Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 70383 times)

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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #950 on: April 14, 2019, 02:55:41 am »
Have you seen this video about KVL/Faraday's law and non-conservative field?

https://youtu.be/pUsdiIl1Kyg

Unfortunately this guy's "intuition" is taking him to the wrong conclusions.

Although he started with the right assumptions, he was giving various hints that sooner or later along the video he would utter some serious bullshit. And he does that exactly @9:36 when he writes elp - iR = 0.

And he concludes, very pseudo-scientifically that "This is a mix, actually, of Kirchhoff and Faraday law, because you don't see this [elp] as a discrete component here in the circuit but it is hovering around and inducing this voltage in the loop".

Note how the people who like to assert that Kirchhoff law holds for non-conservative fields give each one a different explanation for the incongruence between their claims and the results that the phenomenon is proving right in front of their noses.

He said  what he said that because he knows that the only law capable of explaining the phenomenon of induction is Faraday's law. And on top of that, there's no component on the circuit to justify Kirchhoff's EMF. But he can't accept that Kirchhoff doesn't hold for non-conservative fields.

So let's correct his mistakes.

1) elp - iR = 0 violates Kirchhoff's law, because Kirchhoff is adamant about stating that the EMF (elp) has to happen along the path of the circuit at a different place where the voltage drops occur. The EMF cannot happen inside the wire, because the wire (if considered ideal) won't allow the existence of an electric field inside it (its resistance is zero). So it happens inside the resistor, which does allow the existence of an electric field inside it.

Following Kirchhoff's instructions on how to verify his law, we have:

i*0 + i*R = elp != 0. 

¡Hasta la vista, Kirchhoff!

Now let's jump to @21:30

2) The voltage on an impedance is ONE no matter how you measure it.

Impedances are measured where there's no varying magnetic flux, so the way you measure it counts.

3) In non conservative circuits both Kirchhoff and Faraday Laws come into effect and need to be included in the analysis.

Nope. In this case Kirchhoff fails and Faraday holds. Kirchhoff is just a special case of Faraday's law when the varying magnetic flux is zero.

4) When calculating and/or measuring voltages in a non conservative circuits, care should be taken not to include non relevant magnetic flux changes.

All magnetic flux changes that happen in the area bounded by the circuit count. All the magnetic flux changes that happen outside the area bounded by the circuit don't count. Faraday's law. As simple as that.

The four equations that explain the behavior of any electromagnetic phenomenon are Maxwell's (which include Faraday's law). They have been proven time and again for more than a 100 years. So there's no place for "intuition" here, i.e., "alternative" explanations.

Everyone who tries to get around them ends up being cursed.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #951 on: April 14, 2019, 08:13:35 am »
1) elp - iR = 0 violates Kirchhoff's law, because Kirchhoff is adamant about stating that the EMF (elp) has to happen along the path of the circuit at a different place where the voltage drops occur.

Here we go again    :palm:

All the wire of Lewin's circuit is EMF source, thus "wiring" is zero length point where EMF-generating wire is connected to resistor. Resistors are infinismall as well - because their dimensions are *not* specified. You agree that there is actual voltage drop on the resistor (IR) that is equal to EMF voltage created in the loop (elp), yet you say that elp - iR = 0 is bullshit. Come one... Before you decide to participate in the discussion about electromagnetic induction, make sure to learn fundamental laws of physics such as law of conservation of energy.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #952 on: April 14, 2019, 09:50:34 am »
Here we go again    :palm:

Yep. While there will be people spreading pseudo-scientific claims, it is our duty to dispel them. This is not a forum of flat-earthers.

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All the wire of Lewin's circuit is EMF source, thus "wiring" is zero length point where EMF-generating wire is connected to resistor. Resistors are infinismall as well - because their dimensions are *not* specified.

The wires in Lewin's circuit do have a length and so do the resistors. There's nothing "infinitesimal" there.

What amazes me about the people who claim that Kirchhoff holds for varying magnetic fields, and that Lewin's circuit is wrong, is the creativity of the wording of their explanations. Every day is a different explanation. Each one has their own. None of them makes sense, none of them agrees with each other, but they are entertaining, nevertheless.

