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(D) Let the two identical capacitors have capacitance $C$ and initial potentials $V_1$ and $V_2$. Initial energy $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2

Vedclass · Accepted Answer

All of the above.

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(D) Let the two identical capacitors have capacitance $C$ and initial potentials $V_1$ and $V_2$. Initial energy $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2$. When connected in parallel,the common potential is $V = \frac{C V_1 + C V_2}{C + C} = \frac{V_1 + V_2}{2}$. Final energy $U_f = \frac{1}{2}(2C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1^2 + V_2^2 + 2V_1V_2)$. Comparing $U_i$ and $U_f$,we find $U_f Charge is conserved,so the net charge $Q_{net} = Q_1 + Q_2$ remains constant. The final potential difference across each capacitor is $V = \frac{V_1 + V_2}{2}$,which is not equal to $V_1 + V_2$. Thus,all statements are correct.

When two identical capacitors are charged individually to different potentials and connected in parallel to each other,after disconnecting them from the source:

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