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Author Topic: MH's ideal coil and voltage question  (Read 477565 times)

poynt99

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Re: MH's ideal coil and voltage question
« Reply #435 on: May 15, 2016, 05:08:17 PM »
I wasn't planning on going much further MH.

There are enough clues now to fully and correctly answer the question, as long as the "infinite current" and "shorted inductor" barriers are no longer in the way.

verpies

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Re: MH's ideal coil and voltage question
« Reply #436 on: May 15, 2016, 06:20:31 PM »
I do not agree with this. The energy stored in the magnetic field cannot just disappear. That energy has to be dissipated before the opposite magnetic field can be induced
Yes the energy stored in the magnetic field cannot just disappear but it can be transferred elsewhere without dissipation.
P99's current graph is correct so far.

Magluvin

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Re: MH's ideal coil and voltage question
« Reply #437 on: May 15, 2016, 10:47:27 PM »
Just got to sit and go over it all, and there is more I see.

Have to run out for a while. Life.

Just a quicky....

In an ideal world, Im sure that the windings of the coil would be made to perfection as would be the conductors themselves. Zero flaws. Made in a coil shape, not just bent into shape. Ill go further on that when I get back. Would a single ideal wire allow current to flow?  Until I get a grip an all that has been posted since yesterday, Im still looking at the possibility that current wouldnt flow. Just thoughts, not saying you guys are wrong.

Mags

tinman

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Re: MH's ideal coil and voltage question
« Reply #438 on: May 16, 2016, 01:35:10 AM »
The energy in the inductor does not "disappear", some is returned to the source during that period.

If the source is ideal,and the voltage is fixed at -3v,then the flow of energy is in opposition to that stored in the inductor loop/magnetic field.
So this energy that is stored cannot return to the source,as the energy from the source is flowing in the wrong direction. This would be like trying to force a water flow through a pump in opposition to the water already flowing out of the pump.


Brad

tinman

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Re: MH's ideal coil and voltage question
« Reply #439 on: May 16, 2016, 01:41:44 AM »
  There are others that were interested in this and I suggested they brainstorm.  Magluvin may also give it a shot when he comes back.



Quote
I honestly disagree with the hand-holding at this point.

Lets all be honest here MH.
You would never have stated in your answer(if you gave one),that the current would have continued to flow during the 0 volt portion of your question. You would have depicted a sloped drop of current,and then onto the negative voltage protion of the question.

This is why you never answered your question--you needed others input.

But of course,we will never know,as you did not answer your own question.

Quote
The big "breakthrough" is apparently Brad now accepts that the current ramps up to 2.4 amps after three seconds.  What about the other people that disagreed vehemently about this, what do you have to say?

No i do not.
I am going through the process using your understanding.
The coil is ideal,and so should be the CEMF.
I am yet to see any reason posted why the CEMF is also not ideal.


Brad

Magluvin

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Re: MH's ideal coil and voltage question
« Reply #440 on: May 16, 2016, 02:01:11 AM »
Something that keeps coming to mind as i think....

When we see the demo of a magnet being suspended above a superconductor, It is said that as we put the magnet in place, I suspect a particular height above the super conductive object, that as the field of the magnet sets up lenz currents in the superconductor that are set in motion indefinitely and it sets up an opposing field against the mags field and it floats. So in that case the lenz field is even stronger than the mags field at float as to hold up the weight of the magnets mass, let alone just be equal to the field of the magnet. So is it not possible that this balancing act could be inherent in the ideal perfect inductor? Still thinking on it all. Which is good I suppose.  Have some work to do tonight. Will see what I figure on it all

Mags

poynt99

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Re: MH's ideal coil and voltage question
« Reply #441 on: May 16, 2016, 02:01:14 AM »
Lets all be honest here MH.
You would never have stated in your answer(if you gave one),that the current would have continued to flow during the 0 volt portion of your question. You would have depicted a sloped drop of current,and then onto the negative voltage protion of the question.

This is why you never answered your question--you needed others input.

But of course,we will never know,as you did not answer your own question.

