Derivation of transformer equations I've learnt in high school that in an ideal transformer, $$\frac{V_s}{N_s} = \frac{V_p}{N_p}$$ I looked for derivation for this formula, and in every source I look, the argument goes thus:
$$|V_p| = N_p \frac{d\Phi}{dt}$$ $$|V_s| = N_s \frac{d\Phi}{dt}$$ rearrange the equations, and we have the identity.
What bothers me is that doesn't Faraday's Law describe voltage induced by a changing magnetic flux? Since we have a voltage source in the primary circuit supplying Vp, how can we say
$$|V_p| = N_p \frac{d\Phi}{dt}$$ then? None of the sites I came across explains this, not even Hyperphysics.
I feel like I'm missing something obvious and fundamental here since I haven't had to do any physics for years. Please, enlighten me.
 A: 
What bothers me is that doesn't Faraday's Law describe voltage induced
  by a changing magnetic flux? ... isn't Vp coming from the voltage
  source, while Faraday's Law applies for induced voltage?

It's a simple application of KVL.  Assuming ideal circuit elements, if there is a voltage source $V_{AC}$connected to the primary, KVL yields
$$V_p = V_{AC}$$
But it is also the case that
$$V_p = N_p \frac{\mathrm d\Phi}{\mathrm dt}$$
Thus, it must be that
$$\frac{\mathrm d\Phi}{\mathrm dt} = \frac{V_{AC}}{N_p}$$
Whence
$$V_s = N_s \frac{\mathrm d\Phi}{\mathrm dt} = V_{AC}\frac{N_s}{N_p}$$


Doesn't KVL just say that the voltage across the coil is equal to the
  supply voltage? Vp still is coming from the voltage source, not
  induction, and I wonder why Vp/Np = d(phi)/dt holds

Both equations must hold.  Since the voltage source fixes the voltage across the primary, by Faraday's law, the (rate of change) of flux is fixed by the voltage source.
This is no different, in principle, from the case of a voltage source $V_S$ across a resistor.  By KVL we have
$$V_S = V_R$$
But, by Ohm's law, it also the case that
$$V_R = R I$$
Thus, it must be that
$$I = \frac{V_S}{R}$$
Just as the resistor current is not an independent variable when the voltage across the resistor is fixed by the voltage source, the (rate of change) of transformer flux is not an independent variable when the voltage across the primary is fixed by the (AC) voltage source.
A: Transformers only work for AC voltages. As you have also pointed out if the flux does not change no induction current will occur on the other coil.
Edit:
1- The supplied primary voltage ($V_{p}$) generates an oscillating magnetic flux ($\phi _{p}$). One can calculate the time derivative of the flux using the formula given in question (that is, the Faraday Law). 
2- The flux generated by the primary coil is "felt" by the secondary coil, that is $\phi _{p}=\phi _{s}$.
3- The felt flux generates a voltage ($V_{s}$) on the secondary coil. One can calculate the value of the voltage using the formula given in the question (that is, the Faraday Law).
A: I would recommend following derivation, wich I've learned in school:
$$\frac{V_{p}}{V_{n}}=\frac{N_{p}}{N_{n}} $$
$$V_{p}=V_{n}\frac{N_{p}}{N_{n}}$$
and with
$$V=\frac{d\Psi}{dt}=N\frac{d\Phi}{dt}$$
follows:
$$V_{p}=N_{p}\frac{d\Phi}{dt}$$
The main part of understanding Transformers is to understand that one is using alternating current/voltage. So there are two ways of using Transformers:


*

*If you supply alternating voltage to a transformer, a changing magnetic flux will be induced by the primary coil according to Faraday's Law.
This changing Flux induces alternating voltage in the second coil.

*According to Ampére's Law supplying AC to the secondary coil will induce AC in the primary coil. 
