How can current flow in an AC-inductor circuit if the source voltage is always exactly countered by the inductor's back-EMF?

Question

I'm referring to a circuit consisting of an inductor and an AC power source. I found this post here asking the same question, but I wasn't satisfied with the answer; even though the back-EMF is produced by (the changing magnetic field produced by) the changing current, doesn't the fact the KVL must hold in the circuit at all instants of time along with the fact(?) that there needs to be a net voltage in the circuit for current to begin flowing mean that no current can flow in such a circuit, because it cannot begin to flow? I understand how an existing (changing) current can flow while satisfying KVL:

\begin{align*} V(t)-\epsilon&=0\\ V(t)&=\epsilon\\ V(t)&=-L\frac{dI}{dt}\\ \implies I(t)=\frac{V_{\text{max}}}{L}\int\sin (\omega t+\phi)\,dt&=\frac{V_{\text{max}}}{\omega L}\, \cos (\omega t+\phi ), \end{align*}

but I don't understand how a current can begin to flow if KVL guarantees that the source is immediately countered by the back-EMF from the moment of connection, unless the statement that "there needs to be a nonzero net voltage in a circuit to initiate current flow" is not true...

The e.m.f. that opposes the applied voltage is the result of a changing magnetic field in the coil, but the magnetic field in the coil is only changing because the current through the coil is changing. If there were no current there would be no e.m.f. and so nothing to oppose the applied voltage, which means there would be lots of current (so this is inconsistent and cannot be the solution). The differential equation gives the correct answer. — Tony
– Tony, Commented Apr 10 at 2:35
@Tony My question is about how these changing magnetic fields can exist in the first place. They arise due to changing currents in the coil, and one of the first things one is taught about electricity is that current flows because of a potential difference. If the EMF and applied voltage always exactly counter each other, the PD between any point on the circuit and the positive terminal of the source is the negative of the PD between that point and the positive of the coil, meaning net current is zero. How, then, can current start flowing? (Deleted previous comment because of mistake) — cloud
– cloud, Commented Apr 10 at 2:54
Well it is a /differential/ equation. Current will initially be very tiny, infinitesimal even but that's enough to get things started. From the same calculus that brought you zero-limit-thing divided by zero-limit-thing is finite (derivative) and infinite sum of infinitesimal is finite (integral). — Tony
– Tony, Commented Apr 10 at 2:58
Voltage is induced across an inductor only when the magnetic field is changing. If there is no change in the magnetic field, no voltage will be induced across the inductor. The magnetic field changes only when the current is also changing. Therefore, to generate voltage across an inductor, the applied current must vary. If the current remains constant, no voltage will be induced, regardless of its magnitude. In other words, the very existence of induced voltage depends on the fact that current changes, and it must change. — G36
– G36, Commented Apr 10 at 14:16

MarkU · Accepted Answer · 2025-04-10 04:25:15Z

Current is not "only" motivated by voltage across an ohmic conductor; current is also motivated by changing magnetic field/flux (right hand rule, Maxwell's equations, etc.)

Practical inductors are made of wires which have resistance, so there is a \$V=IR\$ term as well as \$V=L\frac{dI}{dt}\$. The changing magnetic field initially develops due to the ohmic current flow through the practical inductor's distributed series resistance. As the magnetic field builds, the back EMF voltage reduces the \$V=IR\$ current flow.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The transient effect you're trying to understand really can't make sense in a perfect, ideal inductor that has no series resistance. I think that ends up similar to the conundrum of what happens when two ideal voltage sources are connected in parallel: it's incomprehensible due to limitations of the lumped-constant model, unless we add some resistors to model the effects of the metal parts of the components and wires.

Note: The topic of magnetically induced current and back EMF is not introduced in DC circuits theory, because there's plenty of new and very abstract material a student has to master (KVL, KCL, and Ohms Law, reading and writing circuit language, recognizing nodes and meshes.) Algebra is good enough for solving systems of equations and learning strategies for DC steady-state analysis.

The more complicated energy storage behavior of capacitors and inductors requires transient analysis and calculus, so that's why the Ldi/dt current is not introduced in DC circuits classes.