Brad, you really are being silly with what you said there. I know you and MH don't seem to be getting along well, but he deserves more benefit of doubt than you are affording him here on this electronics question. MH did know the answer to the question, and could easily have plotted out the current trace as I have encouraged you to do, but he wanted you and others to work it through yourselves.

MH's question is thought-provoking and well thought out, although the wording could have been geared more towards the experimenters here rather than the techies. At any rate, I can assure you that MH knew all along exactly how the circuit currents behave, given the input he specified. After all, he is the one that formulated the question.

poynt99

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Re: MH's ideal coil and voltage question
« Reply #442 on: May 16, 2016, 02:45:07 AM »
If the source is ideal,and the voltage is fixed at -3v,then the flow of energy is in opposition to that stored in the inductor loop/magnetic field.

So this energy that is stored cannot return to the source,as the energy from the source is flowing in the wrong direction.
Sounds to me like you are agreeing with me.

The source is trying to drive the opposite polarity energy into the inductor. When the source is at -3V, it offers up a place for the inductor to dump some of its stored energy. Given more time at -3V, the inductor would fully lose its "positive" energy, and begin to energize in the "negative" direction.

Since it isn't long enough for that to occur, the energy in the inductor simply decreases to a lower "positive" level, and there is only one place for that lost energy to go. You can even see it in the sim.

Magluvin

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Re: MH's ideal coil and voltage question
« Reply #443 on: May 16, 2016, 03:47:17 AM »
Had a thought while drilling holes for this sub box that I wanted to bring up.

While thinking about a physical ideal inductor, I visualized current through the conductors.  If the conductor were very small, very fine wire, would there be a limit to how much 'current' by definition could flow through that ideal wire? Make the wire half the dia. Any limits there?  Cut the dia in half again. Any limits?  Lets think that there may be limits in the superconductive ideal conductors.  Like if we kept increasing current, would there end up being some of it in skin effect?

If we have enough current through a given short piece of wire that ALL the electrons of the conductor(the ones that can)are in motion(Bill says they move at the same speed when they move, and I agree) with the direction of current. All moving in sync. Can more electrons be packed in to consider having more current flow than full up flow?

Just something to think on. You guys may not find these things as interesting possibilities. Maybe some of you do. Along side of what you may think are crazy, loony thoughts, well we are talking artificial, pretend environments. Im just looking at possible details that may alter what we can think when it comes to the legitimacy of the ideal world components. because we are taking them pretty seriously here, sooo.  Example. If the ideal cap follows every capacitor rule except for resistance and inductance, then we could not design a perfect inductor in that environment. And likewise the other way around..


Anyway..  back to making holes. Its a bass reflex box using 2 Kicker 15" for my boss's daughter. Tuned at 30hz.

Mags

wattsup

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Re: MH's ideal coil and voltage question
« Reply #444 on: May 16, 2016, 06:47:12 AM »
@tinman

I put that video here to show what happens when any factor goes above the physical limits of the wire.

The only thing about ideal voltage is that it will not sag like if an ideal car battery is shorted, the voltage will not drop for a few seconds, it will stay at 12 volts and in the case of my video and the copper wire would have popped immediately and not after a few seconds.

But I'll play the game in my OU logic terms....... the math I will leave to others.

Ideal voltage holds the voltage steady while the current CAN rise to infinity, but only if infinity is called for by the inductor. So for current infinity to happen, the inductor would also need to be infinite in either or both wire length or thickness and with or without the resistance the ideal voltage would not care.

But here you have a great disconnect in the logic because current from the ideal voltage that could rise to infinity is now countermanded by an ideal inductor that has no resistance but has an inductance limit to 5H. So the two limiting factors are 4 volts with as much current as required and 5H with no resistance. Figure also that coil impedance will be minimal because most wires have resistance and little reactance to start with so if there is no resistance then only the reactance = impedance = peanuts.

Since the ideal voltage is ideal, it cannot vary but like any DC source, you can add to it, in or out of phase will not matter, but again because its ideal you cannot subtract from it having any less on those two ideal power terminals. Given this is ideal DC, the positive terminal will always be 4 volts (p4) and the negative terminal will always be 0 volts (n0).

