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Author Topic: Professor Walter Lewin's Non-conservative Fields Experiment  (Read 253037 times)
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How about a response Ex?  ;)

Before going further, I would like to understand what is your goal. Is it to affirm that, when a constant voltage is abruptly applied to a solenoid, there is an induced emf whose the mean voltage is a not null DC?

The equation is simple:
L*di/dt + R*i(t) = V * h(t)
L is the inductance, R its resistance, V the voltage and h(t) the Heaviside step function h(t)=0 for t<0 and h(t)=1 for t>=0.
Just before t=0, i=0.
Just after:  L*di/dt + R*i(t) = V whose the solution for i is well known:
i = V/R * (1 - e-t/τ) where τ=R/L L/R

Only the variation of i produces emf. If we suppose a perfect coupling with the solenoid: emf = k * di/dt.
di/dt =d/dt (V/R * (1 - e-t/τ)) = V/R * d/dt (- e-t/τ) = V/R * 1/τ * e-t/τ = V/L * e-t/τ

As  e-t/τ is always >0, so is di/dt and emf. Your goal is correct.
I must admit that my affirmation pretending the contrary made an implicit assumption that I should have given: it is true only when the initial and final conditions are the same (for example i=0, or i=anything but the signal is periodic) which was not the case here.

Do you agree?

I just still ask me a final question about the t<0 to t=0 transition, as modelized in the equation. Is it physical? I think it is but I have some doubt.

« Last Edit: 2012-05-11, 16:14:55 by exnihiloest »
   

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It's not as complicated as it may seem...
My goal is to establish what your understanding is (and if it is the same as mine) of what would be observed on an oscilloscope with the leads placed across nodes D and A using a decoupled method. The input condition is a step function (step and hold) from 0 to some value that produces an induced emf of 1Vp in our 100/900 Ohm loop. Please plot the response across A and D and the loop current.

Read the post again for details of what I am asking if you are uncertain.
   
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It's not ill will but I'm a bit confused. I didn't follow all the story and there are now hundreds posts. I would appreciate that you synthesize the question and put links to what is precisely on the stack for you, schematics and so on... otherwise I will be unable to reply because I could not reread the hundreds of posts of this thread for guessing which are of help.
   

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It's not as complicated as it may seem...
Huh?  ???

Just re-read post #575 above (your quote) and my response and question. That should be sufficient for you to draw out the wave forms. Nothing has changed, it's the same experiment we've been discussing for a while now.
   
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Huh?  ???

Just re-read post #575 above (your quote) and my response and question. That should be sufficient for you to draw out the wave forms. Nothing has changed, it's the same experiment we've been discussing for a while now.

Hello Houston - We have a problem.
Which schematics is "V(D-A)" referring to? Is it the first post? post #17? Which "same experiment"? You produced so many problems with variants of measurement. You would like that I follow a step by step procedure but I don't know why...   ???
Communication problem, we are not on the same wave length. Hello, hello, Poynt, I don't hear you, re-tune, clarify, please.

   

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It's not as complicated as it may seem...
The Lewin experiment is not much different than a standard air-core transformer, so all the same theories apply.

Have you forgotten what our argument was about over only the last several postings in this thread?

It applies to transformers too, so if this makes it easier for you, simply plot out the voltage across the terminals of a 1k Ohm terminated air-core secondary when the primary is excited by a step and hold input voltage. 1Vp is induced in the secondary.

You DO remember what our argument was about I trust? I'm asking you to plot this out to see if you have the same understanding as me, because your responses thus far (excluding your recent ones claiming you don't know what I'm referring to) indicate that you don't.

Stop playing silly-bugger Ex, and just answer the question.  C.C
   
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I will reply later if I can. ADSL here is out of order due to cables theft in my area (for re-selling the copper). The repair is announced for next Monday. I'm using now a cellular phone as modem but it's very expensive...

   
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The Lewin experiment is not much different than a standard air-core transformer, so all the same theories apply.

Have you forgotten what our argument was about over only the last several postings in this thread?

It applies to transformers too, so if this makes it easier for you, simply plot out the voltage across the terminals of a 1k Ohm terminated air-core secondary when the primary is excited by a step and hold input voltage. 1Vp is induced in the secondary.

