Doubt from capacitor

I stucked while writing charge as function of time. Help anyone.



Find work done by battery(W=qV) and change in potential of capacitor.Heat loss=W-( change in U)

I have its soln
Could you pls tell from where 3/2 came.

Initially, charge was CE/2 in 1st capacitor and finally both capacitor has 2CE charge therefore total charge flown from battery is 3CE/2

1 Like

Thanks bro.
I got it.
I make silly mistake by considering the first capacitor to be fully charged initially.

1 Like