n identical condensers are joined in parallel | vedclass

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Question

(b) According to energy conservation, energy remains the same
$==>$ ${U_{parallel}} = {U_{series}}$ $==>$ $\frac{1}{2}(nC){V^2} = \frac{1}{2}\l

Vedclass · Accepted Answer

Energy remains same and potential difference is $nV$

Vedclass · Answer

(b) According to energy conservation, energy remains the same
$==>$ ${U_{parallel}} = {U_{series}}$ $==>$ $\frac{1}{2}(nC){V^2} = \frac{1}{2}\left( {\frac{C}{n}} \right)\,V{'^2}$ $==> V' = nV$
($V'$ $=$ potential difference across series combination)

$n$ identical condensers are joined in parallel and are charged to potential $V$. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be

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