Stop playing silly-bugger Ex, and just answer the question.  C.C

I don't play. I have not to "just answer the question", without knowing why I should answer. It is surprising that you refuse to give the reasons of your question. In any case, see the attached picture for the answer.

I shall not answer any other question that doesn't make sense for me.

Quote from: exnihiloest
ADSL here is out of order...
It should work only this evening. I am exasperated by the delay, so I post from my office.

   

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It's not as complicated as it may seem...
I don't play. I have not to "just answer the question", without knowing why I should answer. It is surprising that you refuse to give the reasons of your question. In any case, see the attached picture for the answer.

I shall not answer any other question that doesn't make sense for me.
It's strange that you're most likely the only one in this thread that did not know what the question was about or why it was being asked, especially since the pertinent discussion was between you and I.  ???

It all started with my statement HERE regarding the loop response as seen by the scope.

Then your refutation 2 posts later, HERE.

Then I elaborate further in an attempt to clarify, HERE. Quoted:
You're splitting hairs here Ex. Reference the Lewin experiment, a DC source is used to pulse the coil. Consequently, there is a corresponding unipolar response on the loop (polarity dependent on meter connection). i.e. the current does not change direction, it rises, then falls back to zero.

Which you then attempt to refute again, HERE. To quote you:
I'm not splitting hairs, I justified what I said. I say that "+" or "-" are not relevant for signals with a null dc component and I explain why it is the case here.
An induced current never rises and fall back to zero. It rises and fall back below zero at some time, otherwise there would be a dc component. If transformers could pass DC, it would be known for a long time!

So now look at your nice graph (which is correct) you just posted. Do you see your error now and that I was correct in what I had said? It should also be clear to you now why I asked you to graph out the voltage and current.
   
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...
Do you see your error now and that I was correct in what I had said?

I had already said it in post #577, with the math!

"As  e-t/τ is always >0, so is di/dt and emf. Your goal is correct [to affirm "an induced emf whose the mean voltage is a not null DC"]. I must admit that my affirmation pretending the contrary made an implicit assumption that I should have given: it is true only when the initial and final conditions are the same (for example i=0, or i=anything but the signal is periodic) which was not the case here."

And I asked you: "Do you agree?". No reply. I understand now that you missed the point of my answer 577 and asked again the same question until you get a picture, while both of us agreed. Waste of time.


   

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It's not as complicated as it may seem...
Ex,

To put it plainly, you said: "An induced current never rises and fall back to zero".

With the step function input (as per the subject matter of this thread from the beginning), the secondary most certainly does.

AND, there ARE two polarities of voltage as measured according to the experimental setup.
   

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I watch this thread with interest and now I see a question pertaining to applying DC to a transformer - and leaving it connected?

Surely it is known that once the current is removed the current will fall below zero and oscillate above an below for a short period? The same goes for primary and secondary.

If you apply a steady DC current to the primary the secondary will a rise in current and then a gradual fall to zero - and not beyond - until the switch is opened.

In any case, the complete cycle of on-off will produce AC. Even the least reduction of primary applied DC, after the applied peak it reached, will cause the secondary to see a reverse in current.

Why the argument?



---------------------------
"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   

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It's not as complicated as it may seem...
In terms of the Lewin experiment, the pertinent interest is what happens from t=0 to the point e-induced falls to zero. Even this latter "falling" part is of limited interest, but showing the complete "half-cycle" makes things look nice and provides a sense of what is going on in terms of the input stimulus and how induction works.

Seems there was simply a misunderstanding of what I said which started this all off.
   
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...
To put it plainly, you said: "An induced current never rises and fall back to zero".

Yes I did. It was in post #557.

But I just told you that in my post #577 I admitted that it was a mistake in the general case, including our particular case of this thread, by stipulating that "an induced current never rises and fall back to zero" is true only when the start and final values of the primary current are the same. It should now be definitely clear.


   
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It is surprising that the time constant depends only on the resistance of the primary, not on the load of the secondary provided that it is much larger. Also surprisingly, the greater the resistance, the lower the time constant.
With a primary circuit resistance of 0.001 Ohm, the time constant is on the order of s. With 1 ohm resistance, it is on the order of ms.
(The input voltage rises to 1v at t=0 and is kept constant after).



