Verified Solution • 2025–26 Session moderate

QJEEM22Ph0d1f: JEE Main 2022 Physics - Class 12 JEEM22Ph0d1f Solution: A coil of inductance 1 H and resistance is connected to a…

A coil of inductance 1 H and resistance 100 Ω100 \,\Omega is connected to a battery of 6 V. Determine approximately :

(a) The time elapsed before the current acquires half of its steady - state value.

(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given ln⁡2=0.693,e−3/2=0.25\ln 2=0.693, \mathrm{e}^{-3 / 2}=0.25 )

Multiple Choice Options

  • A t = 10 ms; U = 2 mJ
  • B t = 10 ms; U = 1 mJ
  • C t = 7 ms; U = 1 mJ
  • D t = 7 ms; U = 2 mJ

Step-by-Step Verified Solution

i(t)=VR(1−e−Rt/L)i(t) = {V \over R}(1 - {e^{ - Rt/L}}) ...... (1)

LR=1100s⇒LR=10 ms{L \over R} = {1 \over {100}}s \Rightarrow {L \over R} = 10\,ms ...... (2)

V2R=VR(1−e−Rt/L){V \over {2R}} = {V \over R}(1 - {e^{ - Rt/L}})

⇒e−Rt/L=12⇒t=LRln⁡2=6.93 ms\Rightarrow {e^{ - Rt/L}} = {1 \over 2} \Rightarrow t = {L \over R}\ln 2 = 6.93\,ms

U=12Li2=12[1−e−15/10]2[6100]2U = {1 \over 2}L{i^2} = {1 \over 2}{[1 - {e^{ - 15/10}}]^2}{\left[ {{6 \over {100}}} \right]^2}

=12[1−0.25]2×36×10−4= {1 \over 2}{[1 - 0.25]^2} \times 36 \times {10^{ - 4}}

=1 mJ= 1\,mJ

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