JEE Main 2019PhysicsAlternating CurrentMediumMCQ

JEE Main 2019 — Alternating Current Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

A 20 H inductor coil is connected to a 10 Ω resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B 2ln2

Step-by-step explanation

Rate of dissipation of energy in resistor =i2R
Rate of energy stored in inductor =ddt12Li2=Lididt
i2R=Lididt
didt=iRL      …i
In L-R  circuit:

i=i01-e-tτ              (∴ τ=LR=2)
didt=i0τe-t/τ
From equation (i),

i0τe-t/τ=i01-e-tτRL
e-t/τ=1-e-t/τ
e-tτ=12  ⇒t=τ ln2
=2 ln2

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About this question

This is a previous-year question from JEE Main 2019, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.