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:
Show 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.
• 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
\[
U_P
=
\frac{1}{2}C
\left(\frac{V}{2}\right)^2
\]
\[
U_P
=
\frac{CV^2}{8}
\]