Let resistances be R1,R2,R3 and R4 and I1 current is passing through R4 as shown in figure ∴I2=(6−I1) is passing through R2 As, same current is flowing through R4 and R3. ∴R4 and R3 are in series.
and series equivalent resistance, Req =R4+R3 ∴Req =2+2=4Ω Voltage through Req and R2 will be same. ⇒I1Req=I2R2⇒I14=(6−I1)2 ⇒2I1=6−I1⇒I1=2A