Q 12-02-143JEE MainJEE Main 2021 (27 Aug, Shift 1)Medium
Calculate the amount of charge on capacitor of $4\ \mu$F. The internal resistance of battery is $1\ \Omega$:
Answer: (B) $8\ \mu$C
In steady state no current flows through the capacitor branch. The current flows through the battery and the $4\ \Omega$ resistor:
$$I = \frac{5}{4 + 1} = 1\ \text{A}$$
The $4\ \Omega$ resistor (and hence the capacitor branch) has $4$ V across it; no current means no drop across $6\ \Omega$.
The two $2\ \mu$F capacitors in parallel make $4\ \mu$F, in series with the $4\ \mu$F capacitor: $C = 2\ \mu$F.
Charge (same on each series capacitor) $= 2\ \mu\text{F}\times4\ \text{V} = 8\ \mu$C.
Solution by Sreeraj P, M.Sc Physics