WAEC Physics 2011 Paper

Concept:For capacitors in series, the charge on each is the same, so the total charge can be used to find the voltage across the parallel combination.Explanation:The two parallel capacitors each have capacitance $4\,\mu\mathrm{F}$.So, their equivalent capacitance is $C_p = 4\,\mu\mathrm{F} + 4\,\mu\mathrm{F} = 8\,\mu\mathrm{F}$.This parallel combination is in series with a $2\,\mu\mathrm{F}$ capacitor.Total capacitance of the circuit is $\frac{1}{C_T} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$.Hence, $C_T = \frac{8}{5} = 1.6\,\mu\mathrm{F}$.Total charge, $Q = C_T V = 1.6\,\mu\mathrm{F} \times 10\,\mathrm{V} = 16\,\mu\mathrm{C}$.Since the parallel branch is in series, it carries the same charge, $Q = 16\,\mu\mathrm{C}$.Therefore, potential difference across the parallel capacitors is $V = \frac{Q}{C_p} = \frac{16\,\mu\mathrm{C}}{8\,\mu\mathrm{F}} = 2.0\,\mathrm{V}$.Answer:The potential difference across the capacitors in parallel is $2.0\,\mathrm{V}$.