The current of primary and secondary windings

[fawk3s]

So I was just reading about transformers from my textbook, and the textbook seems to state that when the resistance on the secondary winding circuit is infinite and no current runs through it thus the power on it P=VI is zero, then, according to the energy conservation law, no current runs through the primary winding as well.
But I can't really grasp the idea. If the resistance on the secondary winding is infinite, shouldn't the transformer just act as a simple inductor for the primary winding? Current still ought to go through it, like in a simple powersource-inductor circuit.
I can totally see that the power on it is zero due to only the reactance of the transformer/inductor, but shouldn't still current pass through it?

Please tell me where I am wrong.

 

[russ_watters]

You're not using conservation law right. A real transformer has one input and TWO outputs. The second output is heat due to inefficiencies.

 

[fawk3s]

I can't really see what you're saying. Are you saying that if the resistance on the secondary winding is infinite, then the energy released is in the form of heat on the transformer/primary winding?

 

[russ_watters]

No "if": there is always heat loss on the primary.

 

[sophiecentaur]

I think that you are not talking about the resistance. You are, I think, talking about the Reactance, due to the Inductance of the windings. A primary coil with infinite resistance would never let any current through it!
The primary inductance would normally be chosen to be high enough to reduce the primary current to an 'acceptably low value' -not zero - with the secondary unloaded.

 

[fawk3s]

[PLAIN]http://img202.imageshack.us/img202/239/transformeru.png

Im not sure if you understood my problem correctly, so I'll try and explain again.
In the picture above, P is the primary coil circuit and S the secondary. The secondary is open, or unloaded, giving it infinite resistance. So no current goes through it.
Now what my textbook states is that when there's no current in the secondary circuit, there is no current in the primary either. And I can't understand why that is.
What I would understand is that there is current going through the primary, but since no work is done on the secondary coil, the transformer would act only as an inductor for the primary circuit. Leaving the resistance of the wire beside, there would only be reactance in that circuit and thus the current would be, well, very big. But certainly not 0 like in the secondary.

 

[sophiecentaur]

Why would the current be big?
If you have enough turns on the primary, the current can be arbitrarily small, surely.
I=V/Z
Lots of current in the coil you've drawn, though! LOL

 

[fawk3s]

The primary inductance would normally be chosen to be high enough to reduce the primary current to an 'acceptably low value' -not zero - with the secondary unloaded.

Ofcourse !
How could have I been so slow ! Oh my!

At last it struck me, while I was taking a bath. The energy in the coil is determined by the inductance and current. The bigger the inductance, the smaller the current needs to be for the same energy value.

Thank you so much for saying that sentence mate! This thing has been giving me a headache since last night.

 

[sophiecentaur]

Any time!