When two charged conductors having different capacitances and different potentials are joined together, show that there is always a loss of energy.

Explanation

Let we have two capacitors of capacitance `C_(1)` and `C_2`
`V_(1), V_(2)=` Potential difference across capacitors `C_(1)` and `C_(2)`
Before connecting the two capacitors :
Energy stored in first capacitor `= U_(1) = 1/2 C_(1)V_(1)^(2)`
Energy stored in second capacitor `=U_(2) = 1/2 C_(2)V_(2)^(2)`
Total energy stored in the two capacitors = U and `U = U_(1) + U_(2)`
`U =1/2 C_(1)V_(1)^(2) + 1/2 C_(2)V^(2)`
When two capacitors are connected in parallel with each other Then total charge = `q=q_(1) + q_(2) = C_(1)V_(1) +C_(2)V_(2)`
Total capacitance of two capacitors`=C_(1) + C_(2)`.
So total energy of the capacitors, after they are connected, is-
`U.=1/2 q^(2)/C =1/2(q_(1)+q_(2))^(2)/(2(C_(1) + C_(2)))`
`U. = 1/2(C_(1)V_(1) + C_(2)V_(2))^(2)/(C_(1) + C_(2))`
Now, `U- U. =(1/2 C_(1)V_(1)^(2) + 1/2 C_(2)V_(2)^(2)) -1/2(C_(1)V_(1) + C_(2)V_(2))^(2)/(C_(1) + C_(2))`
`=(C_(1)V_(1)^(2) (C_(1) + C_(2)) + C_(2)V_(2)^(2)(C_(1) + C_(2))-(C_(1)V_(1) +C_(2)V_(2))^(2))/(2(C_(1)+ C_(2))`
`U - U. = (C_(1)C_(2)(V_(1)^(2) + V_(2)^(2) -2V_(1)V_(2))/(2(C_(1) + C_(2))))`
`U- U. = (C_(1)C_(2) (V_(1)-V_(2))^(2))/(2(C_(1) + C_(2)))`
U-U.=a positive quantity which is only possible when `U gt U` and it clearly shows that there is loss of energy on sharing the charges by the two capacitors.