Source: High school physics (Chinese)
Problem
As shown in the figure, a source with EMF $\varepsilon = 10$ V has negligible internal resistance. The resistors are $R_1 = 20\ \Omega$, $R_2 = 10\ \Omega$, $R_3 = 10\ \Omega$, and the capacitance is $C = 1\ \mu\text{F}$. The source is in series with $R_2$, after which the circuit branches between node $A$ and the bottom rail: $R_4$ connects $A$ directly to the bottom rail, while $R_1$ leads from $A$ to node $B$, from which $R_3$ connects to the bottom rail. The capacitor $C$ is connected between node $B$ and the bottom rail, in parallel with $R_3$.
In steady state, no current flows through the capacitor branch, so the circuit behaves as $R_2$ in series with the parallel combination of $R_4$ and $(R_1 + R_3)$.
The parallel resistance is
$$R_p = \frac{R_4(R_1 + R_3)}{R_4 + R_1 + R_3} = \frac{5 \times 30}{35} = \frac{30}{7}\ \Omega.$$Total resistance: $R = R_2 + R_p = 10 + \frac{30}{7} = \frac{100}{7}\ \Omega$.
Total current from the source:
$$I = \frac{\varepsilon}{R} = \frac{10}{100/7} = 0.7\ \text{A}.$$Voltage across the parallel section (between $A$ and the bottom rail):
$$U_{AB'} = \varepsilon - I R_2 = 10 - 0.7 \times 10 = 3\ \text{V}.$$Current through the $R_1$–$R_3$ branch:
$$I' = \frac{U_{AB'}}{R_1 + R_3} = \frac{3}{30} = 0.1\ \text{A}.$$Since $C$ is in parallel with $R_3$, the voltage across $C$ equals the voltage across $R_3$:
$$U_C = I' R_3 = 0.1 \times 10 = 1\ \text{V}.$$Charge on the capacitor: $Q = C U_C = 1\ \mu\text{F} \times 1\ \text{V} = 1\ \mu\text{C}.$