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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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In the given circuit shown : $L= 200\;mH;$ Emp of battery $E= 12 V$ $R_1=R_2=2 \Omega$ the internal resistance of battrey is negligible . The potential drop across the inductance L after time t.

$(a)\;6\;e^{-5t}\;V \\ (b)\;\frac{12}{t} e^{-3t} \;V \\(c)\;6 [1-e^{t/.2}]V \\(d)\;12\;e^{-10\;t}V $

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Let Emf of battery be E.
Current in the inductance branch of the circuit is
$i =\large\frac{E}{R_2} \bigg[1- e^{-R_2 t/l}\bigg]$
$\large\frac{di}{dt} =\frac{E}{R_2} \frac{R_2}{L} \bigg[e^{-R_2 t/l}\bigg]$
$\qquad= \large\frac{E}{L} e^{-R_2 t/l}$
Drop across $L\bigg[ \large\frac{E}{L} e^{-R_2 t/L}\bigg]$$\times L$
$\qquad= 12 e^{\Large\frac{-2t}{200 \times 10^{-3} }}$
$\qquad=12\;e^{-10\;t}V $
Hence d is the correct answer.
answered Mar 24, 2014 by meena.p

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