Physics - Capacitor Question with Solution | TestHub

The given circuit is a closed loop of five identical capacitors, each of capacitance $C$, forming a pentagon.

To find the equivalent capacitance between points $P$ and $R$ ($C_{PR}$):
The circuit consists of two parallel branches between $P$ and $R$.
Branch 1 (path $P\to Q\to R$) has two capacitors in series. Its equivalent capacitance is $C_1=\frac{C}{2}$.
Branch 2 (path $P\to T\to S\to R$) has three capacitors in series. Its equivalent capacitance is $C_2=\frac{C}{3}$.
Therefore, $C_{PR}=C_1+C_2=\frac{C}{2}+\frac{C}{3}=\frac{5C}{6}$.

To find the equivalent capacitance between points $P$ and $Q$ ($C_{PQ}$):
The circuit consists of two parallel branches between $P$ and $Q$.
Branch 1 (direct path $P\to Q$) has one capacitor. Its capacitance is $C_3=C$.
Branch 2 (path $P\to T\to S\to R\to Q$) has four capacitors in series. Its equivalent capacitance is $C_4=\frac{C}{4}$.
Therefore, $C_{PQ}=C_3+C_4=C+\frac{C}{4}=\frac{5C}{4}$.

The ratio of $C_{PR}$ to $C_{PQ}$ is:
$\frac{C_{PR}}{C_{PQ}}=\frac{\frac{5C}{6}}{\frac{5C}{4}}=\frac{5}{6}\times \frac{4}{5}=\frac{4}{6}=\frac{2}{3}$.

Answer: 2:3