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.
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.
If I show you the equations you don't understand.
The adage that any piece of wire is an inductor is a myth.
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.
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?
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.
You clearly have issues.
Anyway thank you, I will try harder
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.
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
Take a moment and look from different angles on what I just said.
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.
QuoteBtw 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?
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?
Since it starts with an F, it is Michael Faraday!
You say, the EMF is generated IN the resistor ! Hmmm. The resistor being the seat of EMF? You are sure?
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?
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 ?
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).
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.
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.
In fact, none of the (duly debunked) "debunkers" of Lewin managed to measure this voltage in any of their circuits.
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.
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.
I admit that last one made me chuckle.
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.
You say, the EMF is generated IN the resistor ! Hmmm. The resistor being the seat of EMF? You are sure?
Absolutely sure.QuoteImagine 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.
QuoteNah, 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.
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.
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?
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?
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.
You'll understand electromagnetism sooner than those who thanked you.
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.
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?
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.
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?