During the exercise when you get to the zero volts reading this does not mean there is zero across the inductor since the voltmeter is also a differential reading it just sees the same voltage on each side of the inductor as would a scope that would then produce a flat line. If it shows zero with an ideal voltage, then you have to have a combination present at the two terminals like p4/n4 or add 1 from the inductor to p4 to get p5 the inductor would then have to add 5 to the n0 would mean you have p5/n5 which is still zero. At the -3 volts reading the p4 and n0 still cannot change but the inductor can add a -7 to the n side so you can have p4/-n7 and show -3 on the voltmeter. This does not mean the ideal voltage is dropping in voltage but that since it is DC, you can add to it with a positive or a negative value and your volt meter will see this change at the terminals but again it does not imply the ideal voltage changed.

A debate could be had here in terms of can an ideal voltage which also has zero resistance can also receive return voltage from the inductor as I had read here someone implying that at the -3 mark, power is being returned to the voltage source. If that's just another "acceptance" to be able to play the game then OK but I would think not since the ideal part of the voltage should prohibit this. Besides, take a 9 volt and 12 volts battery and connect them in parallel and what do you get. This does not mean the 12 volt battery is feeding anything to the 9 volts battery. This just means you will see 21 volts on your volt meter.

What all parties, especially @MH and company need to understand is that when you ask a question literally, it is taken literally. When you say ideal voltage, then indicate 4 volts, there is no turning back from that. Regardless if the voltage has zero resistance or 300% resistance or any other other factor you wish to equate, that ideal voltage is ideal for only one reason, to stay stable, unchanging, otherwise you just need to call it a standard DC power supply (PS) and then we could have all gone home a long time ago. If it was a regular PS we all know we can adjust the voltage where we want and the current will fix itself as per the inductive requirement. A regular PS is not an ideal voltage because depending on what you run with such a PS, yes you can see on the instrument voltage and current meters that voltage and current will vary with the pulsing or relaxation or inductive kick back that the DUT provides. But as an ideal voltage source that is not the case. Being ideal, one should consider there is a little Jennie inside the power source that magically holds the applied voltage steady under all circumstances. Come hell or high water she shall not flinch.

So to start you need to calculate the first 3 seconds of 4 volts hitting a 5H inductor that measures 4 volts after three seconds. These guys already calculated it to arrive at 2.4 amps of current in the coil. What voltage will that produce in the inductor? Take that apply it to your p4/n0 and go from there.

Anyways I am sure if I asked 1000 students who just graduated from university how they perceived the ideal voltage ideal inductor question, the grand majority will answer "Don't really know, I just faked it and passed". hahahahahaha

But let me make at least one thing very clear with all of you. Regardless of what each holds as truth to a fictitious ideal question, do not have the slightest presumption that this explains how a coil works. This will not tell you how a coil works, it only gives you a presently accepted and measured depiction of how a coil responds. This is far from knowing how it works and for me as an OUer looking to develop a better understanding of the logic behind how a coil actually WORKs, all EE discourse becomes purely presumptive. Now once you have all eaten all this ideal stuff and have washed the dishes and cleaned up the kitchen and then if you really want hard questions of logic, let me know. I have a case full that could keep you busy for a few years. One of them was the last question I asked here but again, better you wait till the kitchen is cleaned.

Went on long enough. As usual, sorry for long post. I even cut it down to half. hahaha

wattsup

verpies

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Re: MH's ideal coil and voltage question
« Reply #445 on: May 16, 2016, 08:49:51 AM »
Besides, take a 9 volt and 12 volts battery and connect them in parallel and what do you get.
Very high current. Equivalent to a 3V shorted battery.

However the same does not apply to a 9V battery inserted into an ideal inductor, because the inductor is not a voltage source - it is a current source. 
The internal impedance of an ideal current source is infinite, while the internal impedance of an ideal voltage source is zero.

