190330-1016 EDT
To add to my comments on measurements.
A transformer can, and is often times, modeled as a series resistor and inductor in the primary of an ideal transformer, plus a series resistor and inductor in each secondary. This also means that for a two coil transformer, a primary and one secondary, that the model can lump all the series impedance in either the primary or secondary. For a multiple secondary transformer the distribution should be between both the primary and secondaries.
In my measurements where the sum of the two individual secondary voltages exactly equaled the line to line measurement there could easily have been readings that might have differed by possibly +/-0.1 V or so from the digital meter alone, and of course there is always the power company voltage fluctuation. Thus, I was quite surprised at the exact match.
For simplicity I will model the transformer as two identical voltage sources connected in series (identical frequency and voltage, and a fixed phase relationship so as to add), and no internal impedance.
For this equivalent circuit I will label the terminals as P1, N, and P2.
P1 to N is 100 V, N to P2 is 100 V, and based on the above assumptions P1 to P2 is 200 V.
Assume three distribution wires from the three terminals. Each wire has a resistance of 0.1 ohm. And a test load of 9.8 ohms is used.
Label the ends of the distribution wires as P1L, NL, and P2L.
Open circuit P1L to NL is 100 V, P2L to NL is 100 V, and P1L to P2L is 200 V.
Place the 9.8 ohm test load from P1L to NL, and the current thru the test load is 10 A, and the voltage drop along P1 to P1L is 1 V. So there is also a 1 V drop along NJ to N. The load voltage is 98 V.
There is no load on P2L, and thus no voltage drop along P2L to P2. Thus, the voltage from NL to P2L is 102 V. The voltage from P1L to P2L is 100-1+100 = 199 V.
Change the neutral resistance from 0.1 ohm to 1.0 ohm. Keep no load on P2, and load P1L to NL with the 9.8 ohm resistance. Load current becomes 100/(0.1+9.8+1.0) = 100/10.9 = 9.174 A. The voltage across the test load is 9.174*9.8 = 89.91 V. The voltage from NL to P2L is 100+9.174 = 109.174 V.
The voltage from P1L to P2L is 89.91+109.174 = 199.08 V. This added to 0.9174 = 199.9997 or 200 V. Thus, a check.
Had there been more drop in the P1 to P1L line it would have shown up in these measurements.
Had there been a high resistance in the P1L to P2 wire it would not show up here unless greater than possibly 10,000 ohms. Thus, the test load test needs to be applied to both phases.
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