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In an AC circuit with only an inductor, we have a current that lags behind the EMF. So, at time 0 we have zero emf and max current. But in realty, how is this possible? How can we have a current without an EMF?

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  • $\begingroup$ The only way you can have current without EMF is if all of your circuit elements are superconductors. When we talk about circuit design, we often are talking about theoretical, ideal components. An ideal inductor and ideal wires have zero resistance. Real, non-superconducting wires have resistance, and real, non-superconducting inductors can have substantial resistance. You have to build that resistance in to your model if you want to accurately predict the behavior of the circuit. $\endgroup$ – Solomon Slow Jun 12 '17 at 22:24
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This is equivalent to asking how can something have a velocity as it passes through the origin.

Because EMF is related to current through a derivative ($V = L \frac{dI}{dt}$ in circuits), it is possible for one to be zero and the other not since the derivative relating the two is nonzero.

Note that in a resistor, which has no time derivatives or integrals in its $I-V$ relation ($V=IR$), it is not possible to have a nonzero current with zero voltage/EMF, or vice-versa. However, because a derivative relates current and voltage in inductors (and capacitors) the value of current or voltage doesn't alter the other variable. However, the change of current or voltage WILL alter the other variable.

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How can we have a current without an emf?

In the absence of resistance, there can be electric current without an associated emf. For example, a superconducting ring has zero resistance and can have a constant current circulating without any emf to 'drive' the current.

An ideal inductor has zero resistance and so there can be a constant inductor current without any emf. If there is an emf, the inductor current must be changing with time.

In the AC (sinusoidal) case, note that when the inductor emf is (instantaneously) zero, the inductor current isn't (instantaneously) changing, i.e., the inductor current is either a maximum or minimum where the 1st time derivative vanishes.

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In an AC circuit with only an inductor, we have a current that lags behind the EMF.

This statement implies to me that the emf you are talking about is the emf of a voltage source connected to the inductor.

The variation of the emf (voltage) of the supply and the current are shown below with the voltage of the source being $90^{\circ}$ ahead of the current.

enter image description here

The inductance $L$ of inductor is defined as $L = \dfrac {\Phi}{I}$ where $\Phi$ is the magnetic flux linked with the inductor and $I$ is the current passing through the inductor.

If the flux thorough the inductor changes as a result of the current through the inductor changing Faraday says that there is an induced (back) emf, $\mathcal E_{\rm b}$, produced which is in opposition to the change producing it.

$\mathcal E_{\rm b}= - L \dfrac{dI}{dt}$

So if there is no change in current then there is no back emf which means that a steady (dc) current can flow through an inductor and no back emf is induced.

If an alternating voltage source is connected across an inductor the voltage source will try and change the current through the inductor and the inductor will oppose that change by inducing a (back) emf.

Now when the voltage of the source is momentarily zero it is not trying to change the current in the circuit and so momentarily the change in the current is zero and consequently the back emf produced by the inductor is zero.
This is your zero emf with a current flowing condition.

Note that if the graph is completed it shows that the back emf produced by the inductor is always equal and opposite to that of the voltage source and so the total emf in the circuit is always zero assuming an ideal circuit having no resistance or capacitance.

enter image description here

This is to be expected because over a quarter of a cycle the voltage source is delivering electrical energy and the inductor is storing it as its magnetic field increases and in the next quarter cycle the magnetic field of the inductor is getting smaller and the inductor is delivering electrical energy to the voltage source.
Overall the net energy dissipated as heat by the circuit is zero.

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