Yes according to the math there are two voltages across A and B, but this is not a voltage that can be used for anything until wires are run to it to connect it to something(Such as a voltmeter). Once that is done the path is defined completely and the resulting voltage is only a single well defined voltage appearing across the voltmeter you connected it to. What sort of EMF the voltmeters wires experience is totally up to you, but usually its most usefully that there is zero EMF because you can then consider the wire as being an ideal connection between two circuit nodes.
What is the math for the above example and what will those two different voltages be ? -0V and +0V
From my understanding of physics in this universe (going to assume there is no parallel universe with same experiment running at the exact same time) there will always be a single voltage between any two points at a fixed moment in time no mater if there is a measurement device in there or not.
This is the reason why its so confusing.
This is the reason why its so confusing.
First of all this is not a weird quantum mechanics thing like Schrodingers cat thought experiment that the cat is both dead and alive simultaneously until you look at it. Instead it is actually the fault of how voltage is defined in textbooks.
The more common way of thinking about voltage is that more electrons there are in one place the more negative that point is, connect this area with lots of electrons to an area with few electrons and you get current between them as the charges want to even out.
The way voltage is actually formally defined is "An integral of all forces working on charges along a path between two points". These forces include the electric field that bunched together electrons create, but it also includes the magnetic forces pushing electrons around (Any charged particle is affected by a changing magnetic field). This force is dependan't on where you are in the magnetic field. This results in a different result of the integral of the force and hence a different voltage.
In Dr. Lewins example you get the magnetic EMF adding to the resistors voltage if you go around one way, but subtracting from the resistors voltage if you go around the other way. Hence the integral is different and there is different voltage. Its just how the math ends up working out. The actual electron charge density at the points A and B is always a single well defined number. The voltage you measure by connecting a voltmeter as you shown will measure this electron density. Hence why the voltmeter shows one voltage.
The two different voltages result from Dr. Lewins example could be sort of a incomplete result of the voltage so far, you need to include the rest of the circuit to properly define the voltage. Think of the two voltages as sort of like complex number math, they don't necessarily exist in the real world but they make the math work.
I do not disagree with anything you mentioned in you last replay. Correct measurement will read a single value and that is the real 0V no multiple voltages at a single defined point in time.
Of course if you move the measurement point you get a different reading that is normal in any circuit not just this particular case.
I don't know if someone has previously posted this link. Its worth watching.
Ah okay you want to know the voltages. Given that the total EMF is 1V the result is -0.5V on one side and + 0.5V on the other side. This is because wires show up as having no voltage while the resistors show a voltage according to Ohms law. Since the current trough the loop is identical everywhere this means both resistors have to show the same drop (given they are the same value). Sine the current is flowing upwards in one resistor and down the other you get opposite voltage polarity. So you do get two voltages.
Well, in case of the first circuit supplied by batteries with no varying magnetic flux, you agreed with me that all five voltmeters would read the same, i.e. the real position of the voltmeters didn't matter. But let's forget that for the moment.
Because I agree with you that in this particular case, any misplacement of the voltmeter will reflect on the precision of the measuring.
But my question remains unanswered.
How do I know that this imprecision is due to the voltmeter not being exactly in the place it should be, and not due to a resistor mismatch?
You know, resistors change their values, either with temperature, or age, or both, etc.
My examples where meant to get rid of the confusion of having wires and separate resistors but any example will work the same.
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You can use KVL to find out voltage between any two points on those example rings.
Not sure what sort of point you want to make.
The discussion here is that there is only one voltage between the two defined points at a fixed moment in time and that KVL applied correctly will give you the correct result.
Of course real circuits are not perfect and that is why you have tolerances but you add those tolerances in your calculation and your result will have a margin of error proportional with those tolerances.
Ok, let's take a loop made of two big resistors - physically big - and some copper wire. Let's say the resistors are shaped into an arc spanning 45 degrees. One is 0.1 ohm, the other one is 0.9 ohm. Emf in the loop is 1V.
What is the real and only voltage across the resistors?
What is the real and only voltage across the remaining two portions of wire (say it's copper)?
