A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2 C and potential 2 V. Then the loss of energy is x/3 E, where x is _______.

Solution

Physics sir · 4 steps

Joining like plates wastes energy: use the series capacitance and the square of the voltage gap.

1. Energy loss formula. Joining two charged capacitors, positive plate to positive plate, always wastes some energy as heat. The loss uses the series combination of C₁ and C₂ and the square of the voltage difference.

   Δ U = 1/2 (C₁ C₂)/(C₁ + C₂) (V₁ - V₂)²

2. Put in numbers. Here (C₁ C₂)/(C₁ + C₂) = (C × 2C)/3C = 2C/3 and V₁ - V₂ = V - 2V = -V. So Δ U = 1/2 × 2C/3 × V² = 1/3CV².
3. Convert to E. The first capacitor has E = 1/2CV², so CV² = 2E. Then Δ U = 1/3 × 2E = 2/3E. As a check, the energy goes from 9/2CV² to 25/6CV² at the common potential 5V/3, and the difference is 1/3CV².
4. So. The loss is 2/3E, so x = 2.

The common slip: 1. Equating the loss 1/3CV² to 1/3E forgets that E = 1/2CV², and that gives 1 instead of 2.

The idea to keep. For two capacitors joined like plate to like plate, the heat lost is 1/2C₁C₂/(C₁+C₂)(V₁-V₂)². If opposite plates are joined, use V₁ + V₂ instead.

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