Re: Faraday "magnetic" induction without a magnetic field.



"blackhead" <larryharson@xxxxxxxxxxxx> wrote in message news:9c568efc-2365-4ae8-a21c-7c88e19f3e36@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 7 May, 06:14, " Don Kelly" <d...@xxxxxxx> wrote:
"blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

news:46fc322f-33af-47c5-b7b1-87ae50e74ee8@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 6 May, 05:43, " Don Kelly" <d...@xxxxxxx> wrote:





> "John Polasek" <jpola...@xxxxxxxxxx> wrote in message

>news:aqc1051i94973qkukn6m0uokde7v5bqspg@xxxxxxxxxx

> > On Tue, 5 May 2009 14:06:50 -0700 (PDT), Benj <bjac...@xxxxxxxxxxx>
> > wrote:

> >>On May 5, 9:30 am, John Polasek <jpola...@xxxxxxxxxx> wrote:

> >>> The sum of the voltages around the loop will be zero. The "hidden"
> >>> batteries" belong not to Lewin, but only to Thevenin.

> >>The "hidden' batteries belong to Thevenin, but also to Faraday. One is
> >>used to thinking that a piece of wire connecting two resistors puts
> >>the ends at the same potential, but of course it doesn't because of
> >>the integral around the loop.
> > I don't know what you're explaining to me, but it would be real noble
> > if you retracted your slanderous accusation of Lewin surreptitiously
> > using hidden batteries to further his claim.

> > I already pointed out that instead of 2 resistors with two taps as
> > shown, we have in reality a 3-tap circuit after inserting Thevenin's
> > volts, so sensibilities are once again restored.

> > So Faraday works 100% and Kirchoff is likewise affirmed.
> > John Polasek

> ------------------------
> John,
> I came into this late because my computer had been down. I agree with > your
> final conclusion.

> However, there is no need to even consider "Thevenin" .
> If the resistors were non-linear, the same result would occur but
> Thevenin's
> theorem would be invalid.

> If you consider the circuit as having a string of n (you choose n) > series
> resistors- all but two being 0, then the total voltage across these two
> individual resistors would be 1V and distributed the same way that they
> would be distributed with the battery in circuit- even if non-linear. > The
> video measured voltage across the 100 ohm resistor and battery in the
> first
> case and the voltage across the 100 ohm resistor in the second case- > with
> a
> lot of hype in between.

> The demo is comparing apples and oranges- the voltage across the 100 ohm
> resistor plus a physical battery is compared to the voltage across the
> resistor alone.

> This sure brought out a lot of argumentation for a simple problem- I
> wonder
> if any of Lewin's student's caught on.

> --
> Don Kelly
> d...@xxxxxxxxxxxx
> remove the x to reply- Hide quoted text -

> - Show quoted text -

I don't think you've understood the point of the video: Two resistors
are in parallel, yet havehave different voltages across them.

I understood the point, It was a con job and it got you.

What was he hiding if it was a con job?
------------------
Nothing but he wasn't telling you everything- and stated things deliberately in such a way that many fell for it. He actually wanted someone to think and point out the fallacies involved.
------------------------

In one case the voltage across a resistor and a battery (1-0.1) was measured
and compared to the voltage across the other resistor (0.9)= agreement.
In the second case the voltage across one resistor(0.1) and the second
resistor (0.9) were measured . The `distributed battery`` was not included
in the measurement.

He calculated the *measured* voltage across the resistors. He was
comparing *measured* voltages across the resistors that you would
observe in practise.
----------
Right and then he took the "measured" voltage across the battery plus the 100 ohm resistor (0.9V) which he compared to the 0.9V "measured" across the 900 ohm resistor. These agreed. Then with the battery out of circuit, he took as you indicate, the calculated voltages across the resistors (and leads) only and got -0.1V vs 0.9V -the previously calculated IR drops and implied that that was what the meter if it existed would read.!!!!!
These would not be the voltages that would be observed in practice.

Implicit in this and not mentioned is that there was no induced voltage in either the left or right side of the circuit -There is no reason to believe that to be true. We actually have no information about the distribution of incremental voltages around the circuit- just that the total was 1V. The distribution in a real circuit (not a blackboard model as this was) depends on the geometry of the loop.
-----

Secondly, go back to the definition of series and parallel circuits. A
series has a single current common to all elements of the circuit and
voltages across elements differ. A parallel circuit has a common voltage
across all elements and currents in the elements differ (implying a
splitting of current between paths). In this primitive circuit we have a
common (single) current and different voltages across elements in the
circuit.- A single loop satisfies the criterion for a series circuit. It
doesn`t satisfy the criterion for a parallel circuit.

