Basics question of Inductor to powering high voltage lamp

[sophiecentaur]

There is no capacitance shown in the schematic, so I relegated it to secondary effects. You can picture as much or as little capacitance as you choose. A perfect switch has no capacitance, and the same goes for the rest of the circuit. If capacitance was to be a consideration, I would expect it to be shown or dotted in or alluded to in the accompanying text.

The question I first addressed concerned the polarity of the flyback voltage. This is a very basic question.

The polarity question is straightforward, of course. The timing is another matter. It can either be calculated by knowing a lot more information than is supplied OR by using the numerical data supplied.
I just went to the trouble of plotting the data and is clearly is a plot of the volts when the switch is Closed and shows the exponential decay in current build up . The ratio of those volt readings, as a fraction of the next, is about 2.7, throughout the period - so it really does have the time intervals in units of time constant.
Problem is that that time constant has nothing to do with the time constant involved when the switch is opened. One thing I was right about was the advice to draw a graph before trying to come to conclusions. - but I did hope the OP would have done that.
Even more , I have just found that (power point) link and read it in full. Quantitative data is only shown for the switch on situation and they are totally non-committal about the timing involved between when the switch is opened and the neon strikes. They are not actually saying 'instantaneous' and nor are they saying what causes delay or how to work it out.A perfect switch has no capacitance and will instantly (on a timescale compared with other effects in a practical / inductive circuit) go to infinite resistance. Any inductor will have self capacitance and it is this that introduces a finite delay and subsequent 'ringing'. This can always be seen on relay windings, driven by a transistor switch, for instance. After the switch goes open circuit, NO current can flow 'around the circuit' but charge flows into the self capacitance of the coil (or anywhere else it can find; the switch capacitance is unlikely to be more than a very few pF, however). The time constant involved is much much shorter than the time constant for energising the coil, though. Any simpler treatment of the topic has to involve the words 'instantaneous' and 'infinite' - which are not very helpful. Neither is that PP presentation very helpful because it peters out without giving any quantitative help about what actually makes the neon strike.

Annoyingly, they quote a conventional car ignition system as an example of this effect. We all know that, without a suitable 'condenser' in place, it just won't work properly. Same goes for a Rumkorf Induction coil.

 

[davenn]

I really don't understand NascentOxygen's responses

lets start at the start and some one clearly define each step please

1) switch closed to A ... current flowing from battery through inductor, resistor and back to battery

2)switch opens from A and closes to B
He's saying there is still current from the battery that's flowing through the coil for a brief time

3) I see the neon being an open circuit till the voltage is high enough to strike the neon ... prior to striking, the neon is a significant capacitor ?
So WHEN does the neon strike ?

a) -- is the opening of the switch contact generating the voltage spike that lights the neon but before the switch closes to B position? or...
b) is it the collapsing magnetic field through the coil and the spike is that EMF that's generated and with current flowing in opposite direction and with the switch closed to B position that "reverse direction " current flow is striking the neon?Dave

 

[Jay_]

Sophiecentaur

"Firstly, what does the table in the diagram refer to? I don't think the heading "time constant" means what it says. Is it supposed to be 'time'? Could you explain what it all means?"

The time constant of an inductor is T = L/R. So it basically shows the values of voltage across the inductor at T, 2T, 3T etc. At the 0T (beginning), it says the value of voltage across the inductor is the highest = 10 V (battery voltage). After that point it keep reducing.

Now when the switch is moved from A to B, there are two immediate scenarios.

1. Switch is between A and B, not connected to either.

2. Switch is closed with B.

What exactly happens to the voltage across the inductor in these moments and why?

"You don't know those losses so you don't know when the volts will reach the 75V you need for your neon to strike. Neither do you know the value of the parasitic C but you could work out what you want if you knew (or could measure) the R and C values."

