# Homework Help: DC motor - power calculations

1. Dec 21, 2015

### Name15

1. The problem statement, all variables and given/known data
A permanent magnet DC motor has an armature resistance of 0.4 Ohms. When a voltage of 110 V is applied to the motor it reaches a steady-state speed of rotation of 126 rad/s and draws 39 A.
(a) Find the power loss in the armature
(b) the power input to the motor

I really don't know what i'm doing. I just used the equations I already know. What is the significance of steady state speed being reached? What do i do with the 126 rad/s?

2. Relevant equations
P=IV (power = current x voltage)
P=(I^2)(R) (power = current squared x resistance)

3. The attempt at a solution
(a) P = (39^2) x (0.4) = 608.4 W
(b) P = 39 x 110 = 4290 W

2. Dec 21, 2015

### Hesch

The armature current will be steady, thus the self induction of the armature can be ignored.
The motor constant, Km, can be calculated.
Back EMF = 110V - 39A * 0.4Ω = 94.4V →
Km = 94.4V / 126 rad/s = 0.749 Vs.
You are just not asked about that, but maybe the the missing questions could be:

3. Dec 22, 2015

### Name15

But why would I find Torque and current at 100 rad/s and not 126 rad/s?
Continuing from your calculations: I find Torque = (0.749 Vs)x(39A) = 29.2 Nm. So using P=Tw; Power = 29.2 x 126 rad/s = 3681 W.
Power input to motor = (39 x 110) = 4290 W. So power loss is (4290 - 3681) = 609 W

Is this correct?

4. Dec 23, 2015

### Hesch

That's just an example, how the exercise could be continued.
Yes, that's correct.