Ajay T.s., Meritnation Expert added an answer, on 31/12/10

Hi Sanal!

The solution to the problems is as follows:

(i)Charge on each capacitors:

The capacitors C_{1} and C_{2} are in series. Therefore the equivalent capacitance of this combination is

As you know, charge Q= CV.

Therefore charge on C_{1}= charge on C2=3µF×12V=36µC.

Charge on C_{3}=6µF×12V=72µC.

(ii)Equivalent capacitance of the network:

The equivalent capacitance of the series combination of C_{1} and C_{2} is 3µF. This is in parallel with C_{3.} Therefore, the equivalent capacitance of the network C_{eq}=3µF+ 6µF =9µF.