#562 – Electroboom!

[jesuscf]

The funny thing about this whole "energy conservative" stuff is, there has to be a source of energy in the first place, for the circuit to consume it. This source of energy can be either lumped or not. If the physical circuit allows for it, you can lump the energy source into a two-terminal component and put it actually into the circuit. In other circuits, like in Dr. Lewins experiment, you can not do it. You're forced to admit that the energy source is external to the circuit and has no discrete source, and this is where KVL stops working.

The reason is simply because such circuit is not energy conservative. The energy source is not IN the circuit. This breaks the very foundation on which KVL is based.

I see you are a lot more confused that I thought you were.  Let me try to explain, in the simplest possible terms what is going on:

This is how Lewin calculated (correctly by the way) the loop current through the circuit:

\$I=\frac{emf}{R_{1}+R_{2}}\$

Or, rearranging a bit, so that you can clearly see what is going on:

\$I^{2}R_{1}+I^{2}R_{2}=emf\cdot I\$

Can you read what the equation above is telling you?  If you are not happy with the units of the equation, multiply both sides by some arbitrary time to find the energy added or consumed by the circuit during that time.

[thinkfat]

The funny thing about this whole "energy conservative" stuff is, there has to be a source of energy in the first place, for the circuit to consume it. This source of energy can be either lumped or not. If the physical circuit allows for it, you can lump the energy source into a two-terminal component and put it actually into the circuit. In other circuits, like in Dr. Lewins experiment, you can not do it. You're forced to admit that the energy source is external to the circuit and has no discrete source, and this is where KVL stops working.

The reason is simply because such circuit is not energy conservative. The energy source is not IN the circuit. This breaks the very foundation on which KVL is based.

I see you are a lot more confused that I thought you were.  Let me try to explain, in the simplest possible terms what is going on:

This is how Lewin calculated (correctly by the way) the loop current through the circuit:

\$I=\frac{emf}{R_{1}+R_{2}}\$

Or, rearranging a bit, so that you can clearly see what is going on:

\$I^{2}R_{1}+I^{2}R_{2}=emf\cdot I\$

Can you read what the equation above is telling you?  If you are not happy with the units of the equation, multiply both sides by some arbitrary time to find the energy added or consumed by the circuit during that time.

Just tell me where in the circuit of Dr. Lewin you find "emf" and we're good. Note: you're not allowed to change the measurement results.

[ogden]

And it is conservative because the magnetic field outside the solenoid is precisely dick.

Solenoid EM fields are miraculously nonexistant for voltmeter leads running next to the test circuit parts. Right. 

[jesuscf]

The funny thing about this whole "energy conservative" stuff is, there has to be a source of energy in the first place, for the circuit to consume it. This source of energy can be either lumped or not. If the physical circuit allows for it, you can lump the energy source into a two-terminal component and put it actually into the circuit. In other circuits, like in Dr. Lewins experiment, you can not do it. You're forced to admit that the energy source is external to the circuit and has no discrete source, and this is where KVL stops working.

The reason is simply because such circuit is not energy conservative. The energy source is not IN the circuit. This breaks the very foundation on which KVL is based.

I see you are a lot more confused that I thought you were.  Let me try to explain, in the simplest possible terms what is going on:

This is how Lewin calculated (correctly by the way) the loop current through the circuit:

\$I=\frac{emf}{R_{1}+R_{2}}\$

Or, rearranging a bit, so that you can clearly see what is going on:

\$I^{2}R_{1}+I^{2}R_{2}=emf\cdot I\$

Can you read what the equation above is telling you?  If you are not happy with the units of the equation, multiply both sides by some arbitrary time to find the energy added or consumed by the circuit during that time.

Just tell me where in the circuit of Dr. Lewin you find "emf" and we're good. Note: you're not allowed to change the measurement results.

First let me clarify that in the equation above both the emf and the current 'I' are a function of time, so in order to find the correct amount of energy added to the circuit, one must add the instantaneous energy values (or find the integral) over that time.  Nevertheless, a dimensional analysis as I did in the quote above, in my opinion could be very useful to understand what is going on.

