Find the potential difference between the points `A` and `B`.
(a)
If `V_A - V_B = 5 V`, find voltage across each capacitor.

Explanation

(a)
Here cells are in series and opposing.
Net emf `= 24 - 12 = 12 V`

`Q = 2C xx 12 = 24 C`

`V_A - V_B = (Q)/(3 C) = (24 C)/(3 C) = 8 V`
(Positive plate is at higher potential)
Alternatively

`V_1 = (6 C)/(3 C+ 6 C) xx 12 = 8 V = V_A - V_B`
(b)
Then cells are in series, opposing
Net emf `= 24 - 12 = 12 V`

`V_1 = (6)/(6 + 3) xx 12 = 8 V = V_B - V_A`
`V_A - V_B = -8V`
( c)
The cells are in series
Net emf `= 24 - 12 = 12 V`
The capacitors are in series
`(1)/(C_(eq)) =(1)/(8 C) +(1)/(12 C) +(1)/(24 C) = (1)/(4 C)`
`C_(eq) = 4 C`

`Q = 4 C xx 12 = 48 C`
Now return to previous diagram, in series charge on each capacitor is same.

Now move from `A` to `B`
`V_A + (Q)/(8 C) + 12 = V_B`
`V_A + (48 C)/(8 C) + 12 = V_B`
`V_A - V_B = -18 V`
(d)
`V_1 = (3C)/(3C + 6C) xx 10 =(10)/(3) V = V_A - V_B`
( e)
Move from `A` to `B` :
`V_A -(Q)/(1)+10 -(Q)/(2) = V_B`
`V_A - V_B + 10 = Q + (Q)/(2)`
`5 + 10 = (3 Q)/(2)`
`Q = 10 mu C`

`V_1 = (Q)/(1) =(10)/(1) = 10 V`
`V_2 = (Q)/(2) = (10)/(2) = 5 V`.