Uh, nevermind. A-C will show slightly higher resistance even with a delta. Not sure if it comes out exactly equal or not, but I’d give it better than odds that it does.
am77494’s post reminds me of another old (but simpler) puzzle: You have a black box with two contacts, which contains either a 1-amp constant current supply in parallel with a 1-ohm resistor, or a 1-volt constant voltage supply in series with a 1-ohm resistor, but you don’t know which. Using any measurement equipment outside of the box, how can you tell the difference?
[spoiler]Well–I’d say “short it, and if the box heats up then it’s a CV supply.”
But if this is really an idealized black box, then we can’t necessarily guarantee that. If it has some kind of super-battery, and an internal heat reservoir, then it can keep the external heat flux to whatever it wants by making the internal reservoir hotter. Alternatively, and more practically, you could have a variable heater in both boxes that keeps both of them emitting a constant flux.
'Course, you didn’t say the boxes had this other stuff… but you also didn’t specify the power supply or anything else, so there’s a lot that has to be inferred.[/spoiler]
[spoiler]You don’t even need to short it, because the constant-current box is already heating up at 1 W even without the short (and will stop heating when you short it).
In any event, it’s a clear proof that Thevenin doesn’t cover everything, so you might be able to tell the difference from the heating.[/spoiler]
And it feels like the OP’s problem should be solvable using an oscilloscope, but I’m too far out of practice on AC to figure out the details.
Here is something I tried:
-
At room temperature, measure resistance between leads A & B, A & C, and B & C.
-
Stick a voltage source between A and B. (Leave C disconnected.) Leave it hooked up for a few minutes.
-
Repeat step 1. Calculate ratios.
If it is a Y generator, two of the leads will heat up. The third will not.
If it is a delta generator, all three leads will heat up (though not equally).
I modeled both the Y and delta generators in Excel. I assumed a certain temperature rise with the tempco of copper at 0.393%/degree C.
The Y and delta generators had the same ratios. 
Ah, yes, I see… The trouble is that, on a Y generator, one winding won’t heat up, but you can’t measure the resistance of a single winding. You should, in principle, be able to tell the difference using calorimetry (the delta will produce more total heat), but that would be very difficult to do in practice.
A coworker has been trying to come up with all kinds of clever schemes using external power supplies, switches, diodes, etc. And each time, my analysis shows that the approach cannot differentiate between a delta and Y configuration.
And again, this assumes you do not know the winding resistance or winding inductance. You only know that all three windings are the same. (If the winding resistance or inductance is known, the problem becomes trivial.)
I am beginning to think it is not possible; there is no solution.
That is only true if I know the resistance of the windings. But I don’t.
A Y will easily produce more total heat than a delta if its winding resistance is very low compare to the delta (and you use a voltage source).
However, on a Y, one winding would remain at room temperature. On a delta, none of the windings would remain at room temperature. So if you could peer inside and take temperature measurements, it might be doable…
So far I think all that has been shown is that the classic star-delta transformation will always stifle anything that tries to measure the simple impedances present. The point being that there is a transform that exactly transforms a delta to a Y. Even one with unbalanced legs. (It might have been that the transformation with unbalanced legs left you with a different result, but it doesn’t 
)
Whether there is some second order effect that allows you to tell, I’m not sure. But it isn’t going to be something that you can tell looking for an asymmetry between delta and Y, because there isn’t one.
Why do you need to peer inside ? The terminals themselves get hot.
Lurker here, sort of understanding, learning:
Is friedo correct or not?
I assume the separate windings are thermally isolated, at least in the very short term?
Quickly pass a large current between terminals A and B. After shutting off the current, immediately measure the resistance between A and B, and the resistance between A and C.
By my calculations, the change in resistance between A and B should be exactly twice the change in resistance between A and C for a Y generator.
For a Delta generator, it is more complicated. For small temperature changes, you must assume one coil (the one between A and B) will receive twice the current of the other two coils, and heat up twice as much, and consequently its resistance will change twice as much. My calculations show that in this circumstance the change in resistance between A and B will be approximately 1.4 times the change in resistance between A and C.
