First the applied voltage is being supplied to the motor using PWM (Pulse Width Modulation). Consequently you are applying a constant voltage and current in pulses whose width represent watts. Consequently when you are feeding an excess number of watts to the motor above its rated wattage and the inductive resistance that is in proportion to the actual RPM is exceeded then you get a current spike and a voltage dip due to insufficient inductive resistance to limit current.
(I think that's right but honestly I have not had breakfast or lunch and was working on another project very late last night so its hard for me to actually no what I'm writing until I read it again.)
Inductive impedance. Resistance (R) on the order of 0.5 ohm and inductive impedance (XL) for inductance of perhaps around 1H ( XL=2(PI)FL )
Inductive impedance is called the imaginary portion since it does not actually dissipate power but only stores it. The resultant impedance is the square root of the sum of the squares or SQRT(0.5^2+XL^2) at an angle of SIN(XL/R).
In an AC circuit the angle of the impedance has to be included in the current calculation as it will not be in phase with the voltage the way it would be in a resistor.
In an AC motor, the instantaneous current is a bit more complicated since you have the angle and magnitude of the CEMF to remove from the voltage before you can procede to calculate the power lost in the windings.
I had been trying to keep things simple for you ...
Back to the simplified version: Motor CEMF is directly proportional to motor RPM.
From the HBS-36 performance curve at no load we have 308RPM and at full load we have 216RPM. from the ratio of these two numbers we can calculate the amount of voltage developed on the winding resistance for that change in current at the supplied 36V:
winding voltage = 36V - 36V x 216/308 = 10.75V
The no load and full load currents are 1.19A and 20.46A, respectively. From the difference between these we can calculate the rotor resistance:
rotor resistance = V/(Ifl-Inl) = 10.75/(20.46-1.19) = 0.56 ohms
Dan