From the figure,
The 3 capacitors are in paralle.
The equivqlent capacitance of those are:
[tex]C^{\prime}=7+9+2=18\mu F[/tex]
This is in series with the fourth capacior.
Thus the equivalent capacitance is:
[tex]\begin{gathered} \frac{1}{c}=\frac{1}{18}+\frac{1}{14}=0.126 \\ \Rightarrow C=7.93\text{ }\mu F \end{gathered}[/tex]
Approximating,
The answer is
[tex]7.88\mu F[/tex]
