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To obtain the emf in the secondary of a transformer as a function of $V_p$ (input voltage in the primary), we asume that $V_s=N_s\frac{d\phi}{dt}$ and that $V_p=N_p\frac{d\phi}{dt}$. What I don't understand y why $V_p$ has that equation if it's an input voltage. Shouldn't it be independent of what's happening in the secondary? Is it because the mutual induction balances when the emf induced in the primary equals the input voltage? If so, wouldn't current stop flowing through the primary?

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  • $\begingroup$ Yes the Vp is independent, and Vs only depends on the returns ratio. The flux in the core is common to both because both the windings go around the core. The voltage has to be AC so that the flux changes with time. The Vp equation also implies that there is no flux (infinitely small) if the secondary circuit is open. $\endgroup$ – PhysicsDave Feb 21 at 0:04
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Yes the Vp is independent, and Vs only depends on the returns ratio. The flux in the core is common to both because both the windings go around the core. The voltage has to be AC so that the flux changes with time. The Vp equation also implies that there is no flux (infinitely small) if the secondary circuit is open.

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Self inductance is responsible for the effect.

Assuming that we have two coils with the same inductance, as in a transformer in this case, then we talk about mutual inductance. The magnetic flux due to the primary on the secondary is one part of it. The other is the EMF by its own current that creates the same flux.

Remember that $\phi=BA$, so $\frac{d\phi}{dt}=n_pA\frac{dB}{dt}$. Assuming the area of the coils are the same. Here, $n_p$ is the number of turns in the primary inductor.

When you calculate the EMF, we use $E=-n_s\frac{d\phi}{dt}$.

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  • $\begingroup$ If the coil was just an inductor, it generates back emf at 90 degree phase, are you saying that the phase relation begins to change when a secondary is added and loaded? ( I can't remember this, its been a while). $\endgroup$ – PhysicsDave Feb 21 at 0:09
  • $\begingroup$ Thanks for the answer. I understand how to calculate the EMF on the secondary. The part that I don't get is the EMF on the secondary. if $V_p$ is the input voltage, wouldn't the resulting voltage on the source be $V_p-N_p\frac{d\phi}{dt}$? $\endgroup$ – jrglez Feb 21 at 9:42
  • $\begingroup$ Yes that is correct. $\endgroup$ – KV18 Feb 21 at 13:12
  • $\begingroup$ You mean to say "equal to" $-N_pd\phi/dt$ $\endgroup$ – KV18 Feb 21 at 13:35
  • $\begingroup$ No, there was no typo in what I wrote. If the input is $V_p$ ant the induced $-N_p\frac{d\phi}{dt}, why isn't the total voltage the addition of both? $\endgroup$ – jrglez Feb 21 at 23:03

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