Understanding the Relationship Between Speed and Current: Expert Explanation

[Winner]

Quick concept question!

Ok,

How does the current relate to the speed.

So let's say everything is normal and if I reduced the speed by 1/3 what would happen to my current?

Thanks!

 

[vsage]

I may have missed something, but are you referring to *electrical* current as it relates to the *average speed* with which electrons move through a metal wire? Current is proportional to velocity, if that's what you are talking about. Please specify!

 

[Winner]

Ok, yea I was referring to say a motor and current required to operate it. Say the speed of the motor was reduced by a third, what happens to the current drawn?

 

[Winner]

Any ideas anyone?? Thanks.

 

[dextercioby]

"The speed of the motor" sounds terribly vague.Vsage referred to drift velocity (in a metal),which is very precise.

So i really doubt anyone would help,unless you 'd be more precise.

Daniel.

 

[Winner]

So electromagnetic induction and Faraday's laws, transformers, I=V-E/Resistance etc, that topic thereabouts. I mean say for example I have R=6, V=120 E=90 I=2.0Aunder normal conditions. Then what would happen if to the current if the speed was reduced by a third.

 

[dextercioby]

Speed of what??What does this formula $I=\frac{V-E}{R}$ mean...?

Daniel.

 

[Winner]

I=V-E/R is the back emf of a motor...and the speed is the motor speed after it starts up. All I want to know, is there a relationship between the current and speed in this case?

 

[SpaceTiger]

Well, if you're talking about an electric motor, then it should go roughly as follows. An electric motor works by spinning a pair of electromagnets inside of a permanent magnet (see Howstuffworks for a more detailed description). The electromagnets in this setup will have a magnetic field which depends on the properties of the solenoid:

$$B=\mu in$$

where B is the magnetic field, $$\mu$$ is the permeability of the object around which the wires are wrapped, i is the current through the solenoid, and n is the number of turns per unit length. This magnetic field will vary linearly with the current, so cutting the current by a third should cut the magnetic field by a third.

As for the velocity, I would imagine you can approximate the two solenoids as a magnetic dipole in an external field created by the permanent magnet. The potential energy of such a setup would be given by

$$U=-\mu_{magnet} \bullet B_{ext}$$

where this time mu is the magnetic dipole moment of the electromagnet. The maximum kinetic energy that can be extracted from such a setup is just equal to the maximum magnitude of the potential, so you'd expect

$$v_{max} \propto \sqrt{U} \propto \sqrt{\mu_{magnet}} \propto \sqrt{i}$$

This is only a very rough argument, however, and I doubt that things are so nicely linear in practice.

 

[Winner]

Hmm yea, that's a bit more complicated than I thought. How does it relate to this formula though I=V-E/R?

 

[SpaceTiger]

Can you define your variables? That formula makes some sense if it's the voltage change across a resistor. Also, are you sure it's the speed vs. current you're looking for? It would make more sense to talk in terms of the power delivered by the motor. The speed depends upon the details of its setup.

 

[Winner]

Ok, here's the exact question: A motor is designed to operate on 117V and draws a current of 12.2A when it firsts starts up. At its normal operating speed, the motor draws a current of 2.30A. Obtain a) the resistance of the armature coil, b)the back emf developed at normal speed, and c)the current drawn by the motor at one-third normal speed.

As you can see, I'm stuck at c)

a was easy, since it's just started we can say E(emf) is 0. So 12.20=117-0/R and solve for R, R=9.59 ohms.

in b) we just go 2.30=117-E/9.59, and E=95V.
c)and finally I don't know... lol

 

[SpaceTiger]

c)and finally I don't know...

The back emf generated by a motor is proportional to its speed, so for the current of the underlying circuit, you'll want to instead assume that one-third speed corresponds to one-third back emf. Then plug into your equation:

$$i=\frac{V-E}{R}$$

and you're done.

 

[Winner]

huh, that easy eh. Well, great stuff, thanks!