Going back to my simplified example...

Given:
L=5H
V=+4V

What happens:
At t<0, XL=0
At t=0, XL=infinity, and the current rises from 0A linearly at 0.8A/s.

At t=1, XL=5 Ohms
At t=2, XL=2.5 Ohms
At t=4, XL=1.25 Ohms
i.e. with every doubling of "t", "XL" goes to 1/2 of previous value. XL never reaches 0 Ohms.

At t>0 when I_L begins rising, an induced cemf of -4V is produced, based on Vcemf=L x di/dt. The cemf remains steady at -4V as long as di/dt stays at 0.8A/s. This of course occurs simultaneous with the application of 4V and the rise of I_L.

The question is, if emf=cemf, would it not makes sense that this result is created through an equalizing process? The amps/s rate ultimately being determined by the applied voltage and the inductance values.

The equalization I refer to comes about via the simultaneous process of an applied emf that wants to drive a current, vs. a reactionary process that wants to lower that current. To me this is very much like a negative feedback mechanism commonly used in linear amplifiers. Your amplifier may have an inherent gain of 1000, but through the application of negative feedback, the gain is reduced to some desired level, such as 100. In the case of our self-inductance, the negative feedback mechanism is the self-induced current and B field, which happens to oppose the B field resulting from the applied voltage. It is not quite an exact analogy, but conceptually similar. It all happens in real time, simultaneously, and only the end result is observable.

Perhaps it comes down to an applied emf, vs. an induced E field. We know the applied emf is 4V, but can we break the induced cemf (equivalent to the E field) down any further? Well, for a multi-turn inductor we can divide the cemf by the number of turns to obtain the induced cemf per turn. To me this would represent the actual value of the E field circulating around the inductor. So if we have a total cemf of 4V, and a 1000-turn coil, the actual induced E field would be 4/1000 = 4mV. This does not sound like much, but with R=0, 4mV could drive a significant current in a single loop, and each loop would carry the same current. I'm sure the induced current can be derived from the E field (or B field) and rate of change.

Some things to perhaps think about anyway. Sorry there are no definitive answers here. Still thinking about this.

and the current rises from 0A linearly at 0.8A/s.

So we have a coil that is ideal--free from resistive losses.
We have a current that is going to follow a straight linear rise at 800mA/s
Our current dose not follow an exponential curve,as it would with a coil that has a resistance value.

This being true,it must also be true that the magnetic fields change it time would remain a constant value to that of the linear current rise,and that value would be the value it was at T=0--the instant the voltage was placed across the coil.
If the magnetic field is increasing/changing in time at a constant value,then the CEMF must also be doing the same. If the CEMF value is the same as the applied EMF,and the coil has no resistance,and cannot dissipate energy,then the reverse current flow must also be the same as the induced current flow.

The only reason we get an exponential current curve,is because real world inductors have a resistance value,and so some of the energy is dissipated as heat,and so the difference between the EMF and CEMF value. But an ideal coil has no resistance,and there for cannot dissipate energy,and so there is no loss associated with the CEMF value as there is in real world coils.

I really hope you have a closer look at this,because as i stated long ago,things are not the same when the coil/inductor is void of resistance.


Brad

Perhaps consider the following:

a) emf = cemf = the E field times the number of turns.
b) If an inductor has 1000 turns, then the effective cemf to emf ratio is 1/1000.
c) per turn, the cemf is therefore much less than the applied emf.

And how can that be,when the EMF also see the same amount of turns.
Also,if this were true,dose this mean an ideal 1 turn coil will have an EMF to CEMF ratio of 1:1,and no current can flow.

I do not think that is right.
If the ratio was 1/1000,then the current would rise much faster.

There is also the fact that,the more turns you have,with the same current flow value,the greater the magnetic field. This results in a greater magnetic field,but the rate of change over time is still a linear constant-along with the induced current.

Are we skipping equal and opposite reactions,even though there is no dissipated power?


Brad

So we have a coil that is ideal--free from resistive losses.
We have a current that is going to follow a straight linear rise at 800mA/s
Our current dose not follow an exponential curve,as it would with a coil that has a resistance value.

This being true,it must also be true that the magnetic fields change it time would remain a constant value to that of the linear current rise,and that value would be the value it was at T=0--the instant the voltage was placed across the coil.
If the magnetic field is increasing/changing in time at a constant value,then the CEMF must also be doing the same. If the CEMF value is the same as the applied EMF,and the coil has no resistance,and cannot dissipate energy,then the reverse current flow must also be the same as the induced current flow.

The only reason we get an exponential current curve,is because real world inductors have a resistance value,and so some of the energy is dissipated as heat,and so the difference between the EMF and CEMF value. But an ideal coil has no resistance,and there for cannot dissipate energy,and so there is no loss associated with the CEMF value as there is in real world coils.

