Find the equivalent capacitance across $A$ and $B$......$\mu F$
Diffcult
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It is clear $\frac{ C _1}{ C _3}=\frac{7}{2}$ and $\frac{ C _2}{ C _4}=\frac{7}{2}$
So balanced W.B., we can remove $13 \mu F$
$\frac{1}{7}+\frac{1}{35}=\frac{6}{35}$
$\frac{1}{10}+\frac{1}{2}=\frac{1+5}{10}=\frac{6}{10}$
$C_{\text {eq }}=\frac{35}{6}+\frac{10}{6}=\frac{45}{6}=\frac{15}{2}$
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