Keep them coming.

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You agree that there is actual voltage drop on the resistor (IR) that is equal to EMF voltage created in the loop (elp), yet you say that elp - iR = 0 is bullshit.

Yes. Surely. The EMF is NOT equal to the voltage drop. The EMF IS the voltage drop. There is just one voltage there. That's why the sum of the voltages in the circuit is different from zero and Kirchhoff doesn't hold.

Now it's up to you to discover how the conservation of energy holds for such a circuit (hint: Maxwell's equations).

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Come one... Before you decide to participate in the discussion about electromagnetic induction, make sure to learn fundamental laws of physics such as law of conservation of energy.

I said before, I always expect you to teach me fundamental physics. It'll be fun.
« Last Edit: April 16, 2019, 05:13:53 pm by bsfeechannel »
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #953 on: April 14, 2019, 11:41:50 am »
The wires in Lewin's circuit do have a length and so do the resistors. There's nothing "infinitesimal" /there.

Are you serious or just trolling? I repeat: all the wire of the contraption is secondary winding, there's no "interconnection wire" that would not generate EMF. So we assume that it's (connection wire) length is zero. It is there only in the circuit *schematics*, not in real world. Dr.Lewin ignores EMF voltage generated *in* the resistors, so he ignores their dimensions, so they are assumed as zero. Is it so hard to comprehend?
 
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Now it's up to you to discover how the conservation of energy holds for such a circuit (hint: Maxwell's equations).

Could you explain please? For simplicity there is single 1Ohm resistor, EMF is 1V, 1A is flowing for 1 sec. Please consider equation in form: power_generated - power_dissipated_in_the_resistor = 0.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #954 on: April 16, 2019, 04:18:54 am »
Are you serious or just trolling?

I might be brutally honest, reveal shocking truths and bust your myths. But you know that I never troll.

Trolls want to see the world burn. They try to destroy the quality of the discussions and hence the value of the forum. I do exactly the opposite. I voluntarily spend my private time to raise the bar of the debate. Trolls don't do that.

Learn the difference, please.

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I repeat: all the wire of the contraption is secondary winding, there's no "interconnection wire" that would not generate EMF.

No static wire in relation to a frame of reference can generate EMF, but I'll interpret your statement as the wire of Lewin's experiment encloses an area with a varying magnetic field and that we don't have terminals in the vicinity of which we don't have this field.

This is precisely the reason why it is impossible to lump-model this circuit.

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So we assume that it's (connection wire) length is zero. It is there only in the circuit *schematics*, not in real world.

The wire represented in the schematic is real wire. He even shows that @44:32 of his lecture 16:

I have a circuit that's exactly what you have here [points at the circuit drawn on the blackboard]: nine hundred ohms and a conducting copper wire here and a hundred ohms here, and here is the solenoid [...] [points at the components of the "contraption" on the table].[/i]

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Dr.Lewin ignores EMF voltage generated *in* the resistors, so he ignores their dimensions, so they are assumed as zero.

Cyriel Mabilde trying to "debunk" Lewin says the following in his video "Walter Lewin: electromagnetic induction=not for the birds(lecture 16)" @3:55:

[Lewin] draws an easy conclusion that an EMF cannot exist in the wires, but only, and only, in the resistors. I quote his own words: "The induced EMF is exclusively in the resistors"[...]

So Lewin does not ignore "the EMF generated *in* the resistors". Quite the contrary.

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Is it so hard to comprehend?

Sometimes it is difficult to decipher the pseudo-scientific talk because it is always contradictory in nature.

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Could you explain please? For simplicity there is single 1Ohm resistor, EMF is 1V, 1A is flowing for 1 sec. Please consider equation in form: power_generated - power_dissipated_in_the_resistor = 0.

Before we can address this problem, let's suppose that we live on a planet where g = 10 m/s². We lift an object of 1kg from the ground up to 1m. The potential energy transferred to this object will be U = mgh = 1kg*10m/s²*1m = 10 joules. If we drop down this object, when it hits the ground the potential energy will be converted to kinetic energy, which will be exactly 10 joules. Energy is conserved.

