For the circuit shown calculate `V_(A) - V_(B), I_(1), I_(2)` and I.

Explanation

`E = (4/4 - 6/12)/(1/4 + 1/12) = 3/2 V`
`r = 1/(1/4 + 1/12) = 3 Omega`
The circuit can be reduced to

`I = (-E)/(R + r) = (3//2)/(3 + 6) = 1/6 A`
`V_(A) - V_(B) = V_(A) - 6 xx 1/6 = V_(B), V_(A) - V_(B) = 1 V`
For upper branch
`V_(A) -I_(1) xx 12 + 6 = V_(B)`
`V_(A) - V_(B) + 6 = I_(1) xx 12`
`1 + 6 = I_(1) xx 12, I_(1) = 7/12 A`
For lower branch
`V_(A) + I_(2) xx 4 - 4 = V_(B)`
`I_(2) = 3/4 A`