(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/\tau})$,where $\tau = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.$(a)$

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(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/\tau})$,where $\tau = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.$(a)$ For the current to reach half its steady-state value $(i = \frac{E}{2R})$:$\frac{E}{2R} = \frac{E}{R}(1 - e^{-t/\tau})$$0.5 = 1 - e^{-t/\tau} \implies e^{-t/\tau} = 0.5$$t = \tau \ln 2 = 10\,ms \times 0.693 = 6.93\,ms \approx 7\,ms$.$(b)$ At $t = 15\,ms$,$t/\tau = 15/10 = 1.5$:$i = \frac{6}{100}(1 - e^{-1.5}) = 0.06(1 - 0.25) = 0.06 \times 0.75 = 0.045\,A$.The energy stored is $U = \frac{1}{2}Li^2 = \frac{1}{2} \times 1 \times (0.045)^2 = 0.5 \times 0.002025 \approx 0.001\,J = 1\,mJ$.

Class 12 Physics Electromagnetic Induction R-L D.C. Circuit

Difficult

$A$ 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 $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)

JEE MAIN 2022 All JEE MAIN

A
$t = 10\,ms; U = 2\,mJ$
B
$t = 10\,ms; U = 1\,mJ$
C
$t = 7\,ms; U = 1\,mJ$
D
$t = 7\,ms; U = 2\,mJ$

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Electromagnetic Induction

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R-L D.C. Circuit

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When the switch $S$ is closed at time $t = 0$,what will be the nature of the graphs for the induced $emf$ $e$ across the inductor $L$ and the current $i$ in the circuit?

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