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In my textbook, it says an alternating current is applied to the primary coil which induces an emf in the both the primary and the secondary windings, it then uses the equations $V_p = N_p (\Delta \phi / \Delta t)$ and $V_s = N_s(\Delta \phi / \Delta t)$ and the fact that the flux is the same through both windings, to derive the equation $V_p / V_s = N_p / N_s$. My question is why are we allowed to use the equation $V_p = N_p(\Delta \phi / \Delta t)$ when the primary voltage is not induced. Surely $V_p$ is just the independent voltage coming from the power station. I am confused as to why the maths works out.

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  • $\begingroup$ because we assume that the wires are so tightly wound that only a small amount of the magnetic field leaks outside hence all field lines are shared between the wires of the primary and the secondary $\endgroup$
    – hyportnex
    Commented Dec 27, 2019 at 17:09
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    $\begingroup$ I have submitted LaTeX formatting edits for your post. If I made errors in my interpretation of your equations, then feel free to say so. In the future, please make sure to use LaTeX to format the mathematics in your posts. $\endgroup$ Commented Dec 27, 2019 at 17:12

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an alternating current is applied to the primary coil which induces an emf in the both the primary and the secondary windings

I think the misunderstanding comes from "alternating current".

An alternating voltage is applied to the primary. Current is whatever the transformer will draw from the AC voltage source that is connected to the primary. You can also have current transformers, and you can derive the equations starting from voltage or current... but here you are starting from voltage.

why are we allowed to use the equation Vp=Np(Δϕ/Δt) when the primary voltage is not induced

You have two different voltages here. One is the applied AC voltage, and the other is the counter-EMF from the primary coil, which is induced.

Consider the transformer with no load connected to the secondary. In this case there is no secondary current so the secondary is out of the picture, and the primary is just an inductor.

Since an inductor is just a bunch of wire of resistance R, at DC the current through it is V/R. But at AC this is no longer true. Why?

Current through the coil produces a magnetic flux $ \phi $ , which in turn induces a voltage $ d\phi / dt $ into the coil. This voltage has the same polarity as the AC voltage applied to the coil, so it opposes a change in current (Lenz law).

So your coil connected to say, mains voltage can be modeled by an AC voltage source of value $ N d\phi / dt $ with perhaps some wire resistance in series.

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  • $\begingroup$ So you are saying that there is induced counter emf in the wires which opposes the alternating voltage coming in from the station. My question is which one of these is considered the primary voltage, or is it their sum? $\endgroup$
    – s12013
    Commented Dec 27, 2019 at 21:52
  • $\begingroup$ "Primary voltage" is mains voltage. If you write the equation "Mains voltage = counter_emf" then you can derive flux, current etc equations for your transformer. $\endgroup$
    – bobflux
    Commented Dec 27, 2019 at 21:59
  • $\begingroup$ so the input/primary voltage is referring to the induced counter emf of the primary, my question is what happens to the original alternating voltage? eg if I have an input voltage of 1000v, is that the pd coming from the station or the induced counter emf? $\endgroup$
    – s12013
    Commented Dec 27, 2019 at 22:52

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