In the case of Lewin's circuit the induced emf is uniformly distributed across the elements of circuit: the two pieces of wire, one at the top and and one at the bottom, as well as the two resistors, one to the left and one to the right.  When drawing the equivalent circuit, it is very important to considered the emf contributions of these four elements.  If the dimensions of the resistors are much smaller than the dimensions of the wires, a good approximation would be to considered the emf contribution of the wires only.

[bsfeechannel]

And it is conservative because the magnetic field outside the solenoid is precisely dick.

Solenoid EM fields are miraculously nonexistant for voltmeter leads running next to the test circuit parts. Right. 

KVLers are a lot happier than engineers. They keep laughing when we show that they don't understand electromagnetism.

Watch Lewin's lecture 15 where he probes a solenoid with a calibrated Hall sensor.



So, in Lewin's setup, where he puts the ring midway from the ends of the solenoid, there's no varying magnetic field outside. Therefore there will be no non-conservative electric field that could interfere with the probes connected to the circuit.

Now you can stop laughing and weep together with us for all the unfortunate KVLers out there.

[ogden]

Watch Lewin's lecture 15 where he probes a solenoid with a calibrated Hall sensor.

So, in Lewin's setup, where he puts the ring midway from the ends of the solenoid, there's no varying magnetic field outside. Therefore there will be no non-conservative electric field that could interfere with the probes connected to the circuit.

Seems, you did not get what I was saying. Rephrased sentence: Solenoid EM fields that causes EMF in the test circuit, are miraculously nonexistant for voltmeter leads running next to the test circuit parts.

[edit] Those who are impatient waiting for answer from "Dr.Lewin's science guru", can try to find answer empirically. All what's needed - AC mains transformer, insulated wire and AC voltmeter. Try to measure EMF for 1) single turn - tightly wrapped around core, then 2) big enough loop that would imitate voltmeter leads not receiving any EMF. 3) compare. Sample configuration for supposedly most popular E-E core transformer attached. Note that test wire is going through outer leg of "E". Comments about your test results are welcome. Do not hesitate to make even bigger loops than shown in M2 pic.

[jesuscf]

Watch Lewin's lecture 15 where he probes a solenoid with a calibrated Hall sensor.

So, in Lewin's setup, where he puts the ring midway from the ends of the solenoid, there's no varying magnetic field outside. Therefore there will be no non-conservative electric field that could interfere with the probes connected to the circuit.

Seems, you did not get what I was saying. Rephrased sentence: Solenoid EM fields that causes EMF in the test circuit, are miraculously nonexistant for voltmeter leads running next to the test circuit parts.

[edit] Those who are impatient waiting for answer from "Dr.Lewin's science guru", can try to find answer empirically. All what's needed - AC mains transformer, insulated wire and AC voltmeter. Try to measure EMF for 1) single turn - tightly wrapped around core, then 2) big enough loop that would imitate voltmeter leads not receiving any EMF. 3) compare. Sample configuration for supposedly most popular E-E core transformer attached. Note that test wire is going through outer leg of "E". Comments about your test results are welcome. Do not hesitate to make even bigger loops than shown in M2 pic.

More likely than not, you'll not get an answer from "Dr.Lewin's science guru".  I just got my Fluke 187 working again, so with the help of an small audio transformer, a function generator (sine wave, 1 kHz, 20Vpp), and my Brymen BM869s, I got what is shown in the attached picture.

[thinkfat]

I'm not quite sure what you two are trying to demonstrate there. Why should the size/diameter/area of the loop make any difference, as long as it loops through the core?
How about you make a loop that does not enclose any part of the core and see if it picks up anything?

[ogden]

I just got my Fluke 187 working again, so with the help of an small audio transformer, a function generator (sine wave, 1 kHz, 20Vpp), and my Brymen BM869s, I got what is shown in the attached picture.

Results as expected - close to identical. Surprisingly quick! Respect

Why should the size/diameter/area of the loop make any difference, as long as it loops through the core?

Bingo. Size/diameter/area does not matter in Maxwell-Faraday equation. It also means that two circuits show in attachment are equivalent and will show equal readings on voltmeter. Those who are familiar with Romer's/Lewin's experiment and still believe that it is not "probing error", can take a moment to sink it in. Yes, it comes from Romer's paper, edited using windows paint (lol).

