In the figure shown , after the switch 's' is turned from postion 'A' to postion 'B' the energy dissipated in the circuit in terms of capactance 'C' and total charge 'Q' is :

Answer

8

Explanation

`q_(1)+q_(2)=CE rArr q_(1)=(CE)/(4)rArr q_(2)=(3CE)/(4)`
`(q_1)/( C) =(q_(2))/(3C)`

initial energy `=(CE^2)/(2)`
Final energy `=(q_(1)^(2))/(2C)+(q_(2)^(2))/(2xx3C)`
`=(C^2E^2)/(32)+(9C^2E^2)/(32xx3)=(4CE^2)/(32)`
`H=Detal U =(CE^2)/(2)-(CE^2)/(8)=(3CE^2)/(8)` .