Just realized my error. For a Delta, one coil will receive twice the current, but will receive 4 times the power of each of the other two coils, so temperature and resistance of that one coil will increase by a factor of 4 also. It appears the overall resistance change will match that of a Y generator.
You would have to exploit the non-linearity of the wire’s resistance versus temperature somehow.
nm
Non linearity is the key. The star-delta transform guarantees that any effect that is linear won’t work.
One that does occur to me is to try to use the hysteresis or saturation curve curve of the magnet cores. You should be able to see this with an oscilloscope, but it means driving quite a bit of current. Measured across two terminals a Y should give a single bump as you start to saturate the two coils in series, whereas a delta should show two bumps, as the current through the single coil will be twice that through the pair, it will saturate first, and the series pair saturates at double the current.
The OP explains the provenance of the problem in post #14.
I made a spreadsheet model.
For the Y I did the following:
Resistance of winding between A and center (R[sub]A[/sub]) at 25 °C = 1 Ω
R[sub]B[/sub] at 25 °C = 1 Ω
R[sub]C[/sub] at 25 °C = 1 Ω
And thus the resistance measurements would be
R[sub]AB[/sub] = 2 Ω
R[sub]AC[/sub] = 2 Ω
R[sub]BC[/sub] = 2 Ω
I apply a voltage between A and C and wait a few minutes. I then assume
Temperature of R[sub]A[/sub]= 60 °C
Temperature of R[sub]B[/sub]= 25 °C (assume no conduction here)
Temperature of R[sub]C[/sub]= 60 °C
Given the tempco of copper is 0.393%/°C,
R[sub]A[/sub] = 1.13755 Ω
R[sub]B[/sub] = 1 Ω
R[sub]C[/sub] = 1.13755 Ω
R[sub]AB[/sub] = 2.13755 Ω (6.8775% increase)
R[sub]AC[/sub] = 2.2751 Ω (13.775% increase)
R[sub]BC[/sub] = 2.13755 Ω (6.8775% increase)
13.775/6.8775 = 2
For the delta I did the following:
Resistance of winding between A and B (R[sub]AB[/sub]) at 25 °C = 3 Ω
R[sub]AC[/sub] at 25 °C = 3 Ω
R[sub]BC[/sub] at 25 °C = 3 Ω
And thus the resistance measurements would be
R[sub]AB_meas[/sub] = 2 Ω
R[sub]AC_meas[/sub] = 2 Ω
R[sub]BC_meas[/sub] = 2 Ω
I apply a voltage between A and C and wait a few minutes. I then assume
Temperature of R[sub]AB[/sub]= 38.75 °C
Temperature of R[sub]AC[/sub]= 80 °C
Temperature of R[sub]BC[/sub]= 38.75 °C
(I assumed 4X the temperature rise for R[sub]AC[/sub] compared to R[sub]AB[/sub] and R[sub]BC[/sub]. This is because there’s twice as much current in R[sub]AC[/sub] compared to R[sub]AB[/sub] and R[sub]BC[/sub].)
Given the tempco of copper is 0.393%/°C,
R[sub]AB[/sub] = 3.2 Ω
R[sub]AC[/sub] = 3.65 Ω
R[sub]BC[/sub] = 3.16 Ω
R[sub]AB_meas[/sub] = 2.15948 Ω (7.97% increase)
R[sub]AC_meas[/sub] = 2.31368 Ω (15.68% increase)
R[sub]BC_meas[/sub] = 2.15948 Ω (7.97% increase)
15.68/7.97 = 1.97 Same as Y!!!
That is the point you are screwed. That is a linear assumption. I don’t think anything will work until you find a non-linear relationship. Which is why I suggested the IB curve of the magnets.
Just to add, here is a theoretical view of a method I think will work.
…
And edited, because I found a flaw.
Well, here’s a non-linear effect:
Run a current through two terminals until one of the coils overheats and fails. If it’s a wye, then the current will drop to zero. If it’s a delta, then the low-resistance path will fail first, leaving the high-resistance path–so the current will not drop to zero.
What? You just said I can’t disassemble it…