energy at the rate of 640W when a current of 8A is

Energy stored in an inductor, U=12LI2U = \dfrac{1}{2}L{I}^{2}
⇒64=12L(8)2⇒L=2H⇒ 64 = \dfrac{1}{2}L ( 8{)}^{2} ⇒ L = 2H
and power dissipated in a resistor, P=I2RP = {I}^{2}R
⇒640=(8)2R⇒R=10Ω⇒ 640 = ( 8{)}^{2}R ⇒ R = 10Ω
Time constant of the LRDCLRDC circuit is,
τ=LR=210=0.2sτ = \dfrac{L}{R} = \dfrac{2}{10} = 0 . 2s

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