A capacitor is charged to a potential differe | vedclass

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(D) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Initial energy $U_0 = \frac{1}{2}CV_0^2$,where $V_0 = 100\, V$.Final energy $U

Vedclass · Accepted Answer

$9/25$

Vedclass · Answer

(D) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Initial energy $U_0 = \frac{1}{2}CV_0^2$,where $V_0 = 100\, V$.Final energy $U = \frac{1}{2}CV^2$,where $V = 80\, V$.The energy lost is $\Delta U = U_0 - U = \frac{1}{2}CV_0^2 - \frac{1}{2}CV^2$.The fractional loss of energy is $\frac{\Delta U}{U_0} = \frac{\frac{1}{2}CV_0^2 - \frac{1}{2}CV^2}{\frac{1}{2}CV_0^2} = \frac{V_0^2 - V^2}{V_0^2}$.Substituting the values: $\frac{\Delta U}{U_0} = \frac{100^2 - 80^2}{100^2} = \frac{(100 - 80)(100 + 80)}{10000} = \frac{20 	imes 180}{10000} = \frac{3600}{10000} = \frac{36}{100} = \frac{9}{25}$.

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