Find the current in each wire applying only Kirchhoff voltage law

Explanation

Applying Krichhoff voltage law in loop ABEF
`i_(1)+30+2(i_(1)+i_(2))-10=0`
`3I_(1)+2I_(2)+20=0`
Applying Kirchhoff voltage law in BEDCB
`+30+2(i_(1)+i_(2))+50+2i_(2)=0`
`4i_(2)+2i_(1)+80=0`
`2i_(2)+i_(1)+40=0`
Solving (i) and (ii)
`3[-40-2i_(2)]+2i_(2)+20=0`
`-120-4i_(2)+20=0`
`i_(2)=-25" A and "i_(1)=10A" "therefore i_(1)+i_(2)=-15A`
current in wire AF = 10 A form A to F
current in wire EB = 15 A from B to E
currnt in wire DE=25 A from E to D.