Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 63708 times)

0 Members and 1 Guest are viewing this topic.

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #975 on: April 18, 2019, 01:46:17 am »
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.

The answer is: it is not possible to measure the impedance across the terminals because this is not an impedance. You see, the voltage V and the current I won't establish a constant proportion, which is a requisite for considering anything an impedance.

Connect your ohmmeter to this battery and see what it indicates.

You can however try to determine an equivalent circuit as I showed in the Reply #886, on 18 February.

 
The following users thanked this post: HuronKing

Online Berni

  • Super Contributor
  • ***
  • Posts: 2487
  • Country: si
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #976 on: April 18, 2019, 04:27:28 am »
The answer is: it is not possible to measure the impedance across the terminals because this is not an impedance. You see, the voltage V and the current I won't establish a constant proportion, which is a requisite for considering anything an impedance.

Connect your ohmmeter to this battery and see what it indicates.

You can however try to determine an equivalent circuit as I showed in the Reply #886, on 18 February.



Makes sense.

How about if the voltage of the battery is indeed a sine wave with no bias, defined as:
V(t)=1.5*ei*6,28*t

Is there an impedance then?


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.

Btw i am interested in what exactly are these certain circumstances when a piece of wire suddenly gets the properties of inductance and when it does not.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 2667
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #977 on: April 18, 2019, 01:51:28 pm »
If I show you the equations you don't understand.

You clearly have issues. Anyway thank you, I will try harder :D

Quote
The adage that any piece of wire is an inductor is a myth.

Problem with your reasoning is that I was not explicit about material of wire or magnetic core or absence of it. Ok, fine. Forget about inductor. - You apparently have problems to fathom not only non-conservative versus conservative fields, but true power versus reactive power as well. Let's assume that variable magnetic field just happens and it induces 1V EMF in the loop, EMF equals (eip). Now what is *correct* (mine was incorrect as you say) equation that shows law of conservation of energy between wire loop and resistor? Just try me and produce equation even if in doubt that anybody except you will understand it.

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.

Discussion about some property of experiment or part of it is not nonsense, especially if property is so small that it does not affect accuracy of measurements. Dr.Lewin ignores resistance of wire - it is scientific and accepted by you. When I say that EMF generated in the (by the) resistors are ignored by Dr.Lewin - just because dimensions of resistors are ignored as well, you say it is pseudoscience and nonsense  |O Take a moment and look from different angles on what I just said.
« Last Edit: April 18, 2019, 02:23:57 pm by ogden »
 
The following users thanked this post: Berni

Offline jesuscf

  • Regular Contributor
  • *
  • Posts: 196
  • Country: ca
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #978 on: April 18, 2019, 06:37:23 pm »
Allow me to falsify Dr. Lewin's experiment.  I'll start with two pieces of wire and two ordinary 5% 1/4W resistors one order of magnitude apart, say 10k\$\Omega\$ and 100k\$\Omega\$.  The resistance of the two pieces of wire are about 0.25\$\Omega\$, so I think we can ignore their resistances for this experiment.  I also attached alligator clips to the pieces of wire to allow for easy and fast connections.



Without  the presence of any electro-magnetic fields that are not natural, such as the earth magnetic field, (I didn't put an inductor in the middle of the ring) I proceeded to measure, and this it what I got:



What?  The voltage across the 10k\$\Omega\$ resistor is -0.111mV while the voltage across the 100k\$\Omega\$ resistor is 1.058mV.  Very similar to what Dr. Lewin's got in his experiment.  Then, Kirchhoff's voltage law doesn't work under this mysterious conditions?!  I am not so sure about that, just to be certain allow me to measure the voltage across the two pieces of wire:



As you can see, there is a voltage across the bottom wire! (Second multi-meter from the top).  If you add the voltages of the four multi-meters the total should be zero.  The problem here is that these multi-meters are out of calibration and also have been through horrible things in the last 10 years, so I didn't expect to have 'perfect' results.  Nevertheless the result is illustrative enough.  The point here is that somehow, there is a voltage induced in the bottom wire.  In this case, the voltage is not induced using an external magnetic field.  What I didn't tell you at the beginning, is that one of the wires is made of the alloy chromel while the other wire is made of the alloy alumel which I removed from a k-type thermocouple.  To produce a voltage, I heated up one of wires with a soldering iron as shown below.


« Last Edit: April 18, 2019, 07:23:02 pm by jesuscf »
Homer: Kids, there's three ways to do things; the right way, the wrong way and the Max Power way!
Bart: Isn't that the wrong way?
Homer: Yeah, but faster!
 
The following users thanked this post: ogden

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #979 on: April 18, 2019, 10:37:08 pm »
Makes sense.

Of course it does.

Quote
How about if the voltage of the battery is indeed a sine wave with no bias, defined as:
V(t)=1.5*ei*6,28*t

Is there an impedance then?

Nope. I regret to tell you. The voltage will be defined by the "battery" and the current by whatever load you connect to its terminals, so there won't be a fixed coefficient of proportionality between the amplitude of the voltage and the amplitude of the current.

You can try to calculate or measure it.

Quote
Btw i am interested in what exactly are these certain circumstances when a piece of wire suddenly gets the properties of inductance and when it does not.

