A capacitor in a circuit behaves in two simple ways at the two ends of time, which is all most exam questions need. The in-between is the RC circuit curve. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
Just after switching on, an uncharged capacitor has no voltage: treat it as a wire. Long after, a full capacitor passes no current: treat it as a break. Then $$Q = CV_C, \qquad U = \tfrac{1}{2}CV_C^2,$$ with $V_C$ the voltage of whatever sits beside it.
Worked example: Before and after
A $12\ \text{V}$ battery, a switch and $R_1 = 20\ \Omega$ are in series with $R_2 = 40\ \Omega$ and an uncharged $50\ \mu\text{F}$ capacitor in parallel. Find (1) the current at the start, (2) the current long after, (3) the capacitor's charge.
- (1) The capacitor shorts $R_2$: $I_0 = 12/20 = 0.6\ \text{A}$.
- (2) No current into the full capacitor: $I = 12/60 = 0.2\ \text{A}$.
- (3) $V_C = (0.2)(40) = 8\ \text{V}$, so $Q = (50 \times 10^{-6})(8) = 400\ \mu\text{C}$.
Answer: $0.6\ \text{A}$, $0.2\ \text{A}$, $400\ \mu\text{C}$
Common mistake
Thinking current flows steadily through a capacitor. Charge piles up on its plates, but none crosses the gap; once full, its branch carries nothing. Check: in steady state, the circuit without the capacitor branch must give the same currents.