Transformer question (Theory)

[Hiwatt25]

How come a transformer won't have any current flowing through the primary if there is no load on the secondary?  I'm pretty sure that's what I read in my introduction to electronics book but it doesn't make sense to me.

Isn't the wire wrapped around the wire core just a conductor?  If I plugged a transformer in to the wall without a load on the oppisate end, wouldn't current just flow through the copper windings on the primary side and back to the wall outlet? 

Transformers are new to me so I'm hoping someone can enlighten me. 

[R.G.]

It does have current flowing if there is no secondary load. The magnetizing current flows.

Here's what happens. If there is no load on the secondary, then the primary just looks like an inductor. The current that flows is exactly what you'd expect: I = Vin/Xprimary.

Xprimary is just the wire resistance plus the impedance of the primary coil, which is XL = 2*pi*Lprimary.

What keeps this under control is that Lprimary is big. So the input current I = V/Xprimary is small, trivially small compared to the main load current rating.

A perfect transformer would have an infinitely big primary inductance so the magnetizing current would tend to infinitely small.

[Hiwatt25]

Thanks again R.G. for a fast reply.

Now, those letters on either side of the equal sign, that's some kind crazy math ain't it?  Like where letters really mean numbers?

Just kidding.   

[db]

The coils in a transformer have a mutual inductance which is to say that any alternating current in one coil will induce a voltage in the other in proportion to the amount of current and the mutual inductance.  This works both ways so if the secondary is open circuit and hence the secondary current is zero, then there is no induced voltage in the primary coil and it looks just like an inductor (with some series resistance).

I would imagine that your book probably goes on to use simultaneous equations to solve the complex situation when there is a load on the secondary.