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As voltage is given to primary coil, current flows through coil and consequently magnetic flux changes. Due to change in magnetic flux a back emf is also induced in the primary which opposes the applied voltage. Thus applied voltage =- induced EMF, i.e. $$V_p=E+IR \, ,$$ and if $R=0$ then $V_p=E$. If an equal and opposite back emf gets induced then how does change of flux takes places through iron core? Shouldn't it suppress the applied voltage completely?

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  • $\begingroup$ The magnitude of the back-emf is the cause of the inductance of electronic components. The characteristics vary with component design. In transformers the energy from the back emf is sent on it's way to the other coil by means as you said of going into the magnetic field. $\endgroup$ – JMLCarter Dec 24 '16 at 21:07
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A back EMF is generated but where did you learn that the back voltage from this EMF is exactly "equal" to the applied voltage? The back EMF is proportional to the time rate of change in the magnetic flux and since the magnitude of the magnetic flux and its rate of change depends, for instance, on what material is inside the coil, it's apparent that the back voltage depends on the details of the coil design and cannot always just be exactly equal to the applied voltage, isn't it?

Read what this article says about Lenz' Law:

The direction of current induced in a conductor by a changing magnetic field due to Faraday's law of induction will be such that it will create a field that opposes the change that produced it.

The law doesn't say that the back voltage is equal to the applied voltage, only that the back voltage will act to oppose the increase in magnetic flux (which is proportional to the electrical current).

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  • $\begingroup$ It's written in our textbook. That If R=0 Then Vp=E.(i.e suppose resistance is negligible ) And From this conclusion turns ratio equation has been derived(in our book) in this way: applied voltage =- induced EMF Vp=-(-Np del.(flux)/del.(time) ) Vp=+Np del(flux)/del(time).....(i) Then EMF induced across secondary= Ns del(flux)/del(time)....(ii) Dividing (ii) by (i) we get transformation ratio as Vs/Vp=Ns=Np. $\endgroup$ – FanBoy Dec 25 '16 at 9:53
  • $\begingroup$ @FanBoy - If the secondary coil is attached to a zero resistance (i.e., it is short-circuited, R=0), and IF the transformer is a perfect transformer with 100% efficiency, then I believe that that's correct that the back voltage will be equal to the applied voltage. $\endgroup$ – Samuel Weir Dec 25 '16 at 16:17

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