If a pure inductor is supplied with an AC voltage we will always have AC current and time changing electric field in the circuit. That always changing electric field will cause changing magnetic field which will cause an EMF in an inductor which is always equal and opposite to the changing applied voltage (Lenz law.)

So every time the voltage is changing in the source it will face equal and opposite voltage in the inductor everytime. My question is if there is always equal and opposite voltage in an inductor against the applied voltage then how can the current flow in such a circuit?

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    \$\begingroup\$ Current through an inductor is proportional to the integral of the voltage across the inductor - does that answer your question? \$\endgroup\$ – user253751 Dec 27 '19 at 12:48
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    \$\begingroup\$ Try read this electronics.stackexchange.com/questions/470171/… \$\endgroup\$ – G36 Dec 27 '19 at 15:08
  • \$\begingroup\$ Why am i not able to see uploaded images on this site? \$\endgroup\$ – Alex Dec 27 '19 at 15:50
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    \$\begingroup\$ If you connect an ideal voltage source across an ideal resistor, the voltage across the source and resistor will be equal and opposite, and a current will flow. Now change 'resistor' to 'inductor' \$\endgroup\$ – Chu Dec 28 '19 at 11:01

In a real inductor scenario (i.e one with no losses), the sign of the back emf voltage will always be opposite of the voltage source with the magnitude of the the back emf voltage always different from the one of the source voltage. Thus, current flows. My argument actually follows from the popular "electric machinery" textbook from Fizgerald and Kingsley. See the following extract from page 61 of the 6th edition: enter image description here enter image description here

The text describes the setup of a no load transformer (i.e. one where the secondary winding is open circuited) which is actually an inductor (with an iron core) attached to a voltage source. Equation 2.3 and especially the last paragraph clearly describes what I previously explained, that is, when the ohmic loses of the inductor are minuscule, the back-emf equals the source voltage. In practice, ohmic losses are always present, and so Usource is not equal to Uemf and thus a current i=(Usource-Uemf)/R flows

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    \$\begingroup\$ No, in a pure inductance, there is no in-phase current. The only current is reactive -- 90° out of phase with the voltage. \$\endgroup\$ – Dave Tweed Dec 27 '19 at 14:46
  • \$\begingroup\$ Your edit explains what YOU meant with respect to non-ideal inductors, but the OP is specifically asking about IDEAL inductors, in which a current still flows, but not by the mechanism you're talking about. \$\endgroup\$ – Dave Tweed Dec 29 '19 at 2:13
  • \$\begingroup\$ Ok, my explanation holds just because the inductance of the primary circuit of a no-load transformer happens to be extremely big. For an ideal inductor (where the ohmic losses are approximately zero, just as with the example of the no load transformer I mentioned), the back-emf (-L*dI/dt) is indeed equal in magnitude with the source voltage but has a phase lag of 90 degrees. So at any given time, Vsource+Vbackemf in not zero, so you can't really tell that there is no voltage in the circuit. So a current flows \$\endgroup\$ – nikos chatziathanasiou Dec 29 '19 at 3:36
  • \$\begingroup\$ No, the "back EMF" is in phase with the applied EMF. It's the current that is 90° out of phase with both of them. \$\endgroup\$ – Dave Tweed Dec 29 '19 at 4:50
  • \$\begingroup\$ Correct. The comment by Chu tells it all. \$\endgroup\$ – nikos chatziathanasiou Dec 29 '19 at 6:35

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