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A ${ }^{6 \mu F}$ capacitor is charged from 10 volts to 20 volts. Increase in energy will be
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$9 \times 10^{-1} \mathrm{~J}$
$\begin{aligned} & \Delta E=E_{\text {Final }}-E_{\text {ineial }}=\frac{1}{2} C\left(V_{\text {final }}^2-V_{\text {ntial }}^2\right) \\ & =\frac{1}{2} \times 6 \times\left(20^2-10^2\right) \times 10^{-6} \\ & =3 \times(400-100) \times 10^{-6}=3 \times 300 \times 10^{-6}=9 \times 10^{-4} \mathrm{~J}\end{aligned}$
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