Now let's suppose that we live on a planet where g can vary. We lift the same object to a height of 1m with g = 10m/s² as before. But as soon as we drop the object, g suddenly becomes 20m/s². When the object hits the ground the kinetic energy will be 20 joules.

Where did this excess energy come from?
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #955 on: April 16, 2019, 05:31:40 am »
2) The voltage on an impedance is ONE no matter how you measure it.

Impedances are measured where there's no varying magnetic flux, so the way you measure it counts.

That's a new one in this thread that i don't agree in.

Why would the impedance of a passive component change in the presence of a magnetic field? The component is still passing current in the same dependence to voltage as before it was placed in a field, since this very dependence is called impedance means the impedance has not changed. If you measured it being different in a changing field then your measured quantities are likely affected by the field (Since impedance is measured indirectly from other quantities) rather than the impedance being affected.

That is unless you make your resistor out of a material that shows significant magnetoresistive effects. But as far as i know pure copper or carbon based resistors don't have any significant magnetoresistive properties. Even then they would react to DC fields too, not only AC fields.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #956 on: April 16, 2019, 05:36:07 am »
Quote
Could you explain please? For simplicity there is single 1Ohm resistor, EMF is 1V, 1A is flowing for 1 sec. Please consider equation in form: power_generated - power_dissipated_in_the_resistor = 0.

Before we can address this problem, let's suppose that we live on a planet

Yes, we live on a planet. It is a fact. Are you going to explain conservation of energy using Maxwell's equations or not? Your usual goalpost shifting tactics is not funny anymore, you shall stop it.
« Last Edit: April 16, 2019, 05:39:27 am by ogden »
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #957 on: April 16, 2019, 05:03:49 pm »

    2) The voltage on an impedance is ONE no matter how you measure it.

    Impedances are measured where there's no varying magnetic flux, so the way you measure it counts.

That's a new one in this thread that i don't agree in.

Au contraire. It is explicitly and utterly expressed in the chapter 22 of Feynman's lectures. And we've called the attention to this fact repeatedly along the thread.

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Why would the impedance of a passive component change in the presence of a magnetic field? The component is still passing current in the same dependence to voltage as before it was placed in a field, since this very dependence is called impedance means the impedance has not changed. If you measured it being different in a changing field then your measured quantities are likely affected by the field (Since impedance is measured indirectly from other quantities) rather than the impedance being affected.

An impedance is Z = V/I. If you're under a varying magnetic field, the voltage will be dependent on the path. So the way you measure this voltage will affect the value you get for your impedance. In extreme cases, as Feynman points out, the varying magnetic field overcomes all the internal reactions and the component becomes a generator, which is not an impedance anymore.

Under a varying magnetic field an impedance can not only change depending on the path, but also cease to be.

Which means that, if you want a "stable" impedance, you have to measure it away from any varying magnetic fields, at its terminals, defined by Feynman as the region where you have no varying magnetic fields.

Quote
That is unless you make your resistor out of a material that shows significant magnetoresistive effects. But as far as i know pure copper or carbon based resistors don't have any significant magnetoresistive properties. Even then they would react to DC fields too, not only AC fields.

A lot of things you took for granted changes when you understand electromagnetism. It opens your mind to many new perspectives. That's the beauty of this phenomenon.
« Last Edit: April 16, 2019, 05:09:18 pm by bsfeechannel »
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #958 on: April 16, 2019, 05:24:22 pm »
Quote
Could you explain please? For simplicity there is single 1Ohm resistor, EMF is 1V, 1A is flowing for 1 sec. Please consider equation in form: power_generated - power_dissipated_in_the_resistor = 0.

Before we can address this problem, let's suppose that we live on a planet

Yes, we live on a planet. It is a fact. Are you going to explain conservation of energy using Maxwell's equations or not? Your usual goalpost shifting tactics is not funny anymore, you shall stop it.

