Author Topic: MH's ideal coil and voltage question (Read 487925 times)

picowatt · « **Reply #1485 on:** July 01, 2016, 06:20:11 PM »

Quote from: tinman on July 01, 2016, 03:47:58 PM

I was asking MH for his answer.

I have ask a number of times now,where is the loss?--why is the negative feedback less than that that caused the negative feedback?
So i have ask for clarification a number of times now,and have received no answer.
I will take it that you have misses !on a number of occasions! that i have indeed requested clarification.

So is your issue ONLY with regard to inductors having no resistance, or do you not agree with or understand the concept of an inductor's CEMF as the feedback mechanism that limits the current flow's rate of change with regard to ANY inductor?

PW

tinman · « **Reply #1486 on:** July 01, 2016, 06:49:01 PM »

Quote from: picowatt on July 01, 2016, 06:20:11 PM

So is your issue ONLY with regard to inductors having no resistance, or do you not agree with or understand the concept of an inductor's CEMF as the feedback mechanism that limits the current flow's rate of change with regard to ANY inductor?

PW

OK,well this seems so hard to make any clearer.
If the CEMF voltage is the same as the EMFs voltage,then where is the loss that allows current to flow,when there is no potential difference between the inductor and source.

Here is the thing. If the CEMF voltage is the same as the V/in,then you have just converted all(100%) of the applied EMFs induced current into CEMF. We now have your two batteries hooked in parallel that have the same voltage--but current still flows. You talk about negative feed back,and this feed back is of the same value,but current still flows--so where is the loss on the negative side(the CEMF side) that allows current to flow with no potential difference.
This is about the fifth or sixth time i have asked,and never get an answer.

From all my test,the CEMF is lower than the EMF by over 1/2 a volt,when 12 volts is applied to that inductor. I am yet to see anyone here making claims that the CEMF is equal to the EMF,actually show it to be the same--my results show it is not,and so current can flow.

Blue is voltage across inductor,yellow is supply voltage.

Brad

picowatt · « **Reply #1487 on:** July 01, 2016, 07:12:03 PM »

Quote from: tinman on July 01, 2016, 06:49:01 PM

OK,well this seems so hard to make any clearer.

It would make it much clearer if you would just answer my question as posed. As in when I asked if your issue was only with regard to zero R inductors or all inductors...

Quote

If the CEMF voltage is the same as the EMFs voltage,then where is the loss that allows current to flow,when there is no potential difference between the inductor and source.

Here is the thing. If the CEMF voltage is the same as the V/in,then you have just converted all(100%) of the applied EMFs induced current into CEMF.

This is not correct. CEMF is only generated by a changing magnetic field and it is only the portion of that changing magnetic field that cuts the conductor at right angles that induces a voltage back into the conductor.

Quote

We now have your two batteries hooked in parallel that have the same voltage--but current still flows. You talk about negative feed back,and this feed back is of the same value,but current still flows--so where is the loss on the negative side(the CEMF side) that allows current to flow with no potential difference.

This is about the fifth or sixth time i have asked,and never get an answer.

Again, in order to help you understand this, I have asked if your issue regarding this is only with regard to zero R inductors, or if it is more so that you are having an issue understanding the CEMF's regulating effect with regard to all inductors.

Quote

From all my test,the CEMF is lower than the EMF by over 1/2 a volt,when 12 volts is applied to that inductor. I am yet to see anyone here making claims that the CEMF is equal to the EMF,actually show it to be the same--my results show it is not,and so current can flow.

Blue is voltage across inductor,yellow is supply voltage.

You're going to have to provide more info for me to make any sense of your scope shot.

If you are connecting an inductor directly across a Vsupply, how can the supply voltage and the voltage measured across the inductor not be equal? A schematic or description of your test set up would be helpful...

