I don't know whether this might explain it, bit I'll offer it.Yes, I'm now reading that all over the place, so I suppose it must be true !!
I've never had reason to think in any depth about, or to understand about, motors, but there is clearly some learning for me to do here!
However, even if I forget that it's a motor and just consider it as a 'black box' with two terminals to which I apply an AC voltage, I still have a problem. The one thing I didn't mention before is that the DC resistance of the motor I was testing was exactly 2.0kΩ. If that were purely resistive, with ~240V AC applied to it, that should result in a current of ~120mA, but what I actually get is a current of ~30mA, which equates to an impedance within my 'black box' of around 8kΩ. However, I am also seeing a PF of around 1.0, which implies a net reactance of zero (i.e. current and voltage in-phase) - so where is 'missing' 6kΩ of impedance within my black box coming from?
Kind Regards, John
In an ordinary induction motor, if you measure the DC resistance of the windings it would indicate a much higher current than observed. The reason the current is lower is not because of inductance, it's because the motor produces a back-emf, opposing the applied voltage. If the synchronous motor produces back-emf 180 volt, that would give the current figure seen.
Problem is I think the motor must be turning to produce the back-emf (but maybe not for a synchronous motor). If so the theory doesn't work when it's stalled. What do you think?