So that means this circuit makes no sense without a capacitor? I can understand that by V = L(di/dt), the quick switching from A to B will cause the top of the inductor to have negative voltage. Could you refer to the diagram I attached (again), and tell me why current won't flow in the manner shown?

If a HIGH voltage is suddenly induced when switch closes with B, what makes us control this value? How do we know the switching won't cause a current of 100 V instead of 65V to damage the neon?

" If you want a more useful answer to homework problems then I suggest you make it more clear what it's all about and the post in the right forum"

I just want to know how an inductor works with DC source! Thats all sir! I know it causes an impedance to an AC voltage, but its applications with a DC source is not something I understand which is why I thought I'd ask here.

But all these posts are themselves confusing, if someone (just one person!) could explain the circuit operation step wise (this happens first, this happens second, this happens third etc) I would get a better understanding. Thanks.

 

[Jay_]

NascentOxygen

"The inductor gets 65V across it at no time during the intervals you tabulated. The switch is obviously fixed in position A during all of that time."

Okay, that is understood the voltage reduces gradually until its zero (like a short wire) across the inductor. But what controls the development of voltage across the inductor to be 65V? Also, what about the direction of the current. From the link it seems like the direction is opposite to what I have shown (with black arrow in Paint), but current always goes from positive to negative. I know about the electron flow and all, but by convention its positive to negative, why would we make it different here?

 

[Jay_]

There is something else in the link.

When the switch is at between A and B ('floating'), the inductor has positive polarity on "top". The moment the switch closes with B, it gets a negative polarity on "top".

Is this how it happens? Again, if someone could explain what exactly controls the voltage developed on the inductor (at a particular value like 65 V) and what exactly is the current direction, I could get an understanding and end this thread which has been pulled way longer that I would have thought it to go.

 

[NascentOxygen]

I really don't understand NascentOxygen's responses

Then let's call him back and demand that he explain himself! http://imageshack.us/a/img10/9783/whiponionheademoticon.gif


First, the question seems to come from Circuit Principles 101, and concerns a simple first order system. So it's not meant to invoke a treatise on inductive non-idealities. Further, the circuit appears to have been dreamt up by someone who himself doesn't have a good grasp of inductors. Well, that's my conclusion, because had he constructed the circuit he would have discovered that it has not a hope in hell of doing what he obviously expects it to.

lets start at the start and some one clearly define each step please

1) switch closed to A ... current flowing from battery through inductor, resistor and back to battery

Not much to argue about there.

2)switch opens from A and closes to B

Ha! There's the problem. And it's a big problem. Unless the switch goes from A to B without opening the circuit for even one-squillionth of a second, then the inductor will dump its energy inside the switch and the party is over and we may as well all go home. (Remember, -L.di/dt. We must not interrupt the inductor current.)

So I gave the author the benefit of the doubt by assuming that he really means for the neon to be connected across the inductor at all times, and for the switch to do nothing more than disconnect the battery. Fair enough? Because if we don't assume this then there is not much left for students to analyze.

3) I see the neon being an open circuit till the voltage is high enough to strike the neon ... prior to striking, the neon is a significant capacitor ?

Why would an ideal neon have significant capacitance? The question doesn't say to consider it is a capacitor as well as a neon, so why complicate it? Let's use a tiny ideal neon globe, it has no capacitance.

So WHEN does the neon strike ?

We are using an ideal neon. It [apparently, since the question doesn't say otherwise] strikes at practically its holding voltage, so we have no need to complicate the issue by separately considering the striking and the holding voltages.

a) -- is the opening of the switch contact generating the voltage spike that lights the neon but before the switch closes to B position? or...

It couldn't, for a couple of reasons. One I pointed out above. Another rather major one being that the neon is not even in the circuit until the switch is safely in position B! (But I'm confident that students of Circuit Principles 101 would not notice there's an issue with the switching, so I wouldn't anticipate they would baulk here.)

b) is it the collapsing magnetic field through the coil and the spike is that EMF that's generated and with current flowing in opposite direction and with the switch closed to B position that "reverse direction " current flow is striking the neon?