¡No, señor! You teach me fundamental physics, and I teach you Maxwell. That's the deal. Answer my question and half of the explanation you want about Maxwell will be answered, requiring only to adapt the formalism.

Besides I'm really busy. So I'm outsourcing my reasoning to your brain and the experiment I proposed can be easily performed right here on earth. If you don't know where the extra energy comes from, just say it, there's no shame about that.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #959 on: April 16, 2019, 06:30:54 pm »
Besides I'm really busy. So I'm outsourcing my reasoning to your brain and the experiment I proposed can be easily performed right here on earth. If you don't know where the extra energy comes from, just say it, there's no shame about that.

¡No, señor! You bragged to let me discover how the conservation of energy holds for such a circuit using Maxwell's equations, yet can't produce anything. Instead came-up with new "experiment" that is not even related to discussion. Please do not shift goalposts and explain yourself. Other option: you can admit that you did not know what you were talking about.  There's no shame about that.
« Last Edit: April 16, 2019, 07:26:29 pm by ogden »
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #960 on: April 16, 2019, 07:17:32 pm »
Quote
Why would the impedance of a passive component change in the presence of a magnetic field? The component is still passing current in the same dependence to voltage as before it was placed in a field, since this very dependence is called impedance means the impedance has not changed. If you measured it being different in a changing field then your measured quantities are likely affected by the field (Since impedance is measured indirectly from other quantities) rather than the impedance being affected.

An impedance is Z = V/I. If you're under a varying magnetic field, the voltage will be dependent on the path. So the way you measure this voltage will affect the value you get for your impedance. In extreme cases, as Feynman points out, the varying magnetic field overcomes all the internal reactions and the component becomes a generator, which is not an impedance anymore.

Under a varying magnetic field an impedance can not only change depending on the path, but also cease to be.

Which means that, if you want a "stable" impedance, you have to measure it away from any varying magnetic fields, at its terminals, defined by Feynman as the region where you have no varying magnetic fields.

In that case the component you are describing is no longer just a passive component.

So then could you please explain how would you measure the impedance of a real battery? (Hence it has internal resistance that can't be directly mesured) Lets also put the battery in a place where there are no magnetic fields present, so in that case we should get only one impedance value as a result, right?
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #961 on: April 17, 2019, 04:30:30 am »
In that case the component you are describing is no longer just a passive component.

What is the surprise?

Quote
So then could you please explain how would you measure the impedance of a real battery? (Hence it has internal resistance that can't be directly mesured) Lets also put the battery in a place where there are no magnetic fields present, so in that case we should get only one impedance value as a result, right?

Impedance is a constant coefficient of proportionality between V and I. It is in general complex number and a function of frequency. Can we talk about a constant coefficient of proportionality between voltage and current of a battery?
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #962 on: April 17, 2019, 05:19:38 am »
Impedance is a constant coefficient of proportionality between V and I. It is in general complex number and a function of frequency. Can we talk about a constant coefficient of proportionality between voltage and current of a battery?

Yep i agree. And to keep things from getting complicated lets suppose this battery is built up inside as 1.5V voltage source with a 1 Ohm series resistance. Such a battery would have no reactive impedance component at all frequencies (At least ones low enough to not get affected by the physical size of the battery) and has no chemistry related variation of voltage.

So then what is the correct procedure for measuring the impedance of this battery from its terminals? You can also give acual numbers in each step since all of the batteries properties are defined with values in the paragraph above.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #963 on: April 17, 2019, 05:52:50 am »
Answer my question and half of the explanation you want about Maxwell will be answered, requiring only to adapt the formalism.

So your question was:

Before we can address this problem, let's suppose that we live on a planet where g = 10 m/s². We lift an object of 1kg from the ground up to 1m. The potential energy transferred to this object will be U = mgh = 1kg*10m/s²*1m = 10 joules. If we drop down this object, when it hits the ground the potential energy will be converted to kinetic energy, which will be exactly 10 joules. Energy is conserved.

Now let's suppose that we live on a planet where g can vary. We lift the same object to a height of 1m with g = 10m/s² as before. But as soon as we drop the object, g suddenly becomes 20m/s². When the object hits the ground the kinetic energy will be 20 joules.

