Load on an electric motor vs. current draw

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SUMMARY

The discussion centers on the relationship between load on an electric motor and its current draw. It establishes that a brushless motor with a resistance of less than a milliohm draws significantly less current when spinning under no load, despite being supplied with 10 volts. When a load is applied, the motor experiences reduced back EMF, resulting in increased current draw. The formula for calculating back EMF is confirmed as Back EMF (DC) = Applied Voltage - (Current through the motor x Resistance of the motor).

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According to classical physics, if I take an electric motor with a resistance of 1 ohm, and run 10 volts through it, 10 amps of current will run through it. But through experiment, a lot less current will run through the motor, an average brushless motor with a resistance of less than a milliohm will draw only a few amps when spinning, even at high voltages.
At the same time, if I take the same motor, run the same voltage through it, and put a load on the motor, it will draw a lot more current.
Does this have something to do with the presence of permanent magnets near the coils or is there another phenomenon going on here?
 
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Exactly that. The spinning coils inside the magnet will cause a back emf that reduces the effective voltage on the coil. The full current only flows if the motor is blocked. A load will cause slower spinning, less back emf, and more current drawn. It has to be like this, because there is energy transmitted to the load and this energy has to come from the electrical power supply.
 
So if you generated the back-emf voltage, and subtracted it from the supply voltage, you would be able to find the actual current flow?
What's the easiest way to calculate that voltage?
 
Back EMF (DC) = Applied Voltage - (Current through the motor x Resistance of the motor)
 
That is much simpler than I made it, thanks.
 

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