What will happen if I put a small resistor in series with a large motor and 24V battery?
Based on Ohms law the maximum current going through a 100 kOhm resistor at 24 V is: I = U / R = 24 V / 100 kOhm = 0,24 mA.

Your motor will never run if you put that huge resistor in there.

And a resistor has no knowledge on how much power it is able to dissipate and intelligently let more or less current pass. So it is the duty of the one designing the circuit to make sure it will be able to handle the typical power dissipation expected from the design. And probably include some safety margin for stuff the designer did not expect.

Current sense resistors should be chosen in a way, that the influence on the circuit is small and have a power rating so they won't burn down.



Now to sense 15 A, and don't influence the circuit much, the voltage drop should be lower than 1 V, probably a lot lower, like 150 mV or so.

So the resistor would have R = U / I = 150 mV / 15 A = 10 mOhm.

The power dissipation of that resistor would be P = U*I = 150 mV * 15 A = 2.25 W.



Measuring motor current might have some difficulties with induced noise and stuff, but I don't have any experience in that field. The phase currents might also be higher than the average you calculated there - but again not sure on that.