Motors and Generators

Question

A 12V electric motor is turned on. When the motor is first started, the current is 1.0A. Once running, the current on the motor is 0.24A.
a) What is the resistance of the motor?

b) What is the running voltage for the motor?

c) What is the back-EMF?

Answer 1

Hi Jamie!

 

The principle of the motor is that it converts electrical energy (via electricity) into mechanical energy by running a current (from an external source) through a loop in a magnetic field.  The force of the magnetic field on the current create a torque on the loop which causes it to turn.  With the right connections to the outside world, this turning can be converted into the mechanical motion of things in the outside world, such as wheels or gears.

 

However, once the loop starts turning, Faraday's law tells us that an EMF will be generated in the loop by induction.  This EMF (informally called the "back EMF") will produce a current of its own in the loop, on top of what was originally running through the loop to get it turning.  This induced current will run counter to the original current, and thus reduce the net current in the wire loop as it turns in the field.

 

Right when the motor starts, the only current in the loop is the one coming from the external source.  As the motor runs and the loop turns, the induced current builds to a steady level.  What is implied (in this problem) to stay the same during the rise of the induced current is the resistance.

 

So, if 12 V is meant to be the voltage across the motor when is starts, and the 1.0 A is the original current from the external source, then you can use that to determine the motor's resistance, as you correctly did to answer (a).  This resistance will STILL be the resistance of the motor when it is running (unless the problem tells us otherwise).  So using V = IR in the second case will still work, as long as I = 0.24 A, the running current of the motor (including the induced current).  If this is not correct, then I am not sure what else to do with the information given.

 

The "back EMF" is then the difference between 12.0 V and the answer to Part (b).  This is the amount that the induced EMF reduces the EMF of the battery.

 

I hope that helps some.  If you have more questions, or there are still problems here, just let me know.