The energy of a charged capacitor is U. It is | vedclass

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(C) Initial energy of the capacitor is $U = \frac{Q^2}{2C}$.When the capacitor is disconnected from the battery and connected in parallel to an unchar

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$\frac{1}{9} U , \frac{2}{9} U$

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(C) Initial energy of the capacitor is $U = \frac{Q^2}{2C}$.When the capacitor is disconnected from the battery and connected in parallel to an uncharged capacitor of capacitance $2C$,the total charge $Q$ is redistributed such that both capacitors have the same potential difference $V'$.Since they are in parallel,$V' = \frac{Q_1}{C} = \frac{Q_2}{2C}$.This implies $Q_2 = 2Q_1$.Since the total charge is conserved,$Q = Q_1 + Q_2 = Q_1 + 2Q_1 = 3Q_1$.Therefore,$Q_1 = \frac{Q}{3}$ and $Q_2 = \frac{2Q}{3}$.The new energy of the first capacitor is $U_1 = \frac{Q_1^2}{2C} = \frac{(Q/3)^2}{2C} = \frac{1}{9} \left(\frac{Q^2}{2C}\right) = \frac{1}{9} U$.The new energy of the second capacitor is $U_2 = \frac{Q_2^2}{2(2C)} = \frac{(2Q/3)^2}{4C} = \frac{4Q^2/9}{4C} = \frac{1}{9} \left(\frac{Q^2}{C}\right) = \frac{2}{9} \left(\frac{Q^2}{2C}\right) = \frac{2}{9} U$.Thus,the energies are $\frac{1}{9} U$ and $\frac{2}{9} U$ respectively.

The energy of a charged capacitor is $U$. It is removed from the battery and then connected in parallel to another uncharged capacitor having capacitance twice that of the first one. The energy of the first and second capacitors respectively is . . . . . . .

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