# Trying to get my head (all the way) around inductors and AC

Discussion in 'Physics' started by autodidact, Jun 28, 2012.

1. ### autodidact Thread Starter New Member

Oct 18, 2009
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Hi All,

I've been struggling for a very long time (years, if truth be known) to reconcile my understandings of various aspects of the relationship between AC current and AC voltage across an inductance. I've read innumerable writeups on the topic, some more understandable, and some less, but all to no avail when it comes to completing my understanding. So, in my first post here (after much lurking), I'm turning to you all for help.

Obviously, I'm thinking about part of the puzzle wrong. I'm hoping someone can point out to me where I'm going wrong, and can also tell me the right way to think about it.

First, let me recite the aspects that do make sense to me:

1. I believe I understand why the (counter) emf developed by an inductor is proportional to di/dt . Basically, the more flux lines cut per unit time by a conductor (regardless of whether the conductor is moving past the flux lines, or the flux lines of an expanding or collapsing field are moving past the conductor), the greater the induced voltage.
2. I believe I understand why, by Lenz's law and the right hand rule, the emf developed by the inductor is a counter emf; if it weren't, the magnitude of the current would tend toward infinity.
3. If I only consider the fundamental e = L di/dt formula, and then look at graphical plots of current and voltage, it's fairly obvious that the current wave lags the voltage wave by 90 degrees, and why.
But there's one more aspect that utterly confuses me. To set this up, allow me to quote from a textbook I own:

All of this seems reasonable, too -- at least, until I look at a plot of the applied voltage, the cemf, and the current. Then everything falls apart for me.

What I see in such a plot is, at all points in the 2π radians of the cycle, the applied voltage exactly opposed and balanced by the cemf. Since the net voltage is always exactly 0 (i.e. the push of the applied voltage is always exactly countered by the reverse push of the cemf, whose magnitude is the same but whose sign is different) my intuition is telling me there shouldn't be any current flowing, at any point in the cycle.

But clearly this is contradicted by experience.

What the heck am I missing here? Where's the extra hole in my head that I need to plug?

Thanks,

M

2. ### Wendy Moderator

Mar 24, 2008
21,424
2,951
I think part of the problem is you need to think of the magnetic field as having inertia. It doesn't, of course, but it acts a lot like it. The current can not flow, because the magnetic field is starting at zero, and build as the current starts to flow. For a little while the current is being converted into a magnetic field. Until the magnetic field is fully established it is absorbing the power from the current, which acts a lot like a mass. It may more correct to say the current is being converted into a magnetic field.

Once the magnetic field is fully established, it does not resist the current any more. Of course with AC, the magnetic field continually collapsing and reforming, so the intertia acts a lot like resistance.

One of the things I like about L and C circuits is the symmetry. The capacitor does something very similar with the electric field that the coil does with the magnetic field. Put the two together and you get an electromagnetic field.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
This is simply the elaboration of Kirchoff's laws. At any instant the sum of the potential differences around the closed loop is zero. Replace the inductor with a resistor - the resistor voltage drop is the same as the source at any instant of time. There is a difference - put crudely, the resistor is obeying Ohm's Law whilst the inductor is obeying Faraday's Law. Energy is lost in the case of the resistor.

Bill is correct - the AC source plus inductor has a history which one must assume has led to the conditions noted when observing the circuit at supposed steady state.

I can propose a case in which an ideal AC source is switched on to an ideal inductor at some time past and the average current in the circuit will be half the peak to peak current value [ad infinitum] - rather than zero as one might intuitively expect.

Last edited: Jun 28, 2012
4. ### autodidact Thread Starter New Member

Oct 18, 2009
5
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I'm aware of this analogy, but I'd rather stay away from it for purposes of this discussion. Like most analogies for the behavior of electrical components, it's useful for giving a general idea of the overall effect, but it actually obscures what's really happening when an expanding or collapsing magnetic field generates an emf in a conductor.

