Physics - Capacitor Question with Solution | TestHub

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 $B$ $=C^{\prime }+C^{\prime \prime }=2+2=4\mu \mathrm{F}$