(a) Find charge on capacitors

(b) Find charge on capacitor

(c ) Find energy stored in capacitor.

Explanation

(a)
`R_(eq) = 8+1+4 = 13Omega`
`i= 26/13 = 2A`
`V_(A) = ixx8-ixx1=V_C`
`V_(A)-V_C = 9i=9xx2 = 18V`
Charge on `C_(1)` :
`Q_(1)=C_(1)(V_A-V_C)=5xx18=90muC`
`V_(B) = ixx1-ixx4=V_D`
V_(B)-V_D=5i=5xx2 = 10V`
Charge on `C_(2)` :
`Q_(2) - C_(2) (V_(B)-V_(D)) = 10xx10 = 100muC`

Between `A` and `O`, via `X , 3Omega` and `3Omega` are in series ie., `6 Omega`.
Between `A` and `O` , `6Omega and 3Omega` are in parallel ie.
`(6xx3)/(6+3_ = 2 Omega`
`R_(eq)=2+3 = 5Omega`
`i=15/5 = 3A`
`i_(1) = 3/(3+6)xxi = 1A`
`i=i_(1)=3-1=2A`

`V_(X)-1xx3-3xx3=V_(Y) implies V_(X)-V_(Y) = 12V`
Charge on capacitor `Q = C(V_(X)-V_Y)`
`=3xx12=36muC`
(c ) First determined currents in the branches
`ADC` and `CB` by junction law at junctions `A and B`.

`V_(A) -3xx5-3xx1-1xx2=V_(B)`
`V_(A)-V_(B)=V= 20V`
`p.d`. across capaitor `V = 20V`
the enregy stored in capacior
`U - 1/2 CV^2`
`=1/2 xx 4xx10^(-6)xx(20)^(2)`
`=400xx10^(-6)J`
`=400muJ`.