`n` idential condenser are joined in parallel and are charged tpo potential `V`. Now they are separted and joined in series. Then the total energy and potential difference of the combination will be

Options

A. Energy and potential difference remain the same
B. Energy remainsi the same and potential difference becomes nV
C. Energy increases n times and potential difference is nV
D. Energy increases n times and potential difference remains the same

Answer

B

Explanation

For n condenser in parallel, `C_(P)=nC`
for n condensers in series, `C_(S)=(C)/(n)`
By conservation of energy
`therefore (1)/(2)(nC)V^(2)=(1)/(2)((C)/(n))V^('2)" "therefore nV^(2)=(V^('2))/(n)`
`therefore V^('2)=n^(2)V^(2)" "therefore V'=nV`
thus E remains the same but the P.D. becomes nV.