[edit] All this just to frustrate students and fellow scientists with resistive divider which *itself* receives EMF from EM induction. Obviously KVL holds. EMF = I*(R1 + R2), V1 = EMF * R1/(R1+R2), V2 = EMF * R2/(R1+R2)

[thinkfat]

Why should the size/diameter/area of the loop make any difference, as long as it loops through the core?

Bingo. Size/diameter/area does not matter in Maxwell-Faraday equation. It also means that two circuits show in attachment are equivalent and will show equal readings on voltmeter. Those who are familiar with Romer's/Lewin's experiment and still believe that it is not "probing error", can take a moment to sink it in. Yes, it comes from Romer's paper, edited using windows paint (lol).

[edit] All this just to frustrate students and fellow scientists with resistive divider which *itself* receives EMF from EM induction. Obviously KVL holds. EMF = I*(R1 + R2), V1 = EMF * R1/(R1+R2), V2 = EMF * R2/(R1+R2)

Err, yes.. But that was entirely not the point of Lewin's experiment.

[ogden]

Bingo. Size/diameter/area does not matter in Maxwell-Faraday equation. It also means that two circuits show in attachment are equivalent and will show equal readings on voltmeter. Those who are familiar with Romer's/Lewin's experiment and still believe that it is not "probing error", can take a moment to sink it in. Yes, it comes from Romer's paper, edited using windows paint (lol).

[edit] All this just to frustrate students and fellow scientists with resistive divider which *itself* receives EMF from EM induction. Obviously KVL holds. EMF = I*(R1 + R2), V1 = EMF * R1/(R1+R2), V2 = EMF * R2/(R1+R2)

Err, yes.. But that was entirely not the point of Lewin's experiment.

AFAIK point of Lewin's experiment was to prove that voltages in his test circuit are "path-dependent" and "KVL is for birds". Both statements are simply BS. Old man frustrated himself with overcomplicated experiment. [edit] You can't agree to both - my equations *and* Lewin's equations (attached)

[jesuscf]

Bingo. Size/diameter/area does not matter in Maxwell-Faraday equation. It also means that two circuits show in attachment are equivalent and will show equal readings on voltmeter. Those who are familiar with Romer's/Lewin's experiment and still believe that it is not "probing error", can take a moment to sink it in. Yes, it comes from Romer's paper, edited using windows paint (lol).

[edit] All this just to frustrate students and fellow scientists with resistive divider which *itself* receives EMF from EM induction. Obviously KVL holds. EMF = I*(R1 + R2), V1 = EMF * R1/(R1+R2), V2 = EMF * R2/(R1+R2)

Err, yes.. But that was entirely not the point of Lewin's experiment.

AFAIK point of Lewin's experiment was to prove that voltages in his test circuit are "path-dependent" and "KVL is for birds". Both statements are simply BS. Old man frustrated himself with overcomplicated experiment. [edit] You can't agree to both - my equations *and* Lewin's equations (attached)

Here, I fixed Lewin's blackboard to match reality.

[Sredni]

Bingo. Size/diameter/area does not matter in Maxwell-Faraday equation. It also means that two circuits show in attachment are equivalent and will show equal readings on voltmeter. Those who are familiar with Romer's/Lewin's experiment and still believe that it is not "probing error", can take a moment to sink it in. Yes, it comes from Romer's paper, edited using windows paint (lol).

[edit] All this just to frustrate students and fellow scientists with resistive divider which *itself* receives EMF from EM induction. Obviously KVL holds. EMF = I*(R1 + R2), V1 = EMF * R1/(R1+R2), V2 = EMF * R2/(R1+R2)

Err, yes.. But that was entirely not the point of Lewin's experiment.

AFAIK point of Lewin's experiment was to prove that voltages in his test circuit are "path-dependent" and "KVL is for birds". Both statements are simply BS. Old man frustrated himself with overcomplicated experiment. [edit] You can't agree to both - my equations *and* Lewin's equations (attached)

Voltages ARE path-dependent. What you fail to see is that in order to see the path dependency you have to enclose a variable flux region between the paths. In the case of Lewin's ring


https://i.postimg.cc/VvFWycbH/Voltage-can-be-path-dependent.jpg

you are just considering two green paths which give equal voltages and come to the conclusion that all paths always give equal voltages.
When you cannot physically put your wires inside the dB/dt region, you are left with two possible sets of paths (the green and purple ones) that give two possible and different values.
Here, a few more pictures:

https://electronics.stackexchange.com/questions/551244/what-would-a-voltmeter-measure-if-you-had-an-electromotive-force-generated-by-a

[bdunham7]

Voltages ARE path-dependent. What you fail to see is that in order to see the path dependency you have to enclose a variable flux region between the paths. In the case of Lewin's ring

Yes, the voltage generated in the voltmeter test leads depends on the path--i.e. where you put them. 

