Question

Three identical capacitors \(P\), \(Q\) and \(S\), each of capacitance \(C\), are connected to a battery of voltage \(V\), as shown in the figure. If the potential 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:

• A. \(\frac{1}{6}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{1}{2}\)

Hint:

Energy stored in a capacitor is proportional to \(CV^2\). Identical capacitors in series share voltage equally. Always find equivalent capacitance first. Total energy equals the sum of energies of individual capacitors.

Answer 1

The Correct Option is A

Step 1: Read the network.
Three identical capacitors of capacitance $C$ connect to a battery $V$. Capacitors $P$ and $Q$ are in series, and that series pair is in parallel with $S$.
Step 2: Series pair $PQ$.
\[ C_{PQ} = \frac{C \cdot C}{C + C} = \frac{C}{2} \]
Step 3: Equivalent capacitance.
Adding the parallel $S$,
\[ C_{eq} = C + \frac{C}{2} = \frac{3C}{2} \]
Step 4: Total stored energy.
\[ U_T = \tfrac12 C_{eq} V^2 = \tfrac12 \cdot \frac{3C}{2} V^2 = \frac{3CV^2}{4} \]
Step 5: Voltage and energy in $P$.
The series pair shares the full $V$, so each identical capacitor gets $V/2$. Energy in $P$ is
\[ U_P = \tfrac12 C \left(\frac{V}{2}\right)^2 = \frac{CV^2}{8} \]
Step 6: Take the ratio.
\[ \frac{U_P}{U_T} = \frac{CV^2/8}{3CV^2/4} = \frac{1}{8}\times\frac{4}{3} = \frac{1}{6} \]
\[ \boxed{\dfrac{U_P}{U_T} = \dfrac{1}{6}} \]