Basics question of Inductor to powering high voltage lamp

[Jay_]

I was trying to study the inductor role in Flyback converters and all, they seem to rely on the principle that the inductor is first energized while the switch is closed and then, it acts as a voltage source itself when the switch is opened.

All that is good, but I am confused with the POLARITY.

http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE5803

The simulations here seem to show all the current directions from negative to positive. Are they talking about the flow of electrons?

Either ways, I don't see when the inductor gets 65volts across it to power the neon lamp.

This is what I understood: First the inductor has a voltage of 10 when the switch is closed with A, then it goes reducing to zero. After the switch is closed with B, the voltage is formed across the inductor to power the neon-lamp.

But all the currents shown are negative to positive?

Can someone clear me on this? I am aware of the equations V = L(di/dt). So when the constant current becomes zero, voltage is there across the inductor, right? But the polarity?

 

[sophiecentaur]

Assume a positive supply voltage, a series L, a switch and a load connected to ground.
When you (try to) break the circuit, the voltage across the inductor will change from a small forward voltage (due to the finite resistance of the winding) to a high reverse voltage. This high, inverse voltage will be such as to maintain the current that is flowing out of the inductor into the circuit it feeds. It will provide a (positive) voltage at the switch terminal that is much higher than the original value. If a high voltage neon has been connected between L and ground, the voltage will then be enough to strike the neon. The more current the load takes, the higher the voltage when the switch breaks the circuit - which you said when referring to Ldi/dt

 

[Jay_]

When exactly does the inductor in the link develop a voltage large enough to power a 65V neon lamp, from a 10 V source?

In the page of the time constant, it has 10 V just when the switch is closed. Then it goes reducing till zero as time passes.

After that the next page has these two situations (shown in the attachment). It seems to shift its polarity once the connection is made. Even after that it seems to be showing current going from negative to positive!

So which is the correct polarity of voltage at the inductor, and what about the current direction?

 

[sophiecentaur]

Your circuit is not the one I was discussing. 'My' switch and neon are the other side of the inductor, so my wording was not appropriate.
In your circuit, in order to maintain the same current through the resistor, when the switch is opened, the volts at the 'top' of the inductor will, as you show, go negative with a large amplitude.
Could your confusion about the direction of the current be to do with the fact that the inductor is an energy source (like a battery) whilst the voltage spike is being formed. For current to flow out of it, through the resistor, the potential that end must be maintained and this is achieved by virtue of the emf, generated in the inductor. When the battery is connected, the negative end of the battery is connected to ground. Whilst the inductor is functioning, the ground connection is replaced by the neon , which will only strike when there is 65V across it. Point B will be 65V negative for this to happen. (total volts on inductor will be 75V - note that on my circuit, you only need 55V across the inductor as there is already 10V from the supply)

You also mention the timing of all this. There will be finite capacitance in the circuit (mostly the self capacitance of the coil) and this will introduce a time constant which will cause a finite time for the voltage reversal. If the neon were replaced with a capacitor, you would have a series of oscillations as the capacitor discharges back through the inductor - and so on - until the energy is dissipated in the load resistor.

 

[NascentOxygen]

When exactly does the inductor in the link develop a voltage large enough to power a 65V neon lamp, from a 10 V source?

The high voltage develops the instant that the current begins to change significantly. Because when current changes rapidly, di/dt is large and there's the large voltage you are asking about.

Let's simplify things even more, using just a battery and the inductor. You connect the battery across the inductor, and current flows, small at first but increasing with time. You then decide to disconnect the battery. Instant removal of the battery would cause the current to instantly fall to zero. But that means di/dt would be infinite, so an infinite voltage would appear across the inductor terminals. An infinite (or, at least, very high) voltage here would jump across the gap of air between the inductor terminals and the battery you are valiantly trying to drag off the inductor, this voltage revealing itself as a blue spark. And if you do the experiment, that's exactly what you'll see. The faster you try to remove the battery from the inductor, the bigger the spark that results.