PW

3Kelvin · « **Reply #1488 on:** July 01, 2016, 08:17:02 PM »

The magnitude of Vsourse and Vinduced are the same.
But Vsource is a static source, Vind is a dynamic source.
For my assumption, this is the unsymmetrical between the both magnitudes.
The difference is how to get the magnitude. Maybe is is the time between first and second action.

With the equation I(t) = (Us / L) *t we cant describe the difference.
0.8 A/s = 4V/5Henry. For this equation we get at t=0 no difference,
so that Vs = Vi. But we also know that without a changing current 0.8A/s we wont get the -4V.

So Vs is not equal Vi, but the Magnitude (Skalar) is equal.

L+P 3K

partzman · « **Reply #1489 on:** July 01, 2016, 08:58:56 PM »

Quote from: picowatt on July 01, 2016, 05:55:08 PM

Consider two voltage sources connected in parallel. One Vsource represents the applied EMF, the other Vsource represents the inductor's generated CEMF. The first Vsource, representing EMF, is set to output +4 volts.

What voltage must the second Vsource be set to output in order for their to be no current flow?

PW

Even though I said I would not post until I could show evidence contrary to your beliefs, I will play along.

The second Vsource or Cemf must be +4v of course. And the point is .......?

pm

Edit: PW or anybody, please give an equivalent circuit or model of the Emf/Cemf interaction in a single inductor driven from an Emf source. Or a math derivation would be OK but not just the same repeat of Emf = Cemf please.

picowatt · « **Reply #1490 on:** July 01, 2016, 09:57:46 PM »

Quote from: partzman on July 01, 2016, 08:58:56 PM

Even though I said I would not post until I could show evidence contrary to your beliefs, I will play along.

The second Vsource or Cemf must be +4v of course. And the point is .......?

pm

Edit: PW or anybody, please give an equivalent circuit or model of the Emf/Cemf interaction in a single inductor driven from an Emf source. Or a math derivation would be OK but not just the same repeat of Emf = Cemf please.

The point is that there is no minus sign where you indicated.

Everyone seems to be in agreement that no current will flow when the EMF=CEMF.

1. When the current flowing thru a 5H inductor is changing at the rate of .8 amps per second, a CEMF of 4 volts is generated (CEMF=dI*L/dt).

2. If the applied EMF is 4 also volts, current flow will cease as soon as the CEMF reaches 4 volts because at that point the CEMF equals the applied EMF (CEMF=EMF).

3. However, as the current flow begins to cease, so does the rate of change, causing the CEMF to be less than 4 volts.

4. When the CEMF is less than 4 volts, that is, when the CEMF<EMF, current will again flow until the rate of change again reaches .8 amps per second and the CEMF again equals 4 volts.

Return to step 2 above (continuous loop)

And again, although described in a stepwise fashion, it is a smooth and continuous process similar to the many instances of negative feedback used in all manner of electronic circuits.

It is this process that limits, or regulates, the current flow's rate of change.

PW

partzman · « **Reply #1491 on:** July 01, 2016, 11:19:21 PM »

Quote from: picowatt on July 01, 2016, 09:57:46 PM

The point is that there is no minus sign where you indicated.

Everyone seems to be in agreement that no current will flow when the EMF=CEMF.

No current flow in what? Could you be specific as you reference Emf=Cemf below in your feedback correction example below when dI=.8A/S.

Quote

1. When the current flowing thru a 5H inductor is changing at the rate of .8 amps per second, a CEMF of 4 volts is generated (CEMF=dI*L/dt).

Well I guess that I'm the only one who disagrees so I will try to explain why! Why are you taking liberty to change Cemf=-dI*L/dt to Cemf=dI*L/dt? Are you throwing the Lenz factor away? If yes, why and how? If not, then when you state that Emf = Cemf, to me you are saying to me that dI*L/dt=-dI*L/dt. Do I not understand the terminology?

Quote

2. If the applied EMF is 4 also volts, current flow will cease as soon as the CEMF reaches 4 volts because at that point the CEMF equals the applied EMF (CEMF=EMF).

