How does the current of a 1/3 hp.115 VAC motor change with load due to power factor assuming 6 amps no load?

Question

When used to rotate a DC generator current does not increase at a rate reflecting the wattage produced by the generator, but it increases slightly with the first 100watts and then more quickly with each 100watt load added.

Answer 1

As mentioned by the first respondent.
After initial inrush on an uncoupled squirrel cage motor, the current settles to its no load condition. As the load is increased the power factor rises towards unity. (1)

For small motors on no load a cage motor the power factor can be as low as .3 as the load is increased the power factor rises to between .7 to .85 at full load (depending on the rotor design).

Inrush currents for cage motors are typically 6 times FLC (Full Load Current). Further to this inrush currents are not load dependent --- so the IC is the same at no load and Full Load.

Source: 50 years electrical engineering experience

Answer 2

The amp draw of AC motors doesn't change much with load. So where does the energy all go? Shouldn't something heat up a lot?

Vectors.

Power DOES = volts x amps but only in DC are those scalars.

In AC, they are vectors with both a magnitude and a phase. When they are out of phase, the vector product is small. When they are in phase (under heavy load conidtions), the vector product is large (approaching or even exceeding the 1/3 hp = 250 watt rating of the motor.

A cheap meter can measure AC amps and AC volts. But you need a moderately fancy meter to measure actual AC power (you could pry the one off the side of your house, but the electric utility might get upset at you).

Answer 3

The first answer looks good. If you are refering to instantanious load change, then your amps would jump (although not as much as when starting the motor). Once the rotor has caught back up with the stator field, it would come back down.
Assuming we are talking about your common squirrel cage motor of course.