The spark represents current flow, meaning the current you tried to interrupt and cause to drop to zero is actually not instantly falling to zero — some current manages to continue flowing by jumping across the air gap. Therefore, di/dt is not infinite (and negative), but it's still a large value.

If you have a neon connected across the inductor terminals, current will preferentially flow through the ionised neon gas rather than jump through air.

 

[sophiecentaur]

Here's a question for 'the student' - What happens if there is no spark or 'neon'? (If the insulation is good enough and there clearly can't be infinite voltage.)

 

[NascentOxygen]

Well, this student says that if that's the case then you haven't rapidly interrupted the current, but rather gradually/steadily wound it back to zero.

 

[sophiecentaur]

What was it that limited how fast you could 'wind it back'. ""

 

[NascentOxygen]

What was it that limited how fast you could 'wind it back'. ""

Apparently the lab that day had issued me with a slow switch. The mechanical separation of its contacts was slow (and, indeed, smooth and gentle) enough that di/dt was limited during the separation process to a value that induced no voltage large enough to ionize any air across the separating points.

 

[sophiecentaur]

No one likes a smartypants.

 

[NascentOxygen]

No one likes a smartypants.

Oh, come on now. Enough of that talk! I'm sure lots of people like you. :tongue:

P.S. I didn't say I believed the instructor's claim of the switch generating no spark.

 

[Jay_]

sophiecentaur

My confusion lies with the direction of current.

1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?

2. How is ground replaced? As I see it first the neon lamp was connected between point B and ground, but no circuit was being made (because switch was at A), after the switch closes with B the circuit is made, and ground stays where it was.

The diagram sort of explains what is going in my mind. The arrow in the loop being the current direction

 

[davenn]

sophiecentaur

My confusion lies with the direction of current.

1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?

2. How is ground replaced? As I see it first the neon lamp was connected between point B and ground, but no circuit was being made (because switch was at A), after the switch closes with B the circuit is made, and ground stays where it was.

The diagram sort of explains what is going in my mind. The arrow in the loop being the current direction

The confusion is that you are using conventional current flow positive to negative
where I suspect sophicentaur is using electron current flow which is negative to positive

so when its negative at the top of the inductor, current flow is anticlockwise from top of inductor down through neon through resistor and back to inductor

hopefully I haven't misread that ;)

Dave

 

[NascentOxygen]

My confusion lies with the direction of current.

With the switch set to position A, current flows through the inductor from top to bottom. As expected, the top is + because it's connected to the battery + terminal. We agree on this.

1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?

When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.

Let's look at the elements apart from the inductor. How to get current to flow in a path that takes it DOWN through the resistor then UP through the neon? This can only happen if the top of the resistor is + with respect to the top of the neon. (Remember, current always flows from + to –.) BUT, we can see that the top of the resistor is the same point as the bottom of the coil, and the top of the neon is the same point as the top of the coil.

So, if the coil is to maintain current unchanged when the switch moves to position B, we have shown that the bottom of the inductor must go + and its top must be – . And indeed it does!

What we haven't examined is the magnitude of this flyback voltage. In this discussion all I've considered is its polarity.

Remember: The voltage across an inductor can change instantly. It's the current through an inductor that takes time to change.

 

[davenn]

Have read a number of times and trying to make heads and tails of it ??

...
When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.

is not this delayed current flow still part of that which flowed from the battery ?
if so then after switching to 'B' its magnitude and direction ( for that brief time) is going to roughly be the same ?
therefore you are not going to have current flowing through the neon as the voltage isn't high enough to strike/light the neon ?

Let's look at the elements apart from the inductor. How to get current to flow in a path that takes it DOWN through the resistor then UP through the neon? This can only happen if the top of the resistor is + with respect to the top of the neon. (Remember, current always flows from + to –.) BUT, we can see that the top of the resistor is the same point as the bottom of the coil, and the top of the neon is the same point as the top of the coil.

So, if the coil is to maintain current unchanged when the switch moves to position B, we have shown that the bottom of the inductor must go + and its top must be – . And indeed it does!

