In a circuit, a cell of emf 18 V and internal resistance 1 Ω is connected in series with a 5 Ω resistor to two junctions a and b. Between a and b are…

After a long time no current flows in the capacitor branch, so the steady current flows only through the internal resistance, the $5\ \Omega$ resistor and the $6\ \Omega$ resistor: $I=\dfrac{18}{1+5+6}=1.5\ \mathrm{A}$. The potential difference between $a$ and $b$ equals the drop across the $6\ \Omega$ resistor, $V_{ab}=1.5\times6=9\ \mathrm{V}$. In branch 2, going from $a$ to $b$ we cross the capacitor (voltage $V_C$) and then the $3\ \mathrm{V}$ cell from its positive to its negative terminal, a drop of $3\ \mathrm{V}$. With zero current in that branch, $V_{ab}=V_C+3$, so $V_C=9-3=6\ \mathrm{V}$. The charge is $Q=CV_C=4\times6=24\ \mu\mathrm{C}$. Taking the cell as aiding instead would give $V_C=12\ \mathrm{V}$ and $48\ \mu\mathrm{C}$, which is not the case for the stated polarity.