MH's ideal coil and voltage question

[tinman]

Somewhere buried back in the noise I stated that the best model of an inductor was a conductor in series with a variable Vsource (all conductors being an inductor, and using a conductor removes the redundancy of having an "inductor" inside an "inductor model").

I believe there has been confusion caused by the discussion of CEMF polarity, which was debated for some time using parallel connected Vsources and the like.

   



PW

Regarding your two drawings, I believe we were debating the correct polarity of the CEMF at the time, and my answer regarding that was and is that the topmost drawing is correct with regard to the CEMF polarity when di=.8A/s.  As well, your drawings do not indicate the CEMF Vsource, it was assumed to be internal to the inductor you drew and that the +/- indicators for CEMF were just indicated measurement points with regard to the polarity debate.

Yes-correct on both accounts.
So dose that not resemble Partzman's drawing on the left?

The model I have been discussing all along is Partzman's "inductor in series with a Vsource" model, his right most model.  If he were to make the change to his notation "C)" as I indicated, it would describe exactly what I have been trying to portray.  Faraday, Lenz, the math, and the negative feedback mechanism, all work as discussed given that one change to his drawing notes.

This discussion is about whether or not that is correct,and so what would be the point in just changing "C") to the value you believe is correct. If we did that,then why would we be having this discussion?. If we are simply to believe what some one else is telling us,then we might as well have no discussions at all,and just be one of the sheep in the flock,following the leader.

The whole idea of this forum,is to question everything,and believe nothing until shown absolute proof.

An example of this is the belief that current lag's voltage with an inductor,when today i find-when you look close enough,that there is an initial current draw that rises at the same time as the voltage. As Mags and myself think,we believe this is the capacitance of the windings being charged.

One would think that the same would apply for a capacitor,where it is thought that current leads voltage. But what dose it take before  current can flow?-->is not an electric field needed between the plates,before current will flow?. If so,would that not mean that a voltage must first appear across the plates,before current could flow?.

Anyway-food for thought.

Brad

[partzman]

Regarding your two drawings, I believe we were debating the correct polarity of the CEMF at the time, and my answer regarding that was and is that the topmost drawing is correct with regard to the CEMF polarity when di=.8A/s.  As well, your drawings do not indicate the CEMF Vsource, it was assumed to be internal to the inductor you drew and that the +/- indicators for CEMF were just indicated measurement points with regard to the polarity debate. 

Somewhere buried back in the noise I stated that the best model of an inductor was a conductor in series with a variable Vsource (all conductors being an inductor, and using a conductor removes the redundancy of having an "inductor" inside an "inductor model").

I believe there has been confusion caused by the discussion of CEMF polarity, which was debated for some time using parallel connected Vsources and the like.

The model I have been discussing all along is Partzman's "inductor in series with a Vsource" model, his right most model.  If he were to make the change to his notation "C)" as I indicated, it would describe exactly what I have been trying to portray.  Faraday, Lenz, the math, and the negative feedback mechanism, all work as discussed given that one change to his drawing notes.   



PW

I have family here for the July 4 celebration so it is difficult to find time to respond at the moment.

OK, below is the corrected model as you have requested.  Do you see a problem? When Emf = Cemf and di = .8A/s, how is di supplied and maintained?  This model is flawed as well IMO.

I have a new model to present time permitting but for now just some food for thought.  Since Emf = L*di/dt and -Cemf = L*di/dt, we would be mathematically correct to say Emf = -Cemf. Note that this is |Emf| = |Cemf| not Emf = Cemf as you have been stating. Big difference!

pm

[picowatt]

I have family here for the July 4 celebration so it is difficult to find time to respond at the moment.

OK, below is the corrected model as you have requested.  Do you see a problem? When Emf = Cemf and di = .8A/s, how is di supplied and maintained?

No, I do not see a problem.  I see an inherent negative feedback mechanism and the reason di is limited to and maintained at precisely .8A/s.

Consider what happens upon application of the 4 volt EMF.  Current begins to flow thru the inductor and...

1.  As di reaches a rate of change equal to .8A/s, the CEMF becomes equal to 4 volts.

2.  As per "B)" in the drawing, as the CEMF becomes equal to the EMF, di also reduces towards zero.

3.  But if di becomes less than .8A/s, the CEMF also becomes less than 4 volts which allows di to increase. 

4.  Return to number 1 above (loop forever)

Although described above in a rather step-wise fashion, it is, in reality, a smooth and continuous process and a classic example of a negative feedback loop.

Thanks for taking the time to post the drawings.  By all means, enjoy the 4th!

PW

Added:

Also note that I believe it more appropriate for L2 to be replaced with a simple conductor (wire).  Having an inductor inside a "model" of an inductor is redundant and could lead to confusion.

I would replace L2 with a wire and then place a dotted line box surrounding both the CEMF Vsource and the wire with di indicated.  The entire dotted line box would then be labeled as L2.

(I would also label the value for CEMF1 as X_v or X_v1 because the only time the CEMF=4volts is when di=.8A/s)

[poynt99]

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.

Huh? The rate of change, i.e. A/s is directly proportional to the applied Vin. The higher Vin, the higher is the A/s. If Vin remains constant, so does the current A/s.

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

To correctly measure the voltages in your circuit, you need to make two separate measurements as noted on the annotation on your diagram. The resistor measurement requires a CH1-CH2 MATH function, while the inductor voltage is simply CH2. Most, if not all the voltage at the instant Vin is applied, should appear across the inductor. What is the value of your R and L?

[minnie]

  It's a job for me to believe this keeps going on and on.
  It seems that scientists know the magnetic moment of an electron
  to be -9284.764 X 10 to the minus 27 with an accuracy of
  7.6 parts in 10 to the minus 13.
  I presume people have been looking at these things for a century
  and a half so in my mind there's just one right answer.
       John.