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An inductor (L = 0.03 H) and a resistance $(R = 0.15 k Ω)$ are connected in series to a battery of 15V emf in a circuit shown below. 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 ( Given $e^{5} \equiv 150$ )

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67 mA

0.67 mA

6.7 mA

100 mA

answer is B.

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Detailed Solution

Decay of current in L-R circuit

$i (t) = I_{0} e^{- R t / L}$

$I_{0} = \frac{E}{R} = \frac{15}{0.15 \times 10^{3}} = \frac{15 \times 100}{15 \times 1000} = 0.1 A$

$i (t) = 0.1 e^{\frac{- 150 \times 10^{- 3}}{0.03}} = 0.1 e^{- 5}$

$= \frac{0.1}{150} = 0.67 m A$

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