How does a generator produce AC current?

Cycloned

I just dont understand how it does, because I never noticed the terminals change positions.

I understood how DC generators work, but not AC.

ZapperZ

Take a loop of wire connected to an ammeter, and spin it inside a magnetic field. Look at sign of the current.

Zz.

Naty1

In a dc generator, current is forced to flow in one direction; in an a/c generator is it allowed to flow in two. A/C power is widely used because it can be transmitted much more economically via step up power transformers...reducing current flow and reducing transmission wire size and power losses....

Remove the rectifiers (diodes) which prevent reverse current flow in a typical generator and you change a dc to an ac generator...

See here for a more complete discussion and diagram:

http://en.wikipedia.org/wiki/Alterna...e_of_operation

generators and alternators are slightly different...see the bottom of the WIKI page for further information.

There are different types of generators, see commutators to understand how current flow direction may be governed:

http://en.wikipedia.org/wiki/Commutator_(electric)

Cycloned

I dont have the apparatus.

So the coil changes directions? It starts moving in clockwise, and then changes to anticlockwise?

Im still so confused =/

jtbell

No, in an AC generator, the coil keeps on rotating in the same direction. As the coil rotates, the magnetic flux goes through the coil first in one direction, then the other, as "seen" from the point of view of the coil. The changes in the magnetic flux through the coil alternate in direction, therefore so does the induced current.

Bob S

The simplest ac generator is a rotating permanent (dipole) magnet inside a stationary coil, sometimes called a dynamo. The only requirement is that the permanent magnet axis of rotation is perpendicular to the dipole field, and in the same plane as the stationary coil. If the permanent magnet were stationary, the ac output must be taken off the rotating armature via sliding contacts via brushes on "slip rings". The rotating dipole magnet actually produces an output voltage on the coil (by Faraday induction), and if a load resistance is connected, a current is produced.

Bob S

thedore

The basic principle involves 3 dimensions. If a magnetic field has a direction straight up from a table top and a straight wire is placed in a north south direction on the table top then moving the wire in a east west direction will cause a voltage to appear on the ends of the wire.

A good example of this can be found if you take apart an old hard disk drive. They have a coil of wire inside two very strong magnets. Be carefull handling them as they can pintch fingers and shatter.

Loud speakers are also an example of this. They consist of a coil of wire in a magnetic field. The coil moves when a current passes through it. If the cone is moved they will generate a voltage.

Another example is in a old mechanical meter movement. Again a coil of wire moves when a current is passed through it.

Bob S

Actually not. High quality loudspeakers have moving voice coils in a uniform magnetic field. In this case, a current in the coil moves the speaker cone, but moving the coil will not generate a voltage.

Bob S

Phrak

He's correct, Bob. Moving the coil in the magnetic field will generate a potential across the coil.

Phrak

Well, that opens up a can of worms. What transformation does the magnetic field undergo in a rotating reference frame?

Andy Resnick

Are you perhaps asking about the slip rings/commutator brushes?

Bob S

Phrak-
Moving a coil in a uniform magnetic field does not produce a voltage, because the voltage induction process is based on the Faraday induction law in integral form:

V = -d/dt[N∫B·n dA] = -N·A·dB/dt = -N·A·(∂B/∂x)(dx/dt)

where A and N are the area and # of turns in the coil, and dx/dt is the coil displacement velocity. If B is a uniform field, then ∂B/∂x is zero, and the induced voltage V is zero. The loudspeaker works on the principle of the Lorentz vxB force, which does not involve the time derivative dB/dt.

Bob S

Phrak

Oh, right. The field is effectively uniform, and radial.