In the given figure, the capacitance C_1, C_3, C_4 , C_5 have a...

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 $\operatorname{across} A$ 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 $C_5$ in series. Therefore their equivalent capacitance, C^{\prime}=$\frac{C_1 \times C_5}{C_1+C_5}=\frac{4 \times 4}{4+4}=$2 \mu $\mathrm{F}$Similarly, $C_4$ and $C_3$ are in series. Therefore their equivalent capacitance, C^{\prime \prime}=$\frac{C_3 \times C_4}{C_3+C_4}=\frac{4 \times 4}{4+4}=$2 \mu $\mathrm{F} .$Now $C^{\prime}$ and $C^{\prime \prime}$ are in parallel. Therefore effective capacitance between $A$ and $B$ =C^{\prime}+C^{\prime \prime}=2+2=4 \mu $\mathrm{F} \text{. }$