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.
Best Answer
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
$$V_s = N_s \frac{\mathrm d\Phi}{\mathrm dt} = V_{AC}\frac{N_s}{N_p}$$
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.