Glad you asked. Take a linear transformer. The secondary of a transformer will be a piece of wire wound around some core, be it air or whatever. But to simplify our reasoning, let's suppose that the core is vacuum.

Well, a piece of wire wound around a core? You say. It is an inductor, isn't it? Of course it is. Take whatever LCR meter you have and connect it to the terminals of this secondary and it'll confirm that.

You can even calculate and measure its impedance  for whatever frequency you want. |Z| = |V|/|I| = 2πfL.

Now connect the primary to a voltage source. A voltage will appear at the terminals of the secondary, won't it? But wait a minute, nothing is connected to the secondary, so the current is zero. If I take this voltage and divide it by the current I will have |Z| = |V|/0, which is undefined.

Don't pull your hair out yet. Suppose that I now connect a resistive load. The resistor will force the current to be in phase with the voltage. The same current that is traversing the wire of the secondary. But in an inductor shouldn't the current be lagging the voltage by 90°?

Yet you have a coiled piece of wire where you have an alternate current in phase with the voltage at its terminals. So much for our inductor!

So, how can it be? What changed from the condition where the primary was disconnected to when it was connected? Certainly it was not the wire, because we haven't touched it. Something else must be producing this change.

What is it?
 
The following users thanked this post: HuronKing

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #980 on: April 19, 2019, 12:02:47 am »
You clearly have issues.

I have many issues, but electromagnetism is not one of them (at least not in the level we are discussing).

Quote
Anyway thank you, I will try harder :D

The world is grateful.

Quote
Problem with your reasoning is that I was not explicit about material of wire or magnetic core or absence of it. Ok, fine. Forget about inductor. - You apparently have problems to fathom not only non-conservative versus conservative fields, but true power versus reactive power as well. Let's assume that variable magnetic field just happens and it induces 1V EMF in the loop, EMF equals (eip). Now what is *correct* (mine was incorrect as you say) equation that shows law of conservation of energy between wire loop and resistor? Just try me and produce equation even if in doubt that anybody except you will understand it.

Your problem is that you think that electricity and magnetism are properties of circuits. You are a circuit head, a circuity guy. You are forever condemned to think electricity and magnetism in terms of circuits and that's crippling your brain. You treat non conservative fields as if they were conservative, and that's why you think that the energy must have come from some component in the circuit, the wire, for instance.

Ditch that bullshit right away.

Here, lemme help you. In the "goalpost-shifting" problem that I proposed to you, the 1kg object that had 10J of potential energy suddenly acquired 10J more. You said that the energy was provided by someone who doubled the mass of the planet. That might have taken a lot of energy. However only 10J was added to the object. We haven't touched it, i. e., we haven't moved it to a higher position or pushed it down. What exactly gave the extra energy to the object? It was the? Anyone? Anyone? (Hint: it starts with an F).

Quote
Dr.Lewin ignores resistance of wire - it is scientific and accepted by you.

That's OK. The resistance in the wire is negligible. Who cares?

Quote
When I say that EMF generated in the (by the) resistors are ignored by Dr.Lewin - just because dimensions of resistors are ignored as well, you say it is pseudoscience and nonsense  |O

The EMF cannot be generated BY the resistors. However it is generated IN the resistors. The dimensions of the resistors are absolutely irrelevant. What counts is the path of the circuit. That's Maxwell's equations. End of story.

Quote
Take a moment and look from different angles on what I just said.

The angle of science is just one. The angles of pseudo-science are many. Each one is free to cook up their own contradictory explanations. I can't keep up with all of them.
 
The following users thanked this post: HuronKing

Offline HuronKing

  • Newbie
  • Posts: 4
  • Country: us
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #981 on: April 19, 2019, 01:06:47 am »
I would add something... but you're doing fine bsfeechannel.
 
The following users thanked this post: bsfeechannel

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #982 on: April 19, 2019, 01:54:22 am »
Thank you HuronKing. By all means feel free to add whatever you deem relevant. I want to learn too.
 

Offline HuronKing

  • Newbie
  • Posts: 4
  • Country: us
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #983 on: April 19, 2019, 03:14:11 am »
The only thing I might want to add is to make even more explicit the nature of non-conservative forces. This isn't some weird voodoo-freaky electromagnetism thing. In Classical Electrodynamics, the analogy that led Maxwell to his discoveries were hydrodynamic vortices. Maxwell's initial picture was inaccurate and Heaviside placed them in the vector form we know and love, but it's useful to go back to what techniques are used to solve explicitly non-conservative problems. Like, friction forces:

https://opentextbc.ca/physicstestbook2/chapter/nonconservative-forces/

Curious note, none of these simple example problems move all the values over to one side and set the whole thing equal to zero. That is, it's not 5+3-8 = 0.

