n identical condensers are joined in parallel and are charged to potential V so that energy stored in each condenser is E. If they are seperated and joined in series, then the total energy and total potential difference of the combination will be

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$nE and \frac{V}{n}$

$\frac{E}{n^{2}} and \frac{V}{n^{2}}$

n²E and nV

nE and nV

answer is D.

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Detailed Solution

When n capacitors are connected in parallel.
$E = \frac{1}{2} {CV}^{2}$ (In each capacitor)
When connected in series ; V₁ = nV and
$E^{1} = \frac{1}{2} C_{s} {(V_{1})}^{2}; E^{1} = \frac{1}{2} \frac{C}{n} (nV)^{2} = nE; E^{'} = nE$

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