   

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It's not as complicated as it may seem...
It is surprising that the time constant depends only on the resistance of the primary, not on the load of the secondary provided that it is much larger. Also surprisingly, the greater the resistance, the lower the time constant.

It's not surprising if you know the equations for the time constant "tau":

For resistive/capacitive, tau=RC

For resistive/inductive, tau=L/R
   
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Tks for the recall, Poynt. I wrongly wrote τ=R/L in reply #577 (without consequence). I have corrected.
So now it's not surprising...    :(

   
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It is surprising that the time constant depends only on the resistance of the primary, not on the load of the secondary provided that it is much larger. Also surprisingly, the greater the resistance, the lower the time constant.
With a primary circuit resistance of 0.001 Ohm, the time constant is on the order of s. With 1 ohm resistance, it is on the order of ms.
(The input voltage rises to 1v at t=0 and is kept constant after).


I found that time constant depends on resistance of secondary, however, only at certain resistance.  The lower the resistance, the lower the time constant. lol
frequency change sim


« Last Edit: 2012-05-13, 23:46:05 by GibbsHelmholtz »
   
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I found that time constant depends on resistance of secondary, however, only at certain resistance.  The lower the resistance, the lower the time constant. lol
...

It depends on the order of magnitude of the resistance of secondary compared to that of the primary, and of the mutual coupling coefficient. If the resistance of secondary is of the same order or lower than the resistance of the primary, the former can't be negligible when viewed from the primary through the coupling, and so the apparent resistance of the primary is lower. The lower the resistance, the higher the time constant, according to the simulation and math.

In the simulation, with a coupling=0.9, τ= around 1ms for Rout=10*Rin, 2ms for Rout=Rin, 6ms for Rout=Rin/10, 10ms for Rout=Rin/100.
With a coupling=0.1, τ= around 1.1ms for Rout=10*Rin, 3ms for Rout=Rin, 8ms for Rout=Rin/10, 13ms for Rout=Rin/100.

   
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It depends on the order of magnitude of the resistance of secondary compared to that of the primary, and of the mutual coupling coefficient. If the resistance of secondary is of the same order or lower than the resistance of the primary, the former can't be negligible when viewed from the primary through the coupling, and so the apparent resistance of the primary is lower. The lower the resistance, the higher the time constant, according to the simulation and math.

In the simulation, with a coupling=0.9, τ= around 1ms for Rout=10*Rin, 2ms for Rout=Rin, 6ms for Rout=Rin/10, 10ms for Rout=Rin/100.
With a coupling=0.1, τ= around 1.1ms for Rout=10*Rin, 3ms for Rout=Rin, 8ms for Rout=Rin/10, 13ms for Rout=Rin/100.



OKay, I see what you mean now.  You looking at the time constant for the secondary.  If we take our view broader, you will see that the time constant for the secondary is the opposite to the time constant for the primary. 

My simulation involves a  bit more than primary and secondary.  It also involves time constant for capacitor and condition at resonance.  I've simulate the increase in frequency in primary circuit by changing secondary resistance.  I believe this is the Thane's effect ... simulated:o
   
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Primary and secondary being coupled, there is only one time constant because coil currents are linked to the variation of a unique magnetic flux. E=-dφ/dt and phi is the same for the two coils.
This time constant depends on both resistances.

   
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Primary and secondary being coupled, there is only one time constant because coil currents are linked to the variation of a unique magnetic flux. E=-dφ/dt and phi is the same for the two coils.
This time constant depends on both resistances.



It's difficult to express my view in this matter.  Let me start with the graph below.  The red curve is primary current, blue curve is secondary current. 

The first case where the blue curve go up with the red is where secondary R<<L secondary.  The second case where the blue curve drop as primary current max is where secondary R>>L secondary.  Notice also the rise time for both cases is significantly different. 

We can say that one is in phase and one is out of phase.  The in phase where R is so high has no time constant. 
   
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I have not such graphs. In mines all is happening as expected. I measure the voltage, not the coils current. What do you measure? Could you give us the schematics?

   
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Here's the sim.

Current is observed from primary and secondary resistance.

Circuit Simulator Applet




   
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The simulator applet gives two lines that evolve in unison. I don't retrieve your previous graph with the inverse slopes of the second view.

   
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