Two things I think are worth noting in these alternative explanations to Faraday's Law (beautiful as the law is):
1) The mathematical contradiction inherent in assuming a value of L just randomly exists as you demonstrated aptly.
2) The assumption that if a mass has more kinetic energy when it hits the ground than it's gravitational potential energy MUST mean someone doubled the mass of the planet.  :o

I admit that last one made me chuckle.
« Last Edit: April 19, 2019, 03:17:04 am by HuronKing »
 
The following users thanked this post: bsfeechannel

Offline seagreh

  • Contributor
  • Posts: 7
  • Country: mx
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #984 on: April 19, 2019, 03:41:41 am »
bsfeechannel,

Since it starts with an F, it is Michael Faraday!
He had his inductive charger with him and pumped a 10 Joules into that‘s things battery! He hasn‘t touched the thing, he hasn‘t moved it neither.
10 seconds charging only. Smart guy.
Kirchhoff - not being aware about inductive chargers - would have tried to bring Mars and Venus over to our planet Earth, to double its mass.

As You can keep up with all angles - that of science and those of pseudo-science...

You say, the EMF is generated IN the resistor ! Hmmm. The resistor being the seat of EMF? You are sure? Otherwise you state correctly, what counts is the path.
Actually, the path itself is already enough for the electric field, won’t even need a wire. But not good enough for the EMF, needing a ‚wire’ as well!

Imagine a super-conductive ring with a single tiny 10k resistor inserted.
You say the EMF sits in the resistor. Now, removing the resistor, would remove the EMF as well ? Or does the EMF now jump into the air gap. Can vacuum as well be a seat for EMF?
Now, let’s close the super conductive ring - short cutting the resistor. Have we killed the EMF now? Assume the ring sits in an environment where a magnetic flux of 1 Wb changes to 0 Wb within 1 second! All the flux of 1 Wb streams through our ring (don‘t care how much energy it takes to do that). Still 0V EMF, since no resistor ? Or could this violate the definition of 1 Wb?

Let‘s assume the EMF sitting in the resistor. Wouldn‘t it then compensate with the voltage drop within the resistor - the resistor being a closed system with energy sourcing part and energy dissipating part in itself? Wouldn‘t this result in 0V at the resistor terminals ? Actually fitting the 0V along your superconductive ring ?

Are you still 100% sure, this is not a pseudo scientific angle of view?
Many people have a hard times to understand the difference between an EMF and a voltage drop. The voltage drop represents the energy dissipating part, the EMF the energy sourcing part. The EMF can be easily identified by transporting charges from negative terminal to positive terminal, resulting in current flow in opposite direction of voltage drop....
I hear you saying - come on, don‘t try to teach me basics...
But on the other hand you say the EMF is NOT equal to the voltage drop. The EMF IS the voltage drop you say! Hmmm.
As I said before, many pople have hard times to understand the difference between EMF and voltage drop!

And maybe Kirchhoff could help you here :-)
As you read his original paper, you must have seen, Kirchhoff‘s intention was not to summarize whole complete world and ‚zerorize‘ it! He simply summed up voltage drops on one side of the equation and summed up EMFs on the other side of the equation. There was no zero in his equation. Now, EMFs on the right side could be coulomb EMFs and non-coulomb EMFs (e.g. Faraday‘s - dPhi/dt).

Hence Prof. Sam Ben-Yaakov is right -

EMF (being -dPhi/dt) = iR (being the line integral of E dl).

Left side of the equation being the energy sourcing part (EMF) being equal to energy consuming part (right side).
 

Offline jesuscf

  • Regular Contributor
  • *
  • Posts: 196
  • Country: ca
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #985 on: April 19, 2019, 06:34:01 am »
This is from chapter 13 of "Engineering Electromagnetics" by Hyat, third edition.

THE LAWS OF CIRCUIT THEORY

To indicate how Maxwell’s equations, the potential definitions, and the concepts of resistance, capacitance, and inductance combine to produce the common expressions of circuit analysis, consider the configuration shown in Fig. 13.1 Between the two points 0 and 1 an external electric field is applied.  These terminals are very close together, and a sinusoidal electric field may be assumed.  Perhaps we could visualize a microscopic transistor oscillator, a rotating machine the size of a pinhead (complete with prime mover), or even a cooperative flea moving a mouthful of charge alternately toward point 1 and point 0.  Whatever the nature of the source, it continues to produce an electric field between these two terminals that is independent of any currents that may consequently flow.  Between points 2 and 3 there is a region of lossy material of cross-sectional area \$S_R\$, length \$d_R\$, and conductivity \$\sigma\$.  At points 4 and 5 are located two capacitor plates of area \$S_c\$, with separation \$d_c\$ and dielectric of permittivity \$\epsilon\$.  These several points are connected as shown by a filamentary perfect conductor of negligible cross section.  Between points 6 and 8 the filament is wound into a helix of N turns having a very fine pitch.



It is probably obvious that we are now going to develop the familiar circuit equation

(1) \$V_{10}  = IR + L_{ind}\frac{{dI}}{{dt}} + \frac{1}{C}\int\limits_{ - \infty }^t {Idt}\$

from Maxwell’s equations.   As we do so, it is interesting to watch how each of these terms arises and what assumptions we have to make in the process.  Our beginning point is the integral from of Faraday’s law,

(2) \$\oint {E \cdot dL = }  - \frac{\partial }{{\partial t}}\int\limits_S {B \cdot dS}\$

We shall see that the right side of this equation provides us with only one term in (1), that involving the inductance.  The other three terms all arise from the closed line integral.

