Let's first find the current in the circuit.
I=2A ....given
I1=I2=I/2=1A
I3=I4=I1/2=0.5A
Finding net resistance between point AD
In AB and BC the currents are :
I1 flows from AB and I1-I3 flows from BC have different values so hence they are connected in parallel...
2*2/2+2=1Ω
Now ,
BC and CD have same values of current so we can take it as Series resistance.
1+2=3Ω
Now we consider CD and BD Their currents are same so they are in series
3+2=5
Νοω ιη ΒD and AD values of current are different so hence they are in parallel combination.
5×2/5+2=10/7=1.428~1.43
Hence I3=0.5 and Req is nearer to 1.25 Value..hope this helps ..
Regards,,,
An aspirant who is learning through lots of mistakes ...
But it's okay to not be okay ...,!