Which of the following statement is NOT true, when two capacitors charged to different potential are connected in parallel by a conducting wire?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The potential lost by one is equal to the potential gained by the other
When two charged capacitors are joined in parallel, charge flows from the one at the higher potential to the other until both are at the same potential. The total charge is conserved:
\( \displaystyle V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \)
The charge \( q \) that moves changes the potentials by \( q/C_1 \) and \( q/C_2 \). These are equal only if \( C_1 = C_2 \), so in general the potential lost by one is not equal to the potential gained by the other.
Option (1) Some energy is lost — True. Energy is lost as heat in the connecting wire (and as radiation) while the charge flows.
Option (2) The charge lost by one is equal to the charge gained by the other — True. This is conservation of charge: the charge that leaves one capacitor goes to the other.
Option (3) Both the capacitors acquire a common potential — True. Charge stops flowing only when both are at the same potential.
Option (4) The potential lost by one is equal to the potential gained by the other — NOT true. The changes \( q/C_1 \) and \( q/C_2 \) differ unless the capacitances are equal.
Note: A similar question was also asked in CUET (UG) 2023 on 05-06-2023 (Slot 2).