[Sredni]

Voltages ARE path-dependent. What you fail to see is that in order to see the path dependency you have to enclose a variable flux region between the paths. In the case of Lewin's ring

Yes, the voltage generated in the voltmeter test leads depends on the path--i.e. where you put them. 

Well, I would have drawn that purple, to match the picture below but... what do you think you have accomplished pointing that out?
The circuit below already showed path dependent voltages. That you have drawn is  a 0V path - it is measuring the voltage along the transformer coil. I simply reused a previously linked figure but if you wish you can draw ten more 0V path. What's your point? As I wrote above, in order to see path dependency you need to enclose the dB/dt region inside your paths.

[bsfeechannel]

Those who are familiar with Romer's/Lewin's experiment and still believe that it is not "probing error", can take a moment to sink it in.

OK. Cool. Since you guys are quick to answer and even do experiments. Help me here, please. What will V1 read?



EDIT: Awwww! Now I get it. 

I need time not to see the probing error, but to absorb the impact of the stupidity of KVLers.

Non, mon chouchou! That's not how you check if there's interference with your probing. Let me show you:



I can't believe I have to teach you basic practical engineering.

If you guys had some formal training in the field, you must have found your degree in the dumpster.

[bsfeechannel]

Here, I fixed Lewin's blackboard to match reality.

Here, I corrected your blunder.



We have to thank Lewin and his superdemo. He really made the dilettantes unhappy.

[ogden]

EDIT: Awwww! Now I get it. 

I need time not to see the probing error, but to absorb the impact of the stupidity of KVLers.

Right. You do it by looking incredibly stupid yourself. Voltmeter leads of the circuit below are part of loop receiving EMF or not?

Hint: yes indeed. EM induction does not care - wire is resistor or voltmeter leads. It does cause EMF in *BOTH*. There is no magical "path directivity", just same old good Faraday's law in action Lewin's preachers pretend to know so well.

[ogden]

Voltages ARE path-dependent. What you fail to see is that in order to see the path dependency you have to enclose a variable flux region between the paths. In the case of Lewin's ring

Yes, the voltage generated in the voltmeter test leads depends on the path--i.e. where you put them. 

Well, I would have drawn that purple, to match the picture below but... what do you think you have accomplished pointing that out?
The circuit below already showed path dependent voltages. That you have drawn is  a 0V path - it is measuring the voltage along the transformer coil.

So you are saying - to measure voltage for given path, one shall route voltmeter leads along that path? In case I have only one voltmeter, I need to place voltmeter and it's leads on right side of Lewin's experiment to measure voltage on right side resistor, as soon as I move voltmeter and leads to left side - I suddenly measure voltage on left side resistor? This is not because of electromagnetic induction but "path dependency"?

I wonder - how one can practically demonstrate path-dependency using transformer with let's say, 100 turns of secondary? He shall wind 100 turns of voltmeter leads on transformer core or what? Would be good to get description of experiment so those who are interested, can repeat. Thank you.

[jesuscf]

Voltages ARE path-dependent. What you fail to see is that in order to see the path dependency you have to enclose a variable flux region between the paths. In the case of Lewin's ring

Yes, the voltage generated in the voltmeter test leads depends on the path--i.e. where you put them. 

Well, I would have drawn that purple, to match the picture below but... what do you think you have accomplished pointing that out?
The circuit below already showed path dependent voltages. That you have drawn is  a 0V path - it is measuring the voltage along the transformer coil.

So you are saying - to measure voltage for given path, one shall route voltmeter leads along that path? In case I have only one voltmeter, I need to place voltmeter and it's leads on right side of Lewin's experiment to measure voltage on right side resistor, as soon as I move voltmeter and leads to left side - I suddenly measure voltage on left side resistor? This is not because of electromagnetic induction but "path dependency"?

