A capacitor of capacitance 900,mu F is charge | vedclass

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(D) Given: Capacitance $C = 900\,\mu F = 900 	imes 10^{-6}\,F$,Voltage $V = 100\,V$.Initial charge on the capacitor: $Q = CV = 900 	imes 10^{-6} 	i

Vedclass · Accepted Answer

$225$

Vedclass · Answer

(D) Given: Capacitance $C = 900\,\mu F = 900 	imes 10^{-6}\,F$,Voltage $V = 100\,V$.Initial charge on the capacitor: $Q = CV = 900 	imes 10^{-6} 	imes 100 = 9 	imes 10^{-2}\,C = 90\,mC$.Initial energy stored: $U_i = \frac{1}{2} CV^2 = \frac{1}{2} 	imes (900 	imes 10^{-6}) 	imes (100)^2 = 4.5\,J$.When connected to an identical uncharged capacitor in parallel,the charge redistributes. Since the capacitors are identical,the final potential $V_f$ across each is $V_f = \frac{Q}{C_1 + C_2} = \frac{90\,mC}{900\,\mu F + 900\,\mu F} = 50\,V$.Final energy stored: $U_f = 2 	imes \left( \frac{1}{2} C V_f^2 ight) = 900 	imes 10^{-6} 	imes (50)^2 = 900 	imes 10^{-6} 	imes 2500 = 2.25\,J$.Loss of energy: $\Delta U = U_i - U_f = 4.5\,J - 2.25\,J = 2.25\,J$.Given $\Delta U = x 	imes 10^{-2}\,J$,so $2.25 = x 	imes 10^{-2} \implies x = 225$.

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