Find the current in each branch of the cirucit shown in figure

Answer

A

Explanation

Loop 1
`-42-6(i_1-i_2)-5i_1-i_1=0` ………i
Loop 2
`-4i_2-10-8(i_2-i_3)+6(i_1-i_2)=0`……….ii
Loop 3
`8(i_2-i_3)=-16i_3+4=0`……….iii
Solving these equations we get
`i_1=4A, i_2=1.0A` and `i_3=0.5 A`