It's possible one of the primary windings has gone open circuit.
It should still be working if only one primary had gone OC. I imagine you'd have to be quite unlucky to OC both primaries on 115VAC.
[/quote]You're right of course. It should still work if one of the primaries had gone open circuit, if they were connected in parallel and it's highly unlikely they would both go open circuit.
It's possible one of the primary windings has gone open circuit. If you connected it to the mains as you described in your first post, which is opposite to schematic posted by ArthurDent, I suspect it's blown. The fact there was no smoke doesn't surprise me as the thermal fuses would have tripped to protect against fire.
Measure the DC resistance of all of the windings with a multimeter. The primary windings should be similar and the secondary much lower. It's difficult to estimate the rough expected values, without knowing the power rating of the transformer. I suspect this is a fairly small transformer? Had it been over several hundred VA, then the circuit breaker in the house would have tripped.
I really owe y'all an apology. I misstated in my original post. Since then, I have returned the toroid so I can't double check it but I think that I had it correct according to Arthur's chart.
Sheesh!
The dots indicate the transformer's phasing and should both be connected together.
I did a bit of Googling and found the technical brochure, which contains links to a clear data sheet.
https://4donline.ihs.com/images/VipMasterCap/Capacitor/AVEL/AVELS00030/AVELS00030-1.pdfI think it would have blown the breaker, if you got the phasing wrong. It doesn't give the primary series resistance, but it does specify the copper losses and the resistance of the primary can be estimated from that. Here are the calculations:
The efficiency is 95.1% and the transformer is rated to 500VA.
P
IN = 500W/0.951 = 526W
I = P/V = 526/120 = 4.38A
The copper losses are specified as 23W, assuming half of the power is lost in the primary and half in the secondary:
P
LOSS(PRIMARY) = 23/2 = 11.5W
So we have a primary current of 4.38A and power dissipation in the primary's resistance of 11.5W, so we can calculate the resistance, using Ohm's law:
R = P/I
2 = 11.5/4.38
2 = 0.6Ω
Of course half the current flows through each of the primary windings which are connected in parallel, so the resistance of each winding will be double that, but they're connected in parallel, so that's the resistance they'll present to the mains, if connected out of phase.
I = 120/0.6 = 200A.
200A will cause the main circuit breaker to pop very quickly.