Physics - Capacitor Question with Solution | TestHub

Rearranging the circuits, we get the following circuit.
$\therefore$ equivalent capacitance between $A$ and $B$, $C_{AB}=\frac{4\times 4}{4+4}=2\mu F$ and equivalent capacitance between $C$ and $D$, $$\begin{array}{l}{C}_{CD}=\frac{8\times 8}{8+8}=4\mu F\\ \therefore C_{ab}=2\mu F+4\mu F=6\mu F\end{array}$$