a a capacitor is charged by a battery. the battery is removed and other identical answers capacitor is connected in parallel. the total electrostatic energy of resulting system is
Answer (1)
22nd Dec, 2019
Hii,
The solution this problem is as follows:
Charge on capacitor is
q = CV
when it is connected with another uncharged capacitor then
Vc = q1+q2/ C1+C2 = q+0/ C+C
Vc = V/2
intial energy
Ui = 1/2 CV^2
final energy
Uf = 1/2 C(v/2)^2 + 1/2 C(V/2)^2
= (CV^2/4)
Loss of energy = Ui - Uf
= CV^2/4
Therefore total electrostatic energy of resulting system will decrease by a factor of 2.
Hope this helps.
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