So if I understand correctly, the bottom line is that this circuit in our lumped-element model is impossible, correct? You mention that current is not only motivated by voltage but also changing magnetic fields; one could argue that the changing magnetic fields only serve to generage a voltage and since current flow is always accompanied by a voltage, the latter is a necessity fot the former. Besides, my question is about how these fields can exist in the first place if we begin with no voltage and no changing fields to motivate current. — cloud
– cloud, Commented Apr 10 at 10:45
MarkU, I am confused by part of your answer, and downvoted. Please see my answer, which addresses my confusion about your answer. I will remove my downvote if/when my concerns are resolved. It is perhaps a misunderstanding. — Math Keeps Me Busy
– Math Keeps Me Busy, Commented Apr 10 at 13:04

Andy aka · Accepted Answer · 2025-04-10 07:49:23Z

An inductors back-emf is an induced voltage that can, should you wish, be seen across another winding if you made one. For a pure ideal inductor, that back-emf is exactly equal to the applied voltage. No need to concern ourselves with series resistance or even the type of supply (DC or AC). That's the way it is.

However, that back-emf is a generated value that is in series with the winding i.e. each turn of the inductor produces a little bit of the final voltage. So, we have an equivalent circuit of a back-emf in series with an inductor.

The back-emf is the same as the applied voltage hence, across that series inductor has a net voltage of zero. There can't be any net voltage because the applied voltage equals the back-emf.

So, if we have zero net voltage across the inductor can we automatically say that the current through it has to be zero for ever and a day?

Of course we can't say that.

Inductor current is defined by time and applied voltage so, if there were 10 amps flowing and, we reduced the applied voltage to zero, 10 amps would continue to flow.

The upshot of this is that you can't use back-emf as an argument for there never being current flow. We don't seem to have this problem with resistors do we? Consider two 1 kΩ resistors in series across a 2 volt DC supply. Either resistor must have 1 volt across them if we used a meter but do we say that there must be zero current flow.

Bottom-line is; forget about back-emf when it comes to inductor current. It doesn't help and confuses too many people. Stick to the basic inductor formula: -

$$V = L\dfrac{di}{dt}$$

Math Keeps Me Busy · Accepted Answer · 2025-04-10 22:56:11Z

How can current flow in an AC-inductor circuit if the source voltage is always exactly countered by the inductor's back-EMF?

The voltage across the terminals of an inductor can be divided into two parts. One part is the inductive EMF, and the other part is the voltage drop due to the winding resistance. Let's call these two components \$\mathscr{E}_L\$ and \$V_R\$. Let's also call the terminal voltage cross the inductor \$V_T\$ Then

$$V_T = \mathscr{E}_L + V_R$$

When one first applies a voltage to an unenergized inductor, the current through the inductor is 0, so \$V_R = 0\$. So, the terminal voltage is equal to the inductive EMF. Since the inductive EMF is non-zero, the current through the inductor must be changing according to the equation

So, when a voltage is first applied, it is both the case that the current is 0, and the current is changing. Therefore, a moment after voltage is first applied, the current will be non-zero.

One might incorrectly assume that current must first be flowing through the conductor before an EMF will be established, but this is incorrect.

One might also incorrectly assume that the difference between the terminal voltage and the inductive EMF must be non-zero for current to flow or for it to be changing. Both of these options are also incorrect. The difference between the terminal voltage and the inductive EMF is the voltage drop due to the winding resistance.

If the winding resistance were 0, either as an ideal component, or as a superconductor, and a constant voltage were applied across the inductor, the current would increase linearly. In the case of a superconductor, I believe the magnetic field around the superconducting wire would eventually reach a point where the winding ceased to be superconducting.

In the case of a real non-superconducting inductor, as the current increases in the inductor, \$V_R\$ will increase. If \$V_T\$ is held constant, then \$\mathscr{E}_L\$ will diminish over time, approaching 0.

I don't understand how a current can begin to flow if KVL guarantees that the source is immediately countered by the back-EMF from the moment of connection, unless the statement that "there needs to be a nonzero net voltage in a circuit to initiate current flow" is not true...