That's why a current source connected to a voltage source does not result in any infinities.

tinman

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Re: MH's ideal coil and voltage question
« Reply #446 on: May 16, 2016, 10:46:40 AM »
MH did know the answer to the question, and could easily have plotted out the current trace as I have encouraged you to do, but he wanted you and others to work it through yourselves.

MH's question is thought-provoking and well thought out, although the wording could have been geared more towards the experimenters here rather than the techies. At any rate, I can assure you that MH knew all along exactly how the circuit currents behave, given the input he specified. After all, he is the one that formulated the question.

Quote
Brad, you really are being silly with what you said there. I know you and MH don't seem to be getting along well, but he deserves more benefit of doubt than you are affording him here on this electronics question.

You mean like the benefit of doubt that he gave me regarding the ICE having resonant systems?. Guessed you missed all that. This is the very same situation. He told me i had no idea what i was talking about,when Internal combustion engines are my forte--my area of expertise.
The difference is,i backed up my knowledge with provided fact's,and this is something no one here can do with an actual test,as we are talking about ideals we do not have. What we are doing is placing a theory based around !best guesses!.

So i have given MH no more than he has given me,and in fact,i have never used the foul language he has toward me.
Not once did i see you,or any other EE guy here tell MH to calm it down when the roll was reversed,but i see you are quick to jump on me when i do the same that has been done to me.

I have seen this very same thing with other members that disagree with MHs analogy.
It's an !agree with me! or your wrong attitude MH has--plane and simple.
As i said in the other thread,i will now treat him as he treats me.

As i said,there is a pattern that is followed on this forum,and that is the EE guys stick together--bar one,that being (as i have always said) verpies. I would also put vortex1 in there with verpies,but he dosnt frequent this forum much-sadly.
As verpies said in reply to this question,Quote : The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
verpies-Just because most of the world does it wrong does not mean that we have to.


Brad

tinman

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Re: MH's ideal coil and voltage question
« Reply #447 on: May 16, 2016, 10:56:38 AM »
Sounds to me like you are agreeing with me.

The source is trying to drive the opposite polarity energy into the inductor. When the source is at -3V, it offers up a place for the inductor to dump some of its stored energy. Given more time at -3V, the inductor would fully lose its "positive" energy, and begin to energize in the "negative" direction.

Since it isn't long enough for that to occur, the energy in the inductor simply decreases to a lower "positive" level, and there is only one place for that lost energy to go. You can even see it in the sim.

Well this is how i see it Poynt.

We have a loop of water pipe that represents our ideal inductor loop. In that loop of water pipe we have a pump. Our pump is our voltage source,and the water in that pipe is our current. We start the pump,and this puts pressure(our voltage) on the water(our current) the water starts to flow around our loop. We have a bypass valve in the pump,so as when we switch the pump off(0 volts),the water can still flow in the direction it was(our current is now flowing with the pump off). While the water is flowing,we start the pump up so as it spins in the opposite direction. The pump(our now reversed voltage) wants to now push the water in the opposite direction to that of which it is already flowing. The energy in the moving water is not added to,or stored in the pump--it is working against the pump,and the pump will draw more current to stop the flow of water that is flowing in the wrong direction, before it can start to move the water(current) in the right direction to that of what the pump wants to move it in.

That is how i see it.

Brad

verpies

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Re: MH's ideal coil and voltage question
« Reply #448 on: May 16, 2016, 12:04:35 PM »
@Tinman

That's a mechanical/hydraulic analogy.  This type of mechanical thinking is a long lost skill among physicists and will lead you to the correct results but the analogy must be precise.  Unfortunately yours is not precise enough.

We have a loop of water pipe that represents our ideal inductor loop. In that loop of water pipe we have a pump.
Let's make it a positive displacement pump- like a Lobe Pump.
I chose the Lobe Pump for this analogy because in such pump, the direction and speed of the impeller has 1:1 correspondence to the direction and speed of the water current.

Our pump is our voltage source and the water in that pipe is our current.
Corrections:
The water in the pipe symbolizes electric charge.  The motion of this charge symbolizes electric current.
BTW: The mass/inertia of the water symbolizes inductance.

