In the circuit shown in figure ,
a. find the current is each branch
b. find the potential difference `V_(ab)` of point a relative to point b.

Explanation

a. `I_(1)` to the left thorugh the 10V battery `I_(2)` to the right through
5V battery , and `I_(3)` to the right thorugh to `10Omega` resistor.
upper loop: anticlockwise
`10V - (2Omega +3 Omega)I_(1)`
`-(1Omega + 4Omega)I_(2) - 5V = 0`
or `5V - (5Omega)I_(1) - (5Omega)I_(2) =0`
or `I_(1)+I_(2) =1A`
Lower loop : antilockwise
`5V + (1Omega +4Omega)I_(2) (10Omega)I_(3) = 0`
or `5V +(5Omega)I_(2) - (10Omega)I_(3) = 0`
or `I_(2) - 2I_(3) = -1A`
Along with `I_(1) = I_(2) +I_(3),` we can solve for th three curents
and find `I_(1) = 0.8A, I_(2) = 0.2 A, I_(3)= 0.6A`
b. `V_(ab) = -(0.2A)(4Omega) - (0.8A) (3Omega) = 3.2V`