It isn't quite like that...
The idler is working with severely unbalanced phases (power coming in on one phase, going out on two phases). This causes a large counter rotating field or, if you prefer, the normal three phase rotating field plus a large oscillating field. As a result of this, the rotor windings are always carrying a large 60Hz induced current, even when the idler is up to speed. The outer copper winding will carry this current in preference to the deeper aluminium winding because the high frequency as you said, but it is carrying that current all the time, not just on startup.
regards
mark
That's not quite what was said...............
The deep winding WON'T carry as much on startup. AND a lower resistance winding won't give as much starting torque.....but it works a bit better for action as an RPC....
So what is done is to add the higher resistance "surface" winding, and bury the other one. What that does is to allow the idler to start better, AND still behave well as an RPC....
When the rotor is slow at startup, the slip frequency is the difference between the back EMF frequency and the line frequency.... A buried winding is shielded from that to some degree, and so the higher resistance surface winding is dominant, giving decent starting characteristics , preferably without the need for a big bank of starting capacitors. (My Arco has NO start caps, and NO switch, but others do)
Once the rotor is up to speed, the lower slip frequency can penetrate to the buried winding, inducing currents in it as is normal for any motor. There are motors with buried rotor bars and NO surface bars..... it's a design decision, perhaps for higher speeds, to retain the bars better.
The fields created by the currents persist as long as the currents do, and as the rotor turns, generate back EMF in other windings. As it happens, the windings connected to the generated phase's wire have little correctly phased exciting voltage on them, and so that induced voltage is almost all effective in producing output current, and a counter-torque is produced on the rotor.
The rotor torque slows the rotor and yes it does increase the input current by decreasing the back emf. Of course there are large rotor currents, that's how it works.
However, since the two windings share a net field, if the same volts per turn are applied to two coils, and one has higher resistance, it will carry less current.
So there is a transfer of current from predominately surface winding at startup, to a sharing with a lot of current carried by the lower resistance buried windings as slip frequency decreases.
The motor is producing an electrical output (which has a perfectly good shaft reaction), of 1/3 of the load motor's rating, max. After all it is generating a current on only one wire, which, if taken as the wye equivalent, is providing 1/3 of the total power delivered to the load motor. Consequently, 2/3 must be due to current coming in directly from the single phase source.
But, the load motor is conventionally 2/3 or less of the idler motor rated power. So, the power put through the idler is 1/3 of 2/3 of the idler rating. Absent a very considerable set of losses, it seems difficult to get to the idler FLA as input current.
So, I don't believe, according to my figuring and hand-waving above, that the input current to the idler is normally ever equal to the idler FLA on the two input wires, which was (or appeared to be) the original contention in the post to which I replied.