Let us consider the surface integral on the right side of (2).  Since the configuration of the circuit does not change with time, the partial derivative may be replaced with the ordinary derivative.  Also, the filament between points 6 and 8, an N-turn helix, serves to produce a much larger magnetic field within the helix than in any other region along the filament.  If we assume that a total magnetic flux \$\Phi \$ links all N turns, the surface integral becomes  \$ - \frac{{d\Phi }}{{dt}}\$, or  \$ - L_{ind}\left( {\frac{{dI}}{{dt}}} \right)\$ by the definition of inductance, and

\$\oint {E \cdot dL =  - L_{ind}\frac{{dI}}{{dt}}} \$

where I is the filamentary current in each turn of the helix.

The closed line integral is taken along the filament, directly between the capacitor plates and points 0 and 1, as indicated by the dashed line.  The contribution from the perfectly conducting filament is zero, because tangential E must be zero there; this includes the helix, surprising as that may be.  We therefore have

\$\oint {E \cdot dL = \int\limits_0^1 {} }  + \int\limits_2^3 {}  + \int\limits_4^5 {} \$

The first integral on the right is the negative of the voltage between points 1 and 0,

\$\int\limits_0^1 {E \cdot dL =  - V_{10} }\$

This integral is a function only of the external source and does not depend on the configuration shown in figure 13.1.  The path is directly between the adjacent terminals, and since we are more used to considering an external source as a voltage than as an electric field intensity, we usually call \$V_{10} \$ an applied voltage.

The second integral is taken across the lossy material, and we apply Ohm’s law in point form and the definition of total resistance R,

\$\int\limits_2^3 {E \cdot dL = } \int\limits_2^3 {\frac{J}{\sigma } \cdot dL = } \int\limits_2^3 {\frac{{JdL}}{\sigma } = } \frac{{Jd_R }}{\sigma } = \frac{{Id_R }}{{\sigma S_R }} = IR\$

The same total current I is assumed, and this justified only when two conditions are met.  There can be no displacement currents flowing from one point of the filament to another (such as from point 3 to 8 ), because we require the continuity of conduction plus displacement current density to be satisfied by conduction current alone.  In other language, we are assuming that stray capacitances are neglected.  Also the dimension of the filamentary path must be small compared to a wavelength.  This will be applied in the final section of this chapter, but our study of wave motion should indicate the complete reversal which may occur in afield over a distance of one-half wavelength.  Here we wish to avoid radiation, but later in the chapter it will provide our main item of interest.

The third integral is evaluated across the region between the capacitor plates where conduction current is zero but displacement current is equal to the current I, as we assumed earlier.  Here we may represent the integral by

\$\int\limits_4^5 {E \cdot dL = } \int\limits_4^5 {\frac{D}{\varepsilon } \cdot dL = } \frac{{Dd_C }}{\varepsilon } = \frac{{Qd_C }}{{\varepsilon S_C }} = \frac{Q}{C}\$

or

\$\int\limits_4^5 {E \cdot dL = } \frac{1}{C}\int\limits_{ - \infty }^t {Idt} \$

where we assume that there is no charge on the capacitor at \$t =  - \infty \$.

Combining these results, we have

\$ - V_{10}  + IR + \frac{1}{C}\int\limits_{ - \infty }^t {Idt}  =  - L_{ind}\frac{{dI}}{{dt}}\$

or

\$V_{10}  = IR + L_{ind}\frac{{dI}}{{dt}} + \frac{1}{C}\int\limits_{ - \infty }^t {Idt}\$

which is the familiar equation for an RLC series circuit that we hoped we would obtain.

« Last Edit: April 19, 2019, 03:35:24 pm by jesuscf »
Homer: Kids, there's three ways to do things; the right way, the wrong way and the Max Power way!
Bart: Isn't that the wrong way?
Homer: Yeah, but faster!
 
The following users thanked this post: Berni

Online Berni

  • Super Contributor
  • ***
  • Posts: 2487
  • Country: si
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #986 on: April 19, 2019, 06:41:54 am »
Nope. I regret to tell you. The voltage will be defined by the "battery" and the current by whatever load you connect to its terminals, so there won't be a fixed coefficient of proportionality between the amplitude of the voltage and the amplitude of the current.

You can try to calculate or measure it.

This was exactly my point. It doesn't make sense to talk about a voltage source as purely being an impedance.

When analyzing the circuit it makes more sense to think of it as an equivalent circuit. This separates the voltage source from the impedance part, allowing both to remain the same no matter the load.

Or is the use of equivalent models somehow forbidden in circuit analysis?

Quote
Btw i am interested in what exactly are these certain circumstances when a piece of wire suddenly gets the properties of inductance and when it does not.

Glad you asked. Take a linear transformer. The secondary of a transformer will be a piece of wire wound around some core, be it air or whatever. But to simplify our reasoning, let's suppose that the core is vacuum.

Well, a piece of wire wound around a core? You say. It is an inductor, isn't it? Of course it is. Take whatever LCR meter you have and connect it to the terminals of this secondary and it'll confirm that.

You can even calculate and measure its impedance  for whatever frequency you want. |Z| = |V|/|I| = 2πfL.