That. (But I'm uneasy with your emphasis on reversed current flow. I thought we agreed that the inductor current does not reverse? Ever — this being a first order system. No current in any element here reverses.)

Hope that has cleared some things up.

 

[NascentOxygen]

what controls the development of voltage across the inductor to be 65V?

That is wholly determined by the characteristic of the neon. We can infer from the data that the author of the question is using a neon which maintains 65V across its terminals regardless of the current though it. Were you to replace it with a 95V neon, then the inductor would increase its voltage to 95V, at which point the neon would prevent the inductor from increasing the voltage further. The inductor's voltage just keeps increasing until current punches through the neon.

Also, what about the direction of the current. From the link it seems like the direction is opposite to what I have shown (with black arrow in Paint), but current always goes from positive to negative.

In wires and resistors, yes. But that does not apply to inductors. The current (magnitude and direction) in an inductor is unrelated to the voltage at that moment. You know v=L.di/dt , so it's the derivative of the current that determines an inductor's terminal voltage.
Or, expressing current in terms of voltage, i(t) = (1/L)⋅_∫[/size]v(t).dt

 

[sophiecentaur]

I really don't understand NascentOxygen's responses

lets start at the start and some one clearly define each step please

1) switch closed to A ... current flowing from battery through inductor, resistor and back to battery

2)switch opens from A and closes to B
He's saying there is still current from the battery that's flowing through the coil for a brief time

3) I see the neon being an open circuit till the voltage is high enough to strike the neon ... prior to striking, the neon is a significant capacitor ?
So WHEN does the neon strike ?

a) -- is the opening of the switch contact generating the voltage spike that lights the neon but before the switch closes to B position? or...
b) is it the collapsing magnetic field through the coil and the spike is that EMF that's generated and with current flowing in opposite direction and with the switch closed to B position that "reverse direction " current flow is striking the neon?


Dave

There is no certain answer to this. The only data available refers to energising the coil and not discharging it.
But you can make an educated guess about it. Assume that the demo needs to be impressive, so let's assume you have 1J stored. The energy stored is given by
W = l²L/2
With 10A flowing, that requires an L of 20mH.
Including the capacitances of the switch, neon and self capacity of the coil (the majority), would give a total C of around, say 100pF.
Two reactances will give a resonant frequency of
f =1/2π√LC
which gives around 100MHz and this means that the neon would strike at or before the first peak of oscillation, which would mean a time of around 2.5ns. I can't help thinking that the self capacitance would be greater than this - which would make the time correspondingly longer.
Once the neon has struck, everything changes, of course.

The value of the original R would be 1Ω. (10A 10V) and the time constant for this and the 20mH inductor would be L/R, which = 20ms.
Interesting to compare the two times involved - even if they are only back of fag packet calculations. Feel free to find gross errors in my arithmetic.

 

[NascentOxygen]

There is something else in the link.

Which link would that be?

When the switch is at between A and B ('floating'), the inductor has positive polarity on "top". The moment the switch closes with B, it gets a negative polarity on "top".

My contention is that, had the author here intended that the top of the inductor spend any time floating, even for the most fleeting of moments, then his schematic would clearly show that. The way to show this would be a 3 position switch, one having an intermediate position interposed between A and B, let's designate it position N. In position N the top of the inductor would be connected nowhere. The switching would be from position A, to position N, then to position B. Such a switching sequence is not shown, so I can only conclude that in the operation he envisaged it doesn't happen that way.

Is this how it happens? Again, if someone could explain what exactly controls the voltage developed on the inductor (at a particular value like 65 V)

Note that the neon's 65V embraces the inductor and that easily-overlooked current limiting resistor. Without that resistor the inductor would probably dump all its energy by overloading the neon in a bright purple flash. The resistor slows down the delivery of energy by limiting the loop current and allows the neon to operate as as neon globe should.

and end this thread which has been pulled way longer that I would have thought it to go.

