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(B) $1$. To find the equivalent capacitance between $P$ and $R$ $(C_{PR})$: The path $P-Q-R$ consists of two capacitors in series,giving $C/2$. The pa

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$2:3$

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(B) $1$. To find the equivalent capacitance between $P$ and $R$ $(C_{PR})$: The path $P-Q-R$ consists of two capacitors in series,giving $C/2$. The path $P-T-S-R$ consists of three capacitors in series,giving $C/3$. These two branches are in parallel. Thus,$C_{PR} = C/2 + C/3 = 5C/6$.$2$. To find the equivalent capacitance between $P$ and $Q$ $(C_{PQ})$: The path $P-Q$ is one capacitor $C$. The path $P-T-S-R-Q$ consists of four capacitors in series,giving $C/4$. These two branches are in parallel. Thus,$C_{PQ} = C + C/4 = 5C/4$.$3$. The ratio $C_{PR} / C_{PQ} = (5C/6) / (5C/4) = 4/6 = 2/3$.

Five capacitors,each of capacitance $C$,are connected as shown in the figure. The ratio of the equivalent capacitance between $P$ and $R$ to the equivalent capacitance between $P$ and $Q$ is:

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