Now connect the primary to a voltage source. A voltage will appear at the terminals of the secondary, won't it? But wait a minute, nothing is connected to the secondary, so the current is zero. If I take this voltage and divide it by the current I will have |Z| = |V|/0, which is undefined.

Don't pull your hair out yet. Suppose that I now connect a resistive load. The resistor will force the current to be in phase with the voltage. The same current that is traversing the wire of the secondary. But in an inductor shouldn't the current be lagging the voltage by 90°?

Yet you have a coiled piece of wire where you have an alternate current in phase with the voltage at its terminals. So much for our inductor!

So, how can it be? What changed from the condition where the primary was disconnected to when it was connected? Certainly it was not the wire, because we haven't touched it. Something else must be producing this change.

What is it?

Oh we are already at transformers from inductance, good to hear. Yep we all agree that a coil of wire is indeed an inductor.

What you describe is called mutual inductance in circuit analysis. As soon as inductors start sharing magnetic fields they also start sharing there inductance too. The total inductance of the coil stays the same so as mutual inductance increases the self in distance decreases. In transformers this self inductance is actually called leakage inductance. Here is a quick summary of how this works: https://physics.stackexchange.com/questions/119638/choosing-sign-for-kvl-mutual-inductance

So now how does the transformer secondary not have 90 degree phase shift if its an inductor? Because the voltage is all coming from mutual part of inductance and this inductance only cares about the current in the primary. The current trough the load resistor actually affects the voltage of the primary side. Once all of these currents and voltages are summed up neither the primary or secondary have 90 degree phase shifts anymore. In an ideal 1:1 transformer with no leakage the currents and voltages are actually all perfectly in phase (If the load is a resistor like in your example) or 180 out of phase depending on what way around you connect the coil.

All my hair is still in its place as there is no need to pull it out over a concept that works just fine, or is the use of mutual inductance somehow forbidden?

So then back to the original question. Under what circumstances is a wire not an inductor and when it is? Or does only self inductance count? If not then at what coupling factor (k) does it stop counting as an inductor? 0.5? 0.1? 1E-6?

EDIT: Oh and btw and the ideal 1:1 transformer above also has infinite inductance. For a real transformer with a finite inductance the primary inductance still takes a current according to its impedance, this is the magnetizing current for a transformer and its what limits the lowest usable frequency for the transformer. This current is indeed lagging 90 degrees, but once the load current joins it then the total sum current has a phase shift less than 90.(Tho the currents don't literally sum due to where the leakage inductance is)
« Last Edit: April 19, 2019, 08:02:09 am by Berni »
 
The following users thanked this post: ogden

Offline jesuscf

  • Regular Contributor
  • *
  • Posts: 196
  • Country: ca
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #987 on: April 19, 2019, 04:03:29 pm »
So then back to the original question. Under what circumstances is a wire not an inductor and when it is? Or does only self inductance count? If not then at what coupling factor (k) does it stop counting as an inductor? 0.5? 0.1? 1E-6?

This wire inductance calculator may be useful for the person this question was directed:

https://www.eeweb.com/tools/wire-inductance

Homer: Kids, there's three ways to do things; the right way, the wrong way and the Max Power way!
Bart: Isn't that the wrong way?
Homer: Yeah, but faster!
 

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #988 on: April 19, 2019, 04:14:05 pm »
Since it starts with an F, it is Michael Faraday!

That's a good one. But nope. It's not Faraday.

Quote
You say, the EMF is generated IN the resistor ! Hmmm. The resistor being the seat of EMF? You are sure?

Absolutely sure.

Quote
Imagine a super-conductive ring with a single tiny 10k resistor inserted.
You say the EMF sits in the resistor. Now, removing the resistor, would remove the EMF as well ? Or does the EMF now jump into the air gap. Can vacuum as well be a seat for EMF?

The resistor happens to be where the EMF is. If you remove the resistor, the EMF stays in the same place. You don't need a superconductor to prove this. Just connect a voltmeter to the secondary of a transformer capable of providing a sufficiently large current. Then connect a 10k resistor. The voltage won't change.

Quote
Let‘s assume the EMF sitting in the resistor. Wouldn‘t it then compensate with the voltage drop within the resistor - the resistor being a closed system with energy sourcing part and energy dissipating part in itself? Wouldn‘t this result in 0V at the resistor terminals ? Actually fitting the 0V along your superconductive ring ?

Sensational! I was waiting for someone to come up with exactly that "explanation". The EMF and the voltage drop cancelling each other inside a resistor. You have a current induced through the resistor and its voltage is zero.  It is pseudo-science in all its splendor. You can't blame them for their creativity. I love it.

The EMF is path-dependent. So following the path inside the conductor it is zero. Following the path along the resistor it is non zero. Faraday's law. There's no way around it.

Quote
As I said before, many pople have hard times to understand the difference between EMF and voltage drop!

Electromagnetism, the fundamentals of our trade, is not intuitive. You have to use reason to understand it. Tell these people to study.

Quote
And maybe Kirchhoff could help you here :-)
As you read his original paper, you must have seen, Kirchhoff‘s intention was not to summarize whole complete world and ‚zerorize‘ it! He simply summed up voltage drops on one side of the equation and summed up EMFs on the other side of the equation. There was no zero in his equation. Now, EMFs on the right side could be coulomb EMFs and non-coulomb EMFs (e.g. Faraday‘s - dPhi/dt).

