The torque-speed characteristics of a DC motor have been determined, as shown in the figure below. Note that u is the voltage applied to the motor armature, m t

Question

 

 

(a) Obtain the motor constant, torque constant, and armature resistance of the motor.

(b) When the motor is rotating at 100 radians per second, what is the back emf voltage induced at the armature?

(c) When the motor is producing 2x10-3 Nm of torque, how much power is dissipated as heat at the motor windings?

 

Given: 4 x 10-3 Nm u = 24 volts u = 20 volt 160 r/s 200 r/s

Answer 1

Using the torque–speed data, the no-load condition shows the motor constant Ke equals 24 divided by 200, giving 0.12 volt-seconds per radian, and in SI units this is also the torque constant Kt of 0.12 newton-meters per ampere. From the stall torque of 4 × 10⁻³ newton-meters, the stall current is torque divided by Kt, which gives 0.0333 amperes, and using the applied voltage of 24 volts the armature resistance is about 720 ohms. At a speed of 100 radians per second, the back emf is Ke times speed, which equals 12 volts. When the motor produces 2 × 10⁻³ newton-meters of torque, the current is 0.0167 amperes, and the power dissipated as heat in the windings is i squared R, which is approximately 0.20 watts.