Oh, I guess you must be new around here? Welcome to Physics Forums!

You'll always get your answer, in the end. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon10.gif

 

[sophiecentaur]

Note that the neon's 65V embraces the inductor and that easily-overlooked current limiting resistor. Without that resistor the inductor would probably dump all its energy by overloading the neon in a bright purple flash. The resistor slows down the delivery of energy by limiting the loop current and allows the neon to operate as as neon globe should.

Oh, I guess you must be new around here? Welcome to Physics Forums!

You'll always get your answer, in the end. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon10.gif

I agree that the switch action needs to be instant - a justifiable approximation. This is why I said the switch doesn't need to be a changeover and the neon could just as easily be permanently wired to make the analysis easier.

It's a 1Ω resistor! (10V 10A) and will obviously act as an energy loss mechanism but it will hardly affect the rise time of the first cycle of oscillation. It's the neon that dissipates the energy significantly - being the higher resistance - as W = I²R and its on resistance will be a fair bit higher higher than 1Ω. So it will limit its own current and damp the resonance once it has struck
I just wish somebody would acknowledge that the Capacitance of the circuit is vital if you want a quantitative clue about its operation and that the coil has the most C. Presumably at least some readers have been involved with some practical effects of self resonance of inductors.

 

[Jay_]

NascentOxygen

Note that the neon's 65V embraces the inductor and that easily-overlooked current limiting resistor. Without that resistor the inductor would probably dump all its energy by overloading the neon in a bright purple flash. The resistor slows down the delivery of energy by limiting the loop current and allows the neon to operate as as neon globe should.

Okay. If the resistor didn't exist, would it keep the neon from preventing the inductor to increase its voltage at the neon's rated value, like you mentioned? If so, what is the value of this resistance?

That is wholly determined by the characteristic of the neon. We can infer from the data that the author of the question is using a neon which maintains 65V across its terminals regardless of the current though it. Were you to replace it with a 95V neon, then the inductor would increase its voltage to 95V, at which point the neon would prevent the inductor from increasing the voltage further. The inductor's voltage just keeps increasing until current punches through the neon

So if my neon was like 1000 V and required a lot of current to punch through it. In that case would the inductor's voltage keep increasing till it punches through the 1000 V neon? Where exactly would the inductor get all this energy from?

----

The link I was mentioning is the one at the OP. In that clearly, at the N position (as you have described), the voltage at the top of the inductor is shown as positive.

----

Question on switching

Does the voltage developed across the inductor depend on how quickly we switch from point A to B? Or is it still as NascentOxygen says keep increasing till the neon prevents it from increasing?

 

[Jony130]

I strongly advise you to read this
http://www.elsevierdirect.com/samplechapters/9780750679701/9780750679701.PDF (from page 22 Understanding the Inductor)


sample

Why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument because we know Ohm’s law (V = I × R) all too well. But an inductor has (almost) no resistance it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to “hold-off” any voltage across it?
In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor, we are not very clear!
A mysterious electric field somewhere inside the inductor! Where did that come from?
It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of ‘induced voltage.’ This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor, it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed) is the ‘induced voltage.’
So let us now try to figure out exactly how the induced voltage behaves when the switch is closed. Looking at the inductor charging phase, the inductor current is initially zero. Thereafter, by closing the switch, we are attempting to cause a sudden change in the current. The induced voltage now steps into try to keep the current down to its initial value(zero).
So we apply ‘Kirchhoff’s voltage law’ to the closed loop in question. Therefore, at the moment the switch closes, the induced voltage must be exactly equal to the applied voltage, since the voltage drop across the series resistance R is initially zero (by Ohm’s law).
As time progresses, we can think intuitively in terms of the applied voltage “winning.” This causes the current to rise up progressively. But that also causes the voltage drop across R to increase, and so the induced voltage must fall by the same amount (to remain faithful to Kirchhoff’s voltage law).
That tells us exactly what the induced voltage (voltage across inductor) is during the entire switch-closed phase.
Why does the applied voltage “win”? For a moment, let’s suppose it didn’t. That would mean the applied voltage and the induced voltage have managed to completely counter-balance each other — and the current would then remain at zero. However, that cannot be, because zero rate of change in current implies no induced voltage either! In other words, the very existence of induced voltage depends on the fact that current changes, and it must change.
We also observe rather thankfully, that all the laws of nature bear each other out. There is no contradiction whichever way we look at the situation. For example, even though the current in the inductor is subsequently higher, its rate of change is less, and therefore, so is the induced voltage (on the basis of Faraday’s/Lenz’s law). And this “allows” for the additional drop appearing across the resistor, as per Kirchhoff’s voltage law!

