Q 12-02-129JEE MainJEE Main 2021 (27 Jul, Shift 1)Easy
A capacitor of capacitance $C = 1\ \mu\text{F}$ is suddenly connected to a battery of $100$ V through a resistance $R = 100\ \Omega$. The time taken for the capacitor to be charged to get $50$ V is: (Take $\ln 2 = 0.69$)
Answer: (C) $0.69\times10^{-4}$ s
Charging: $V = V_0\left(1 - e^{-t/RC}\right)$. With $V = \dfrac{V_0}{2}$: $e^{-t/RC} = \dfrac12 \Rightarrow t = RC\ln2$.
$RC = 100\times10^{-6} = 10^{-4}$ s, so $t = 0.69\times10^{-4}$ s.
Solution by Sreeraj P, M.Sc Physics