Find the potential difference `V_(a)-V_(b)` between the points (1) and (2) shown in each part of Fig.

Explanation

(a) We first distribute the charges on different capacitors and branches keeping in mind Kirchhoff's junction rule .We can start from any battery. In this case, We start from battery `epsilon_(1)`. Let battery `epsilon_(1)` supply a charge `q` The charge on the left plate of capacitor `C_(1)` wil be `q` as this plate is receiving the charge. Now charge `q` reches junction `b` where it is divided into two Paths. Let charge `x` go to capacitor `C_(3)`, then the remaining charge `q-x` will go to capacitor `C_(2)`. In juncation `a`, we can verify Kirchoeff's juncation rule, i.e., the incoming charge is equal to the outgoing charge. Now we write the loop equations.
For loop `1`
`2 - (q)/(2) - (x)/(C) = 0`
For loops `2`
`2 + (x)/(C) - ((q-x))/(2) = 0` (i)
or `2 + (x)/(C) - (q)/(2) + (x)/(2) = 0` ......(ii)
From `(i)` and `(ii)`
`(2x)/(C) + (x)/(2) = 0`
or `x ((2)/(C) + (1)/(2)) = 0`
or `x = 0`(since `(2)/(C) + (1)/(2) ne 0`)
No charge will go to the capacitor connected across `(1)-(2)`. A there is no charge in the capacitor connected across `(1)` and `(2)`, the potential difference across `(1)` and `(2)` should be zero.

(b) In this case also, we will first distribute the charges. Let us start from the `24 V` battery. Let this battery supply a charge `q`, which is divided into two paths after reaching juncation `(2)`. Let `x` charge move toward the `2 muF` capacitor, then the reminaning charge `q - x` will move toward the `4 muF` capacitor, as per Kirchhoff's juncation rule. Now we write the loop equations.

In the upper closed loop
`+6+((q-x))/(4)-(x)/(2)-12=0` or `(q)/(4)-x=6`....(i)
In the lower closed loop
`12+(x)/(2)+(q)/(1)-24=0` or `q + (x)/(2) = 12`.....(ii)
From `(i)` and `(ii)`,
`x = (-8)/(3) mu C`
Moving from `(1)` to `(2)` through middle path
`V_(a) + 12 - (8//3)/(2) = V_(b)`
or `V_(a) - V_(b) = (4)/(3) - 12 = (-32)/(3) V`