Hence Prof. Sam Ben-Yaakov is right -

EMF (being -dPhi/dt) = iR (being the line integral of E dl).

Left side of the equation being the energy sourcing part (EMF) being equal to energy consuming part (right side).

The problem has nothing to do with where you place the terms of the equation. This is just a convention. The problem lies in the fact that Kirchhoff demands that EMFs and voltage drops must be found in separate places IN the circuit. Using the convention Kirchhoff did, the sum of all voltage drops must be equal to the sum of all EMFs. Or, using our modern convention, all the voltages FOUND IN THE CIRCUIT, i.e. directly measured, (EMFs and drops) must add up to zero.

Prof. Sam Ben-Yaakov says, to the add more confusion to the mess:

"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".

If Kirchhoff held in this case, then elp should be found in the circuit as a separate entity. But it obviously isn't. And he even declares that ("you don't see", "it is hovering around").

So he contradicts Kirchhoff, who demands the voltage to be explicit.

Walter Lewin is a vehement critic of this kind of people espousing those ideas for obvious reasons: it not only causes all kinds of misconceptions but is also completely pseudo-scientific. In short it is WRONG.

Conventionally today the voltages (emfs or voltage drops) found in the circuit are placed to the left of the equation. To the right is the account of the effect of induction. If there's no induction, all the voltages add up to zero and KVL holds. If there is induction, the voltages will add up to the calculation indicated by the right side, and Kirchhoff doesn't hold anymore.

The calculation to the right side is not a voltage that you'll find in the circuit (as Prof. Ben-Yaakov confirmed), which means that you cannot conventionally place it to the left side as if you'd find that with your voltmeter. This is simply stupid. (And it is even "stupider" to look at the zero left to the right of the equation after this mathematical bodge and declare because of this that Kirchhoff holds.)

In fact, none of the (duly debunked) "debunkers" of Lewin managed to measure this voltage in any of their circuits.

Νot even this unfortunate professor.
 

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #989 on: April 19, 2019, 04:16:40 pm »
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.

ogden needs to agree with nature, not with me.

I would have no problem drinking beer with anyone in the forum. I have no hard feelings against ogden or anyone else. Since Shahriar said that the next thing you could do to a person besides loving them is teaching them, I think that by this time I must be madly in love with ogden (Berni comes second).
 

Offline jesuscf

  • Regular Contributor
  • *
  • Posts: 196
  • Country: ca
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #990 on: April 19, 2019, 04:37:36 pm »
Conventionally today the voltages (emfs or voltage drops) found in the circuit are placed to the left of the equation. To the right is the account of the effect of induction. If there's no induction, all the voltages add up to zero and KVL holds. If there is induction, the voltages will add up to the calculation indicated by the right side, and Kirchhoff doesn't hold anymore.

The calculation to the right side is not a voltage that you'll find in the circuit (as Prof. Ben-Yaakov confirmed), which means that you cannot conventionally place it to the left side as if you'd find that with your voltmeter. This is simply stupid. (And it is even "stupider" to look at the zero left to the right of the equation after this mathematical bodge and declare because of this that Kirchhoff holds.)

In fact, none of the (duly debunked) "debunkers" of Lewin managed to measure this voltage in any of their circuits.

Νot even this unfortunate professor.

I just want to preserve this for posterity in case the original poster decides to erase it once he realizes how massively incorrect it is.
Homer: Kids, there's three ways to do things; the right way, the wrong way and the Max Power way!
Bart: Isn't that the wrong way?
Homer: Yeah, but faster!
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 2667
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #991 on: April 19, 2019, 07:04:12 pm »
In fact, none of the (duly debunked) "debunkers" of Lewin managed to measure this voltage in any of their circuits.

Every AC transformer in operation by itself is proof that you are mistaken.

The resistor happens to be where the EMF is. If you remove the resistor, the EMF stays in the same place. You don't need a superconductor to prove this. Just connect a voltmeter to the secondary of a transformer capable of providing a sufficiently large current. Then connect a 10k resistor. The voltage won't change.

Seems, you accidentally disproved yourself - because in case you just described, KVL holds.

2) The assumption that if a mass has more kinetic energy when it hits the ground than it's gravitational potential energy MUST mean someone doubled the mass of the planet.  :o
I admit that last one made me chuckle.

Your self-embarrassment could be avoided if you would read original post carefully meaning all the provided info, not just last words:

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.
« Last Edit: April 20, 2019, 04:57:17 am by ogden »
 

Online Berni

  • Super Contributor
  • ***
  • Posts: 2487
  • Country: si
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #992 on: April 19, 2019, 07:28:00 pm »
I do sometimes end up at these electronics trade shows in Germany. Always up for a chat if the timing is right, the accompanying beer is even free if you find the right booth handing it out.

Tho lately I'm finding it a bit hard to even tell what everyone is even arguing about in this thread. It feels like people are contradicting each other simply for the sake of contradicting them. Tho surprisingly my explanation of mutual inductance hasn't been contradicted yet, but its probably just a matter of time until someone disagrees with at least part of it.

Still fun thread to read tho.
 

