I remember reading somewhere that someone said that if you keep the electric set up the same (i.e. same volts etc) but increase the resistance (prop) then you also increase the amp draw. That makes sense because the battery is being strained more, but if you look at the simple physics V=IR then if R goes up and V stays the same , then I (amp) must go down... Could someone explain me to what actually happens and which one is correct
thanks

The resistance in the formula is electrical resistance not mechanical. It's the overall resistance/inductance of the motor windings, which has not changed in your example.

Changing the prop would increase the load, or the "power" required to spin the prop at the same speed. The formula for power is P=IV. Using arbitrary numbers if you were to assume it takes 100W to spin a particular prop at a certain speed with 10V then your current is I=P/V or 10A. Now if you put on a larger prop that would require 200W of power to spin at the same RPM then your current draw is 200/10 or 20A.

More load means more power required which means more current draw.

Or something like that.

Jack

Hi Racer.... i think you're possibly confusing the larger props physical additional rotating resistance with more ohm's law electrical motor resistance because of the bigger prop... The motors electrical operating resistance actually goes to a lower numerical ohm resistance number as the larger prop causes the motors operating RPM to run slower than with a smaller prop.... When you plug in the lowered R value into the ohm's law formula, the Amps resultant should get larger, consistant with your "more motor loading intuitional" thoughts... You are familar with the load reduced RPM causing less motor counter EMF against the battery's EMF causing more foward motor amps concept, right ???? ..... kw