Where did this excess energy come from?

Energy was added to the system by one who brought extra mass to the planet. Now please stop playing challenge games and be so kind - explain law of conservation of energy using Maxwell's equations. I am looking for equation in form: power_generated - power_dissipated_in_the_resistor = 0.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #964 on: April 17, 2019, 07:01:20 am »
Energy was added to the system by one who brought extra mass to the planet.

Thank you.

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Now please stop playing challenge games and be so kind - explain law of conservation of energy using Maxwell's equations. I am looking for equation in form: power_generated - power_dissipated_in_the_resistor = 0.

Alright. Google for Poynting's theorem. Or go directly to chapter 27 of volume II of Feynman's lectures. Good read.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #965 on: April 17, 2019, 07:13:01 am »
So then what is the correct procedure for measuring the impedance of this battery from its terminals? You can also give acual numbers in each step since all of the batteries properties are defined with values in the paragraph above.

Is a battery a component of an AC circuit? Do you understand in what context we can talk about impedances?  If you are in doubt I recommend you read the title of chapter 22 of Feynman's lectures, Volume II.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #966 on: April 17, 2019, 07:27:18 am »
Quote
Now please stop playing challenge games and be so kind - explain law of conservation of energy using Maxwell's equations. I am looking for equation in form: power_generated - power_dissipated_in_the_resistor = 0.

Alright. Google for Poynting's theorem. Or go directly to chapter 27 of volume II of Feynman's lectures. Good read.

Poynting's theorem does not explain anything about resistor. Please come back to earth from your arrogance heights and explain w/o sending to google. Others reading this... khm... debate could be interested in **your** wizdom as well.

[edit] Ok..

As energy generated by wire loop is equal to energy dissipated in the resistor, we conclude that at any instant moment of time power delivered by wire loop is equal to the power dissipated by resistor. Now we will express what I just stated using equation: L(di/dt)*I = R*I^2. Do you agree?

Then we  divide it all by I:

L(di/dt) = R*I

It results in what I was looking for,

L(di/dt) - R*I = 0
« Last Edit: April 17, 2019, 10:02:49 am by ogden »
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #967 on: April 17, 2019, 07:40:47 am »
Is a battery a component of an AC circuit? Do you understand in what context we can talk about impedances?  If you are in doubt I recommend you read the title of chapter 22 of Feynman's lectures, Volume II.

Yes batteries are perfectly valid in AC circuits because AC current is capable of flowing trough them. It just doesn't have any frequency dependence. Just like a resistor.

I never found any mention in that particular Feynman lecture about impedance not being applicable to components that act the same in AC and DC.

So do we get a method for measuring impedance of a battery? Or is such a thing impossible to do?
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #968 on: April 17, 2019, 09:09:20 am »
So do we get a method for measuring impedance of a battery? Or is such a thing impossible to do?

It is possible or not - depends on convenience of your debate opponent.

He takes sentences and words out of context, twist meaning as he pleases to suit his agenda of combative debate. Example: I say "Dr.Lewin does not account for EMF voltage generated *in* the resistors", his argument is that Dr.Lewin said: "EMF cannot exist in the wires, but only, and only, in the resistors". Dr.Lewin did not say that resistors generate EMF, he said that EMF generated by the loop of wire can be observed on the terminals of resistor. Absolutely different meaning, but hey - most readers of this thread will not even notice, right?  :-DD
 

Offline Zucca

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #969 on: April 17, 2019, 09:51:35 am »
bsfeechannel and ogden you should meet and drink a beer together.
I guess after some time you will agree, probably only after just the first beer.

In forums it's easy to escalate for no real reasons.
Can't know what you don't love. St. Augustine
Can't love what you don't know. Zucca
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #970 on: April 17, 2019, 10:04:31 am »
bsfeechannel and ogden you should meet and drink a beer together.
I guess after some time you will agree, probably only after just the first beer.

