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:

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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.
Updated On: Jul 5, 2026
  • \(\frac{1}{6}\)
  • \(\frac{2}{3}\)
  • \(\frac{1}{3}\)
  • \(\frac{1}{2}\)
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The Correct Option is A

Solution and Explanation

Concept:

• Energy stored in a capacitor is \[ U=\frac{1}{2}CV^2 \]

• Capacitors in parallel have the same potential difference.

• Capacitors in series carry the same charge.

• Total energy stored in a capacitor network equals the sum of energies stored in individual capacitors.

Step 1: Identify the effective combination
From the circuit, capacitors \(P\) and \(Q\) are connected in series. Therefore, \[ C_{PQ} = \frac{C\times C}{C+C} = \frac{C}{2} \] This series combination is connected in parallel with capacitor \(S\).

Step 2: Find the equivalent capacitance of the network
\[ C_{\text{eq}} = C+\frac{C}{2} \] \[ C_{\text{eq}} = \frac{3C}{2} \]

Step 3: Calculate total energy stored in the system
\[ U_T = \frac{1}{2}C_{\text{eq}}V^2 \] \[ U_T = \frac{1}{2} \left( \frac{3C}{2} \right)V^2 \] \[ U_T = \frac{3CV^2}{4} \]

Step 4: Determine the voltage across capacitor P
The series combination \(PQ\) is connected across the battery. Hence total voltage across the pair is \[ V \] Since the capacitors are identical, \[ V_P=V_Q=\frac{V}{2} \]

Step 5: Calculate energy stored in capacitor P
\[ U_P = \frac{1}{2}C \left(\frac{V}{2}\right)^2 \] \[ U_P = \frac{CV^2}{8} \]

Step 6: Find the required ratio
\[ \frac{U_P}{U_T} = \frac{\dfrac{CV^2}{8}} {\dfrac{3CV^2}{4}} \] \[ = \frac{1}{8}\times\frac{4}{3} \] \[ = \frac{1}{6} \] \[ \boxed{\frac{U_P}{U_T}=\frac{1}{6}} \]
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