I really hope you have a closer look at this,because as i stated long ago,things are not the same when the coil/inductor is void of resistance.


Brad

Tinman,

Are you stating that you believe that the only amount of current that can flow thru an inductor with zero DC resistance is the amount of current that flows at T=0?

PW

Tinman,

Are you stating that you believe that the only amount of current that can flow thru an inductor with zero DC resistance is the amount of current that flows at T=0?

PW

The CEMF induced current is determined by the rate of change of the magnetic field in time--yes?
If as Poynt says,that the current will rise at a steady linear rate of 800mA/s,then it must also be true that the rate of change of the magnetic field in time will also change at that steady linear rate.That being the case,then the value of the CEMF,and there for the self induced current,must also rise at that steady linear rate--the joys of having no exponential current curve--no time constant.

As there is no time constant--the process is infinite,then the determined action/reaction is determined at T=0. The coil is also ideal,and there for dose not dissipate any energy,and so none is lost such as it is in a coil with resistance.

The answer given for MHs question is not correct,as the fact that there is no time constant ,was simply ignored--and it should not have been.
Having no time constant results in a steady linear climb in EMF induced current,and there for must also result in a steady climb in self induced current that is in opposition to that which created it.


Brad

The CEMF induced current is determined by the rate of change of the magnetic field in time--yes?
If as Poynt says,that the current will rise at a steady linear rate of 800mA/s,then it must also be true that the rate of change of the magnetic field in time will also change at that steady linear rate.That being the case,then the value of the CEMF,and there for the self induced current,must also rise at that steady linear rate--the joys of having no exponential current curve--no time constant.

As there is no time constant--the process is infinite,then the determined action/reaction is determined at T=0. The coil is also ideal,and there for dose not dissipate any energy,and so none is lost such as it is in a coil with resistance.

The answer given for MHs question is not correct,as the fact that there is no time constant ,was simply ignored--and it should not have been.
Having no time constant results in a steady linear climb in EMF induced current,and there for must also result in a steady climb in self induced current that is in opposition to that which created it.


Brad

Again, to be clear, are you stating that you believe that the only amount of current that can flow thru an inductor with zero DC resistance is the amount of current that flows at T=0?

A simple yes or no would be most helpful.

PW

As long as the current is rising at a steady rate of 0.8A/s, the cemf will be a steady 4V, it will not rise at a steady linear rate.
It is very unlikely anyone agrees with you. How are you going to prove this?

Ok -now we are getting somewhere.
If at T=0,4 volts is placed across the coil. The CEMF is also 4 volts as you state the moment current starts to flow.
So the outcome is?.-remembering that we have agreed (with the exception of MH-who seems to have his own ruels)that the CEMF has to be lower than the applied EMF in order for current to flow.


Brad

Sorry, but you sound incredibly stupid.  You have been playing with electronics all this time, for years, and you can't understand what a bloody voltage drop is?


MileHigh

\

.

As long as the current is rising at a steady rate of 0.8A/s, the cemf will be a steady 4V, it will not rise at a steady linear rate.

The answer given for MHs question is not correct

It is very unlikely anyone agrees with you. How are you going to prove this?

You and others are going to do it for me . I only hope you  and others will take the time to read all i have gathered,as i have taken the time to put it altogether.

Quote post 1218-Poynt

I think you have the basic concept, yes. Again, the fundamental frequency and the harmonic content influences how the inductor reacts. The higher the inductance, the higher the induced cemf for a given frequency. At some point (either relatively large L or high frequencies) the cemf will equal the applied voltage (or it may be more correct to say the induced current will equal the applied current) and the net resulting current will be minimal.

Quote post 1227-MH

So even though the battery is imposing its voltage on the coil, you need to be able to shift your perspective and go "inside" the coil and realize that the coil is pushing back with the same CEMF. The applied EMF and the CEMF from the coil must add up to zero. Therefore, the CEMF must be equal and opposite to the EMF.

Post 1231--Poynt

The voltage across the coil terminals does not change, it is determined by the voltage source. But the induced cemf is in series opposing with the voltage source Vin, and its value is determined by the frequency of Vin and the inductance L.
So, from this perspective the induced cemf is usually not equal to the applied emf (Vin). It is usually lower.

Then there is this confusing one by MH-post

I think that you are just going to confuse Brad with that diagram because he is going to see the CEMF being opposite the EMF

Then at the bottom of the same post,he say this

If you agree with this, then the CEMF is always equal and opposite to the applied EMF

So im lost with that one,as he says that i will get confused and think that the CEMF is opposite the EMF,and then go's on to say that it is

Post 1233--Poynt

I don't think he will assume that. I believe he knows that even though the induced emf (cemf) is opposite in polarity to that of the applied voltage, it will almost always be less, and therefore there will still be a net applied emf and resulting current.

Post 1282--Hoptoad,in reference to my statement--> if the CEMF was equal and opposite to the EMF,then the total voltage across the inductors terminals would be 0v,and no current would flow.

I agree. No potential difference, No current flow.

Post 1284 from PW

As well, it would seem that any device capable of producing a CEMF exactly equal to an applied EMF would prevent current flow.Consider two identical voltage sources connected in parallel (positive to positive, negative to negative).  One Vsource represents EMF and the other Vsource represents CEMF.  As long as both sources produce identical voltage, there will be no current flow.

Post 1297 from Hoptoad. I am going to post the link in that reply.

https://www.wisc-online.com/learn/career-clusters/stem/ace5903/an-inductor-opposing-a-current-change
You will see,if you follow the test pages,this statement.
Quote:  At 1 time constant,the CEMF cannot be sustained,and after the first time constant,the counter voltage reduces by 63%.
As our coil is ideal,we have no time constant,and the counter voltage will remain at a steady 4 volt,s--as you stated Poynt.

Post 1301-Citfta

If your claim that the CEMF equals the EMF were true then no current would flow

Post 1308 from Weby-a link attached

https://books.google.com/books?id=9dsWAAAAQBAJ&pg=PA268&lpg=PA268&dq=Is+a+resistance+a+CEMF?&source=bl&ots=OmxEsMZ6B2&sig=21Hvbdp3lsZo50GzV2t_DlJF7Bc&hl=en&sa=X&ved=0ahUKEwiRq4-Tm77NAhVC9GMKHegBCcgQ6AEIMjAI

A quote from that book regarding a DC motor.
Quote: If the BackEMF is equal to the applied EMF,then no current will flow through that motor. As we know,an increase in BackEMF means a decrease in current draw from the motor--this i showed in my last video very clearly.

Post 1313 by Hoptoad

If the cemf was a steady value, all other factors would also be steady.

Post 1331-Poynt

One is called cemf because that is precisely what it is; i.e. it is a generated voltage in this case. Going around the loop is simply confirming KVL, and it always holds.

Post 1332-Poynt

Since the amps/sec is constant, the induced cemf should be steady.See post 1313 above from Hoptoad

Post 1333 from PW

As I responded to Tinman, if it were somehow possible to cause all the magnetic flux created by a current flowing thru an ideal conductor to be confined to, and cut thru, that conductor in such a way as to make the inductor's CEMF be equal to the EMF, I believe that inductor would have infinite inductance.  To avoid the "chicken or egg paradox" in answering whether current could flow thru such an inductor, I stated that I believed that an infinitely small current would flow over an infinitely long period of time.

Post 1359--Loner

Again, if the CEMF were EXACTLY equal to the applied EMF, the "Rate of change" would be equal to 0,

Post 1363-Hoptoad

What will happen in an 'ideal' inductor is great debating material, however, almost all explanations could be considered equally valid simply because the ideal doesn't exist (except possibly - inductors made with superconductors?) and therefore any hypothesis relating to it is (currently) unfalsifiable. But 'ideal' hypothesis do give the brain matter something to chew on.

And post 1446-Poynt

As long as the current is rising at a steady rate of 0.8A/s, the cemf will be a steady 4V,

At T=0 seconds the coil connects to the ideal voltage source. For three seconds the voltage is 4 volts

Post 1450-Poynt

The mechanism of exactly how the cemf equals the emf while still permitting current flow is still somewhat of a mystery.

Do i know what CEMF is?
To quote PW post 1399

 

Although a bit on the loose side where I pointed it out, his definition of CEMF was, for the most part, correct.

I believe that i have provided enough information by all here,that it is agreed upon,that if the CEMF value is equal to that of the applied EMF,then no current could flow.
It has also been concluded that the CEMF is Equal to the applied voltage at T=0.

In your last post Poynt(quoted above),you state that how the current can flow,when the CEMF value is equal to that of the applied EMF,is still a mystery.

So i ask-how is an accurate and correct answer gained from something yet unknown(the mystery),and also go's against all that most of us here believe to be true-even your self and PW ?.
You ask how am i going to provide proof to back up what i believe?. It would seem to me,that most of you here,are trying to find the proof that go's against all that has been stated in this thread-with the exception of MH,who provided a formula base around inductors/coils that have a resistance value.

I have spent the last 2 1/2 hours gathering all this information throughout this thread,and i only hope it dose not fall on death ears.


Brad