Q 12-06-024NEETJEE MainMedium
In an LR circuit connected to a battery at $t = 0$, the current reaches half its final value after a time
Answer: (B) $\dfrac{L}{R}\ln 2$
$I = I_0(1 - e^{-Rt/L}) = \dfrac{I_0}{2} \Rightarrow e^{-Rt/L} = \dfrac{1}{2} \Rightarrow t = \dfrac{L}{R}\ln 2$.
Solution by Sreeraj P, M.Sc Physics