A coil of inductance 300 mH and resistance 2 Ω is connected to a source of voltage 2 V. The current reaches half of its steady state value in (a) 0.1 s (b) 0.05 s (c) 0.3 s (d) 0.15 s

Answer: (a). Growth of current: I = I₀(1 - e-t/τ), τ = L/R = 0.3/2 = 0.15 s I = I₀/2 ⇒ e-t/τ = 1/2 ⇒ t = τln 2 t = 0.15 × 0.693 ≈ 0.1 s So the answer is (a).

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Combination of Inductors, LR & LC Circuits

A coil of inductance $300\text{ mH}$ and resistance $2\,\Omega$ is connected to a source of voltage $2\text{ V}$ . The current reaches half of its steady state value in

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Answer: (a)

Growth of current: $I = I_0(1 - e^{-t/\tau})$ , $\tau = L/R = 0.3/2 = 0.15\text{ s}$
$I = I_0/2 \Rightarrow e^{-t/\tau} = \frac{1}{2} \Rightarrow t = \tau\ln 2$
$t = 0.15 \times 0.693 \approx 0.1\text{ s}$
So the answer is (a).

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