MH's ideal coil and voltage question

[Magluvin]

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]

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]

@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]

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]

Tinman,



PW

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.

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.

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