A permanent magnet DC motor has an armature resistance of 0.5 Ω and when a voltage of 120 V is applied to the motor it reaches a steady-state speed of rotation of 20rev/s and draws 40A. What will be:

(a) the power input to the motor?

(b) the power loss in the armature?

(c) the torque generated at that speed?

Question

Engineering

Maestro

18 February, 12:13

A permanent magnet DC motor has an armature resistance of 0.5 Ω and when a voltage of 120 V is applied to the motor it reaches a steady-state speed of rotation of 20rev/s and draws 40A. What will be:

(a) the power input to the motor?

(b) the power loss in the armature?

(c) the torque generated at that speed?

+1

Answers (1)

Know the Answer?

Jeffrey Moyer · Answer 1

12 N-m

Explanation:

The dc motor is operating at 24 V that is its terminal voltage V = 24 V

Armature resistance = 0.2 ohm

No load speed = 240 radian / sec

For motor we know that as the motor is on no load so so

Power developed in the motor

Now we know that power = torque * angular speed

So