Physics - Capacitor Question with Solution | TestHub

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 $\mathrm{across}A$ 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 $B$ $$=C^{\prime }+C^{\prime \prime }=2+2=4\mu \mathrm{F}\text{. }$$