Current drawn by a motor operating at half speed

[BOAS]

Hello,

i'm unsure of how to complete the final part of this question;

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I have completed parts a and b.

a) $V = IR$

$R = \frac{V}{I} = \frac{118}{12} = 9.83 \Omega$

b) $2A$ flow through the armature with resistance $9.83 \Omega$.

$V = IR = 19.66V$

Supply voltage = back emf + voltage across armature

back emf = 118 - 19.66 = 98.34V

But how do I figure out c?

Thanks!

 

[CWatters]

Perhaps start by working out the back emf at half the normal speed?

 

[BOAS]

Perhaps start by working out the back emf at half the normal speed?

I can't find any material in my textbook that deals with the relationship speed has with other known quantities.

Do you know of any places I should look?

 

[BOAS]

I can't find any material in my textbook that deals with the relationship speed has with other known quantities.

Do you know of any places I should look?

Bump, I still haven't found anything.

If the load stays constant, then decreasing the voltage will decrease the current in the armature and hence the speed of the motor, but I don't know how to quantify these ideas.

 

[haruspex]

Isn't it important to know why it's at half normal speed? I.e. is it because the applied voltage has been lowered or because the load has been increased?
If you knew which, I think you could use consideration of power to obtain an answer.