JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Capacitance

The target $\frac{6}{13}\mu$F must be assembled from seven $2\mu$F capacitors. Placing three of them in parallel gives $6\mu$F, and this group is put in series with the remaining four taken one after another. Adding reciprocals along the series chain, $\frac{1}{C_{eq}}=\frac{1}{6}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{1}{6}+2=\frac{13}{6}$, so $C_{eq}=\frac{6}{13}\mu$F. The arrangement of three in parallel followed by four in series delivers this value.