A 5, mu F capacitor is charged fully by a 220 | vedclass

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(A) Initial energy stored in the capacitor: $U_i = \frac{1}{2} C_1 V^2 = \frac{1}{2} 	imes (5 	imes 10^{-6}) 	imes (220)^2 = 0.121 \, J$.When conne

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$4$

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(A) Initial energy stored in the capacitor: $U_i = \frac{1}{2} C_1 V^2 = \frac{1}{2} 	imes (5 	imes 10^{-6}) 	imes (220)^2 = 0.121 \, J$.When connected to an uncharged capacitor $C_2 = 2.5 \, \mu F$,the charge $Q = C_1 V = 5 	imes 10^{-6} 	imes 220 = 1.1 	imes 10^{-3} \, C$ redistributes.The common potential $V'$ is given by $V' = \frac{Q}{C_1 + C_2} = \frac{1.1 	imes 10^{-3}}{(5 + 2.5) 	imes 10^{-6}} = \frac{1100}{7.5} = \frac{440}{3} \, V$.Final energy stored: $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} 	imes (7.5 	imes 10^{-6}) 	imes (\frac{440}{3})^2 = 0.08066 \, J$.Energy change $\Delta U = U_f - U_i = 0.08066 - 0.121 = -0.04033 \, J$.Given $\Delta U = -\frac{X}{100} \, J$ (considering magnitude of loss),we have $\frac{X}{100} = 0.04033$,so $X \approx 4$.

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