In the arrangement shown in in the switch `S` is , find

the final charge on the `6 mu F` capacitor

Options

A. `12 muC`
B. `24 muC`
C. `32 muC`
D. `48 muc`

Answer

C

Explanation

In the close loop,
`(q_(0)- Deltaq)/C_(1)-(Deltaq)/C_(2)-(Deltaq)/C_(2)=0`
.
`q_(0)/C_(1)=Deltaq[1/C_(1)+1/C_(2)+1/C_(3)]`
`Deltaq=q_(0)/(C_(1)[1/C_(1)+1/C_(2)+1/C_(3)])=(192)/([1+3+2)]=32 muC`
Potential defferece across `C_(1)= (Deltaq)/(C_(2)) = 8 V`
Potential defferece across `C_(1)=((192-32))/(12)=(160)/(12)=(40)/3V`.