Inductors & DC Transients (Mixed) question from JEE Main 2020 (4 Sep, Shift 2)

Answer: (b)

1. $U = \frac{1}{2}Li^2$, so $U = \frac{U_{max}}{n}$ means $i = \frac{i_0}{\sqrt{n}}$.
2. Growth: $i = i_0(1 - e^{-Rt/L}) \Rightarrow 1 - e^{-Rt/L} = \frac{1}{\sqrt{n}}$
3. $e^{-Rt/L} = \frac{\sqrt{n}-1}{\sqrt{n}} \Rightarrow t = \frac{L}{R}\ln\left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$
4. So the answer is (b).