Q 12-06-130JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
A coil of self inductance $10\ \text{mH}$ and resistance of $0.1\ \Omega$ is connected through a switch to a battery of internal resistance $0.9\ \Omega$. After the switch is closed, the time taken for the current to attain $80\%$ of the saturation value is: [$\ln5 = 1.6$]
Answer: (D) $0.016\ \text{s}$
Total resistance $1\ \Omega$, so $\tau = \dfrac LR = 0.01\ \text{s}$.
$$0.8 = 1 - e^{-t/\tau} \;\Rightarrow\; t = \tau\ln5 = 0.01\times1.6 = 0.016\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics