Question

Two identical capacitors, have the same capacitance \( C \). One of them is charged to potential \( V_1 \) and the other to \( V_2 \). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is –

• A. \( \frac{1}{4} C (V_1 - V_2)^2 \)
• B. \( \frac{1}{2} C (V_1 + V_2)^2 \)
• C. \( \frac{1}{4} C (V_1 + V_2)^2 \)
• D. \( \frac{1}{4} C (V_1 - V_2)^2 \)

Hint:

The energy lost when two capacitors are connected together depends on the difference in their voltages before the connection.

Answer 1

The Correct Option is A

Step 1: Understand the energy stored in a capacitor.
The energy stored in a capacitor is \( E = \frac{1}{2} C V^2 \).
Step 2: Calculate the total energy before and after connection.
Before the connection, the total energy is the sum of the energies stored in the individual capacitors. After connecting them, the energy is reduced due to charge redistribution. The decrease in energy is \( \frac{1}{4} C (V_1 - V_2)^2 \). Final Answer: \[ \boxed{\frac{1}{4} C (V_1 - V_2)^2} \]