What we haven't examined is the magnitude of this flyback voltage. In this discussion all I've considered is its polarity.

Remember: The voltage across an inductor can change instantly. It's the current through an inductor that takes time to change.

isnt the voltage/current spike going to be caused by the collapse of the magnetic field
and its going to flow in the opposite direction ?
its that spike that's going to be high enough to strike/light the neon ?

just want to clarify this for myself too :)

Dave

 

[NascentOxygen]

The voltage across an inductor is no indication of the direction or magnitude of current through that inductor. To determine the current through an inductor you need to know its history.

 

[davenn]

The voltage across an inductor is no indication of the direction or magnitude of current through that inductor. To determine the current through an inductor you need to know its history.

not sure who or what particular comment that's aimed at ??

I agree, and we do know the history :) so not sure what point you are making ?

Dave

 

[NascentOxygen]

is not this delayed current flow still part of that which flowed from the battery ?

Um, delayed current? That is your word, and I don't know what you mean. I have never mentioned delayed current. Current doesn't get delayed, or bank up, or anything like that.

if so then after switching to 'B' its magnitude and direction ( for that brief time) is going to roughly be the same ?

Yes, inductor current never changes instantly, for to do so would require infinite voltage.

therefore you are not going to have current flowing through the neon as the voltage isn't high enough to strike/light the neon ?

I think this is a question you should answer. You agree that there is an interval of time after the battery has been disconnected when current continues flowing through the inductor. A fundamental principle is that current always flows in a closed loop. You say no current is flowing in the neon. We agree the battery is now out of the picture, so we can forget the battery. So can you describe the closed loop that you perceive the inductor current completing during this time.

isnt the voltage/current spike going to be caused by the collapse of the magnetic field

Yes.

and its going to flow in the opposite direction ?

Are you suggesting inductor current instantly reverses direction? It doesn't. Inductor current continues as it was, before starting to decrease.

its that spike that's going to be high enough to strike/light the neon ?

Yes.

 

[NascentOxygen]

not sure who or what particular comment that's aimed at ??

That was a follow on to what I'd written. I hadn't noticed your post.

 

[Jay_]

I am lost in all this discussion of so many posts. Can someone just clarify if my logic in the diagram in my previous post is right?

Also, at the various time constants the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?

I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?

 

[sophiecentaur]

I am lost in all this discussion of so many posts. Can someone just clarify if my logic in the diagram in my previous post is right?

Also, at the various time constants the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?

I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?

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? This looks more like a h/w question which supplies the information that you need to solve it. Are you expected to plot a graph of the volts across the L and extrapolate to the high volts you need (75V) after the graph has crossed the Y axis and intercepts -75. It's a strange circuit because it requires more volts from the L than necessary - if it were connected as I described a few days ago - you'd only have to get to 55V. But the question setter was after an answer from what he gave you, I suppose. Plot the graph and see what you get. When you are given data which makes no sense to you, a graph can often help.

To work out the "when" question (precisely), we need to know more than the information given. As I mentioned way back, the rate of change of current will be due to the Inductance value and also the parasitic Capacitance involved. You basically have a resonant circuit which has current flowing through it and, ignoring any other paths in the circuit, when you open the switch, the current in the Inductor will divert into charging the capacitance and the resonant circuit will ring at a frequency given by 1/2∏√(LC). There will be losses which will cause the oscillation to decay, of course and which will limit the peak voltage. 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.
If you wanted to, you could put a capacitor across the L which would swamp the parasitic one and then you would have an idea of the (much lower resulting) resonant frequency and that would tell you how long before the peak was reached.
But all this may not be relevant to the answer you want. 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 - where you won't find people batting your problem about in quite the same way. You can see a "no homework" message at the top of the list of topics in each forum. It tells you what to do.

 

[davenn]

Um, delayed current? That is your word, and I don't know what you mean. I have never mentioned delayed current. Current doesn't get delayed, or bank up, or anything like that.

poor choice of words ... was referring to the brief period of time the current still flows

Yes, inductor current never changes instantly, for to do so would require infinite voltage.

