# Quantifying the current drawn by a motor at half speed

I have a h/w question regarding the current in the armature of a motor when it is operating at half it's normal speed. The question is;

A motor is designed to operate at 118V and draws a current of 12A when it first starts up. At it's normal operating speed, the motor draws a current of 2A. Compute:

a) the resistance of the armature of the coil.

b) the back emf developed at normal speed.

c) the current drawn by the motor at half the normal speed.

I have done parts a) and b) but I don't know how to complete c).

My reasoning goes like this;

Voltage is a measure of how 'hard' you push the current around a circuit, so decreasing the voltage will decrease the amount of current in the circuit (since current is coulomb's per second) which will in turn result in a smaller magnetic force on the armature and hence the motor will be running at a lower speed. But I don't know how to quantify this - All I have managed to find online is that angular velocity is directly proportional to voltage...

I would really appreciate a general discussion of the concepts I need to understand to answer this question.

Thanks!

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This seems like a valid homework question, or is it? Moderators? – Parth Vader Apr 9 '14 at 13:43
I provided the actual question for context - I would really like to understand the concepts so I can go away and do it myself. – Jacobadtr Apr 9 '14 at 13:45

I've seen this question before. :) You can model a DC motor as an ideal resistor, an ideal inductor and a back-emf in series with the voltage source. When you are talking about steady state operation like your question is, you can treat the inductor as a short, leaving you with just the resistance and the back-emf.

From this circuit, you can see that $V = e + RI$, where $I$ is the current in the armature. It is also important to note that the back-emf, $e$, is proportional to the angular velocity of the armature, $e = K*\dot{\theta}$. $K$ here (sometimes called $K_v$ or $K_e$) is the back-emf constant. One more equation I should mention ... the torque developed in a DC motor is directly proportional to the current, $T = K_t*I$, where $K_t$ is the torque constant.

The reason that your stated reasoning above is wrong is that you are treating the motor as a purely resistive device when it isn't. But notice that the reason you can calculate the resistance of the motor is because you were given the voltage and current at $\dot{\theta} = 0$. That means that $e = 0$ and in this case you do have a purely resistive circuit.

So, given all this, what happens when the speed of the motor is cut in half? Well, you can see that if the speed is decreased then the back-emf, $e$, is decreased. When this happens, one of two things needs to happen to keep the equivalent circuit balanced. Either the supply voltage, $V$, needs to decrease or the current in the armature, $I$, needs to increase (or some combination of the two). The question doesn't specify why the speed is being cut in half, but as you can see, if the speed decreases it must be because the supply voltage was decreased or because the current was increased (which implies the torque was increased).

I think this should give you the concepts you need to determine the answer to your homework. I'll note again that I don't think the question gives you enough information to answer the question. If I had to guess, I would say the question is probably asking what happens if the voltage stays the same and the torque (and therefore current) is increased to cause the speed to go to half speed.

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