Okay, sure; this more or less characterizes what happens when you connect a DC voltage source to the terminals of an inductor. But we can describe what happens in this situation without resorting to the inertia analogy at all, like this:

When the inductor gets hit by the near-instantaneous rise in current, the inductor drops the full voltage of the source back in the opposite direction, opposing the applied voltage (i.e., generates the counter emf). This happens because the magnetic flux lines are moving rapidly across the inductor as it begins to charge. But whatever resistance is present in the circuit imposes an upper limit on the current that can flow through it, so the expansion of the magnetic field slows asymptotically (and along with it, the counter emf decays asymptotically) until, after 4 or 5 time constants, there's virtually no counter emf.

Hmm... wouldn't it be fair to say that "the inertia (actually the counter emf) acts a lot like resistance" regardless of whether the applied voltage is AC or DC?

No, on second thought, I don't want to go there. Except for the edge-case of initial application of DC to an inductor -- where it acts like an infinite resistance, i.e., no current flow in that very first moment -- I don't think that the "inertial" property of an inductor is anything at all like a resistance. Current is in phase with voltage across a resistance; not so with an inductor. Resistors dissipate power as heat; inductors store energy, and return it to the circuit when they discharge. I think these differences are significant enough that I'd prefer to avoid muddying the waters with this particular analogy.

Okay. Faraday's law says that the magnitude if the counter emf is proportional to the time rate of change of the magnetic flux vector. That's fine, I've stipulated to that from the very start. But there's another salient feature of the difference you mention -- and that's what's causing me all the heartache.

In the case of the resistance, there's a single voltage waveform, and it and the current waveform are in phase with one another. This is easy to understand.

On the other hand, in the case of the inductance, there are two voltage waveforms 180° out of phase with one another, and having equal and opposite magnitudes. Now, if I were to sum those two voltage waveforms, I would get 0 volts everywhere, which should mean (naively) no current flow. Clearly that's not what happens, because empirically we know that AC currents do in fact flow though inductors (attenuated by inductive reactance).

So why does summing those two voltage waveforms not correctly characterize the current flow?

If I were to connect two AC sources with identical voltage waveforms, but 180° out of phase, in series, I believe I would see no net current flow in the resulting circuit. The graph of the two voltages would look just like the graph of the two voltages in the inductive circuit with the single applied AC voltage, but the graphs of current in the two circuits would be radically different. Why?

I think I may see what you're getting at here. But then, what do those graphs of voltage and current between initial power-up and steady state actually look like?

All I've ever seen are the asymptotic graphs that illustrate DC across an inductance (along with the whole time-constant exposition), and the graphs of steady state AC voltage and current through an inductor. (I have looked at an animation on YouTube which purports to show what happens when AC is applied to an inductor, but between the unfamiliar visual conventions they use to represent the properties of the waveforms, and the caption windows rolling across the screen every couple of seconds, I found it pretty much useless.)

If you know of a graph that illustrates the initial-conditions to steady-state progression that you're proposing as the explanation, and if that graph makes it clear how the steady-state current waveform can be as it is, despite the perfect balance of forward and counter emf, please share!

Thanks,

M

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If the inductor counter emf was an [independent] active source then your supposition would be correct. But the inductor counter emf isn't an active source - it is produced in response to a primary excitation - in this case the actual AC source driving the inductor current.

6. ### autodidact Thread Starter New Member

Oct 18, 2009
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I'm not quite sure what to do with this. Because it's not an active source, it:

• ... doesn't actually "push" in opposition to the applied voltage?
• ... isn't a true voltage in same way that the applied voltage is?
• ... somehow "doesn't count" (in some other way)?
How am I to understand it as something other than a real difference in electrical potential?

M

7. ### Wendy Moderator

Mar 24, 2008
21,424
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The whole point of counter emf is it is a direct result of the primary emf, it is a reaction, not an active source. If you increase the primary emf, the counter emf will increase in response.

In rejecting the analogy you missed several key points. Current is being converted into a magnetic field. Magnetic fields take a finite time to reach full capacity. This is where the intertia like response is coming from. You can not just create the magnetic field full blown, it has to grow and is defined by rate of change.

Analogies exist to help you wrap your head around concept that may not be obvious. Good analogies will offer insites to other behaviors besides. Many concepts in electronics are taught wrong at first, such as AM Modulation, then retaught once you get comfortable with the core concepts.

