So i connected a 12v motor on a pretty big car battery the other day (12 volt,70Ah). When the motor was rotating with its shaft free (no load) the voltage across its terminals would be 12V and the current drawn 200mA. When i squeezed the shaft with my fingers (a significant amount of load applied) to the point where the motor was ready to stop, the voltage would be 8V and the current drawn would be 1A.
Could you please explain this to me in terms of both the power law (Power = Voltage x Current) and Ohm's law (Voltage = Current x Resistance)?
Is there such a thing as lower resistance for motors? (if the equation V=IxR is applied, in the first case [no load] the resistance is 60 ohms which then drops to 8 ohms in the second case [full load])
Why did the voltage drop from 12volts to 8 volts? (i was using a pretty huge source of stable DC energy)
Am i right on this? : "The only two factors determining the current drawn by something is the voltage supplied to that something and the resistance of that something. If the voltage is steady (or drops), the resistance has to also drop in order for the current drawn to increase "
Any "light" to my question is well appreciated. Thank you in advance :)