JEE Main 2015 · 4 Apr · Shift 1 · Physics Q22

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An inductor (L = 0.03H) and a resistor (R = 0.15 kΩ) 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 = 1ms, the current in the circuit will be : ( $e^5\cong150$ )

The 0.03 H inductor and 0.15 kΩ resistor in series on the top branch; key K₂ in the middle branch shorts them; the 15 V battery with key K₁ is on the bottom branch.

100 mA

67 mA

6.7 mA

0.67 mA

Answer and worked solution

Official answer: D

Before switching, $I_0=\frac{15}{150}=0.1$ A. Then the current decays through L and R with $\tau=\frac LR=\frac{0.03}{150}=2\times10^{-4}$ s. At $t=1$ ms $=5\tau$ :
$I=0.1\,e^{-5}=\frac{0.1}{150}\approx0.67\text{ mA}.$

Topics:LR circuits

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 22