Find the charge on the three capacitors shown in figure.
(a)
(b)

Explanation

(a)
The cells is parallel can be replaced by a single cell.
`E = (C_1 E_1 + C_2 E_2)/(C_1 + C_2) = (3 xx 10 + 6 xx 40)/(3 + 6) = (30 + 240)/(9) = 30 V`
`C = C_1 + C_2 = 3 + 6 = 9 mu F`

`V' = ((9)/(9 + 18)) xx 30 = 10 V`
`V_A - V_B = 10 V`
Now return to previous diagram :

Branch `A xx B`
`V_A - (Q_2)/(3) - 10 = V_B`
`V_A - V_B - 10 = (Q_2)/(3)`
`10 - 10 = (Q_2)/(3) rArr Q_2 = 0`
Branch `A Y B` :
`V_A -(Q_1)/(6) - 40 = V_B`
`V_A -V_B - 40 = (Q1)/(6)`
`10 -40 = (Q_1)/(6) rArr Q_1 = - 180 mu C`
`Q_1 + Q_2 = -180 mu C`

Charge on capacitor
`3 mu F : 0 , 18 mu F: 180 mu C, 6 mu F : 180 mu C`.
Alternatively by Kirchoff's rules
Assume charge in branches, balancing at junctions.

Left loop :
`V_A - (Q_2)/(3) -10 -((Q_1 + Q_2))/(18) = V_A`
`-Q_1 - 7 Q_2 = 180`
`Q_1 + 7 Q_2 = -180`....(i)
Right loop :
`V_A -(Q_1)/(6) - 40 -(Q_1 + Q_2)/(18) = V_A`
`4 Q_1 + Q_2 = - 720`
Solving (i) and (ii)
`Q_1 = -180 mu C, Q_2 = 0`
(b)
`E = (20 xx 12 - 10 xx 6)/(12 + 6) = 10 V`
`C = 12 + 6 = 18mu F`
`V' = ((80)/(18 + 9)) xx 10 = (20)/(3) V`
charge on `9 mu F : 9 xx (20)/(3) = 60 mu C`

Branch `AXB` :
`V_A - 20 + (Q_1)/(12) = V_B`
`(Q_1)/(12) = V_B - V_A + 20 = -(20)/(3) + 20 = (40)/(3)`
`Q_1 = 160 mu C`
Branch `AYB` :
`V_A + 10 + (Q_1)/(6) = V_B`
`V_A - V_B + 10 = - (Q_2)/(6)`
`(20)/(3) + 10 = (-Q_2)/(6)`
`Q_2 =- 100 mu C`
Check : `Q_1 + Q_2 = 60 mu C, O.K.`

( c)
`E = (30 xx 2 +15 xx 4 + 20 xx 6)/(2 + 4+ 6) = 20 V`
`C = 2 + 4 + 6 = 12 mu F`

`V_B - V_A = 20 V`

`AXB` :
`V_A + 30 - (Q_1)/(2) = V_B rArr V_A - V_B + 30 = (Q_1)/(2)`
`-20 + 30 = (Q_1)/(2) rArr Q_1 = 20 mu C`
`AYB` :
`V_A + 15 -(Q_2)/(4) = V_B rArr V_A - V_B + 15 = (Q_2)/(4)`
`-20 + 15 = (Q_2)/(4) rArr Q_2 = -20 mu C`
Check `AZB` :
`V_A + 20 -(Q_1 + Q_2)/(6) = V_B`
`V_A - V_B + 20 = (Q_1 +Q_2)/(6)`
`-20 + 20 = (Q_1 + Q_2)/(6)`
`Q_1 + Q_2 = 0`.