Offline seagreh

  • Contributor
  • Posts: 7
  • Country: mx
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #993 on: April 19, 2019, 10:00:12 pm »
You say, the EMF is generated IN the resistor ! Hmmm. The resistor being the seat of EMF? You are sure?

Absolutely sure.

Quote
Imagine a super-conductive ring with a single tiny 10k inserted.
You say the EMF sits in the resistor. Now, removing the resistor, would remove the EMF as well ? Or does the EMF now jump into the air gap. Can vacuum as well be a seat for EMF?

The resistor happens to be where the EMF is. If you remove the resistor, the EMF stays in the same place. You don't need a superconductor to prove this. Just connect a voltmeter to the secondary of a transformer capable of providing a sufficiently large current. Then connect a 10k resistor. The voltage won't change.

I did your experiment. At the begin of the experiment the EMF must have been in the air gap between both secondary terminals. However, as soon as I connected the voltmeter, I got the feeling the EMF jumped into my voltmeter (as it is 10MOhm resistive. As you say EMF is path dependent). With the 10k resistor I was expecting it to sit now in the Voltmeter AND in the resistor at the same time - according to your thesis.

And I always thought, the EMF is defined as the tangential force per unit charge in the wire integrated over length, once around the complete circuit/loop! But not at all, it was hiding in the resistor and my voltmeter!

But you confused me a bit later, when talking about the [elp]. The [elp] being the EMF (I guess the Israeli Prof meant [Epsilon loop] with this abbreviation). You tell me it’s invisible and can’t be measured? Although you told me before it’s sitting in the resistor and how to measure it?

On the other hand you say today’s convention is to put voltage drops AND EMFs on the left side. But insist at the same time to keep the -dFlux/dt on the right side? -dFlux/dt being the EMF.
Looks like you don’t consider -dFlux/dt as EMF, but rather the EMF as part of the line integral E ds?
 
The following users thanked this post: jesuscf, ogden

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 186
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #994 on: April 20, 2019, 12:52:42 am »

Quote
Nah, I simply decided not to contribute to this blog anymore than it is necessary to get my potential future questions answered. You had someone who could provide content (and believe me, I had a lot more material to share), now you have a leech.

That's a shame. But you are the one who deleted all your posts, not us.

Not all of them.

Quote from: EEVblog
We do care that you removed your content, that's a real shame. But it's you who made that decision, I hope you are not accusing us of somehow forcing you to do that?

When people are banned their posts are NOT deleted, so no "hard work" is lost, the content still remains for others to enjoy, learn from and discuss. You and you alone decided to delete all your posts and your hard work.

Again, not all of them. The reasons I deleted my posts was twofold: one I explained in my previous post, and the other was to give you a taste of your medicine. When you ban someone when he's writing a post, that someone finds out to have been banned only when he hits the "Post" button.
Result: the post he has just written is lost, wasted, deleted.
Now, if the ban was rightful, I could tolerate that. But I did not see any explanation for my ban, except for the one I assumed. That is, puncturing the ego of someone who DID NOT QUALIFY AS A MODERATOR (to me that was some bloke cracking a joke, and I treated him as such), and made no specific requests as to what we should have done. Stop posting? Saying the counterpart was right? Take a bow?
Do you even know how the posts in this forum are rendered in an Android browser? There is no "moderator" tag.
So, learn to do your 'job' before wasting other people's time.

A site like this exists because of the content people contribute. If you show you do not care about the effort they put into it, why should they care themselves?

This is feedback, in case you are wondering.

All that said, let me add that I have no hard feelings toward the site. As a matter of fact, apart from this 'incident', I believe it's one of the most relaxing places where to talk about electronics. But the moderator screwed up, and you lost a contributor.

I am still leeching, tho.


All instruments lie. Usually on the bench.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 2667
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #995 on: April 20, 2019, 04:43:43 am »
But I did not see any explanation for my ban, except for the one I assumed.

You got your ban because you insulted people, including moderator. How about such explanation?

Quote
That is, puncturing the ego of someone who DID NOT QUALIFY AS A MODERATOR (to me that was some bloke cracking a joke, and I treated him as such), and made no specific requests as to what we should have done. Stop posting? Saying the counterpart was right? Take a bow?

Seems you got gigantic hole in your ego as well and it is still there.  :popcorn:
 

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #996 on: April 20, 2019, 04:00:51 pm »
I did your experiment. At the begin of the experiment the EMF must have been in the air gap between both secondary terminals.

Yep.

Quote
However, as soon as I connected the voltmeter, I got the feeling the EMF jumped into my voltmeter

Yep.
 
Quote
(as it is 10MOhm resistive.  As you say EMF is path dependent).

Under a varying magnetic field it is. Where the magnetic field doesn't vary, i.e. outside the transformer, it is not.

Quote
With the 10k resistor I was expecting it to sit now in the Voltmeter AND in the resistor at the same time - according to your thesis.

I do not have a thesis. Maxwell does. You need to get along with him more often.

Quote
And I always thought, the EMF is defined as the tangential force per unit charge in the wire integrated over length, once around the complete circuit/loop! But not at all, it was hiding in the resistor and my voltmeter!

