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 $\dfrac{U_P}{U_T}$ is :

From the given circuit diagram, capacitors $P$ and $Q$ are connected in series, and their combination is connected in parallel with capacitor $S$.

The equivalent capacitance of the series combination of $P$ and $Q$ is:
$C_{PQ} = \dfrac{C \times C}{C + C} = \dfrac{C}{2}$

The total equivalent capacitance of the system is:
$C_{eq} = C_{PQ} + C_S = \dfrac{C}{2} + C = \dfrac{3C}{2}$

The total energy stored in the system is:
$U_T = \dfrac{1}{2} C_{eq} V^2 = \dfrac{1}{2} \left(\dfrac{3C}{2}\right) V^2 = \dfrac{3}{4} C V^2$

Since capacitors $P$ and $Q$ are identical and connected in series across the battery of voltage $V$, the potential difference across each of them is equal. Thus, the voltage across capacitor $P$ is:
$V_P = \dfrac{V}{2}$

The energy stored in capacitor $P$ is:
$U_P = \dfrac{1}{2} C_P V_P^2 = \dfrac{1}{2} C \left(\dfrac{V}{2}\right)^2 = \dfrac{1}{8} C V^2$

The ratio of the energy stored in capacitor $P$ to the total energy stored in the system is:
$\dfrac{U_P}{U_T} = \dfrac{\dfrac{1}{8} C V^2}{\dfrac{3}{4} C V^2} = \dfrac{1}{8} \times \dfrac{4}{3} = \dfrac{1}{6}$

Answer: $1/6$