An inductor $(L = 0.03 H)$ and a resistor $(R = 0.15 k Ω)$ are connected in series to a battery of $15 V E M F$ in a circuit shown. The key $K_{1}$ has been kept closed for a long time. Then at $t = 0, K_{1}$ is opened and key $K_{2}$ is closed simultaneously. At $t = 1 m s$ , the current in the circuit will be $(e^{5} ≅ 150)$

67 m A

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6.7 m A

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0.67 m A

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100 m A

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Solution

The correct option is C $0.67 m A$
$A t t i m e t = 0, I n d u c t o r b e h a v e s l i k e a s h o r t c i r c u i t . S o, I o = \frac{E o}{R} = \frac{15 V}{0.15 k Ω} = 100 m A W h e n K_{2} i s c l o s e d, i n d u c t o r s t a r t s d e c a y i n g c u r r e n t A t t i m e t = 1 m s, c u r r e n t, I = I_{0} e^{\frac{- R t}{L}} = 100 \times 10^{- 3} e^{\frac{0.001 \times 15 \times 10^{3}}{0.03}} = 0.1 \times e^{\frac{15}{0.03}} = 0.67 m A$

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