I wonder - how one can practically demonstrate path-dependency using transformer with let's say, 100 turns of secondary? He shall wind 100 turns of voltmeter leads on transformer core or what? Would be good to get description of experiment so those who are interested, can repeat. Thank you.

You make a very good point here.  Sredni's "path dependency" is less and less noticeable the more turns on the secondary.  I did the experiment with a three turn secondary and posted the results in this forum.  Sredni almost calculated the voltage between nodes A and D, V_AD but suddenly realized that if he did the calculation he would show that KVL works perfectly, and backed up quickly...  Cyriel Mabilde in Youtube did a similar experiment too with a five turn secondary and also demonstrated that KVL works perfectly.  Now I am waiting for some insight from Sredni of what would happen if we replace the resistors in the loop with capacitors, but he is MIA...

[Sredni]

You make a very good point here.  Sredni's "path dependency" is less and less noticeable the more turns on the secondary.  I did the experiment with a three turn secondary and posted the results in this forum.  Sredni almost calculated the voltage between nodes A and D, V_AD but suddenly realized that if he did the calculation he would show that KVL works perfectly, and backed up quickly... 

You clearly have a very fuzzy recollection of those events.
I computed all the values in your silly circuits in less than 15 minutes and got the results right. Including the voltage along a diameter in the case of perfectly circular and concentric geometry.
You still cannot understand that voltage IS path dependent.

Cyriel Mabilde in Youtube did a similar experiment too with a five turn secondary and also demonstrated that KVL works perfectly.  Now I am waiting for some insight from Sredni of what would happen if we replace the resistors in the loop with capacitors, but he is MIA...

Mabilde, from the depth of his garage, is another KVLer who cannot imagine a path dependent quantity. And this is an old movie that is being rerun over and over. The KVLers propose their 'killing' experiments that should make us "Armchair Nobel prize physicists" fly away to another galaxy. Then we post the solutions according to classical ED (it's not 'our' theory, it's plain old classical electrodynamics) and you fade to silence for a while, except coming back with muddy recollections of events.
It happened with the 'two secondaries is series', it happened with the straight partial coil, it happened with the multiturn coil, it happened with that sentence by Belcher (Jesse is still touting it in his boilerplate answer on his channel and he is forced to ban users who do not agree with him to make them 'fly away to another galaxy')...


...and now we are at the ring with two capacitors. As if these capacitors could change something.

So, here, are the results for the following values of capacitors

    C1 = 4.7 uF , C2 = 22 uF
    freq = 50 Hz
    emf = 374 mV

I get - from simulation and without even invoking MEAS, just by eyeballing the plots

    VcapL = 308.5 mV,           VcapH = 65.5 mV

Guess what I measure with a true RMS multimeter?

    VcapL = 308 mV,            VcapH = 66 mV

And nothing, ok almost nothing, in the copper joining the caps.

So, what are the revolutionary results that you said were bad news for 'team Lewin'?

[Sredni]

So you are saying - to measure voltage for given path, one shall route voltmeter leads along that path?

I am saying that when I measure a voltage I want to make sure there is not a variable magnetic flux region inside my measurement loop.

When I measure from the outside of the ring, since the dB/dt is well inside the ring, there is no dB/dt region inside my measurement loop, so if we look at it in 2D in the area enclosed by the loop voltage is not path-dependent and the value along the branch I test is ALSO the value across the branch I test which is ALSO the value across the probes and voltmeter, which is ALSO the value shown by the voltmeter.

[ogden]

When I measure from the outside of the ring, since the dB/dt is well inside the ring, there is no dB/dt region inside my measurement loop, so if we look at it in 2D in the area enclosed by the loop voltage is not path-dependent and the value along the branch I test is ALSO the value across the branch I test which is ALSO the value across the probes and voltmeter, which is ALSO the value shown by the voltmeter.

I wonder - how dB/dt discerns between wires of resistor and voltmeter leads? You think that EM induction miraculously stops at first wire it encounters, acts only on circuit but not on voltmeter leads? Then please explain how multi-turn transformers work? Also you did not answer my 2nd question: how one can practically demonstrate path-dependency using transformer with let's say, 100 turns of secondary.

