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A coil of resistance 40 Ω is connected across a 4.0 V battery. 0.10 s after the battery is connected, the current in the coil is 63 mA. Find the inductance of the coil.
3 years ago

Kevin Nash
332 Points
Sol. R = 40 Ω, E = 4V, t = 0.1, i = 63 mA

i = i0 – (1 – e^tR/2)
⇒ 63 × 10^–3 = 4/40 (1 – e^–0.1 × 40/L)
⇒ 63 × 10^–3 = 10^–1 (1 – e^–4/L)
⇒ 63 × 10^–2 = (1 – e^–4/L)
⇒ 1 – 0.63 = e^–4/L ⇒ e^–4/L = 0.37
⇒ –4/L = ln (0.37) = –0.994
⇒ L = -4/-0.994 = 4.024 H = 4H
3 years ago
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