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(D) $1$. Capacitors $P$ and $Q$ are in series. Their equivalent capacitance $C_{PQ}$ is given by $\frac{1}{C_{PQ}} = \frac{1}{C} + \frac{1}{C} = \frac

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(D) $1$. Capacitors $P$ and $Q$ are in series. Their equivalent capacitance $C_{PQ}$ is given by $\frac{1}{C_{PQ}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$, so $C_{PQ} = \frac{C}{2}$.$2$. The voltage across the series combination of $P$ and $Q$ is $V$. Since $P$ and $Q$ are identical, the voltage across each is $\frac{V}{2}$.$3$. The energy stored in capacitor $P$ is $U_P = \frac{1}{2} C (\frac{V}{2})^2 = \frac{1}{2} C \frac{V^2}{4} = \frac{CV^2}{8}$.$4$. Capacitor $S$ is in parallel with the combination of $P$ and $Q$. The voltage across $S$ is $V$. The energy stored in $S$ is $U_S = \frac{1}{2} CV^2$.$5$. The total energy stored in the system is $U_T = U_P + U_Q + U_S$. Since $P$ and $Q$ are identical, $U_Q = U_P = \frac{CV^2}{8}$.$6$. Thus, $U_T = \frac{CV^2}{8} + \frac{CV^2}{8} + \frac{1}{2} CV^2 = \frac{CV^2}{4} + \frac{2CV^2}{4} = \frac{3CV^2}{4}$.$7$. The ratio $\frac{U_P}{U_T} = \frac{CV^2/8}{3CV^2/4} = \frac{1}{8} 	imes \frac{4}{3} = \frac{1}{6}$.

Three identical capacitors $P$, $Q$ and $S$, each of the capacitance $C$, are connected to a battery of voltage $V$ as shown in the figure. If the energy stored in the capacitor $P$ and total energy stored in the system are $U_P$ and $U_T$, respectively, then the ratio $\frac{U_P}{U_T}$ is:

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