A 60 p F capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is…

Solution

${\mathrm{V}}_0=20\mathrm{}\mathrm{V}$
Heat loss $={\mathrm{U}}_{\mathrm{i}}-{\mathrm{U}}_{\mathrm{t}}$
$=\frac{1}{2}\mathrm{C}{\mathrm{v}}_{\mathrm{o}}^2-2\left[\frac{1}{2}\mathrm{C}{\left(\frac{{\mathrm{V}}_0}{2}\right)}^2\right]$
$=\frac{\mathrm{C}{\mathrm{V}}_{\mathrm{o}}^2}{4}$
$=\frac{\left(60\times {10}^{-12}\right){\left(20\right)}^2}{4}\mathrm{}\mathrm{J}$
$=6\times {10}^{-9}\mathrm{J}=6 \mathrm{n}\mathrm{J}$

Asked in: JEE Main 2020 (07 Jan Shift 2)

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