Wilkinson Divider Current Density

If I simulate a Wilkinson Power Divider current density without the isolation resistors, will I get current density minimums at the locations where the isolation resistors belong? I'm running AWR/Microwave Office.

Tnx,
GE

Where is your feed? If you feed from the common port, the structure is symmetric and there is no current through the resistor anyway.

The feed could be from either direction depending on how the divider/combiner is used. If I feed from the divider ports (not common port), I guess that I will need to excite both divider ports in-phase.

Yes, the resistor is only needed for isolation (asymmetric currents).

Yes, the resistor is only needed for isolation (asymmetric currents)[/I][/B].....
Not only, The resistor is also needed to provide the matching to Z0 system of the output ports.