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JEEMAIN and NEET
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Physics
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Class12
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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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