Mesh Analysis for Current Source in between "blocks"
You can see it as a current divider consisting of the 2&Omega; resistor and the other two resistors that add up to 1&Omega; + 3&Omega; = 4&Omega;.

So the proportion of the conductances and thus the proportion of currents of both branches are1/4 : 1/2  or  1:2,
i.e. 1/3 of the total 6A are going one way and 2/3 the other way, i.e. 2A are goind through the left branch and 4A through the right branch.

If 2A are going through a 1 &Omega; resistor the voltage across it is 2V.