A capacitor stores 60 mu C charge when conne | vedclass

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(C) Initial state: Charge $Q_i = 60 \ \mu C$,Capacitance $C_i = 2 \ \mu F$. The battery voltage $V = Q_i / C_i = 60 \ \mu C / 2 \ \mu F = 30 \ V$.Fina

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(C) Initial state: Charge $Q_i = 60 \ \mu C$,Capacitance $C_i = 2 \ \mu F$. The battery voltage $V = Q_i / C_i = 60 \ \mu C / 2 \ \mu F = 30 \ V$.Final state: Additional charge $120 \ \mu C$ flows from the battery,so final charge $Q_f = Q_i + 120 \ \mu C = 60 \ \mu C + 120 \ \mu C = 180 \ \mu C$.The final capacitance $C_f = Q_f / V = 180 \ \mu C / 30 \ V = 6 \ \mu F$.Initial energy $U_i = \frac{1}{2} C_i V^2 = \frac{1}{2} 	imes 2 \ \mu F 	imes (30 \ V)^2 = 900 \ \mu J$.Final energy $U_f = \frac{1}{2} C_f V^2 = \frac{1}{2} 	imes 6 \ \mu F 	imes (30 \ V)^2 = 2700 \ \mu J$.Work done by the battery $W_b = (Q_f - Q_i) V = 120 \ \mu C 	imes 30 \ V = 3600 \ \mu J$.Heat produced $H = W_b - (U_f - U_i) = 3600 \ \mu J - (2700 \ \mu J - 900 \ \mu J) = 3600 \ \mu J - 1800 \ \mu J = 1800 \ \mu J$.

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