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If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to
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The correct answer is:
$\frac{1}{2} \mathrm{nCV}^2$
$\frac{1}{2} \mathrm{nCV}^2$
$E=\sum \frac{1}{2} C V^2=\frac{1}{2} n C V^2$
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