Self inductance and mutual inductance on transformers?

[inmyblood]

So I know transformers work by mutual inductance. A first loop that is connected to AC current produces an alternating B which produces a changing flux. As a result an induced Emf is formed. This Alternating B is also alternating in the second coil as well to produce same emf (if same number of loops) . right ?

and that the relation of the emf2 produced in second coil is= -N2(change in flux)/Time ... and emf1 in first coil is= N1(Change in flux)/time...

We end up with relation Emf2/N2 = Emf1/N1 .. So the emf in the first or second coil only depends on number of loops only?

BUT wait, isn't there also an Emf added in the first loop due to the AC generator itself that produces an induced emf ? doesn't that mean that the first loop has an emf from AC generator as well as another emf due to self inductance by that (AC current) ?

While the second loop only has an emf produced due to the changing flux of the first loop (mutual inductance ) and does not get the emf from the generator...

Here is a general idea of my understanding of the concept:

Generator produces-->induced emf1--->induced changing current 1 ---> induced Changing B1--->Changing in flux--> causes another induced emf2 in first loop ---> Also Change in B1 causes change in flux on loop 2---> which causes same induced emf2 in second loop.

Loop 2 has induced emf due to changing B1.

Loop1 has induced emf1 due to generator,and emf2 due to changing B1...

So how can both emf be equal even when the number of loops are the same ?
I don't know if there is something wrong in my question but i am really confused about this.

 

[Vailanor]

ThanksA:The transformer is an example of a linear circuit, so you can use the superposition theorem to examine what's going on. The emf in the second coil is due to two sources:the alternating current from the generator (this does not depend on N2 at all)the changing flux in the first coil (this does depend on N2).If you assume that the flux in the first coil is completely determined by the primary current, and you do the math, then you get the result that you mentioned:$$\frac{e_2}{N_2} = \frac{e_1}{N_1}$$However, this is a simplification, and in reality the flux in the first coil is affected by both the primary current and the secondary current. In that case, the result you mentioned no longer holds; the emf in the second coil depends not only on the number of turns, but also on the load connected to it.