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A series L-R circuit with a battery of emf V is switched on at t = 0. Find the time at which the energy stored in the inductor reaches 1/n times its maximum (final) value.

1. (L/R)*ln[sqrt(n)/(sqrt(n) - 1)]
2. (L/R)*ln[(sqrt(n) - 1)/sqrt(n)]
3. (L/R)*ln[sqrt(n)/(sqrt(n) + 1)]
4. (L/R)*ln[(sqrt(n) + 1)/(sqrt(n) - 1)]

Correct answer: (L/R)*ln[sqrt(n)/(sqrt(n) - 1)]

Solution

The current grows as i = i0(1 - e^(-Rt/L)), i0 = V/R, and Umax = (1/2)L i0². The energy ratio U/Umax = (i/i0)². Setting this to 1/n gives (1 - e^(-Rt/L))² = 1/n, so 1 - e^(-Rt/L) = 1/sqrt(n), hence e^(-Rt/L) = 1 - 1/sqrt(n) = (sqrt(n) - 1)/sqrt(n). Taking logarithm: Rt/L = ln[sqrt(n)/(sqrt(n) - 1)], so t = (L/R)*ln[sqrt(n)/(sqrt(n) - 1)].

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