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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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In the circuit given the switch is closed at time $t=0$ What is the amount of charge passing through the battery in are time constant ?

$(a)\;\frac{eR^2E}{L} \\ (b)\;E \bigg(\frac{L}{R}\bigg) \\(c)\;\frac{EL}{eR^2} \\(d)\;\frac{eL}{ER} $

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1 Answer

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The current flowing in LR circuit at a given time t is
$i= i_0 (1-e^{-t/T})$
$i_0=\large\frac{E}{R}$ and $T=\large\frac{L}{R}$
Charge flowing in time $t=T$
$q=\int \limits_0^T idt$
$\quad= \int \limits_0^r i_0 (1-e^{-t}{r})dt$
$\qquad= \large\frac{i_0T}{e}$
$\qquad= \large\frac{\Large\frac{E}{R} \frac{L}{R}}{e}$
$\qquad= \large\frac{EL}{R^2e}$
Hence C is the correct answer.
answered Mar 24, 2014 by meena.p
 

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