Q 12-03-285JEE MainJEE Main 2017 (2 Apr)Medium
In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance $C$ will be:
Answer: (D) $CE\dfrac{r_2}{r + r_2}$
In the steady state no current flows in the branch containing the capacitor, so no current flows through $r_1$ either. The current circulates through $E$, $r$ and $r_2$:
$$I = \frac{E}{r + r_2}$$
The capacitor branch is connected across $r_2$ (the same two junctions), and with no current in $r_1$ the capacitor voltage equals the voltage across $r_2$:
$$V_C = Ir_2 = \frac{Er_2}{r + r_2},\qquad Q = CV_C = CE\frac{r_2}{r + r_2}$$
Solution by Sreeraj P, M.Sc Physics