Electric Motor Starting: Increase or Decrease Current?

[Jimmy25]

Homework Statement

Does the current drawn by an electric motor increase or decrease as the motor starts from rest?

Homework Equations

The Attempt at a Solution

I can't find an answer anywhere. I would think it would decrease because less power would be required to keep the motor spinning. Is this correct or is there more to this question?

 

[Stonebridge]

As soon as the motor starts to spin, it behaves like a coil rotating in a magnetic field. This causes an induced emf in the coil, and this emf is governed by Lenz's Law.
So there is more to it...

 

[jrive]

the spinning motor creates a voltage (back electromotive force, EMF) which counteracts the applied voltage. Thus the current is less since the "back emf" increases as the spinning speed increases and there is a greater voltage drop across the motor.

 

[Watchman2012]

The EMF generated by Faraday's law of induction due to relative movement of a circuit and a magnetic field is the phenomenon underlying electrical generators.

When a permanent magnet is moved relative to a conductor, or vice versa, an electromotive force is created.
If the wire is connected through an electrical load, current will flow, and thus electrical energy is generated, converting the mechanical energy of motion to electrical energy.

Although Faraday's law always describes the working of electrical generators,
the detailed mechanism can differ in different cases.
When the magnet is rotated around a stationary conductor, the changing magnetic field creates an electric field, as described by the Maxwell-Faraday equation, and that electric field pushes the charges through the wire.
This case is called an induced EMF
On the other hand, when the magnet is stationary and the conductor is rotated, the moving charges experience a magnetic force (as described by the Lorentz force law), and this magnetic force pushes the charges through the wire.
This case is called motional EMF.

An electrical generator can be run "backwards" to become a motor. (AND VICE VERSA !)

The induced voltage depends on the diameter (area) of the coil and the number of turns of the coil.
The power is determined as well by the resistance of the coil.
Faraday's law will give you the induced emf in the coil:

V= (dΦ/dt)= NA(dB/dt)


N=the number of turns in the coil
A=area
B is the magnetic field
dt=time
ΦB is the magnetic flux in Webers or Tesla ( Tesla is the SI unit)

The current will be I = V/R. OR:
E=N((dΦB)/dt)

A complete discussion on the topic of windings, power output
and an example of an actual motor is on my website:

http://www.watchman2012.com/articles/spout24.shtml [Broken]

The page is quite popular among those building wind turbines to calculate the output of
ANY given motor they build, to see if what they are building is adequate.

I worked out the maths with a real physicist over a two month collaboration period
and quite pleased at the resulting paper. (the vector calculus was a bear !)
the first of it's kind information on the web.
and it is all boiled down to one simple plug-in formula !

I started from the beginning with particle physics thru faraday etc
to have a complete understanding, it was FUN, then again if I was a spice girl i'd be geeky-spice ;)
enjoy
-Watchman

 

[voko]

This can be answered without looking at the details of operation.

When a motor starts, it has to accelerate its rotor (+ its load, if any) and overcome the friction force. When it runs steadily, it only needs to overcome the friction force.

 

[NascentOxygen]

This can be answered without looking at the details of operation.

When a motor starts, it has to accelerate its rotor (+ its load, if any) and overcome the friction force. When it runs steadily, it only needs to overcome the friction force.

True, but when it is running at full speed, the fluid friction forces (air resistance) are much greater than when it's turning slowly.