0
\$\begingroup\$

I mean, if we know that the motor consumes 100A to move a 10kg load, will it need 10A to move a 1kg load?

\$\endgroup\$
4
  • \$\begingroup\$ Only approximately. You may find it takes 5 or 10A to move itself (0 load) and then 10A per additional kg. \$\endgroup\$
    – user16324
    May 17, 2022 at 20:44
  • \$\begingroup\$ The torque is a linear with the current (within the rated range of course). \$\endgroup\$
    – Eugene Sh.
    May 17, 2022 at 20:49
  • \$\begingroup\$ It depends on tons of things. The only thing we can say for sure is that moving a 10 kg load will require less than or equal to the same current as the 100 kg load. If you are moving a weight up a hill on a cart that ways 1kg, then yes, it will probably take only 10 A to move it up the same hill. But if the cart weighs 1000 kg, then the load doesn't matter. And if you are moving at 100 kph, then the weight probably doesn't matter (it is all wind resistance). \$\endgroup\$
    – user57037
    May 17, 2022 at 22:53
  • \$\begingroup\$ You need to define what you mean by "move". Newton says an object in motion stays in motion. In the absence of things like friction or drag any tiny application of force will eventually accelerate an object of any mass to a given speed. The easiest example to think of spooling up a very, very heavy, but balanced wheel. The current of a DC motor is about proportional to the TORQUE, but the TORQUE applied is not the same thing as "moving". \$\endgroup\$
    – DKNguyen
    May 17, 2022 at 23:10

2 Answers 2

1
\$\begingroup\$

The mass of the load doesn't matter, it's the energy expended. If you're moving it downhill, it will take virtually no energy and, with some motors, might give some back.

Let's assume you mean "lift" as in moving it directly upward from the surface of the earth. Assuming they're both lifted to the same height, the larger weight will have a proportionally larger energy requirement. Assuming further that they're lifted at the same rate, the power used during that time will be proportional to the energy expended.

This is where it gets tricky. Power is the product of voltage and current. If you're feeding the motor a constant voltage, the current will increase with the square of the power. However, if you're using voltage as a means of controlling the motor speed, the current will be affected by the voltage/torque curve.

Many DC motors are controlled with pulse width modulation (PWM) so the power is the product of the instantaneous power and the duty cycle.

This analysis assumes ideal power conversion and omits wasted energy in friction, deformation, heat, etc.

\$\endgroup\$
0
\$\begingroup\$

A DC motor has a commutator that will put the stator and rotor fields at the same angle, so the resulting force is sort of the product of the field strengths. Now some DC motors use a permanent magnet for one of the two fields; in this case the resulting force (and summarily torque) will be in first approximation linear to the flowing current. Other DC motors use a solenoid for both: in this case the resulting force (and summarily torque) will be quadratic to the flowing current, assuming the respective solenoids are in series. At slow rotation speeds, the resistance of the coils is essentially ohmic in nature, so if they are wired in parallel, again the torque will be quadratic to the flowing currents which are both linear to the applied voltage.

The dynamic (speed-dependent) behavior will be more complex.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.