The energy stored in four capacitors, each of capacitance ' \(C^{\prime}\), placed in parallel configuration…

The energy stored in four capacitors, each of capacitance ' \(C^{\prime}\), placed in parallel configuration and connected to a ' \(\mathrm{V}^{\prime}\) volt source is
  1. CV
  2. \(\mathrm{CV}^{2} / 2\)
  3. \((1 / 8) \mathrm{CV}^{2}\)
  4. \(2 \mathrm{CV}^{2}\)

Solution

As the capacitance of all the capacitors is equal(C) so, total capacitance in parallel, \(\mathrm{C}^{\prime}=\) \(\mathrm{nC}\)
(\(\mathrm{n}\) = number of capacitors in parallel)
So, \(C^{\prime}=4 C\)
Total energy stored \(=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{2}\)
$\begin{aligned} &=\frac{1}{2}\left(4 CV^{2}\right) \\ &=2 CV^{2} \end{aligned}$

Asked in: JEE Mains - Capacitance - Chapter Test

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