Question

A parallel plate capacitor of capacitance 'C' is connected to a battery and charged to a potential 'V'. Another capacitor of capacitance '\(3C\)' is charged to a potential '\(3V\)'. The charging battery is then disconnected and both the capacitors are connected in parallel to each other such that positive terminal of one is connected to negative terminal of the other. The final energy of the configuration is

• A. \(1.5\,CV^2\)
• B. \(6.5\,CV^2\)
• C. \(8.0\,CV^2\)
• D. \(18.0\,CV^2\)

Hint:

Opposite polarity means the charges subtract; then find the common voltage and the stored energy.

Answer 1

The Correct Option is C

Step 1: Charge Conservation:
Total signed charge $=9CV+(-CV)=8CV$, shared between plates of total capacitance $4C$.

Step 2: Energy Directly From Charge:
$U=\dfrac{q^2}{2C_{eq}}=\dfrac{(8CV)^2}{2\cdot4C}=\dfrac{64C^2V^2}{8C}=8CV^2$.

Step 3: Compare With the Start:
Initially the energy was $\tfrac12CV^2+\tfrac12(3C)(9V^2)=14CV^2$. The drop to $8CV^2$ is energy lost as heat during the charge flow. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } 8.0\,CV^2} \]