MH's ideal coil and voltage question

[poynt99]

As i said-at what point were actual real world measurements substituted  for math that just followed the trend of those actual measurement taken?.

If we look at what nature shows us,we can go from a liquid to a solid when we go from a value of 1 down to 0. So we know drastic changes are indeed possible in one small change-one value less,and we went from a liquid to a solid.

I am remaining undecided until such time that some one can give a solid reason as to how current can flow from a source to the coil,when that coil has the same voltage across it-->apparently


Brad.

So the 1 Ohm current trace is ok with you, you have no problem with that. But you can't (or won't) see the progression to 1p Ohm and the associated trace and make a mental connection?

Nor will you conclude or acknowledge from that data that a 158 thousand year time constant in a 3 second time frame (the first 3s) can be considered an ideal straight line for the current trace?

Is there anyone else here that also can't or won't make the connection? Are there others that remain "undecided" and or unconvinced?

Anyone else feel that the simulation is only correct for the 1 Ohm trace, but not the 1p Ohm trace? Anyone care to give an explanation why one is correct, while the other is not? I am curious to see the logic here, because this doe not make any sense at all. Are we cherry picking the results of the simulation so that it is only correct when it is in alignment with our own preconceptions (or perhaps misconceptions)? Is that good science? I think not.

I am using the simulation program within the limits of its capabilities, so I fail to see why the results are being scrutinized, especially when one result is considered ok, while the other (tweak to R value) is not.

Please enlighten me, as this is becoming discouraging.

[poynt99]

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

PW has explained in detail how/why the current is limited and that the cemf=Vin.

It is a step-wise process that does not happen in steps; it is an instantaneous process. It can be understood however by breaking it into steps as he has posted a few times already.

A lot of time and effort by some here is essentially wasted when the offered explanations are not even challenged on a technical level. I did not see any attempt to analyse and/or debate from a technical perspective PW's descriptions of the process. And the same has been done to a number of my own posts as well. It seems that the explanations are rejected or dismissed out of hand and the same question repeated, indicating to me that either the explanation was not understood, or not agreed to. That is fine, but out of respect, if it is not understood, ask for clarification, if it is not agreed to, explain in technical terms, why.

In summary, it is a self-regulating process facilitated by negative feedback caused by self-induction and Lenz's law.

How does a governor work an a small engine? Similar in concept, but the process with the inductor begins as soon as current flows. The regulation is achieved via the self-induction process and the fact that more current produces more cemf, so it regulates itself to a fixed rate of A/s.

The inductor "produces" a cemf equal to Vin because all the input voltage is always fully across the inductor; there is no resistor to trade off the voltage with. From the simulations, even the 1 Ohm case clearly shows that all the input voltage is across the inductance for the fist part of the trace, and that has not been acknowledged by you either, and this is KEY. It crushes the argument that current can not flow if the cemf equals Vin.

[tinman]

No, it has absolutely nothing to do with the way Australians speak English and everything to do with your own personal problems with language.  Don't play straw man.  There is never going to be a build-off of anything, so fantasize away.

I have no problem with the way i speak MH,nor dose anyone else.
It seems that you are the only one with the problem here--another failed attempt at misdirection from the question.

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

Brad

[tinman]

In summary, it is a self-regulating process facilitated by negative feedback caused by self-induction and Lenz's law.

PW has explained in detail how/why the current is limited and that the cemf=Vin.

I was asking MH for his answer.

It is a step-wise process that does not happen in steps; it is an instantaneous process. It can be understood however by breaking it into steps as he has posted a few times already.

A lot of time and effort by some here is essentially wasted when the offered explanations are not even challenged on a technical level. I did not see any attempt to analyse and/or debate from a technical perspective PW's descriptions of the process. And the same has been done to a number of my own posts as well. It seems that the explanations are rejected or dismissed out of hand and the same question repeated, indicating to me that either the explanation was not understood, or not agreed to. That is fine, but out of respect, if it is not understood, ask for clarification, if it is not agreed to, explain in technical terms, why.

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.

How does a governor work an a small engine? Similar in concept, but the process with the inductor begins as soon as current flows. The regulation is achieved via the self-induction process and the fact that more current produces more cemf, so it regulates itself to a fixed rate of A/s.

A good point.
A governor works by regulating the RPM at which an engine will run. It will not allow the RPM to double every second,unlike the current value dose !apparently! in our ideal coil.
If the governor was set to kick in at the point at which the engine crank started revolving,then the engine would not revolve at all. If there was a non ideal governor,and there was some loss in that governing system,then the engine crank will rotate.
So i will ask once again--where is the loss in the negative feed back system?.

The inductor "produces" a cemf equal to Vin because all the input voltage is always fully across the inductor; there is no resistor to trade off the voltage with. From the simulations, even the 1 Ohm case clearly shows that all the input voltage is across the inductance for the fist part of the trace, and that has not been acknowledged by you either, and this is KEY. It crushes the argument that current can not flow if the cemf equals Vin.

And so,the question remains--where is the loss?
Also,are you saying that there is no voltage drop across the series resistor at T=0/s ?
If this is the case,then it should not matter as to what value resistor we use as our series resistor,in order to see the CEMF value--we should see no voltage drop across the resistor at T=0/s-->time of connection--correct?.

Brad

[poynt99]

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?

Asking this question indicates that you do not understand the explanation.

There is no loss; cemf=Vin at all times with the ideal inductor.

A governor works by regulating the RPM at which an engine will run. It will not allow the RPM to double every second,unlike the current value dose !apparently! in our ideal coil.
If the governor was set to kick in at the point at which the engine crank started revolving,then the engine would not revolve at all. If there was a non ideal governor,and there was some loss in that governing system,then the engine crank will rotate.

I see RPM as the voltage and torque as the current. The governor I would be referring to (more or less equal to the inductor current limiting process) would limit torque not voltage. So it would be like an centrifugal clutch that could adjust itself to produce a constant rise in torque.

Also,are you saying that there is no voltage drop across the series resistor at T=0/s ?
If this is the case,then it should not matter as to what value resistor we use as our series resistor,in order to see the CEMF value--we should see no voltage drop across the resistor at T=0/s-->time of connection--correct?.

Brad

For a non-ideal inductor, that is correct. At the instant Vin is connected, unless R is large relative to the L value, the resistor will have very little voltage across it compared to the inductance. So for normal non-ideal inductors, this holds true. You can confirm this on your bench to a certain degree. Obviously real inductors already have a series resistance, but you will have to introduce another so you can measure the voltages across each when the connection is made.