OK can accept that

I think this is a question you should answer. You agree that there is an interval of time after the battery has been disconnected when current continues flowing through the inductor. A fundamental principle is that current always flows in a closed loop. You say no current is flowing in the neon. We agree the battery is now out of the picture, so we can forget the battery. So can you describe the closed loop that you perceive the inductor current completing during this time.

No I'm asking you to clarify as I cannot follow your expalanation :)

there is NO closed loop, its an open cct at the neon, but you said the current is flowing up through the neon...

When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.

So if that "residual" current is still flowing, and we know its not enough to strike/light the neon so as said I believe its an open cct.

I want you to tell me how you have current flowing in an open cct please :)

Are you suggesting inductor current instantly reverses direction? It doesn't. Inductor current continues as it was, before starting to decrease.

no I am not, we have already established that

Dave

 

[Windadct]

The time constant discussion was for the "charging" of the inductor, as the current is building up from 0 to Max Amps.
When the switch is moved from A to B - this is now a V= L * di/dt discussion, not based on the time constant. It is also interesting to note - that you are discharging energy from the inductor - so a faster switch may see a bigger arc ( higher arc Voltage) but faster decrease in current ( higher V - Lower I).
Nevertheless - the 2 questions are not yet understood - Why does the V across the inductor "reverse" and how does the V get to 65 V.
Since the Inductor current can not change instantly - when the switch is opened the inductor will "push" current in the came direction, to do this the polarity at the bottom as shown will be +.
As for getting to 65V - it will be best to "do the math" ... so throw a few more data points in there. Make L=50mH and R=50Ohms. ... I max (Switch in position A for more than 5 TC... then I = 0.2A. - Now - the key here is that there is no "perfect" switch, the better the switch ( how fast and cleanly it opens) the higher the voltage will be created by the inductor. Let's consider the point when the contacts in the switch begin to separate - at first they are separated by a very small gap - for example enough that 1V can jump the gap... also once you start the arc, the ionized air is more conductive. Assume we know or can control - how long it takes for the switch to extinguish this arc ( 1nS - 1 mS ? How "fast" will the switch need to be to get the 65+V to ignite the Neon Lamp.

I would not let the diagrams - current representation throw you off too much. The current convention is meaningless unless you agree on a convention or do real math ( since again the Physicists like to think of the literal electrons ( negative charges flowing in the negative direction) and EEs - prefer the + to - Current flow abstraction.

 

[sophiecentaur]

The time constant discussion was for the "charging" of the inductor, as the current is building up from 0 to Max Amps.
When the switch is moved from A to B - this is now a V= L * di/dt discussion, not based on the time constant. It is also interesting to note - that you are discharging energy from the inductor - so a faster switch may see a bigger arc ( higher arc Voltage) but faster decrease in current ( higher V - Lower I).
Nevertheless - the 2 questions are not yet understood - Why does the V across the inductor "reverse" and how does the V get to 65 V.
Since the Inductor current can not change instantly - when the switch is opened the inductor will "push" current in the came direction, to do this the polarity at the bottom as shown will be +.
As for getting to 65V - it will be best to "do the math" ... so throw a few more data points in there. Make L=50mH and R=50Ohms. ... I max (Switch in position A for more than 5 TC... then I = 0.2A. - Now - the key here is that there is no "perfect" switch, the better the switch ( how fast and cleanly it opens) the higher the voltage will be created by the inductor. Let's consider the point when the contacts in the switch begin to separate - at first they are separated by a very small gap - for example enough that 1V can jump the gap... also once you start the arc, the ionized air is more conductive. Assume we know or can control - how long it takes for the switch to extinguish this arc ( 1nS - 1 mS ? How "fast" will the switch need to be to get the 65+V to ignite the Neon Lamp.

I would not let the diagrams - current representation throw you off too much. The current convention is meaningless unless you agree on a convention or do real math ( since again the Physicists like to think of the literal electrons ( negative charges flowing in the negative direction) and EEs - prefer the + to - Current flow abstraction.