 

[sophiecentaur]

That quotation, above, is dodgy because Kirchoff 2 doesn't apply for non conservative fields. Also, phrases like
"The induced voltage now steps in to try to keep the current down " are anthropomorphic and don't help a proper understanding.

The only way to predict what will happen in any circuit is to write the differential equation out and solve it! There is no 'explanation' that ignores all relevant components and parasitic components - including the dominant resistances and capacitances. Any arm waving description involves a 'bootstrapping' argument which is not very useful.

 

[Jay_]

Jony130

It says the page cannot be displayed.

Sophiecentaur

But Kirchoff's laws are used in RLC circuits too - with the laplace taken and then, converted back to time domain form in the end right?

----------

Another thing I don't understand - isn't an inductor as good as a short circuit when dealing with a DC source. So why would there even be 10V across the coil in the beginning, shouldn't it just be zero? Or does it go to say that the DC source in reality is actually passes a zero current first and then its value increases and so a di/dt exists?

 

[Jony130]

Try this link
http://www.elsevierdirect.com/samplechapters/9780750679701/9780750679701.PDF

And inductor in DC circuit act just like a piece of wire (short) in steady state condition.
Not in the transient condition. So after the time equal to t = 5*R/L . The coil acting more like a short circuit.

And before we proceed to AC analysis kept in mind that when you connect the coil to the DC voltage source, at first step the voltage across the coil reach its maximum value equal to Vsource, but the current is equal zero amps. And when current in the coil reach his maximum value the voltage across the coil reach zero volts.
So we can say that current in the coil lags the voltage.

This imagine will help you understand behavior of the coil using this equation:

V = L * dI/dt

when we replace DC source with AC source.

So as you can see "the larger" the rate of change of a AC current (AC signal Frequency).
The "larger" the voltage drop across the coil. So less voltage is seen by the load.

 

[sophiecentaur]

Jony130

It says the page cannot be displayed.

Sophiecentaur

But Kirchoff's laws are used in RLC circuits too - with the laplace taken and then, converted back to time domain form in the end right?

----------

Another thing I don't understand - isn't an inductor as good as a short circuit when dealing with a DC source. So why would there even be 10V across the coil in the beginning, shouldn't it just be zero? Or does it go to say that the DC source in reality is actually passes a zero current first and then its value increases and so a di/dt exists?

Firstly, look at this link, which will tell you why K2 is not always applicable. I think this applies in your inductor case.

The data given tells you there is zero volt drop across the L when there is 10V across the R.

 

[NascentOxygen]

Okay. If the resistor didn't exist, would it keep the neon from preventing the inductor to increase its voltage at the neon's rated value, like you mentioned? If so, what is the value of this resistance?

Some series resistance must be present, otherwise the inductor will dump all of its energy in a brilliant flash and likely exceed the neon's rating.

So if my neon was like 1000 V and required a lot of current to punch through it. In that case would the inductor's voltage keep increasing till it punches through the 1000 V neon? Where exactly would the inductor get all this energy from?

