Two capacitors, `C_(1) and C_(2)`, charged to a potential `V/2` and V respectively with polarities as shown, are connected in a circuit with an ideal battery of EMF V. The capacitances are `C_(1)=2C` and ``aC_(2)=C`. At `t=0`, the switch S is closed. A long time after S is closed,

Options

A. The upper plate of `C_(1)` has positive charge CV
B. Total charge that has flowed through battery is 3CV
C. Total loss in heat dissipation is `(9)/(4)CV^(2)`
D. Potential energy stored in `C_(1)` is same as potential energy stored in it at t = 0

Answer

A

Explanation

Between t=0 and the steady state,
`+(3)/(2)CV` charge flowed into upper plate of `C_(1)`,
`+CV` charge flowed into upper plate plate of `C_(2)`
So, `+(5)/(2)CV` charge flowed out of positive terminal of battery
Heat dissipation`=U_(i)+W_("battery")-U_(f)`
`=((1)/(2)C((V)/(2))^(2)+(1)/(2)(2C)V^(2))+((5)/(2)CVxxV)-(1)/(2)CV^(2)-(1)/(2)(2C)V^(2)=(7)/(4)CV^(2)`.