But all you did was split the ring in to 4 equal parts
and have 4 resistors two with equal low resistance (the copper wires) and the other two with higher resistance 0.1Ohm and 0.9Ohm
you will measure say +650mV across the 0.9Ohm resistor and then -150mV across the 0.1Ohm resistor
But all you did was split the ring in to 4 equal parts
Not equal (360 - 2x45 = 270; 270/2 = 135 != 45), but that's immaterial.Quoteand have 4 resistors two with equal low resistance (the copper wires) and the other two with higher resistance 0.1Ohm and 0.9Ohm
Correct. So, let's make the resistors span 90 degrees instead of 45 so we can use your numbers.Quoteyou will measure say +650mV across the 0.9Ohm resistor and then -150mV across the 0.1Ohm resistor
The only and true voltage across the 0.9 ohm resistor is .65 V. Ohm would say there's a current of .65/.9 = .72 amps
The only and true voltage across the 0.1 ohm resistor is .15 V. Ohm would say there's a current of .15/.1 = 1.5 amps
Now, .65 + .15 = .80 V. EMF is 1V, I suppose you want to put that into 'modified KVL', so did you bungle the calculation or are you assuming there are 0.20 voltage drop on the copper wires? If we assume 0 resistance that would mean ohm would think there's infinite current.
But let's leave ohm alone for the moment because you might want to put some EMF here and there.
Can you put KVL into a formula with numbers? Please make the (possibly corrected, if required) numbers of all the 'true' voltage drops in the loop and show that KVL balance.
Mehdi posted a follow-up video:
The formula is very simple each quarter of the ring (90 degree) will see a quarter of that 1V so 0.25V
Thus depending on the direction of the current you have +0.9V across the 0.9Ohm resistor but you subtract 0.25V thus +650mV
On the other side of the ring (again my assumption the other resistor is on the other half) you have 0.1V on the resistor and subtract 0.25V so -150mV
The copper wires are treated the same as the resistors so a 0.25V source and whatever resistance those wires have say is 1mOhm for each quarter segment in series
If you disagree with this please provide your numbers and how you got to them.
I do not believe in a 'true' voltage (actually it's not a belief, the formulas tell me). In this case it all depends on how you measure it - I can get your numbers by suitably partitioning the disk with the probes, or many other values. But I tell you how I can balance Faraday here:
EMF = 1V, total loop resistance 1 ohm, current 1 amp
integral of E dl in 0.9 ohm + integral of E dl in 0.1 ohm + nearly nothing in copper = EMF
0.9 + 0.1 + 0 = 1
In your case, well, what is the field inside the copper parts, if you have 0.25 V across each of them?
QuoteIf you disagree with this please provide your numbers and how you got to them.
I do not believe in a 'true' voltage (actually it's not a belief, the formulas tell me). In this case it all depends on how you measure it - I can get your numbers by suitably partitioning the disk with the probes, or many other values. But I tell you how I can balance Faraday here:
EMF = 1V, total loop resistance 1 ohm, current 1 amp
integral of E dl in 0.9 ohm + integral of E dl in 0.1 ohm + nearly nothing in copper = EMF
0.9 + 0.1 + 0 = 1
(signs come about when you consider the correct conventions)
My ohm's law still work. I had to give up uniqueness of voltage between two points, though.
Most importantly, there is practically no field inside the copper conductor, as predicted by Maxwell's equations.
In your case, well, what is the field inside the copper parts, if you have 0.25 V across each of them?
If superconductive, then E field is 0.0V. In this case it is 0.002 V because wire is specified as 0.001 Ohms per segment.
[edit] We don't measure E field or EMF between two points of the circuit. We usually measure potential difference.
So are you saying that voltage across the 0.9Ohm resistor in the current example is 0.9V ?
because that sure is not the case unless the resistor has an infinitely small size
If superconductive, then E field is 0.0V. In this case it is 0.002 V because wire is specified as 0.001 Ohms per segment.
[edit] We don't measure E field or EMF between two points of the circuit. We usually measure potential difference.
Thanks I'm worse on expressing myself so your replay is shorter and more to the point.
I'd like to kindly ask the Kirchhoff experts what tools do Kirchhoff rules give me to calculate the "right" voltage of the loop below. Except for the irregular perimeter, all conditions are the same as for my rectangular loop above.
I need to know the exact points where to connect the voltmeter and its precise location. Any reply will be appreciated.
So, you measure electric field in volts?
You need to brush up on basic physics.
Please, humor me and try again: what is the field inside the copper conductor and how do you justify - with formulas - that there is a 0.25 volts drop across it?
So are you saying that voltage across the 0.9Ohm resistor in the current example is 0.9V ?If I measure from outside, and without crossing the flux varying region, yes.
Note that I did not use 'voltage drop' but the integral of E.dl in my balance.
I sort of feel like I deal with trolls
So, you measure electric field in volts?
You need to brush up on basic physics.
I did not mention any units at all. You need to check your vision.
I sort of feel like I deal with trolls and I hope that is not the case.