Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2 V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

Solution

Physics sir · 4 steps

Find the common voltage from charge conservation, then compare energy before and after.

1. Energy before. Each capacitor stores 1/2CV² of energy. Before joining, the total is 1/2CV² + 1/2C(2V)² = 5/2CV².
2. Common potential. Like plates are joined, so the charges add: CV + 2CV = 3CV. This charge now sits on a total capacitance of 2C, so the common potential is 3CV/2C = 3/2V.
3. Energy after. U_f = 1/2(2C)(3/2V)² = 9/4CV².

   Δ U = 5/2CV² - 9/4CV² = 1/4CV²

4. So. The energy decreases by 1/4CV², which is option (a). The lost energy goes into heat and radiation while charge flows through the connecting wires.

The idea to keep. When two charged capacitors are joined plate to plate, charge is conserved but energy is not. The loss is 1/2C₁C₂/(C₁+C₂)(V₁-V₂)².

Answer from the NTA final answer key. Solution by PrepWiser.