This DC motor has a power range of 9.0W-300W, while its voltage range is from 6 to 20 volts. Does this mean that its internal resistance is somehow dependent on the load? Or does the user determine the actual power with using a resistor in the circuit?
Performance Table
MODEL ---- VOLTAGE ---- -- NO LOAD --- ---- AT MAXIMUM EFFICIENCY ----- ----- STALL ------
OPERATING NOMINAL SPEED CURRENT SPEED CURRENT TORQUE OUTPUT TORQUE CURRENT
RZ-735VA RANGE V r/min A r/min A mN·m g·cm W mN·m g·cm A
9517 6 - 20 18 20400 2.8 17990 20.9 149 1523 281 1265 12895 156
Best Answer
Yes, of course the effective resistance of the motor depends on the load. This is true of all motors.
When a motor is lightly loaded, it generates a high level of back EMF, which opposes the flow of current. You can think of this as a high effective resistance, although the actual resistance of the coils doesn't change. What is actually changing is the net voltage (= source voltage - back EMF) across that resistance.
When a motor is heavily loaded, the back EMF is reduced, allowing more current to flow. This decreases the effective resistance by increasing the net voltage across the internal resistance of the motor.1
The wide range of power ratings for this motor indicate that it is both efficient (low power consumption at low loads) and robust (can handle the current associated with high loads).
The methods for controlling a motor depend on how you're using it. If you're primarily interested in controlling its speed, you regulate the voltage, allowing the current to vary (within limits) with the load. If you're primarily interested in controlling its torque, you regulate the current, and the voltage varies with load. A resistor is NOT a particularly useful way to accomplish either of these.
1With AC (e.g., induction) motors, the situation is a bit more complicated. The magnitude of the back EMF doesn't change as much as its phase relationship to the source voltage. This still has the effect of increasing the net internal voltage.