A coil of inductance 1 H and resistance $100 \,\Omega$ 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 $\ln 2=0.693, \mathrm{e}^{-3 / 2}=0.25$)
Solution
<p>$i(t) = {V \over R}(1 - {e^{ - Rt/L}})$ ...... (1)</p>
<p>${L \over R} = {1 \over {100}}s \Rightarrow {L \over R} = 10\,ms$ ...... (2)</p>
<p>${V \over {2R}} = {V \over R}(1 - {e^{ - Rt/L}})$</p>
<p>$$ \Rightarrow {e^{ - Rt/L}} = {1 \over 2} \Rightarrow t = {L \over R}\ln 2 = 6.93\,ms$$</p>
<p>$$U = {1 \over 2}L{i^2} = {1 \over 2}{[1 - {e^{ - 15/10}}]^2}{\left[ {{6 \over {100}}} \right]^2}$$</p>
<p>$= {1 \over 2}{[1 - 0.25]^2} \times 36 \times {10^{ - 4}}$</p>
<p>$= 1\,mJ$</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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