JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
A capacitor C is fully charged with voltage V₀. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C/2. The energy loss in the process after the charge is distributed between the two capacitors is :
- (a) 1/2CV₀²
- (b) 1/4CV₀²
- (c) 1/3CV₀²
- (d) 1/6CV₀²
Answer and solution
Answer (d) 1/6CV₀²
Solution
Physics sir · 4 steps
Charge is conserved when the capacitors share it, so find the common voltage first, then compare energies.
- Charge is conserved. After the source is removed, the charge Q = CV₀ has nowhere to go. It spreads over the parallel combination of capacitance C + C/2 = 3C/2.
- Common voltage. V = CV₀/(3C/2) = 2V₀/3. Both capacitors end up at this voltage.
- Compare energies. Initial energy is 1/2CV₀². Final energy is 1/2 × 3C/2 × 4V₀²/9 = 1/3CV₀².
Δ U = 1/2CV₀² - 1/3CV₀² = 1/6CV₀²
- So. The energy lost as heat and radiation in the connecting wires is 1/6CV₀², which is option (d). The shortcut Δ U = 1/2C₁C₂/(C₁+C₂)V₀² gives the same result.
The idea to keep. When a charged capacitor shares charge with another, charge is conserved but energy is not. The loss is 1/2C₁C₂/(C₁+C₂)(V₁-V₂)², whatever the wire resistance.
Answer from the NTA final answer key. Solution by PrepWiser.