You clearly thought wrong, I regret. EMFs will never "hide" inside (static) wires (considered as ideal, that is).

Quote
But you confused me a bit later, when talking about the [elp]. The [elp] being the EMF (I guess the Israeli Prof meant [Epsilon loop] with this abbreviation). You tell me it’s invisible and can’t be measured? Although you told me before it’s sitting in the resistor and how to measure it?

It's understandable that you are confused. Electromag is unintuitive. It requires diligent study. You didn't pay attention when I said that the EMF can't be measured as a SEPARATE ENTITY in this circuit.

Once you measure the voltage on the 10Mohm of your meter or the 10kohm load, you won't find this voltage anywhere else in the circuit. All that you'll find is a wire with (ideally) zero volts. This violates Kirchhoff's law which demands that a second voltage be found around the path of the wire to balance the one you found across the resistor.

Quote
On the other hand you say today’s convention is to put voltage drops AND EMFs on the left side. But insist at the same time to keep the -dFlux/dt on the right side? -dFlux/dt being the EMF.
Looks like you don’t consider -dFlux/dt as EMF, but rather the EMF as part of the line integral E ds?

Again you left out the phrase FOUND IN THE CIRCUIT. To the right are the EMFs FOUND IN THE CIRCUIT, i.e. directly measured. -dΦ/dt is an EMF that cannot be found in the circuit as a separate entity, so it must be on the right side of the equation.

But you are on the right track. You are at least questioning, instead of trying to disprove Faraday's law with pseudo-scientific assumptions, which is a good thing.

You'll understand electromagnetism sooner than those who thanked you.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 2667
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #997 on: April 20, 2019, 04:15:01 pm »
Once you measure the voltage on the 10Mohm of your meter or the 10kohm load, you won't find this voltage anywhere else in the circuit. All that you'll find is a wire with (ideally) zero volts.

Wow. How transformers with multi-tap secondary windings work then?

Quote
You'll understand electromagnetism sooner than those who thanked you.

 :-DD

Right. Be prepared to become enlightened. Those who thanked you are lost souls - nonbelievers. They do not believe that Dr.Lewin is always right even when he is not.
« Last Edit: April 20, 2019, 04:35:42 pm by ogden »
 

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #998 on: April 20, 2019, 04:35:33 pm »
All that said, let me add that I have no hard feelings toward the site. As a matter of fact, apart from this 'incident', I believe it's one of the most relaxing places where to talk about electronics.

Yay! Sredni is back!

Quote
But the moderator screwed up, and you lost a contributor.

Sredni, man, can't we leave this moderation imbroglio behind? After all, your access has been reinstated, you were allowed to rant on what happened, the thread is still open for us to discuss the subject and no one is bothering us.

Quote
I am still leeching, tho.

You're not punishing Dave, you are punishing us. Dave and Simon have more important things to worry about than this "electromagnetism rubbish". So leave them be.

If you have something to contribute to the discussion, please do. We are in desperate need of people who can help us destroy bullshit thinking in engineering.
 

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 902
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #999 on: April 22, 2019, 05:27:43 am »
This was exactly my point. It doesn't make sense to talk about a voltage source as purely being an impedance.

A voltage source is NOT an impedance, much less "purely". Read Feynman lectures volume II chapter 22.

Quote
When analyzing the circuit it makes more sense to think of it as an equivalent circuit. This separates the voltage source from the impedance part, allowing both to remain the same no matter the load.

Or is the use of equivalent models somehow forbidden in circuit analysis?

What determining the equivalent circuit of a battery has to do with measuring an impedance in an AC circuit under a varying magnetic field?

Quote
What you describe is called mutual inductance in circuit analysis. As soon as inductors start sharing magnetic fields they also start sharing there inductance too.

So what causes the change in their behavior is the _ _ _ _ _.

Quote
Here is a quick summary of how this works: https://physics.stackexchange.com/questions/119638/choosing-sign-for-kvl-mutual-inductance

So now how does the transformer secondary not have 90 degree phase shift if its an inductor? Because the voltage is all coming from mutual part of inductance and this inductance only cares about the current in the primary.

As you confirm, not only the secondary doesn't behave like an inductor (there's no 90 degree phase shift), but also it is not even an impedance: because the voltage on the secondary is a function of something external to itself, which is in fact the _ _ _ _ _ generated by the current in the primary.

Quote
The current trough the load resistor actually affects the voltage of the primary side.

How can this be, once the primary is connected to a voltage source?

Quote
Once all of these currents and voltages are summed up neither the primary or secondary have 90 degree phase shifts anymore. In an ideal 1:1 transformer with no leakage the currents and voltages are actually all perfectly in phase (If the load is a  resistor like in your example) or 180 out of phase depending on what way around you connect the coil.

So now not even the primary can be identified as an inductor anymore.

Quote
All my hair is still in its place as there is no need to pull it out over a concept that works just fine, or is the use of mutual inductance somehow forbidden?

Forbidden? Why? It perfectly demonstrates what I said before: if you don't understand the underlying physics of electromagnetism you'll be limited in your ability to design and analyze circuits.

Quote
So then back to the original question. Under what circumstances is a wire not an inductor and when it is?

If you read your own words again you'll see that you've already answered your question.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf