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




"John Polasek" <jpolasek@xxxxxxxxxx> wrote in message news:ggs505thqaupa88oqltlrls38079mlfh2n@xxxxxxxxxx
On Wed, 6 May 2009 21:29:59 -0700, " Don Kelly" <dhky@xxxxxxx> wrote:



"John Polasek" <jpolasek@xxxxxxxxxx> wrote in message
news:n28305l35p3adl658mc5mv3aliqsrggu73@xxxxxxxxxx
On Tue, 5 May 2009 21:43:44 -0700, " Don Kelly" <dhky@xxxxxxx> wrote:

"John Polasek" <jpolasek@xxxxxxxxxx> wrote in message
news:aqc1051i94973qkukn6m0uokde7v5bqspg@xxxxxxxxxx
On Tue, 5 May 2009 14:06:50 -0700 (PDT), Benj <bjacoby@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.

Illegal procedure: you have just assumed what needs to be proved:
namely, that the resistor are *in series* when the video plainly shows
them welded together at top and bottom, yet, inexplicably, with
different voltages.
-----------
Sorry, I should have said, "in series with your "virtual battery" or in
series with the actual battery" Single loop on the assumption of perfect
voltmeters.

While there seems no way out, we first should investigate how the
fluxrate, like a ghost, can introduce the known driving voltage. From
Maxwell we have dB/dt causing:
Curl E = -dB/dt (what a genius to think of this)
Using Stokes to deal with the curl
Int curl E = Int E.dL = 1 volt for 1 turn, showing a
continuous gradient all round
Also
1V/1K = 1 mA (= shorted current of Thevenin supply)

We can forsake the inch by inch gradient in favor of lumped constants
by invoking Thevenins theorem which says the loop represents a power
supply of 1 volt in series with 1K ohms. We can put VThev anywhere in
the loop, and we choose the bottom shorting bar. The 1 volt pushes off
rising from C to D, then falls .9 from D to B and another .1 from A to
C.

A____B
| |
| .1 | .9
C_1V_D

You are right in that *if* we can show that the resistors are in
series we don't need Thevenin. The trick is that Thevenin affords us a
'pry bar' to break the parallel circuit into a series of 3 elements.
--------
But Thevenin's theorem (as is true of most circuit theorems) applies only to
linear circuits. The simple use of KVL and KCL works well.
---

This sure brought out a lot of argumentation for a simple problem- I
wonder
if any of Lewin's student's caught on.
The students probably felt justified in not understanding and just
being as chagrined as the observing professors were said to be. (Not
very chagrined, only a 'B').

An engineer knows Thevenin's theorem; a physicist doesn't. (Check
Smythe and Panofsky who don't have a closed circuit in the whole
book). An admixture is best.
How explain without Thevenin? V = N dph/dt yes, but you need
Thevenin's insight to install V as a 'lump' which is the essential
point in the partitioning.

John Polasek

It is convenient to install V as a lump but this owes nothing to Thevenin
but a hell of a lot to Kirchoff.

There is no illegal procedure.
By illegal procedure I meant you initially assumed a series loop where
the diagram clearly shows the two resistors connected in parallel with
each other, being the source of the astonishment and controversy. (Did
you see the video?).
Granted one VM is showing -.1V and the other +.9V, then it would be
the work of a moment to put the prods across the two shorting bars to
reveal the deficit in the equation, which should be fragments of 1
volt and its complement. I brought that up in a prior note.
-----------

I saw the video- Levin could sell the Brooklyn bridge- In this case he proved that bull*** really can baffle brains.

I did not initially assume a series circuit but it is patently obvious that ignoring the meters there is only a single current loop. Hence, by definition, the circuit is (- see below**). However, there is no need to consider the concepts of series or parallel or Thevenin- Just Kirchoff.

All I did was apply the fundamental core of circuit analysis- Kirchoff's equations.
The sum of the voltages around a closed loop =0 KCL
The sum of currents into a junction =0 KVL
In this case- at every connection between elements, or even between segments of the connecting wires, there is a single "in" and a single "out lead so what goes in comes out and into the next segment/element all around the loop and this reduces to a single current everywhere in the loop. (assuming ideal voltmeters).
Considering KVL then we can say
The sum of the voltage rises due to the sources=the sum of the IR voltage drops. and all elements have a common current.


That is all that is needed.
Nothing more.

**However, for your information, I checked a couple of references

"Elements are said to be connected in series when they all carry the same current" Johnson, Johnson, Hilburn,. "Elecric Circuit Analysis "(2nd Edition Prentice Hall, 1992).

Fitzgerald, Higginbotham, Grabel "Basic Electrical Engineering" agrees but is less succinct.

Wikipedia agrees as well

This indicates that this single loop circuit is "series" and not parallel. The way he took "measurements" tended to give the impression of a parallel circuit- (more misdirection).

As to the deficits:
In the battery case he "measured" the voltage at the top with respect to the bottom - "measuring" across the 100 ohm resistor PLUS the battery getting 1-0.1 =0.9V which was equal to the 0.9V across the 900 ohm resistor. Now when he measures the voltages with the flux source, he "measured" the voltage across the 100 ohm resistor and got -0.1V Thus, he "measured" the -0.1 V rise in the direction of the current but not the battery voltage. He is comparing apples and oranges. The -0.1V rise is the same as in the first case. Flip it to get a voltage drop of +0.1V in the direction of the current and add it to the 0.9V drop in the direction of the current in the 900 ohm resistor- the sum is 1V which is the same as the induced voltage. No "residue" left over. KVL satisfied, KCL satisfied, Faraday satisfied.

You have indicated the use of a virtual battery" which could be anywhere in the circuit without affecting the voltage drops across the resistors or the current in the circuit- but affecting the top-bottom "meter" readings depending on location. Levin made no statements as to the location and distribution of the virtual battery but obviously was not including any part of the induced voltage in either the left or right leg.

HOWEVER
Note that there were *no actual measurements* made- all values were calculated for an idealized *blackboard model * with ideal elements.- Naturally there were no discrepancies. Note also the the current was calculated as 1ma by Ohm's law and this required a resistance of 100+900 ohms (Req=R1+R2 for a series circuit).
Cheers,
--
Don Kelly
dhky@xxxxxxxxxxxx
remove the x to reply

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