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The initial value problem governing the current i following in an RL circuit when a step voltage of magnitude $\in$ is applied to circuit at $t=0$ is given by (R, L, t are constants ) $iR= \large\frac{Ldi}{dt} $$=\in, t > 0, i(0)=0$ Determine limiting value of i as $t \to \infty$

$(a)\;\frac{\in}{R} \\ (b)\;\frac{2 \in}{R} \\ (c)\;e^{-\frac{RE}{L}} \\ (d)\;e^{\frac{R}{L}} $

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1 Answer

$L \large\frac{di}{dt}$$=\in -iR$
$\large\frac{di}{\in -IR}$$=\frac{dt}{L}$
$\large\frac{-1}{R} $$ \log \in iR=\large\frac{t}{L}$
$\log t -iR \bigg]_0^1 =\large\frac{-R}{L}t \bigg]_0^t$
$\log \large\frac{t- iR}{\in}= \large\frac{-R}{L}$$t$
$\in -iR=\in e^{\large\frac{-Rt}{L}}$
$i= \large\frac{\in }{R} \bigg(1- e^{-\large\frac{Rt}{L}}\bigg)$
as $ t \to \infty$
$i \to \large\frac{\in }{R}$
Hence a is the correct answer.
answered Feb 6, 2014 by meena.p

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