Now I'm really confused by the above statement because you and others hold to the idea that Emf = Cemf during normal inductor current so again, what current will cease to flow?

Quote

3. However, as the current flow begins to cease, so does the rate of change, causing the CEMF to be less than 4 volts.

OK, let's look at what could cause the current in the inductor to decrease even slightly during any given dt. A), the Emf could drop but so would the assumed Cemf so we still follow the EMF law. B), the inductance could change due to self heating (neglecting any change in the DC resistance) creating a slight physical movement so this could really be ignored, but this too would follow the Emf law. So what does that leave but C), the initial instant that the Emf is applied when supposedly the Emf/Cemf feedback loop corrects until we experience dI=Emf*dt/L.

Quote

4. When the CEMF is less than 4 volts, that is, when the CEMF<EMF, current will again flow until the rate the change again reaches .8 amps per second and the CEMF again equals 4 volts.

Again, apart from any outside influence such as an approaching magnet or any material that would change the permeability, what would cause the required feedback adjustment above other than the instant of applied Emf?

Quote

Return to step 2 above (continuous loop)

And again, although described in a stepwise fashion, it is a smooth and continuous process similar to the many instances of negative feedback used in all manner of electronic circuits.

PW

Well, I can't seem to get past step 1.

Believe me, I understand what you are trying to explain, I just don't agree with it for the above and previously stated reasons.

IMO, your thought model is flawed but I just can't prove it at this point in time.

pm

Magluvin · « **Reply #1492 on:** July 01, 2016, 11:30:48 PM »

Quote from: picowatt on July 01, 2016, 09:57:46 PM

The point is that there is no minus sign where you indicated.

Everyone seems to be in agreement that no current will flow when the EMF=CEMF.

1. When the current flowing thru a 5H inductor is changing at the rate of .8 amps per second, a CEMF of 4 volts is generated (CEMF=dI*L/dt).

2. If the applied EMF is 4 also volts, current flow will cease as soon as the CEMF reaches 4 volts because at that point the CEMF equals the applied EMF (CEMF=EMF).

3. However, as the current flow begins to cease, so does the rate of change, causing the CEMF to be less than 4 volts.

4. When the CEMF is less than 4 volts, that is, when the CEMF<EMF, current will again flow until the rate of change again reaches .8 amps per second and the CEMF again equals 4 volts.

Return to step 2 above (continuous loop)

And again, although described in a stepwise fashion, it is a smooth and continuous process similar to the many instances of negative feedback used in all manner of electronic circuits.

It is this process that limits, or regulates, the current flow's rate of change.

PW

Are we talking about cemf of the inductor or an applied cemf equal to the applied emf?

When we first apply 4v emf, how can cemf be equal to the emf? What happened to create the initial cemf? Did some current flow in the beginning due to emf in order for the cemf to develop in the first place? I mean, I can understand that at T0 that the 4v can be read across the inductor leads, but Im not sure it is due to cemf created by induction unless some initial emf current must have happened in order for cemf to develop in the first place.

Im kinda betting on cemf to be in some form, no matter how small the difference, to be always less than the emf.

Mags

picowatt · « **Reply #1493 on:** July 01, 2016, 11:50:21 PM »

Quote from: Magluvin on July 01, 2016, 11:30:48 PM

Are we talking about cemf of the inductor or an applied cemf equal to the applied emf?

When we first apply 4v emf, how can cemf be equal to the emf? What happened to create the initial cemf? Did some current flow in the beginning due to emf in order for the cemf to develop in the first place? I mean, I can understand that at T0 that the 4v can be read across the inductor leads, but Im not sure it is due to cemf created by induction unless some initial emf current must have happened in order for cemf to develop in the first place.

Im kinda betting on cemf to be in some form, no matter how small the difference, to be always less than the emf.