Immediately the battery is disconnected, the inductor current must find another route. If it were to find no new route, then current in the inductor would cease instantly, making di/dt = infinity, so an infinite voltage would appear across the inductor. But, as you'd expect, any neon will begin conducting well before the voltage reaches near infinity. (If the neon was not included, then you could expect to hear a high voltage discharge sizzle through the air around the inductor.)

The energy stored in the inductor is the energy that the battery delivered to it while the switch was in position A for current to build up in the coil. This stored energy has a value = ½LI², where I is the value of current at the moment the battery is disconnected.

The link I was mentioning is the one at the OP. In that clearly, at the N position (as you have described), the voltage at the top of the inductor is shown as positive.

All I see is position A and position B. It's a 2 position switch.

Does the voltage developed across the inductor depend on how quickly we switch from point A to B?

For the situation I'm discussing, the switch instantly jumps from A to B. It is either in position A, or it's in position B. There can be no intermediate state for this exercise.

Or is it still as NascentOxygen says keep increasing till the neon prevents it from increasing?

That.

 

[sophiecentaur]

@NascentOxygen
Unless you are prepared to use some actual values (other than zero) for the other components in a real circuit then the circuit is not real. This is what I have been attempting to do, in order to answer the OP question of "When?"
That 1Ω resistor is fine, although it would be a bit low if anyone wanted to do an actual experiment / demo. Would you care to choose a different value of energy stored from the 1J I chose in my example? Would you also care to suggest a different value of total parasitic C? There would be an answer for your choices too. You do realize (?) that, without some Capacitance, the circuit is totally unreal and will produce an infinite, instantaneous leap of volts - clearly nonsense.
You can't skirt round the issue of possible component values. There's no point in just talking vaguely about a neon tube. Neons, along with every other components, have VI characteristics and there are examples available in spec sheets. Once it has struck, a real neon will have a resistance value in the order of tens of Ohms (it's not a dead short). Before it has struck, it is a pretty good open circuit.
We have to stop arm waving and start to do something more constructive, with some values assigned to components, if we want proper answers.

 

[Jay_]

Firstly, look at this link, which will tell you why K2 is not always applicable. I think this applies in your inductor case.

The data given tells you there is zero volt drop across the L when there is 10V across the R.

Thanks. That was revealing. But consider this: If I connect both my voltmeter probes at point "d", I am measuring Vd - Vd, which means I am measuring the potential of one point with respect to the very same point.

Logically, a place can have just one potential, yet the difference being non-zero tells me that it has two values?! That's like an object existing in two places wouldn't it?

 

[Jay_]

Jony130

I got the pdf, I will give it a read soon. Thanks :)

NascentOxygen

So let me see if I got this right. If V = L(di/dt) and the switching is instant, so the inductor will try to reach infinite voltage. But the neon gives the voltage a path to flow when it reaches 65, and so the voltage build up on the inductor stops? Would that be right?

"Some series resistance must be present, otherwise the inductor will dump all of its energy in a brilliant flash and likely exceed the neon's rating."

Even a coil has some resistance right? So what I was asking is: Is there a particular resistance value (a minimum) for this condition of "dumping all its energy in a brilliant flash" to not happen?

 

[NascentOxygen]

So let me see if I got this right. If V = L(di/dt) and the switching is instant, so the inductor will try to reach infinite voltage. But the neon gives the voltage a path to flow when it reaches 65, and so the voltage build up on the inductor stops? Would that be right?

That's the right idea. But because of the presence of the resistor, the inductor must develop a bit more than 65V. Some of its voltage is dropped across the resistor, to leave 65V for the neon.

Even a coil has some resistance right? So what I was asking is: Is there a particular resistance value (a minimum) for this condition of "dumping all its energy in a brilliant flash" to not happen?

