show that there is always a loss of energy when two capacitors charged to different potentials share charge
Dear Student ,
Let there be two capacitors Ca and Cb which are charged to different potentials.Let Ca be at a potential Va and Cb be at a potential Vb.
Common potential of the two capacitors, Vc = ( CaVa + CbVb )/ ( Ca + Cb )
Energy that can be obtained from capacitor Ca = CaVa2/2
Energy that can be obtained from capacitor Cb = CbVb2/2
Total Energy that can be obtained from both Ca and Cb before sharing of charges takes place = CaVa2/2 + CbVb2/2
Energy that can be obtained from both the capacitors after sharing of charges has taken place = ( Ca + Cb ) Vc2/2
Thus , Loss of Energy during sharing of charges = Total Energy that can be obtained from both the capacitors before sharing of charges has taken place - Energy that can be obtained after sharing of charges has occured
Energy Loss = ( CaVa2/2 + CbVb2/2 ) - (Ca+Cb)Vc2/2
= (CaVa2/2 + CbVb2/2 ) - [( Ca + Cb )( CaVa + CbVb )2/2(Ca+Cb)2] ......(1)
Solving equation 1 , we get :-
Energy Loss = CaCb ( Va - Vb )2/2(Ca+Cb). This much amount of energy is always lost when two capacitors charged to different potentials share their charges.
Let there be two capacitors Ca and Cb which are charged to different potentials.Let Ca be at a potential Va and Cb be at a potential Vb.
Common potential of the two capacitors, Vc = ( CaVa + CbVb )/ ( Ca + Cb )
Energy that can be obtained from capacitor Ca = CaVa2/2
Energy that can be obtained from capacitor Cb = CbVb2/2
Total Energy that can be obtained from both Ca and Cb before sharing of charges takes place = CaVa2/2 + CbVb2/2
Energy that can be obtained from both the capacitors after sharing of charges has taken place = ( Ca + Cb ) Vc2/2
Thus , Loss of Energy during sharing of charges = Total Energy that can be obtained from both the capacitors before sharing of charges has taken place - Energy that can be obtained after sharing of charges has occured
Energy Loss = ( CaVa2/2 + CbVb2/2 ) - (Ca+Cb)Vc2/2
= (CaVa2/2 + CbVb2/2 ) - [( Ca + Cb )( CaVa + CbVb )2/2(Ca+Cb)2] ......(1)
Solving equation 1 , we get :-
Energy Loss = CaCb ( Va - Vb )2/2(Ca+Cb). This much amount of energy is always lost when two capacitors charged to different potentials share their charges.