Q 12-03-183JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
As shown in the figure, in steady state, the charge stored in the capacitor is ______ $\times10^{-6}$ C.
Numerical value type. Enter your answer.
Answer: 10
In steady state no current flows through the capacitor branch, so the cell drives current only through $R$:
$$I = \frac{E}{R + r} = \frac{10}{110} = \frac{1}{11}\ \text{A}$$
The capacitor branch is in parallel with $R$, and with no current $R'$ has no drop, so
$$V_C = IR = \frac{100}{11}\ \text{V}, \qquad Q = CV_C = 1.1\times10^{-6}\times\frac{100}{11} = 10\times10^{-6}\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics