JEE Main 2020 · 5 September · morning · Single correct · +4 / −1
Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :
- (a) Zero
- (b) 3/2CV²
- (c) 9/2CV²
- (d) 25/6CV²
Answer and solution
Answer (b) 3/2CV²
Solution
Physics sir · 4 steps
Plus joined to minus means the charges subtract. Find the net charge first, then the energy.
- Charges before joining. Q₁ = C × V = CV and Q₂ = 2C × 2V = 4CV.
- Net charge. The positive plate of one touches the negative plate of the other, so opposite charges cancel on each joined pair of plates. What remains is Q = 4CV - CV = 3CV.
- Common potential. In parallel the capacitances add: C + 2C = 3C. The common voltage is 3CV/3C = V.
U = Q²/2Cₙₑₜ = (3CV)²/(2 × 3C) = 3/2CV²
- So. The final energy is 3/2CV², option (b). The starting energy was 9/2CV², so 3CV² is lost as heat in the connecting wires.
The idea to keep. When capacitors are joined, charge is conserved but energy is not. Find the net charge with signs, divide by the total capacitance, then use Q²/2C.
Answer from the NTA final answer key. Solution by PrepWiser.