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When N identical capacitors are connected in series across a battery, the energy stored is U. If the same capacitors were connected in parallel across the same battery, then how much energy would be stored?

NU

N2U

U/N

U

Verified Answer :

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Below is the solution:

capacitors are connected in series 

same q same V1=V/N

U1=1/2 C(V/N)^2

U(total-1)=NU1=CV^2/2N

capacitors connected in parallel

same V

U2=1/2 CV^2

U(total-2)=1/2 NCV^2


U(total-2)/U(total-1)= CV^2/2N/NCV^2/2=>U(total-2)=N^2 U(total-1)