Seven capacitors, each of capacitance $2\ \mu$F, are to be connected in a configuration to obtain an effective capacitance of $\left(\dfrac{6}{13}\right)\mu$F. Which of the combinations, shown in figures below, will achieve the desired value?
Answer: (B) see figure
$\dfrac{6}{13}\ \mu$F is less than $2\ \mu$F, so some capacitors must be in series with a parallel group. Try $n$ in parallel ($2n\ \mu$F) in series with the remaining $7 - n$:
$$\frac1C = \frac{1}{2n} + \frac{7 - n}{2}$$
For $n = 3$: $\dfrac1C = \dfrac16 + 2 = \dfrac{13}{6}$, so $C = \dfrac{6}{13}\ \mu$F.
(Check the others: $n = 5$ gives $\tfrac{10}{11}$, $n = 4$ gives $\tfrac{8}{13}$, $n = 2$ gives $\tfrac{4}{11}\ \mu$F.) So three in parallel followed by four in series.
Solution by Sreeraj P, M.Sc Physics