A 6 micro farad capacitor is charged from 10 volts to 20 volts . Increase in energy will be

Answers (1)

Energy stored in capacitor, E = \frac{1}{2}CV^{2}

Increase in energy, \Delta E = \frac{1}{2}CV_{2}^{2} - \frac{1}{2}CV_{1}^{2}

\Rightarrow \Delta E = \frac{1}{2}C\left (V_{2}^{2} -V_{1}^{2} \right )

\Rightarrow \Delta E = \frac{1}{2}6\times 10^{-6}\left (20^{2} -10^{2} \right ) =9\times 10^{-4} J

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