A DC Motor, when supplied from a 195V power supply is observed to rotate under no load conditions at 1730 rpm. A stall test shows that the winding resistance is 1.66 Ω. The motor is connected to a load whose torque-speed characteristic is Torque = 0.128 × ω.

Find the speed (in rpm) at which the motor will drive the load.

So far, I have tried to use the following method:

$$\ K= \frac{V}{\omega} = \frac{195}{1730*\frac{2\pi}{60}} = 1.0764 $$

$$\ Torque= K*I = 1.0764* \frac{195}{1.66} = 126.4446 $$

And from the question, \$Torque = 0.128 \omega\$

$$\ \omega = \frac{126.4446}{0.128} = 987.85RPM $$

Can anyone tell me why this is wrong?

  • \$\begingroup\$ You are using the stall current in place of the no-load motor current, for one thing. \$\endgroup\$ – Brian Drummond Apr 26 '18 at 8:47
  • \$\begingroup\$ I believe that the torque-speed characteristic of the load is given in N-m vs. radians/sec. \$\endgroup\$ – Charles Cowie Apr 26 '18 at 12:48
  • \$\begingroup\$ You are not considering the back-emf created by the rotating shaft. \$\endgroup\$ – MathieuL Jan 31 at 20:50

There are two errors:

  • the torque you calculate is the torque at stall condition (with a destructive current, if it is maintained for a sufficient long time). When the motor will start rotating, there will be a back electromotive force that will reduce the current;
  • omega is in rd/s and not RPM

What you should do:

  1. calculate first the torque-speed characteristic of the MOTOR (torque as a function of speed). As explained above, you'll have to take into account the effect of the back emf on the current;
  2. combine the MOTOR characteristic with the LOAD characteristic (Torque = 0.128 × ω): the solution will give you both the operating speed and the operating torque
| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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