When two capacitors charged to different potential are connected in parallel the statement which is not true about it is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → potential lost by one is equal to potential gained by the other
When the capacitors \( C_1 \) and \( C_2 \) (at potentials \( V_1 \) and \( V_2 \)) are joined in parallel, charge flows from the higher to the lower potential until both reach a common potential V. The total charge is conserved:
\( \displaystyle V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} \)
Option (1) some energy is lost — True. The energy lost is \( \frac{1}{2}\frac{C_1C_2}{C_1+C_2}(V_1 - V_2)^{2} \), which is positive unless \( V_1 = V_2 \); it appears as heat in the connecting wires.
Option (2) both the capacitor acquire a common potential — True. Charge flows until the potentials are equal.
Option (3) potential lost by one is equal to potential gained by the other — Not true. The potential lost is \( V_1 - V \) and the potential gained is \( V - V_2 \). Equal charge moves (\( C_1(V_1 - V) = C_2(V - V_2) \)), so the two potential changes are in the ratio \( C_2 : C_1 \), and they are equal only when \( C_1 = C_2 \).
Option (4) charge lost by one is equal to charge gained by the other — True. This is conservation of charge.
Note: A similar question was also asked in CUET (UG) 2023 on 22-05-2023 (Slot 3).