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If there are n capacitors in parallel connected to V volt source , then energy stored is equal to

$(a)\;C V\qquad(b)\;C V ^2\qquad(c)\;\large\frac{1}{2n}\;C V^2\qquad(d)\;\large\frac{1}{2}\;n\;C V^2$

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Answer : (d) $\;\large\frac{1}{2}\;n\;C V^2$
Explanation : Each capacitor has voltage difference = V
Number of capacitors = n
Total energy = $\;n\times\;$ energy across one capacitor
$=\large\frac{1}{2}\;n\;C V^2\;.$
answered Jan 30, 2014 by yamini.v

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