If by "net voltage" you mean the difference between the applied voltage and the inductive EMF, then the statement "there needs to be a nonzero net voltage in a circuit to initiate current flow" is not true.

By KVL, the applied voltage is the terminal voltage across the inductor, and the difference between the terminal voltage an the inductive EMF is just the resistive voltage drop of the winding. Whatever the value of that voltage drop, if the the inductive EMF is non-zero, then the current in the inductor must be changing. If the current at time t is 0, but is changing, then a moment later, the current will be non-zero. That is, current flow will be "initiated".

From a comment

I acknowledge that the differential equation is, in reality, accurate even in this case. My question is about why, taking KVL for granted, this equation holds when there is no existing current in the circuit. Specifically, why is the EMF given by $-L\frac{dI}{dt}$ in this case if we know that a current is only motivated by a potential difference, and that the net voltage in the circuit is zero?

I believe here you may be using "potential difference" in the same way you used "net voltage" before, i.e. the difference between the applied voltage and the inductive EMF. But again, this difference is just the voltage drop due to winding resistance. Even if this value is 0, current will be "motivated" to flow if there is a non-zero EMF in the inductor.

Consider a circuit that consists of a 1 V ideal voltage source and a 100 \$\Omega\$ resistor.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

One might say that a 10 mA current flows through the resistor "as a result of the 1 V voltage drop across the resistor".

Now consider a circuit that consists of an ideal 10 mA current source and a 100 \$\Omega\$ resistor.

schematic

^{simulate this circuit}

One might, with equal justification say that the 1 V drop across the resistor is the result of the 10 mA flowing through it. Whether current or voltage drop is a "result" of the other is perhaps not so cut and dried.

Now let's consider a third circuit.

schematic

^{simulate this circuit}

If the switch and wires are ideal, and the switch is closed, there is no voltage across the resistor. Now what happens if the switch were to be opened? We know in reality that 1 V will appear across the resistor. However, if we reasoned that "at the moment that the switch is opened, there is no voltage drop across the resistor, therefore, there is nothing "motivating" current to flow through it", we would fall into error. The current that definite will flow through the resistor, ensures that a voltage drop will appear. As best as I was able to come up with an analogy for what I think is a/the mistake in your conceptualization, the example I just gave is that analogy.

It is much safer to stick with equations that are known to be true, than to second guess them with possibly mistaken ideas about what causes what, and what is the result of what.

The equation

$$\mathscr{E}_L = \frac{dI}{dt}$$

holds. (Possibly with a minus sign). So if there is a inductive EMF present, there must be a changing current. Any theory about cause and effect must conform to that.

The magnetic field in the inductor is proportional to the current through the inductor. So, as the magnetic field increases, so does the IR voltage drop within the inductor. I mention this because another answer seems to say something different, i.e.

The changing magnetic field initially develops due to the ohmic current flow through the practical inductor's distributed series resistance. As the magnetic field builds, the back EMF voltage reduces the \$V=IR\$ current flow.

I'm not sure I understand what is being said. I don't know if "the \$V=IR\$ current flow" is supposed to be anything different from the total current through the circuit. If it is something different, I don't know what. If it is not something different, it seems to in contradiction to the fact that the magnetic field is proportional to the total current, and so the two grow together, i.e. one does not grow while the other is reduced.

The same answer also contains the statement:

The transient effect you're trying to understand really can't make sense in a perfect, ideal inductor that has no series resistance

to which I also object. The behavior of ideal inductors does make sense. The applied voltage is always equal to the inductive EMF. A non-zero inductive EMF implies that the magnitude of the current is changing, according to the formula

$$\mathscr{E}_L = \frac{dI}{dt}$$

I acknowledge that the differential equation is, in reality, accurate even in this case. My question is about why, taking KVL for granted, this equation holds when there is no existing current in the circuit. Specifically, why is the EMF given by $-L\frac{dI}{dt}$ in this case if we know that a current is only motivated by a potential difference, and that the net voltage in the circuit is zero? — cloud
– cloud, Commented Apr 10 at 2:52

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