The resistors are NOT in parallel- either with a battery or with an induced
voltage source. There is no common voltage across the resistors. In fact
there is a voltage source- either a battery or a distributed induced voltage
in a single loop which contains resistors -that is series.
If the source produced a current which was split between two branches- then
that is parallel- this isn`t the case here.

OK, let's call them in series then.
---
According to the definition of a series circuit-
--

A voltmeter across the 100Ohms will measure 0.1V, across the 900Ohms
will measure 0.9V. So even though the voltmeters are physically in
parallel with one another, they measure different voltages! I don't
think you've grasped this point. Put another way, your voltmeter will
read different voltages between A and B depending on where it's
located in a circuit, even though there is no changing magentic field
in the loop the voltmeter is a part of. And that's surprising, at
least for me.

---------
The voltmeter was across the resistance *and the leads* between the terminals where the meter was connected- and assumes (assumption not explicit but obvious as the voltages given were the calculated IR voltages) no induced voltage in these parts , then all the induced voltage would have to be in the top and bottom "shorted" bars (and the induced voltage in these bars depends on the flux and the geometry of the situation, not the fact that the bars are assumed to have 0 resistance). If you wish you can lump the induced voltage anywhere in the loop (or split it into several lumps in different places).
If on the left side- then the meters as connected would read 0.9 on both left and right sides. If on the right side, the meters would read -0.1 V on both sides. In these cases, the voltmeters would be in parallel with each other (this doesn't make the main circuit a parallel circuit). The "results" obtained correspond to the effective induced voltage being in the "shorted" bars and in this case the voltmeters would be measuring different legs of the circuit as they would not be in parallel. Work it out. Consider Vab=0, Vbc=0.9v, Vcd=-1V (source), Vba=100I (or Vab as measured would be -0.1V. Try it with the source moved to some other part of the circuit or split equally between the 4 legs- the IR drops, the source voltage and the current will not change but the "meters will have different readings.

No problems exist in this case. If, as would be the situation in a real circuit- thee induced voltage is distributed somehow, the voltmeters would still read differently because they would be measuring the IR drop + some of the induced voltage. Note that he said nothing about the voltage across the "shorting bars" leaving one to think it is 0 because the resistance is 0. The loop resistance or variations in resistance has no effect on the induced voltage or its distribution in the loop- that is dependent on geometry which is not taken into account in lumped circuit analysis.

So- a con job with a purpose- trying to get people to think- that was his job.

Don Kelly
dhky@xxxxxxxxxxxx
remove the x to reply- Hide quoted text -

- Show quoted text -

.

Relevant Pages

  • Re: transient analysis of linear system
    ... Anyway the circuit is shown above. ... Clearly in steady state it's just a voltage divider of the difference of Vx and Vy. ... The point of all this is to 'see' if the resistors change through the "fog" caused the time varying sources. ... Looks like you have everything wrong, attempting to measuring a circuit parameter that nature is forcing to be constant, meaning you have to measure *current* to detect the resistor changes, the voltage measurements will barely move by ppm and be undiscernible from drift. ...
    (sci.electronics.design)
  • Re: transient analysis of linear system
    ... Anyway the circuit is shown above. ... voltage divider of the difference of Vx and Vy. ... the user but it must be the steady state result. ... The point of all this is to 'see' if the resistors change ...
    (sci.electronics.design)
  • Re: Faraday "magnetic" induction without a magnetic field.
    ... I already pointed out that instead of 2 resistors with two taps as ... If you consider the circuit as having a string of n ... resistors- all but two being 0, then the total voltage across these two ... would be distributed with the battery in circuit- even if non-linear. ...
    (sci.physics.electromag)
  • Re: Faraday "magnetic" induction without a magnetic field.
    ... ignoring the meters there is only a single current loop. ... the circuit is. ... Thevenin's voltage is an easy way that usually ... You have an ideal voltage source in series with two resistors -3 elements with a single common current (Note that Lewin also assumed this when calculating the current by Ohms law. ...
    (sci.physics.electromag)
  • Re: The KISS AMP: a progress report
    ... It would be useful to know the secondary voltage at that transformer ... read your output from the rectifier tube, that would come to some 415V ... or so out of the 5AR4/GZ34, excepting the dropping resistors. ...
    (rec.audio.tubes)