(D) Initial charges on the capacitors are:$Q_1 = C \times V = CV$$Q_2 = 2C \times 2V = 4CV$Since they are connected with opposite polarities (positive

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(D) Initial charges on the capacitors are:$Q_1 = C \times V = CV$$Q_2 = 2C \times 2V = 4CV$Since they are connected with opposite polarities (positive to negative),the net charge available is:$Q_{net} = Q_2 - Q_1 = 4CV - CV = 3CV$When connected in parallel,the equivalent capacitance is:$C_{eq} = C + 2C = 3C$The common potential $V_c$ is given by:$V_c = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$The final energy $U_f$ stored in the configuration is:$U_f = \frac{1}{2} C_{eq} V_c^2 = \frac{1}{2} \times (3C) \times V^2 = 1.5 CV^2$

Class 12 Physics Electric Potential and Capacitance Sharing of Charge in Capacitor Circuit

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Two capacitors of capacitances $C$ and $2C$ are charged to potential differences $V$ and $2V$,respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is $.....CV^2$.

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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