I think we have to leave resistors of any kind out of this,as there is no such thing as an ideal resistor.
If a resistor did not dissipate energy,then it would have a resistive value of 0 ohms,and so,you have no resistor.

A resistor has no inductance,dose not store energy,and has resistance.
An ideal coil has inductance,dose store energy,and has no resistance.

The applied EMF see's the number of turns in the coil. The CEMF also see's the same number of turns in the coil.
As soon as the EMF induced current starts to flow,an equal and opposite CEMF produced current will start to flow,due to the fact that there are no losses associated to R.

Only when the RATE OF CHANGE of the current flowing thru the inductor equals .8 amps per second will the CEMF be equal to the applied EMF (with regard to the 4V applied to 5H).  For an ideal inductor, it does not matter if the actual current flowing is 1 amp or 1000 amps.  It is the RATE OF CHANGE that determines the CEMF.

The only reason the self induced current is of a lesser value to that of the EMF induced current in a real world inductor,is due to the losses associated to R in that inductor.

The resistance of a real world inductor is a source of error that causes the inductor's rate of change of the current flowing thru it to deviate from ideal (linear) and also limits the maximum current that can flow thru the inductor.

The formula for calculating the CEMF of an ideal inductor does not use or require a resistance.

PW

The instant 4V is applied to the inductor, the cemf goes to -4V and current begins to flow.

In the case of an ideal inductor there is no trade-off with a resistor, therefore the cemf remains constant (and equal to Vin) as does the rate of rise of the current.

Are we all in agreement?

I disagree PW.

The instant current begins to flow and rise, it is already rising at 0.8A/s (there is no gradual or "reaching" rise in current to the level of 0.8A/s). But we may be saying the same thing.

As far as the polarity of the cemf, yes it is the same as Vin.

I disagree PW.

The instant current begins to flow and rise, it is already rising at 0.8A/s (there is no gradual or "reaching" rise in current to the level of 0.8A/s). But we may be saying the same thing.

As far as the polarity of the cemf, yes it is the same as Vin.

I was just trying to clarify that the CEMF did not exist prior to, or reach 4 volts, until the rate of change was .8 amps per second, which for all practical intents and purposes, can be considered instantaneous. 

Any rate of change faster or slower than .8 amps per second would generate a CEMF that is more than of less than 4 volts.   

PW

Technically speaking, current would not flow if the CEMF and EMF were exactly equal.  But then what would the rate of change of the current be?  Zero volts per second, so no CEMF would be generated.  With no CEMF, current would again flow until it reached .8 amps per second, at which time the CEMF would again equal the EMF, and the cycle would repeat.  But it does not happen in this step-wise fashion, it is a smooth feedback type process where the CEMF is maintained as equal to the level of the applied EMF (or at the least, very, very close to it, if that helps with the visualization) so that the rate of change is .8 amps per second.

No, the R value is not being forgotten, it just does not come into play with regard to the CEMF.  The CEMF is an induced voltage.  The CEMF equals 4 volts when the RATE OF CHANGE of the current flow is .8 amps per second (4V applied to 5H).  Do not confuse the CEMF, which is a mechanism that determines the rate at which energy can be stored, with the actual amount of energy that is stored.  And do not forget that the inductor is storing energy. 

As soon as the RATE OF CHANGE reached .8 amps per second, 4 volts of CEMF would be generated and effectively regulate/limit the rate of the current's rise to that .8 amps per second.

You are wrong.  Faraday says nothing about a required resistance or required dissipation.  Resistance does not enter into it.  If the rate of change of the current flowing thru the 5H inductor is .8 amps per second, the generated CEMF will be 4 volts.  If the rate of change is less than .8 amps per second, the generated CEMF will also be less than 4 volts.

Consider reading this whole Wiki, but from the Wiki:

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

PW

The assumption is still being made , that the current value will continue to rise at 8 amps per second.
You seem to keep missi g the point i am making,and one that Poynt hast stated is yet to be understood.
At T=1 second,the EMF and CEMF are the same value. If at this point the current continues to rise,then it must be stated as to why the self induced current is of a lesser value than that of the EMF induced current,to allow current to continue to flow, when there is no potential difference across the coil

It would seem to me that making a claim to have a definitive answer,is very premature, when the mechanism is not yet known as to why the self induced current should be less than that of the EMF induced current.