Mags

All thru this thread we have been discussing the 4 volts applied across the inductor from an ideal Vsource ,as being the applied EMF, or just EMF. The EMF is the fixed potential of 4 volts applied to the inductor at T=0.

CEMF refers to the voltage induced in the inductor's windings as the current flowing thru the inductor changes at a speciific rate.

PW

picowatt · « **Reply #1494 on:** July 02, 2016, 12:10:56 AM »

Has everyone at least read the Wiki?

https://en.wikipedia.org/wiki/Inductor

PW

Magluvin · « **Reply #1495 on:** July 02, 2016, 03:26:56 AM »

Quote from: picowatt on July 01, 2016, 11:50:21 PM

All thru this thread we have been discussing the 4 volts applied across the inductor from an ideal Vsource ,as being the applied EMF, or just EMF. The EMF is the fixed potential of 4 volts applied to the inductor at T=0.

CEMF refers to the voltage induced in the inductor's windings as the current flowing thru the inductor changes at a speciific rate.

PW

Well, that is not what I asked. Anyway...

I do know about the 4v, as I did mention that in my post..

When we say "...cemf refers to the voltage induced in the inductors windings......" , I look at it a bit differently..

Say we apply the input and current begins to rise. I, instead of seeing the cemf as an actual reverse voltage potential against the input, I see all of that happening in the magnetic realm where the fields are at odds with each other 'from winding to winding' which determines the current flow, rather than thinking that 2 emf's are actually in opposition in the wound conductor.

Mags

tinman · « **Reply #1496 on:** July 02, 2016, 04:40:30 AM »

Quote from: picowatt on July 01, 2016, 09:57:46 PM

The point is that there is no minus sign where you indicated.

Everyone seems to be in agreement that no current will flow when the EMF=CEMF.

1. When the current flowing thru a 5H inductor is changing at the rate of .8 amps per second, a CEMF of 4 volts is generated (CEMF=dI*L/dt).

2. If the applied EMF is 4 also volts, current flow will cease as soon as the CEMF reaches 4 volts because at that point the CEMF equals the applied EMF (CEMF=EMF).

3. However, as the current flow begins to cease, so does the rate of change, causing the CEMF to be less than 4 volts.

4. When the CEMF is less than 4 volts, that is, when the CEMF<EMF, current will again flow until the rate of change again reaches .8 amps per second and the CEMF again equals 4 volts.

Return to step 2 above (continuous loop)

And again, although described in a stepwise fashion, it is a smooth and continuous process similar to the many instances of negative feedback used in all manner of electronic circuits.

It is this process that limits, or regulates, the current flow's rate of change.

PW

I have highlighted 3 because-->

Being that the coil is ideal,there is no rate of change-change-->the rate of change remains a constant,regardless of the applied voltage value,because the time constant is infinite.

Every time the EMF tries to make a change,the CEMF makes the very same change at an instant--your feed back system.
So the very moment an EMF is applied(the force is applied),the CEMF pushes back with the same force--when you push against a concrete wall,the wall will push back with the same force you applied to it,and there is no motion.

Then there is this-->the equal and opposite to your feed back system.
If the EMF is 4 volt's,and the CEMF is 4 volt's,the only way current can start to flow again,is if either the EMF voltage rises above 4 volts,or the CEMF voltage drops below 4 volt's. In your feed back system,one of these two has to happen in order for current to flow.

As the voltage is ideal,then we can assume that it will not drop below that value. This leaves only the CEMF voltage value. As our rate of change is constant-as our coil is ideal,and free from resistance,i ask !once again!,where is the loss that allows the current to flow?-->how is the CEMFs value reduced when the rate of change is a constant,and the applied voltage is ideal,and will not change from that 4 volts-regardless of load.

So i am asking you(with regards to the original MH question),where is the loss that allows current to flow,if the CEMF value is always that of the !!ideal!! voltages value of 4 volts,and where there is no rate of change in time of the magnetic field.