Yes, unless it is a superconductor. The particular value of resistance is determined by many factors, including the characteristics of the neon. The resistor is not crucial, but the neon may be damaged if the current is not limited to a value it is designed for. If you were worried, you could leave out the neon and let the spark jump across a narrow air gap between two wires.

I'm hoping you are about ready to sketch the waveform (current versus time) of the current in the inductor. The task is already half done for you: in the original video they have sketched the current as it builds up while driven by the battery. Now all you have to do is sketch the current as it winds back down to zero. Don't worry about getting the time scale exact, because you can't. We don't know the value of L or R, so it becomes a generalized sketch.

 

[sophiecentaur]

Jony130

I got the pdf, I will give it a read soon. Thanks :)

NascentOxygen

So let me see if I got this right. If V = L(di/dt) and the switching is instant, so the inductor will try to reach infinite voltage. But the neon gives the voltage a path to flow when it reaches 65, and so the voltage build up on the inductor stops? Would that be right?

"Some series resistance must be present, otherwise the inductor will dump all of its energy in a brilliant flash and likely exceed the neon's rating."

Even a coil has some resistance right? So what I was asking is: Is there a particular resistance value (a minimum) for this condition of "dumping all its energy in a brilliant flash" to not happen?

If you aren't prepared to get down and dirty with some assumed values then you can do nothing but wave arms about. I have been trying to do this but no one seems interested. The 1Ω is fine to be getting on with but, unless you make some assumptions about the energy stored, you might as well say it will take a week for the neon to flash. It's even worse if you choose to ignore the Capacitance involved.
Engineers need to use numbers as well as Calculus.

P.S. What exactly is a "Voltage Path"?

 

[Jay_]

Voltage path is closing the circuit or in other words, not opposing the flow of current to the extend of behaving like an open.

 

[sophiecentaur]

Voltage path is closing the circuit or in other words, not opposing the flow of current to the extend of behaving like an open.

Ummm. Voltage does not follow a path; it is the potential difference between two points, which is independent of the path taken by charges flowing between the points. It is difficult to understand any argument that is using such 'alternative' usage of terms.

You are right when you say that any coil will have finite resistance. But the finite Capacitance in a circuit will also affect its behaviour.

 

[The Electrician]

Yes, unless it is a superconductor. The particular value of resistance is determined by many factors, including the characteristics of the neon. The resistor is not crucial, but the neon may be damaged if the current is not limited to a value it is designed for. If you were worried, you could leave out the neon and let the spark jump across a narrow air gap between two wires.

How can the resistor limit current? It has been pointed out over and over in this thread that at the instant the switch moves from A to B, the current in the inductor will remain unchanged. If there's a resistance in series with the inductor, then the inductor will cause the voltage to rise just the amount needed so that the current is unchanged, as you point out yourself in the post I'm quoting from:

That's the right idea. But because of the presence of the resistor, the inductor must develop a bit more than 65V. Some of its voltage is dropped across the resistor, to leave 65V for the neon.

 

[The Electrician]

Here are some oscilloscope captures of the behavior of an inductor under circumstances similar to the one being discussed in this thread.

I used a 1.734 mH crossover inductor. It is an "air core" inductor, with a DC resistance of 1.82 ohms.

I connected it to a power supply adjusted to cause a current of 3 amps DC to flow in the inductor. I then disconnected the power supply and monitored the voltage across the inductor. There was no neon lamp involved, nor was anything else other than the 100x scope probe connected to the inductor. Here's the voltage across the inductor at the instant of disconnecting the power supply; the two cursors are used to measure the self resonance frequency of about 555 kHz. Even though there's nothing connected across the inductor and theoretically the voltage should rise to infinity, the distributed capacitance of the inductor limits the voltage rise:

I didn't have a neon lamp handy, so I connected an approximately 100 volt zener diode across the inductor, with the polarity arranged to clamp the inductor voltage when the power supply was disconnected. Here is the scope capture of that event:

This iimage shows the leading edge of the rising inductor voltage as it approaches the clamping voltage; we can see that the voltage rises at a rate such that it takes about 50 nS to reach about 100 volts.

