What happens to a motor's amp draw when you decrease the supplied voltage?

[Technoid]

I'm confused about what happens to the amp draw of a DC boat motor if I change the voltage being supplied to it. If I decrease the voltage, does the amp draw *also* decrease (because the prop is spinning slower, thus putting less physical load on the motor), or does the amp draw *increase* (because P=I*V, and the boat and propeller aren't changing)? I can't seem to wrap my head around this, even though I know that P=I*V, V=I*R, and RPM = KV*Volts. Oh yeah, and Power (kW) = Torque (N.m) x Speed (RPM) / 9.5488.

For example, this motor on Amazon has the following specs:
KV(RPM/V): 2250KV
Watts: 1800W
Max Voltage: <22.2V
Max Amps: 80A
https://www.amazon.com/gp/product/B0...?ie=UTF8&psc=1

That all makes sense: P=I*V, so 1800=80*22.2 (roughly). But if I want to run this motor off my 4S battery (14.8 volts), then how does this affect the current draw, and the ESC I will need to spec? I've been assuming that this motor would require an 80-amp ESC, but if I'm only supplying 14.8 volts... do I need a larger ESC, or can I get away with using a smaller one? And *how much* larger or smaller? I have no idea what happens to my current as I change the voltage in this context!

[NativePaul]

If you drop the voltage whilst keeping the same prop the current will drop too. Having sad that an 80A ESC isn't enough, you need at least a 120A and preferably more.

[Speed3]

Like the above reply said once the prop is the same and you reduce voltage the current will also decrease.

As for ESC with a 2250kv an 150 to 180 amp ESC should be suitable for 4s.

I would like to point out, that it is very easy to pull 80amps on 4s with a 2250kv motor.

You would need to use a x440 size or smaller prop with that motor on 4s or 14.8v.