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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A inductor $(L= 50 \;mH)$ and resistor of $R =100\;\Omega$ and a battery , of emf 100 V is initially connected in series. Then the battery is short circuited , the current flowing 1 ms after the short circuit is

$(a)\;0.1\;A \\ (b)\;\frac{1}{e} \;A \\(c)\;e^{-1/2}A \\(d)\;\frac{1}{e^2} \;A $

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1 Answer

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Discharge of coil at $t= 1 ms=10^{-3} S$
$E= 100 V$
$R= 100 \;Ohms$
$L= 50 \times 10^{-3} \;Henry$
$I= \large\frac{E}{R} e^{-R/L t}$
$I=\large\frac{100}{100} e^{\Large\frac{-100}{50 \times 10^{-3}} \times 10^{-3}}$
$\qquad =e^{-1/2}\;A$
Hence c is the correct answer.
answered Mar 21, 2014 by meena.p
 

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