We have all agreed that the applied 4 vots-our EMF is a constant,and will not change unless we change it. You (and some others) have also clearly stated that you believe that the CEMF is always going to be equal to that of the applied voltage,as there is no rate of change to the magnetic field in our ideal inductor.
So,in order for current to flow through our ideal inductor,the CEMF !must! be lower than the applied !ideal! EMF.

It would seem to me PW,that you are using this !feed back! stuff,so as it aligns with current mathematics,and not looking at the situation as it is.

The circuit in regards to the scope shot,is as below.
It would seem to me that the actual value of the CEMF value at T=0(moment EMF is placed across the inductor),would be 11.6 volts,if the average between the EMF and CEMF is 12.2 volts--the blue trace.
I will have to set up a more robust circuit to confirm this,but it is clear that the CEMF is of a lesser value than the applied EMF.

Brad

tinman · « **Reply #1497 on:** July 02, 2016, 04:49:58 AM »

@MH

Could you please post again-in detail as to how the current will flow without a potential difference between coil an source.

This is all i have from you-post 1365
Quote:-->a coil integrates on voltage to give you current just like a shopping cart integrates on force to give you velocity. It's just Mother Nature in action.
All of the stuff in your head about "battling currents" is a model that simply does not work. It's crazy talk. It's like something that you found in a pumpkin patch.

Brad

picowatt · « **Reply #1498 on:** July 02, 2016, 04:52:14 AM »

Tinman,

First, in your scope shot and schematic you are not measuring the CEMF. The difference you are seeing between channels is the Vdrop across R1. You are in effect measuring current.

The CEMF and EMF across the inductor are equal and the only way you can observe the CEMF is by its effect on current flow.

Second, once again I do not know if you are having problems with seeing the inductor's CEMF as a feedback mechanism in ALL inductors or only with regard to inductors of zero resistance. Please let me know if it is one or the other or both.

If your problem is only with regard to zero R inductors, then perhaps an inductor with .1R should be discussed.

PW

tinman · « **Reply #1499 on:** July 02, 2016, 05:09:35 AM »

Quote from: picowatt on July 02, 2016, 04:52:14 AM

Tinman,

PW

Quote

First, in your scope shot and schematic you are not measuring the CEMF. The difference you are seeing between channels is the Vdrop across R1. You are in effect measuring current.

The voltage being measured is the instantaneous voltage-Vmax.
If the CEMF was equal to the EMF at T=0,then there should be no current flow,so why would there be a voltage drop across the resistor at T=0
Remember,we are reading the maximum voltage values on each channel,and so this is the voltage at T=0.

Quote

The CEMF and EMF across the inductor are equal and the only way you can observe the CEMF is by its effect on current flow.

I disagree.
If the CEMF and EMF are equal at T=0,then there should be no current flow at that time,and there for there should be no voltage drop across the resistor.

Quote

Second, once again I do not know if you are having problems with seeing the inductor's CEMF as a feedback mechanism in ALL inductors or only with regard to inductors of zero resistance. Please let me know if it is one or the other or both.
If your problem is only with regard to zero R inductors, then perhaps an inductor with .1R should be discussed.

The difference between the EMF and CEMF i am seeing,can only be due to the coils winding resistance. As we have no winding resistance in an ideal coil,then i would agree that the CEMF is equal to the EMF--and there we have our problem.
This feed back system you talk about,must have a loss in the negative feed back in order for it to no be the same that induced it in the first place. At T=0,every change that the EMF tried to make to the current flow,the CEMF would counteract this change with 100% efficiency-and so no change takes place. As you you have said that the drop in voltage in my scope shot is due to current flow,and that is the two instantaneous voltage value's,then current could only flow instantly if the instantaneous CEMF value was less than that of the applied EMF. As i stated,i believe this to be true when the coil has winding resistance.

I am yet to see(as partzman said),anything that confirms that the CEMF is equal to the applied EMF at T=0. So far,i have seen the opposite.

Brad

Brad

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