 

[sophiecentaur]

Interesting. From your results (the frequency of the oscillation), I conclude that the self capacitance of the circuit must be around 200pF. This is not an inconsiderable capacitance.
Also, the Q of the circuit is round about 2 (Looking at the decay rate on the trace), so that makes the resistance around 1.5k at that frequency. Quite a bit higher than the DC value but not surprising as it's an audio frequency core.
I could always be wrong in my arithmetic but I checked it once or twice.

 

[NascentOxygen]

How can the resistor limit current?

By Ohms Law. The resistor could be omitted, or set to a very low value, but this risks the current rising to a destructive level — exceeding the rating of the inductor while in switch position A, and/or exceeding the rating of the neon when in position B. Were the current to not be limited by resistance, then it would be crucial that the switch not being kept in position A beyond a certain duration. This is an unrealistic expectation to place on manual switching.

The point you raise suggests a further improvement: in addition to modifying the switching and the neon's placement, the resistor could be repositioned to be attached to the battery so it disappears from the circuit when the switch moves off position A. There may be practical considerations why some resistive damping is desirable while the inductor powers the neon, but I think the instructor probably felt that the circuit resistance should be lumped in with the inductor because at least some coil resistance is inescapable.

 

[The Electrician]

Interesting. From your results (the frequency of the oscillation), I conclude that the self capacitance of the circuit must be around 200pF. This is not an inconsiderable capacitance.
Also, the Q of the circuit is round about 2 (Looking at the decay rate on the trace), so that makes the resistance around 1.5k at that frequency. Quite a bit higher than the DC value but not surprising as it's an audio frequency core.
I could always be wrong in my arithmetic but I checked it once or twice.

You've raised an issue that I wasn't going to mention because it's an additional complication to the factors being discussed.

This inductor I used was not the sort of thing one would use in a switcher. It's a multi-layer solenoid wound on a nylon bobbin with 20 gauge wire; it has 10 layers in fact.

This kind of construction leads to considerable proximity effect loss and the real part of the impedance increases dramatically at higher frequencies.

Here's a plot of the magnitude of the impedance (green) and the real part of the impedance (yellow):

The frequency we saw on the scope is the natural resonance frequency, not necessarily the frequency of maximum impedance, and it's lower than what the analyzer shows because of the additional capacitance of the scope probe, etc. Anyway, at the frequency of maximum impedance, the real part is almost 30k ohms.

So, of course, the resistance causing the damping of the free oscillations is not the DC resistance, but a much higher value AC resistance. I wasn't going to mention this.

 

[The Electrician]

By Ohms Law. The resistor could be omitted, or set to a very low value, but this risks the current rising to a destructive level — exceeding the rating of the inductor while in switch position A, and/or exceeding the rating of the neon when in position B. Were the current to not be limited by resistance, then it would be crucial that the switch not being kept in position A beyond a certain duration. This is an unrealistic expectation to place on manual switching.

The point you raise suggests a further improvement: in addition to modifying the switching and the neon's placement, the resistor could be repositioned to be attached to the battery so it disappears from the circuit when the switch moves off position A. There may be practical considerations why some resistive damping is desirable while the inductor powers the neon, but I think the instructor probably felt that the circuit resistance should be lumped in with the inductor because at least some coil resistance is inescapable.

Here's what you said on the same issue in post #39:

"Note that the neon's 65V embraces the inductor and that easily-overlooked current limiting resistor. Without that resistor the inductor would probably dump all its energy by overloading the neon in a bright purple flash. The resistor slows down the delivery of energy by limiting the loop current and allows the neon to operate as as neon globe should"

This sure looks to me like you were talking about limiting the current after the switch moves to B, and is delivering energy to the neon bulb. At that time the resistor can't limit the current.