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

Solution

Physics sir · 4 steps

Energy goes as the square of current, so 1/n of the energy means current at 1/√n of its maximum.

1. Energy to current. The inductor stores U = 1/2Li², so U/Uₘₐₓ = (i/i₀)². Setting this equal to 1/n gives i = i₀/√n, where i₀ = V/R.
2. Growth of current. In an L-R circuit switched on at t = 0, the current rises as i = i₀(1 - e^(-Rt/L)). Put i = i₀/√n into it.

   e^(-Rt/L) = 1 - 1/√n = (√n-1)/√n

3. Take the log. Taking the natural log gives -Rt/L = ln((√n-1)/√n). Use the minus sign to flip the fraction, which gives t = L/R ln(√n/(√n-1)). This is positive, as a time must be.
4. So. The energy reaches 1/n of its maximum at t = L/R ln(√n/(√n-1)), which is option (b).

The idea to keep. Whenever a question gives a fraction of the inductor's energy, first convert it to a fraction of the current using U ∝ i². Only then use i = i₀(1 - e^(-t/τ)).

Answer from the NTA final answer key. Solution by PrepWiser.