A 2,mu F capacitor is charged to a potential | vedclass

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(B) Initial charges on the capacitors are:$Q_1 = C_1 V_1 = 2\,\mu F 	imes 10\,V = 20\,\mu C$$Q_2 = C_2 V_2 = 4\,\mu F 	imes 20\,V = 80\,\mu C$Since

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$600$

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(B) Initial charges on the capacitors are:$Q_1 = C_1 V_1 = 2\,\mu F 	imes 10\,V = 20\,\mu C$$Q_2 = C_2 V_2 = 4\,\mu F 	imes 20\,V = 80\,\mu C$Since the positive plate of one is connected to the negative plate of the other,the net charge is $Q_{net} = |Q_2 - Q_1| = |80 - 20| = 60\,\mu C$.The common potential $V$ after connection is $V = \frac{Q_{net}}{C_1 + C_2} = \frac{60\,\mu C}{2\,\mu F + 4\,\mu F} = \frac{60}{6} = 10\,V$.Initial energy $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 = \frac{1}{2}(2)(10)^2 + \frac{1}{2}(4)(20)^2 = 100 + 800 = 900\,\mu J$.Final energy $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2}(2 + 4)(10)^2 = \frac{1}{2}(6)(100) = 300\,\mu J$.Heat evolved $H = U_i - U_f = 900\,\mu J - 300\,\mu J = 600\,\mu J$.

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