A coil of inductance 300 mH and resistance 2 Ω is connected to a source of voltage 2V . The current reaches half of its steady state value in: - Physics MCQ AIEEE 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Current growth in an L R circuit: I = I_{0} (1 - e^{- Rt / L}) .For I = I_{0} /2 : 1/2 = 1 - e^{- Rt / L} ⟹ e^{- Rt / L} = 1/2 Taking natural log: - Rt / L = ln (1/2) = - 0.693 t = \frac{0.693 L}{R} Given L = 300 mH = 0.3 H, R = 2 Ω : t = \frac{0.693 \times 0.3}{2} = \frac{0.2079}{2} \approx 0.104 s Approximately 0.1 s .

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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:

0.05 s

0.1 s

0.15 s

0.3 s

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