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$Initial

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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$Initial energy stored in the capacitors is:$U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 = \frac{1}{2} (2 	imes 10^{-6}) (10)^2 + \frac{1}{2} (4 	imes 10^{-6}) (20)^2 = 100 \mu J + 800 \mu J = 900 \mu J$Since the plates are connected with opposite polarity,the net charge is:$Q_{net} = |Q_2 - Q_1| = |80 \mu C - 20 \mu C| = 60 \mu C$The common potential $V$ after connection is:$V = \frac{Q_{net}}{C_1 + C_2} = \frac{60 \mu C}{2 \mu F + 4 \mu F} = \frac{60}{6} V = 10 \ V$Final energy stored in the capacitors is:$U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (6 \mu F) (10 \ V)^2 = 300 \mu J$Heat evolved in the circuit is the loss in energy:$H = U_i - U_f = 900 \mu J - 300 \mu J = 600 \mu J$

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