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) $\dfrac{L}{R}\ln 2$
Concept. Growth of current: $I = I_0(1 - e^{-Rt/L})$.
Step 1. Set $I = \dfrac{I_0}{2}$. $e^{-Rt/L} = \dfrac12$.
Step 2. Solve. $t = \dfrac{L}{R}\ln2$.
Answer. $\dfrac{L}{R}\ln2$ (Option B).
Solution by Sreeraj P, M.Sc Physics