Current Electricity Part 2

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