I have a 24V motor which is rated output of 250W at 2750 rpm (I calculated torque to be .87 Nm). If I have a wheel which needs 1 Nm of torque to retain a speed of 2750 rpm, I assume that the motor will have a lower rpm. However, will the motor voltage also be lower?

In the case of my wheel, will the power output be greater or less than 250W (how so)? I am new to motors and am trying to figure out their behaviors for a project. Any help would be appreciated. Thank you.


1 Answer 1


If the power supply is able to deliver the extra current that the motor demands, then the motor will draw more current, and deliver more torque. The increased voltage drop in the motor windings will reduce the speed of the motor.

The increased current in the windings is now dissipating more power than is rated for the motor. The extra heating in the motor will be approximately (1/0.87)^2 = 1.32x. This is OK for a while, while the copper absorbs the extra heat by getting hotter. It may take minutes for the motor to reach its rated temperature under this small extra heating.

If left like this while the motor reaches thermal equilibrium, you would expect the eventual temperature rise above ambient of the motor to be 1.32x higher than rated. This might make the motor smoke, and if it didn't destroy it by damaging the insulation, would certainly reduce the life of the motor.

  • \$\begingroup\$ I see. So, in general is current always linearly proportional to torque and voltage to speed? \$\endgroup\$ Commented Nov 13, 2017 at 8:34
  • \$\begingroup\$ Yes, to first order, neglecting losses, voltage is speed, and torque is current. Of course, voltage drop due to current*resistance means the terminal voltage/speed ratio is not exact. Friction and air resistance mean that the shaft torque/current ratio is not exact. These two are the major losses. For small changes in torque and speed, as you suggest in the OP, the proportionality will stay reasonably constant. \$\endgroup\$
    – Neil_UK
    Commented Nov 13, 2017 at 8:44

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