Can someone explain in simple terms how an induction motor with no load can draw a current
Not totally true. The magnetizing current to run an unloaded motor is much less than the real current to power it at full load. You can estimate the relative value of reactive current to real current from the loaded PF of a typical induction motor of about 0.85 to 0.9.Mechanical losses: Friction, windage
Electrical losses: heating, magnetic
An unloaded induction motor draws almost as much current as a loaded one. It is just that the current lags the voltage, so electric bill is smaller...
Yes, but I think the discussion concludes that the reactive (magnetizing) current doesn't change much from no load to full load, not that the total current doesn't change. So, for example, with a PF of .9 and a total current of 10A under load you would have 9A of real current and 4.36A of reactive current, which should also be close to the no-load reactive current. If the PF dropped to say 0.35 at no-load, then the reactive current would still be about 4.36A but the total current would drop to 5.4A with about 3.2A of real current (if my vector math is correct).But the magnitude of the motor current doesn't change much loaded to unloaded.
Good discussion here.