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n capacitors identical to each other joi...

n capacitors identical to each other joined in parallel are charged to a common potential V. The battery is disconnected. Now the capacitors are joined in series. For the new combination:

A

Energy and potential becomes n times.

B

Energy becomes n times, potential difference remains V.

C

Energy will remain same, potential difference becomes nv.

D

Energy and potential, both remains unchanged.

Text Solution

Verified by Experts

The correct Answer is:
A
Energy and potential becomes in n times
Explanation :
`U_("parallel") =(1)/(2)QV`
In series `V=V_(1)+V_(2)+V_(3)+V_(n)=nV`
Hence the energy
`=(1)/(2)(nV)`
`=nE`
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