The force/torque applied to the pump's impeller symbolizes the voltage. 
It is important not to conflate the pump with the voltage, because the pump itself is not the force - it is only a mechanism to transfer the force to the water/charge.

We start the pump,and this puts pressure(our voltage) on the water(our current) the water starts to flow around our loop.
Generaly, I agree.
I would write: "...this puts force/pressure on the water causing its acceleration and motion (current)"

We have a bypass valve in the pump,so as when we switch the pump off (0 volts), the water can still flow in the direction it was (our current is now flowing with the pump off).
Yes, the water can flow even when the force (torque) applied to the impeller is zero, but a bypass valve is not necessary for this flow to continue, as e.g. the impeller of a lobe pump will continue to rotate under the current of water already flowing through this pump.

This "bypass valve" is an extraneous component that should have tipped you off, that the analogy is not precise enough.

While the water is flowing,we start the pump up so as it spins in the opposite direction. The pump(our now reversed voltage) wants to now push the water in the opposite direction to that of which it is already flowing.
Here is where you reap the bad fruit of your analogy ;(
Precisely the cause of the conceptual error is in the phrase: "we start the pump up so as it spins in the opposite direction".
The correct phrase should have been: "we apply a torque to the pump's impeller in the opposite direction".

Note, that the application of opposite force/torque to the impeller, does not immediately result in the reversal of the impeller's direction (nor water's direction).  I remind you of the 1:1 correspondence between the direction of the impeller and the direction of the water, (their speed also).

I am sure that now you have the tools to complete the rest of the analogy by yourself.

tinman

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Re: MH's ideal coil and voltage question
« Reply #449 on: May 16, 2016, 12:35:51 PM »
@Tinman

That's a mechanical/hydraulic analogy.  This type of mechanical thinking is a long lost skill among physicists and will lead you to the correct results but the analogy must be precise.  Unfortunately yours is not precise enough.
Let's make it a positive displacement pump- like a Lobe Pump.
I chose the Lobe Pump for this analogy because in such pump, the direction and speed of the impeller has 1:1 correspondence to the direction and speed of the water current.
Corrections:
The water in the pipe symbolizes electric charge.  The motion of this charge symbolizes electric current.
BTW: The mass/inertia of the water symbolizes inductance.

The force/torque applied to the pump's impeller symbolizes the voltage. 
It is important not to conflate the pump with the voltage, because the pump itself is not the force - it is only a mechanism to transfer the force to the water/charge.
Generaly, I agree.
I would write: "...this puts force/pressure on the water causing its acceleration and motion (current)"


This "bypass valve" is an extraneous component that should have tipped you off, that the analogy is not precise enough.


Note that the application of opposite force/torque to the impeller does not immediately result in the reversal of the impeller's direction (nor water's direction).  I remind you of the 1:1 correspondence between the direction of the impeller and the direction of the water, (their speed also).

I am sure now you have the tools to complete the rest of the analogy by yourself.

Quote
Here is where you reap the bad fruit of your analogy ;)
Precisely the cause of the error is in the phrase "we start the pump up so as it spins in the opposite direction".
The correct phrase should have been: "we apply a torque to the pump's impeller in the opposite direction".

I see no difference in what i said,and what you have stated above.
The motor that drives the positive displacement pump you have used,will still encounter the force of the water moving in the opposite direction to that of the applied torque to the pump.
You have simply separated the pump from the motor. But any force placed upon the pump will be transferred to the motor.

Quote
Yes, the water can flow even when the force (torque) applied to the impeller is zero, but a bypass valve is not necessary for this flow to continue, as e.g. the impeller of a lobe pump will continue to rotate under the current of water already flowing through this pump.

As the motion of the pump/motor combo i was using to represent the voltage(our force),then i included the bypass valve to represent no motion of the motor/pump--the equivalent of the 0 volt level in the question. Having the impeller still moving,would be seen as a resistance against the flow of water,and in our ideal loop,we have no resistance to the flow of the current,and so the bypass valve was included for that reason also.
I believe that this is a close representation to what we have,than having the impeller being rotated by the water,resulting in a resistance to that flow of water.

Brad