Find charge and potential on each capacitor. All capactances are in `muF`.

Explanation

(a)
Series, `(1)/(C_(eq))=(1)/(12)+(1)/(20)+(1)/(5)`
`C_(eq)=3muF`
Now return to previous diagrams.
Now return to previous diagram.

In series, charge on each capacitor is same and equal to charge on equivalent capacitance.
`V_(1)=(45)/(12)=3.75V,V_(2)=(45)/(5)=9V,V_(3)=(45)/(20)=2.25V`

`5muF` is resultant of `2` and `3muF` in parallel. In parallel potential on individual capacitors and equivalent capacitor is same. here potential on `5 muF,2 muF` and `3muF` will be same and equal to `9V`.
(b)
`V_(1)=((10)/(10+40))xx10=2V,V_(2)=10-V_(1)=8V`
Now, return to previous diagram keeping in mind that in parallel, potential is same and in series charge is same.


(c)
`V_(1)=((6)/(3+6))xx6=4V, V_(2)=6-V_(1)=6-4=2V`
`V_(3)=6V`
Return to previous diagram.

Final diagram :

(d)

Now, return to previous diagrams.

`V_(1)=(240)/(12)=20V,V_(2)=(240)/(24)=10V,V_(32)=(240)/(8)=30V`

Final diagram:

(e)
`V_(1)=((20)/(20+10))xx30=20V`
`V_(2)=30-V_(1)=10V`
Now, return to previous diagram.
Final diagram:
`V_(1)=(2)/((2+(16)/(3)))x110=(6)/(22)xx110=30V`
`V_(2)=110-V_(1)=80V`
Now, return to previous diagram. Final diagram :