[Sredni]

When I measure from the outside of the ring, since the dB/dt is well inside the ring, there is no dB/dt region inside my measurement loop, so if we look at it in 2D in the area enclosed by the loop voltage is not path-dependent and the value along the branch I test is ALSO the value across the branch I test which is ALSO the value across the probes and voltmeter, which is ALSO the value shown by the voltmeter.

I wonder - how dB/dt discerns between wires of resistor and voltmeter leads? You think that EMF miraculously stops at first wire it encounters, acts only on circuit but not on voltmeter leads? Then please explain how multi-turn transformers work?

It doesn't.
Maybe one day you will learn about current dividers.
Try to compute the current that will flow in your 100meg (or even 10meg) internal voltmeter resistance when you shunt your voltmeter with the branch you are measuring (900 ohm or 100 ohm in the case of Lewin's ring).

The measurement loop that does not include the dB/dt region will give you the correct voltage (EDIT: with a nanovolt order-of-magnitude load effect).
The measurement loop that include the dB/dt region will give you a measurement that is the voltage of the distance branch plus (or minus, depending on orientation) the linked EMF.

Also you did not answer my 2nd question: how one can practically demonstrate path-dependency using transformer with let's say, 100 turns of secondary.

Another killer question, I suppose.
I did it for a generic autotransformer. You probably have to look in the last 30 or so pages to find it.
When you measure from the outside you can look at the measure in two ways:

the voltmeter shows the actual voltage in the gap between taps
the voltmeter shows the voltage along the portion of filament between the taps (which is the ohmic drop that is nearly zero volts) plus (or minus depedending on orientation) the linked emf (emf of one turn times integer number of turns linked).

[jesuscf]

You clearly have a very fuzzy recollection of those events.
I computed all the values in your silly circuits in less than 15 minutes and got the results right. Including the voltage along a diameter in the case of perfectly circular and concentric geometry.
You still cannot understand that voltage IS path dependent.

The voltage the voltmeter shows depends on the induced emf in the probes of the multimeter.  Is that what you call path dependent?   The voltage at the tips of the voltmeter probes, the ones connected to nodes A and D, V_AD, is not path dependent because the ring circuit is neither changing nor moving!  You don't seem to understand that V_{voltmeter_screen} != V_AD. Unless you get rid of the induced voltage in the voltmeter probes or account for it somehow (using KVL of course) you'll get the wrong value for V_AD.  Nevertheless, that should not prevent you for calculating the voltage V_AD correctly

Mabilde, from the depth of his garage, is another KVLer who cannot imagine a path dependent quantity. And this is an old movie that is being rerun over and over. The KVLers propose their 'killing' experiments that should make us "Armchair Nobel prize physicists" fly away to another galaxy. Then we post the solutions according to classical ED (it's not 'our' theory, it's plain old classical electrodynamics) and you fade to silence for a while, except coming back with muddy recollections of events.
It happened with the 'two secondaries is series', it happened with the straight partial coil, it happened with the multiturn coil, it happened with that sentence by Belcher (Jesse is still touting it in his boilerplate answer on his channel and he is forced to ban users who do not agree with him to make them 'fly away to another galaxy')...

All of these are trivial experiments, that show that KVL works perfectly.  Also, what is wrong about doing experiments in your garage?  Both Mabilde, I, and probably you too, have equipment at home to do this experiments that is several orders of magnitude better than all the equipment the big wigs of physics had in the 19th century when doing similar tests.

...and now we are at the ring with two capacitors. As if these capacitors could change something.

So, here, are the results for the following values of capacitors

    C1 = 4.7 uF , C2 = 22 uF
    freq = 50 Hz
    emf = 374 mV

I get - from simulation and without even invoking MEAS, just by eyeballing the plots

    VcapL = 308.5 mV,           VcapH = 65.5 mV

Guess what I measure with a true RMS multimeter?

    VcapL = 308 mV,            VcapH = 66 mV

And nothing, ok almost nothing, in the copper joining the caps.

So, what are the revolutionary results that you said were bad news for 'team Lewin'?

Everything is bad news for team Lewin, because team Lewin is convinced that the induced voltage in an arbitrary length of wire is zero.  Can you show us your calculations or are you using a circuit simulator?  Did you measure V_AD?  Any pictures of your setup?

Do you think there is an induced voltage in the capacitors via Faraday's law?