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

When a battery is applied across $A and B$, then the points $b and c$ will be at the same potential $\left(\because C_1=C_4=C_3=C_5=4 \mu \mathrm{F}\right)$ Therefore, no charge flows through $\mathrm{C}_2$. $\mathrm{As}, C_1$ and C_5 are 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 $B =C^{\prime}+C^{\prime \prime}=2+2=4 \mu \mathrm{F}$