Physics - Capacitor Question with Solution | TestHub

Between the points $P$ and $Q$, a capacitor $C_1$ and series combination of $C_2,C_3,C_4,C_5$ are connected in parallel. $\therefore \frac{1}{C_s}=\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}+\frac{1}{C_5}$
$$=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{4}{C}$$ or, $C_s=C/4$. The equivalent capacitance between $P$ and $Q$ is $$C^{\prime }=C_1+C_s=C+\frac{C}{4}=\frac{5C}{4}\text{. }$$ Between the points $P$ and $R$ a series combination of $C_1,C_2$ and a series combination of $C_3,C_4,C_5$ are connected in parallel. $$\begin{array}{rl}& \therefore \frac{1}{C_s^{\prime }}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}\text{ or, }{C}_s^{\prime }=\frac{C}{2}\\ & \frac{1}{C_s^{\prime \prime }}=\frac{1}{C_3}+\frac{1}{C_4}+\frac{1}{C_5}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}\text{ or, }{C}_s^{\prime \prime }=\frac{C}{3}\end{array}$$ The equivalent capacitance between $P$ and $R$ is $$\begin{array}{rl}& C^{\prime \prime }=C_s^{\prime }+C_s^{\prime \prime }=\frac{C}{2}+\frac{C}{3}=\frac{5C}{6}\\ & \therefore \frac{C^{\prime \prime }}{C^{\prime }}=\frac{(5C/6)}{(5C/4)}=\frac{4}{6}=\frac{2}{3}\end{array}$$