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Question: Two identical capacitors are connected in parallel across a potential difference V. After they are f...

Two identical capacitors are connected in parallel across a potential difference V. After they are fully charged the battery is removed and the positive of first capacitor is connected to negative of second and negative of first is connected to positive of other. The loss in energy will be-

A

12\frac{1}{2} CV2

B

CV2

C

CV24\frac{CV^{2}}{4}

D

Zero

Answer

CV2

Explanation

Solution

\ loss in energy = (V1 – V2)2

= (V – (–V))2

= × 4V2 = CV2