The rate of change of current is limited, ultimately, by the stray Capacitance.
But the data (table) given in the question is what will yield the answer to it. All the rest is arm waving because we do not know enough of the details about the actual circuit. The slope of the graph gives the rate of change of the volts - i.e. it will give you the rise time of the pulse.

 

[NascentOxygen]

I am lost in all this discussion of so many posts.

I'm sorry to hear that.

Also, at the various [strike]time constants[/strike] times the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?

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.

I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?

The neon gets a high voltage the very instant the switch is moved out of position A. The voltage across the neon will increase to whatever it takes to strike the neon. The 65V here is obviously a characteristic of that particular manufacturer's product. If the neon happened to not be present, then I would expect to see a blue spark leap through the shortest path of air between the electrical ends of the coil.

 

[NascentOxygen]

So if that "residual" current is still flowing, and we know its not enough to strike/light the neon so as said I believe its an open cct.

Current always flows in a closed loop. The instant that it ceases to flow from the battery, it begins to flow in the neon, so there is no moment when there is not a closed path for inductor current.

I don't believe you can claim "we know its not enough to strike/light the neon". Basically, it definitely is enough, because if it isn't then current would instantly stop, and if current in an inductor stops flowing di/dt would be infinity and an infinite voltage would develop. But before it hits infinity the neon strikes, current flows, and di/dt is not infinity, so the infinite voltage never develops and during that crucial time when the switch opens inductor current never ceases.

I want you to tell me how you have current flowing in an open cct please :)

There is no current flowing in an open circuit. There is no open circuit.

 

[sophiecentaur]

I'm sorry to hear that.

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.

The neon gets a high voltage the very instant the switch is moved out of position A. The voltage across the neon will increase to whatever it takes to strike the neon. The 65V here is obviously a characteristic of that particular manufacturer's product. If the neon happened to not be present, then I would expect to see a blue spark leap through the shortest path of air between the electrical ends of the coil.

Read the question with the attached image. Do you not think that the exercise is to draw a graph of the given data and to extrapolate to the 65V? This isn't a practical problem at all - or even much of a theoretical one. If it were theoretical then there would be actual values of inductance etc. given.
All this chat we have been giving is overkill and not relevant to the real question.

 

[sophiecentaur]

Current always flows in a closed loop. The instant that it ceases to flow from the battery, it begins to flow in the neon, so there is no moment when there is not a closed path for inductor current.

I don't believe you can claim "we know its not enough to strike/light the neon". Basically, it definitely is enough, because if it isn't then current would instantly stop, and if current in an inductor stops flowing di/dt would be infinity and an infinite voltage would develop. But before it hits infinity the neon strikes, current flows, and di/dt is not infinity, so the infinite voltage never develops and during that crucial time when the switch opens inductor current never ceases.

There is no current flowing in an open circuit. There is no open circuit.

This is not true. With a perfect switch, there is still some Capacity in the circuit, into which charge will flow as the volts increase. Only when the Neon has struck, can current flow in it. There is always a Delay, due to the LC time constant. Please don't confuse the OP by making unfounded assertions and over-simplifying the situation. He asked for the time involved and, as I have already made clear, the data can be used to find that time. It's a college exercise.

 

[NascentOxygen]

Read the question with the attached image. Do you not think that the exercise is to draw a graph of the given data and to extrapolate to the 65V?

The instructional video? It was followed by 3 questions, and I got all 3 correct. I never saw anything about extrapolating to 65V. Sorry, have no idea what you are referring to.

This exercise must be like poetry, it brings different things to different people!

This isn't a practical problem at all - or even much of a theoretical one. If it were theoretical then there would be actual values of inductance etc. given.

The circuit is not practical. The switch shouldn't switch in the neon, the neon must be kept across the coil at all times or the switch contacts will glow brighter than the neon. But the student wouldn't know that, so picturing the switch as an ideal switch serves the purpose.

 

[NascentOxygen]

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.

 

[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.