Question

two identical capacitors have the same capacitance C. one of them charged to potential V1 and other to V2 . the negative ends of the capacitor are connected together. when the positive ends are also connected then what will be the decrease in the energy of the combined system

Answer 1

∆U=initial energy - final energy
1/2C (V1 - V2) -1/2(2C)(V1 -V2/2)
=1/4C (V1^2 -V2^2)
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Answer 2

Answer:

Explanation:

Initially, the total electrostatic energy stored in two capacitors is Ui =1/2CV²1 + 1/2CV²2.

Initial charges on two capacitors are q1=CV1 and q2=CV2.

Let q′1 and q′2 be the charges on capacitors when the same terminals of two capacitors are connected to each other. By charge conservation

= q′1+q′2=q1+q2

The potential V′ across two capacitors is equal = q′1/C =q′2/C

Thus, q′1 = q′2 =(q1+q2)/2, and V′ = (V1+V2)/2

Thus, final electrostatic energy stored in the two capacitors is -

Uf = 1/2CV² + 1/2CV² = 1/4C ( V²1 + V²2 + 2V1V2)

and loss in energy will be = Ui − Uf = 1/4C(V1−V2)².