8. ### autodidact Thread Starter New Member

Oct 18, 2009
5
0
Okay: so both you and t_n_k are pointing out to me that I can't think of a reactively-produced emf in the same way as I think of an emf produced by an active source.

I'm willing to accept that, but I need some guidance in understanding how I should understand it. Just saying that it's produced in response to the primary emf, but not explaining why or how it acts differently from an emf produced by an active source, doesn't really help.

I believe I said as much when I said that the generation of the counter emf is the result of the movement of the magnetic flux lines across the conductor, which occurs as the inductor begins to charge. The counter emf is what produces the "inertia-like effect" -- is it not?

In the DC example, that effect tapers off and then completely ceases as the coil comes up to full charge (i.e., as the current stabilizes after 4 or 5 time constants, and the magnetic flux lines stop moving).

If I said anything that made it sound like I thought the magnetic field could be created instantaneously, I'd appreciate your pointing it out to me, because that definitely was not my intent!

M

Last edited: Jun 29, 2012
9. ### Wendy Moderator

Mar 24, 2008
21,424
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The point is since the magnetic field is grown, this is where the delay comes from. Just as there is a RC curve, there is a LR curve, and the magnetic field slowly coming into being is what creates the curve, just as the electrostatic field being created in the cap defines the TC.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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There may be a subtle distinction between field and flux. For the inductor to resist change at the instant of application of the voltage one must postulate the existence of 'something' within the inductor physical domain which creates the resistance to change.

As another little thought experiment consider an inductor to which is applied an ideal sinusoidal current source.

That is, at time t=0 an ideal current source Im*sin(ωt) is switched onto an ideal inductor L. How would one distinguish by observation of the circuit parameters any difference between this circuit and a 'dual' circuit comprising the same inductor driven by an ideal voltage source Vm*cos(ωt) at time t=0 - where Vm=ω*L*Im?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I also suspect the concept of considering the circuit condition at some instant of time and reasoning that nothing can change by virtue of a perfect balance of driving emf and counter emf is rather like an example of Zeno's Paradox.

12. ### WBahn Moderator

Mar 31, 2012
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7,207
Let me come at it from a different direction.

Why do you think that no voltage means no current?

Consider a wire ring that has zero resistance. If I could get a current into that ring, why would it ever stop? Yet there would be no voltage drop anywhere.

It wouldn't.

This is not theoretical supposition, this is how superconducting magnets operate when they are in persistence mode.*

So I think you are mentally applying the behavior you expect for a circuit that has resistance to a problem in which it is explicitly assumed there is NO resistance. If there is ANY resistance in the circuit, then the applied voltage and the back-emf do NOT exactly cancel and the necessary voltage to support the current appears across the resistance.

*How persistence mode works is you take a coil of superconducting wire and connect it back on itself. You then wrap a heater around the wire in one place and connect your current injection leads on either side of it. You turn on the heater and then ramp the current into the magnet via the leads. Once you have the current flowing that you want, you turn off the heater. You then reduce the current in the injection leads to zero and disconnect them (up out of the cryogen bath where they exit the dewar).

If you want to see a brutal application of Lenz's Law, make the mistake of turning on the heater without reattaching the injection leads and getting the current to match the magnet current. A coil with dozens (perhaps over a hundred) henries of inductance and something like a hundred amps makes for an awesome show -- if seen from a suitable distance.

13. ### davebee Well-Known Member

Oct 22, 2008
539
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I wonder if the problem with getting the math to work is that given Ohm's law, i=v/r, when you assume a perfect resistanceless circuit, you have assumed a zero in the denominator, setting up a mathematically impossible condition.

Maybe the answer is that since any real circuit will have some resistance, the back emf will not exactly equal the forward voltage, allowing current to flow.

14. ### WBahn Moderator

Mar 31, 2012
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But not all circuits do have resistance. I've just described a superconducing magnet. These things are very real. But Ohm's Law doesn't apply because a resistanceless material is not an ohmic material, so you have to drop back to more general fundamentals.