Brad

The CEMF is only less than the applied EMF of 4 volts when the rate of change of the current flowing the the 5H inductor is less than .8 amps per second.  At .8 amps per second the CEMF would equal the applied EMF.  The generated CEMF is what limits the rate at which the current can rise to .8 amps per second.

It is not an assumption, it is the how and why inductors work like they do.  Have you read the Wiki?  Perhaps that will help.

I do not know how to explain the action of the inductor's CEMF any better than I already have several times now. 

PW

PW
You are still missing the point,and you have not yet explained why current would continue to rise when there is no potential different across the coil.

I understand that in order to have a CEMF valje  equal to that of the applliex EMF value,that the  current must rise at .8 amps per second.
The point is,the current is rising  at that rate from T=0-not just from T=1s
So,as soon as that 4 volts is placed across that inductor,a CEMF of 4 volts exist..From the moment of connection,there is no potential difference across that coil,and so the only way a current could flow,is if the self induced current is of a lower value than that of the EMF induced current.
So-i am asking why the self induced current is a lesser value than the EMF induced current,so as current can flow.

I might also add that Faraday never had a means to determine the outcome of the stated question this thread is about..

 author=picowatt link=topic=16589.msg487380#msg487380 date=1467185200]

I am not missing the point.  Current would not continue to flow if there was no potential difference, but then there would also cease to be any CEMF, so current would again increase until the rate of change was .8 amps per second and the CEMF was again equal to the EMF.  Repeat continuously...  But it does not happen in this step wise manner, it is a smooth analog feedback mechanism.

So,as the CEMF is equal to the applied EMF at the instant the voltage is placed across the resistor,then the outcome is what?-->how have you decided that the EMF wins over the CEMF,when both forces are the same(if we are to stick with voltage as being a force),and acting in opposition.
So thats all im asking--how have you decided that the current can flow,when there is no potential difference?.

A CEMF of 4 volts will exist as soon as the rate of change of .8 amps per second is achieved

Just prior to the connection there is no CEMF.  As soon as the EMF is applied current begins to flow and as soon as the rate of change of .8 amps per second is reached, a CEMF of 4 volts is generated.

PW
The rate of change to the value of .8 amps per second is instant--there is no waiting for this rate of change value to reach .8 amps per second--it is there as soon as there is an EMF--a voltage placed across the inductor.
Quote Poynt: The instant 4V is applied to the inductor, the cemf goes to -4V and current begins to flow.
That statement by point makes no sense when you read it.
We have no potential difference,but current begins to flow

This rate of change and CEMF thing, as well as Faraday and Lenz, is all wrapped up in the very definition and quantification of an inductor.  So yes, it does refer to the question asked.

So which one(Lenz or Faraday) placed an ideal voltage across an ideal coil of 5H,to quantify there law stands with ideal coils/inductors.
Why dose every circuit model of a circuit that includes an inductor,have a series resistor associated  with it?.

Perhaps this will help:
Consider two ideal voltage sources connected in parallel.  If both are set to output 4 volts (EMF=CEMF), no current will flow.
How much less than 4 volts would one of the supplies have to be to allow an infinite amount of current to flow?

Ah,ok.
Well the difference would only have to be minute--> i dont have enough time to place that many 0's before the decimal point,and even then,we would be no where near as low as we needed to be,before the current flow would no longer be an infinite amount.
When you start to talk about infinite amount's,there is no !just a bit less!--math simply cannot deal with divisions,subtraction,additions or fractions of infinite--there simply is no such math.
It's like asking how long it will take to walk around a circle,and reach the end,where there is no end to a continuous loop.

If you are familiar with negative feedback mechanisms, the action of an inductor's CEMF as a regulator of the current's rate of change should be fairly easy to grasp.   If not, after all this, I doubt i would have the patience to explain emitter degeneration or the like.  I know you are capable of understanding this, and I am at a loss as to why you do not.

I have no problem with that ,when dealing with real world devices that dissipate energy.
The problem comes when ideals are involved,and where we have a coil that dose not,and cannot dissipate energy.

Please read my posts again where I describe the action of the CEMF in a step-wise fashion.  The part you appear to be having difficulty grasping is the smooth continuous action of the CEMF as a feedback mechanism that maintains the .8 amps per second rate of change.

No
The part i have a problem with,is how you have come to the conclusion that the EMF wins over the CEMF in this smooth transition step function ,when there is no potential difference.
Here is how i am seeing it.
A water pump has a coiled hose attached to it. Both the pump and hose are ideal.
The pump has a pressure switch that cuts the power to the pump when the pressure reaches 8psi in the hose. The hose now has one end blocked off,and so the pump will stop as soon as the pressure in the hose reaches that which the pump will supply. Every time the pump tries to raise the pressure in the hose,the hose pushes back with just as much pressure,and so no water flows.
We can adjust that pressure switch as much as we want,and as fine as we want,but no water will ever flow--equal and opposite action/reaction.

Consider the CEMF to "hover" between the 4 volt value and just below it if you must to help visualize the feedback mechanism.  Once you can grasp it that way, it will be easier to see the smooth continuous action.

Im simply not going to drop a value to just below it's value,so as something fits with what the books say--that's just no me. I do not bend rules to make things work.
With a non ideal coil,i have no problem with the current rising,as the CEMF value cannot be, and is not the same as that of the applied EMF,as we have a current rise that has an exponential curve,and this curve starts at T=0. And this is why i said that the resistance of the coils windings are the reason that the CEMF is a slightly lower value than the applied EMF. Even that !minute !amount will start a chain reaction when winding resistance is involved,and so the current can flow,due to the coils ability to dissipate energy. But that just is not the case when talking ideals--that is why they are ideal.

On the bench,using non ideal motors,i can show that when the BackEMF equals the EMF,no current will flow. I can do this because i can externally control the amount of BackEMF produced by the motor.

Perhaps we could find the answer using real world inductor's

Lets say we take a 1:1 transformer--any size and value.
We pulse each side of that transformer with the same voltage across each winding continuously.
Now,we know that even though the two coils/windings have the same voltage placed across them,the two coils will still allow a current to flow--why?

One last thing.
What polarity is this CEMF as far as to that of the polarity of the supplied voltage.
A silly question you may think,but draw it out,and see what you see.
Poynt was right--it makes no sense.


Brad

Technically speaking, current would not flow if the CEMF and EMF were exactly equal.  But then what would the rate of change of the current be?  Zero volts per second, so no CEMF would be generated.  With no CEMF, current would again flow until it reached .8 amps per second, at which time the CEMF would again equal the EMF, and the cycle would repeat.  But it does not happen in this step-wise fashion, it is a smooth feedback type process where the CEMF is maintained as equal to the level of the applied EMF (or at the least, very, very close to it, if that helps with the visualization) so that the rate of change is .8 amps per second.

No, the R value is not being forgotten, it just does not come into play with regard to the CEMF.  The CEMF is an induced voltage.  The CEMF equals 4 volts when the RATE OF CHANGE of the current flow is .8 amps per second (4V applied to 5H).  Do not confuse the CEMF, which is a mechanism that determines the rate at which energy can be stored, with the actual amount of energy that is stored.  And do not forget that the inductor is storing energy. 

As soon as the RATE OF CHANGE reached .8 amps per second, 4 volts of CEMF would be generated and effectively regulate/limit the rate of the current's rise to that .8 amps per second.

You are wrong.  Faraday says nothing about a required resistance or required dissipation.  Resistance does not enter into it.  If the rate of change of the current flowing thru the 5H inductor is .8 amps per second, the generated CEMF will be 4 volts.  If the rate of change is less than .8 amps per second, the generated CEMF will also be less than 4 volts.  And do not forget that with regard to an ideal inductor, we are not dissipating energy, we are storing/retrieving energy. 

PW

Makes perfect sense.