Two capacitors C_1 = 4 mu F and C_2 = 2 mu F | vedclass

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(D) Initial charges on the capacitors are $Q_1 = C_1 V = 4\ \mu F 	imes 500\ V = 2000\ \mu C$ and $Q_2 = C_2 V = 2\ \mu F 	imes 500\ V = 1000\ \mu C

Vedclass · Accepted Answer

$A$ and $C$ both.

Vedclass · Answer

(D) Initial charges on the capacitors are $Q_1 = C_1 V = 4\ \mu F 	imes 500\ V = 2000\ \mu C$ and $Q_2 = C_2 V = 2\ \mu F 	imes 500\ V = 1000\ \mu C$. Since they are connected with opposite polarity,the net charge after closing the switches is $Q_{net} = Q_1 - Q_2 = 2000\ \mu C - 1000\ \mu C = 1000\ \mu C$. The equivalent capacitance is $C_{eq} = C_1 + C_2 = 4\ \mu F + 2\ \mu F = 6\ \mu F$. The common potential is $V' = Q_{net} / C_{eq} = 1000\ \mu C / 6\ \mu F = 500/3\ V$. Thus,option $A$ is correct. Initial energy $U_i = 1/2 C_1 V^2 + 1/2 C_2 V^2 = 1/2 (4+2) 	imes 10^{-6} 	imes (500)^2 = 3 	imes 10^{-6} 	imes 250000 = 0.75\ J$. Final energy $U_f = 1/2 (C_1 + C_2) (V')^2 = 1/2 (6 	imes 10^{-6}) 	imes (500/3)^2 = 3 	imes 10^{-6} 	imes 250000 / 9 = 0.75 / 9\ J$. The ratio $U_f / U_i = (0.75 / 9) / 0.75 = 1/9$. Thus,option $C$ is also correct. Therefore,option $D$ is the correct answer.

Two capacitors $C_1 = 4\ \mu F$ and $C_2 = 2\ \mu F$ are charged to the same potential $V = 500\ V$,but with opposite polarity as shown in the figure. The switches $S_1$ and $S_2$ are closed. Determine the correct statement.

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