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Physics 12 Online
OpenStudy (anonymous):

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

OpenStudy (mos1635):

capacitors are connected in series same q same V1=V/N U1=1/2 C*(V/N)^2 U(total-1)=N*U1=C*V^2/2N capacitors connected in parallel same V U2=1/2 C*V^2 U(total-2)=1/2 N*C*V^2 U(total-2)/U(total-1)= C*V^2/2N/N*C*V^2/2=>U(total-2)=N^2* U(total-1)

OpenStudy (anonymous):

\[Ceq=N/C1\] \[U=1/2.(C1/N).V^{2}\] \[U'=N.1/2.(C1.V^{2})\]

OpenStudy (anonymous):

just divide now

OpenStudy (anonymous):

\[U'=N^{2}.U\]

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