Given circuit is simplified and redrawn in right side of figure.
Parallel combination of two 2Ω resistances across points A and B
give equivalent resistance 1Ω . This equivalent resistance 1Ω is in series with 2Ω resistance gives
effective resistance 3Ω which is shown in simplified circuit at right side of figure.
The simplified circuit has two loops marked as (1) and (2) .
Let us consider current distribution i1 , i2 and i3 in the circuit as shown in figure.
Let us apply Kirchoff's voltage law (KVL) to loop-1.
-6 - 11i2 + 8 = 0 i.e., i2 = ( 2/11 ) A
Let us apply Kirchoff's voltage law (KVL) to loop-2.
-8 + 4 i3 = 0 or i3 = 2 A
Current passing through 8Ω resistance is i2 = (2/11) A
Current passing through 4Ω resistance is i3 = 2 A