An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15 V EMF in a circuit shown below. The key K₁ has been kept closed for a long time. Then at t = 0, K₁ is opened and key K₂ is closed simultaneously. At t = 1 ms, the current in the circuit will be: (e⁵ 150) (a) 6.7 mA (b) 0.67 mA (c) 100 mA (d) 67 mA

Answer: (b). Initial current (K1 closed long): I₀ = 15/150 = 0.1 A After switching, the current decays in the L-R loop with τ = L/R = (0.03)/(150) = 0.2 ms At t = 1 ms: I = 0.1e-1/0.2 = 0.1e⁻⁵ = (0.1)/(150) A I ≈ 6.7×10⁻⁴ A = 0.67 mA, so the answer is (b).

An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected… | JEE Main 2015

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Inductors & DC Transients (Mixed)

An inductor $(L = 0.03\text{ H})$ and a resistor $(R = 0.15\text{ k}\Omega)$ are connected in series to a battery of $15\text{ V}\ \text{EMF}$ 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 $t = 1\text{ ms},$ the current in the circuit will be: $\left(e^5 \cong 150\right)$

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Answer: (b)

Initial current (K1 closed long): $I_0 = \frac{15}{150} = 0.1\text{ A}$
After switching, the current decays in the L-R loop with $\tau = \frac{L}{R} = \frac{0.03}{150} = 0.2\text{ ms}$
At $t = 1\text{ ms}$ : $I = 0.1e^{-1/0.2} = 0.1e^{-5} = \frac{0.1}{150}\text{ A}$
$I \approx 6.7\times10^{-4}\text{ A} = 0.67\text{ mA}$ , so the answer is (b).

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