In the given capacitive network the resultant capacitance between point
\(\mathrm{A}\) and \(\mathrm{B}\) is

In the given circuit, \(\mathrm{C}_3\) and \(\mathrm{C}_4\) are in series and
\(
\begin{aligned}
\mathrm{C}_3 & =\mathrm{C}_4=8 \mu \mathrm{F} \\
\therefore \frac{1}{\mathrm{C}_5} & =\frac{1}{\mathrm{C}_3}+\frac{1}{\mathrm{C}_4}
\end{aligned}
\)
\(
\begin{aligned}
\therefore \mathrm{C}_{\mathrm{s}} & =\frac{\mathrm{C}_3^2}{2 \mathrm{C}_3} \\
\mathrm{C}_{\mathrm{s}} & =\frac{\mathrm{C}_3}{2} \\
\therefore \mathrm{C}_{\mathrm{S}} & =4 \mu \mathrm{F}
\end{aligned}
\)
\(\mathrm{C}_5\) and \(\mathrm{C}_6\) are in parallel and \(\mathrm{C}_5=\mathrm{C}_6=4 \mu \mathrm{F}\)
\(
\begin{aligned}
\mathrm{C}_P & =\mathrm{C}_5+\mathrm{C}_6 \\
\therefore \mathrm{C}_{\mathrm{P}} & =8 \mu \mathrm{F}
\end{aligned}
\)
\(\therefore\) Equivalent circuit is as shown in figure.

Now, \(C_2\) and \(C_P\) are in series and their combination in parallel with \(\mathrm{C}_5\)
\(
\begin{aligned}
\therefore C_E & =\frac{C_2 C_P}{C_2 C_P}+C_5 \\
& C_E=\frac{(8)(8)}{16}+4 \\
\therefore C_E & =8 \mu \mathrm{F}
\end{aligned}
\)
Now, \(C_1\) and \(C_E\) are in series,
\(
\begin{aligned}
& \therefore \mathrm{C}=\frac{\mathrm{C}_1 \mathrm{C}_{\mathrm{E}}}{\mathrm{C}_1+\mathrm{C}_{\mathrm{E}}} \\
& \therefore \mathrm{C}=\frac{(8)(8)}{16} \\
& \therefore \mathrm{C}=4 \mu \mathrm{F}
\end{aligned}
\)