I have a power supply that contains a dual primary 330 VA transformer, so I thought that I would do a test. The two primary windings are 2.3 Ω each at DC.
I put two 470 Ω power resistors in series (so 940 Ω), in series with the mains, and tested the transformer with virtually no load in three conditions.
Normal, both winding in series. There was 24 V across the 940 Ω ohms, so about 25 mA
One winding reversed, and both in series. There was 235 V across the 940 Ω, and 1 V across the transformer. The current would have been 235 / 940 = 0.25 A, so the transformer impedance was about 4 Ω, which is all accounted for by the resistance of the windings.
Finally, one winding open, the other connected. There was 139 V across the transformer, 113 V across the 940 Ω. As the windings are 120 V each, the core would have been saturating, which accounts for the current of 113 / 940 = 120 mA. There would have been much less current at 120 V, but I don't have an 120 V supply here.
Anyhow, reversing one winding reduces the transformer inductance to nearly zero. With no safety resistor, on 120 V, the primary current would have been about 30 A, and the transformer heating would have been about 3 kW.
If you don't know the phasing of the transformer primaries, do not guess and then connect straight to the mains.