So we all (including you) meet in Hannover, Germany, during next electronics trade show? ;)
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #971 on: April 17, 2019, 06:28:24 pm »
Yes batteries are perfectly valid in AC circuits because AC current is capable of flowing trough them. It just doesn't have any frequency dependence. Just like a resistor.

I never found any mention in that particular Feynman lecture about impedance not being applicable to components that act the same in AC and DC.

So do we get a method for measuring impedance of a battery? Or is such a thing impossible to do?

Before we do that, I need you to do me a favor.

Please find a battery whose voltage at its terminals varies like this

V(t)=Veiωt

I.e., a battery that has a sinusoidal output. Yes because that's the condition that Feynman imposed to study AC circuits, that the voltages would be sinusoidal. I'm sure you read chapter 22 carefully and paid attention to this detail.

Then I need you to establish the coefficient of proportionality between the current and the voltage of this strange battery at its terminals. You understand that a coefficient of proportionality implies that when the amplitude of the current at the terminals of the battery is zero, the amplitude of the voltage is also zero.

Doesn't need to be numeric. You can come with just an equation or something like that.

After we know what to measure, we can discuss a method to do it. What do you say?
 
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Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #972 on: April 17, 2019, 07:38:08 pm »
Here is the whole model with all values:


If you want it in that form you can take the battery voltage as being V(t)=1.5 + 0*ei*0*t. If such a thing is impossible then feel free to chose different numbers at your liking, can get rid of the DC component too if you think it matters. If a load needs to be placed across the output that is also fine as long as you specify what the load is.

What to measure is this: Whatever quantity you claim is impedance across the output terminals as a numeric value.

In the case the impedance at DC does not have a value, please explain the reason for it.
« Last Edit: April 17, 2019, 07:49:14 pm by Berni »
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #973 on: April 17, 2019, 10:56:29 pm »
Poynting's theorem does not explain anything about resistor. Please come back to earth from your arrogance heights and explain w/o sending to google. Others reading this... khm... debate could be interested in **your** wizdom as well.

That's why you can't have nice things. If I show you the equations you don't understand. If I try to explain it to you without resorting to the fast lane of math you tell me to stop it.

Quote
[edit] Ok..

As energy generated by wire loop is equal to energy dissipated in the resistor, we conclude that at any instant moment of time power delivered by wire loop is equal to the power dissipated by resistor. Now we will express what I just stated using equation: L(di/dt)*I = R*I^2. Do you agree?

Then we  divide it all by I:

L(di/dt) = R*I

It results in what I was looking for,

L(di/dt) - R*I = 0

Assuming that your reasoning were correct, since the current in the resistor is the same in the wire, di/dt = R*i/L. If i= sin(2πft), then we will have 2πfcos(2πft) = R*sin(2πft)/L => cos(2πft) = (R/2πfL) * sin(2πft). Sine and cosine are orthogonal functions, which means that you can't multiply one by a constant and have the other. So we have a problem here.

And the problem arises from the fact that we simply do not have an inductance there. The adage that any piece of wire is an inductor is a myth. A piece a wire can be an inductor under CERTAIN circumstances with which this piece wire does not comply.

So, the energy that's powering the resistor must be coming from somewhere else.

If you let me go on with my "goalpost-shifting experiment", you will understand why this piece of wire is not an inductor and how conservation of energy can be maintained even when KVL fails.
 
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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #974 on: April 17, 2019, 11:19:46 pm »
So do we get a method for measuring impedance of a battery? Or is such a thing impossible to do?

It is possible or not - depends on convenience of your debate opponent.

He takes sentences and words out of context, twist meaning as he pleases to suit his agenda of combative debate. Example: I say "Dr.Lewin does not account for EMF voltage generated *in* the resistors", his argument is that Dr.Lewin said: "EMF cannot exist in the wires, but only, and only, in the resistors". Dr.Lewin did not say that resistors generate EMF, he said that EMF generated by the loop of wire can be observed on the terminals of resistor. Absolutely different meaning, but hey - most readers of this thread will not even notice, right?  :-DD

It's impossible to reason on pseudo-scientific assumptions because they lead you inevitably to contradictions. So, before we can answer anything we need to check if we are not talking nonsense.
 
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