Capacitance of capacitors $C_1, C_3, C_4, C_5 = 4 \mu \mathrm{F}$ each and capacitance of capacitor $c_2 = 10 \mu \mathrm{F}$ . If a battery is applied across A and B , the points b and c will be at the same potential (since $C_1 = C_4 = C_3 = C_5 = 4 \mu \mathrm{F}$ ).

Therefore no charge flows through $C_{2}$ .

We have the capacitors $c_1$ and $\mathrm{C}_5$ in series. Therefore their equivalent capacitance,

$c' = \frac{C_1 \times C_5}{C_1 + C_5} = \frac{4 \times 4}{4 + 4} = 2 \mu F$

Similarly, $C_4$ and $C_{3}$ are in series. Therefore their equivalent capacitance,

$C'' = \frac{C_3 \times C_4}{C_3 + C_4} = \frac{4 \times 4}{4 + 4} = 2 \mu \mathrm{F}$

Now $C'$ and $C''$ are in parallel. Therefore effective capacitance between A and $\mathbf{B} = C' + C'' = 2 + 2 = 4 \mu \mathrm{F}$