@ Bill

I believe this may be the one you are referring to?
From 17:00

https://www.youtube.com/watch?v=imlqEKrfS-k

If so,then yes. Ohms law still hold,as the resistance of the light bulb can be calculated as the heat in the bulb increases.
Not sure why he says that ohm's law dose not hold?.


Brad

Part 1

Part 2

Lewin fails to see or explain the experiment for what it is, and wrongly claims that Kirchhoff's law does not hold, when in actual fact it does. He just does not show how. His lecture is very biased towards Faraday and is quite misleading in my opinion.

So i take it that you know what is happening in that experiment?.

Well i do not,but i would like to know,and i would like to nut it out for myself

Looking at the circuit below--is this correct to that used in the video?.
If so,i do not see how when a voltage potential of 1 volt is placed across the inductor,that the instantaneous voltage across both resistors is also not 1 volt. In fact,if there is enough supply current to maintain that 1 volt across the inductor,then the two resistors (100 & 900 ohm) should always have 1 volt across them. How he managed to get a negative value across the 100 ohm resister has me a little lost ATM . Even his frozen scope shot show's that he dose indeed have a negative value at the 100 ohm resistor ,and a positive value across the 900 ohm resistor.

UMmm

Brad

So i take it that you know what is happening in that experiment?.

I believe I know most of what is going on there, yes.

Well i do not,but i would like to know,and i would like to nut it out for myself

Looking at the circuit below--is this correct to that used in the video?.

Not quite.

First, make it an air-core coil.

Second, the coil is the "primary". The circuit including the two resistors and wire is the "secondary". So there is no electrical connection between the two. Yes, essentially a transformer with a 1-turn secondary.

This is in fact the experiment I encouraged you to perform a number of months ago when we were "discussing" the electric field.

So i take it that you know what is happening in that experiment?.

Well i do not,but i would like to know,and i would like to nut it out for myself

Looking at the circuit below--is this correct to that used in the video?.
If so,i do not see how when a voltage potential of 1 volt is placed across the inductor,that the instantaneous voltage across both resistors is also not 1 volt. In fact,if there is enough supply current to maintain that 1 volt across the inductor,then the two resistors (100 & 900 ohm) should always have 1 volt across them. How he managed to get a negative value across the 100 ohm resister has me a little lost ATM . Even his frozen scope shot show's that he dose indeed have a negative value at the 100 ohm resistor ,and a positive value across the 900 ohm resistor.

UMmm

Brad

Hi Tinman, replace the resistors with diodes, add caps to the diodes , room for 4, the coil will never know it is being tapped.
artv

Yes!!  Exactly.  At 24 minutes in he shows the break down of Ohm's law and why it does not hold.

Thank you for finding this, it would have taken me a long time to do so.  I watched this about 6 years ago or so and MIT has taken the vids down...some of my saved links are to youtube but most were to MIT.  Here he shows the current going up on a curve relative to the voltage and then tapering off and never reaching the level Ohm's law says it should be.

Like I said, I would have thought this was Nobel Prize stuff and would have made many headlines.  This man obviously knows much more about this than I ever will so I will not argue if he is right or not.

Thanks again for digging this one up Brad.  I would be very interested in what .99, TK or MH has to say about Dr. Lewin's conclusions here.

Bill

There is nothing Nobel about this stuff Bill. It's just Lewin making trying to make a big deal out of nothing. There is no "breakdown" as he calls it, just basic physics.

I looked at the current trace for the light bulb being lit at about 25:00 in the clip.  If you ask me, Lewin is getting whackadoo when he says, "a breakdown in Ohm's Law."  All that he is demonstrating is a resistance where the value is a function of time.  Or you could say that it is a resistance where the value is a function of the temperature of the element.

There are some issues relating to deductive logic when you look at things like this.  I remember once reading something where the person talked about scientists "discovering" the value of Pi.  When you look at it logically, there is nothing to discover.  For Ohm's Law, all that it states is that for a two-terminal device you will be able to measure the voltage across the device and the current though the device.  So how can you get a "breakdown" in Ohm's Law?  It's nothing more than an observation of voltage and current for a device under test and defining the term "resistance" as the voltage divided by the current.  Presuming that the voltage and current are always measurable, then there can be no such thing as a breakdown in Ohm's Law.

That one is a lot more complicated than just Ohm's law.

Lewin left out a few important details on that one, resulting in misleading information.

Bottom line is Kirchhoff always holds, regardless of what your measuring tool is telling you.

Wattsup:

I think that many people around here would agree with me when I say that the concept of voltage, what it means, and how you interpret it and work with it is mostly an enigma for you.  So until you have mastered the concept of voltage, you are at a disadvantage when trying to do bench experiments.  If you are going to beat me up for saying that because it isn't politically correct, go ahead.

I will take a stab at your first questions in general terms.  I believe that in the US and Canada, the third-prong ground line is connected to the neutral line at the breaker panel.  Then that ground-neutral connection at the breaker panel is connected to a solid and nearby earth ground.  The big step-down transformer up on a telephone pole also connects it's center tap to a solid earth ground.  Please anybody correct me if I am wrong.

You did not mention if the scope is isolated or not, why is that?

If the scope is grounded (not isolated) then the two bare probes will pick up the hot and neutral potentials.  Naturally this assumes that no device plugged into the power line is misbehaving and there is no current flowing in the ground line.

That begs some questions for you to ponder:  What if between you and the breaker panel there is a coffee maker plugged into the line and you turn it on?    What if there is a coffee maker further down the line and you turn it on?   What will you see on the scope display for the two channels when the coffee maker is ON?

If the scope is isolated (not grounded) then the two bare probes will be floating and yet in all likelihood you will still see some waveforms on the scope display.  Before even discussing the waveforms you see, there is a fundamental problem with the test in this case, what is it?  Moving on, why will you see those waveforms, and what will they be?  How do you explain those waveforms?  What can you say about the impedance of the waveforms that you are looking at?  What about putting the coffee maker between you and the breaker panel and putting the coffee maker further down the line?

Keep in mind that I never did a lot of the "crazy tests" you see discussed around here.  Even though way back when I had a lot of bench experience, I never had any logical reason to do them.

MileHigh