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 $\mathbf{B} = \mathbf{C}' + \mathbf{C}'' = 2 + 2 = 4 \mu \mathrm{F}$