This will Blow your mind!

That did occur to me, but I think stimulating curiosity in a few while confusing the hell out of most is elitist and not A Good Thing.

He schematically shows voltmeters connected near their respective resistors while claiming that they are both connected to points A and D. Schematically, they are. But schematically, a line conventionally represents a node where voltage is everywhere the same. That isn't the case here because EMF is induced in the wire he shows as a line in his schematic. The razzledazzle is in interpreting the schematic differently at different parts of his presentation while never explicitly pointing that out.

Very well put! Succinct & dead on.

The razzledazzle is in interpreting the schematic

Yes. And while stimulating a few who just can't believe it & go on to the the real answer, it will also leave more (?) who will accept what the Great Professor has said & what they have "seen" with their own eyes.

Bob

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Tell me you're joking, Larry.

--Winston

No test - lead induction is necessary to explain the effect.

Turns out the power in R1 *is* 180 degrees phase - displaced in relation to the power in R2. If Doc had used two voltage dividers instead of just two resistors, he could have displayed differences in instantaneous polarity and voltage with a two - channel scope. Instead, he constructed his circuit so that the really low resistance of the connecting wires shorted the metering channels together, blinding us to the effect he was trying to show.

It still isn't D.C. as he kind of implied, and the circuit he drew cannot display any "polarity" or voltage difference.

I now grok how a different circuit would demonstrate what the prof was on about and show what he was *trying* to say, even if he stated it very very poorly.

--Winston

I don't know what that means.

Conventional notation is that lines in a schematic have zero resistance and represent equipotential nodes. You can't have it both ways in the same presentation.

Back on Thursday, Cross-Slide said: > "I believe the leads are physically connected to the same points.. > The leads themselves must be picking up voltage from the induction > coil."

I was just saying that the effect the Prof pointed out does not rely on inductively coupled voltage in the meter leads themselves.

The resistors R1 and R2 (and their leads) do form a closed, center-tapped inductor, however. A.C. converted from the magnetic field surrounding them circulates back and forth within this loop. The voltage dropped between the center tap and ground is a function of the ratios of R1 and R2. No significant voltage will be present between the center tap and ground when the values of R1 and R2 are equal because at any given instant, an equal amount of positive power produced across the R1 R-L pair is phase-canceled by the negative power produced across the R2 R-L pair (and vice versa at some other instant).

As R1 and R2 differ from each other in terms of resistance, the voltage between the center tap and ground will increase.

Doc didn't give us series 'sense resistors' to measure the amount of current through R1 and R2 or else he would have been able to show evidence of the current flowing within the loop just by connecting a scope channel to each sense resistor.

There are special cases though, like an inductor feature on a PCB or as in my case, components and leads that form two half-turns within a changing magnetic field, each of which have 120 mV p-p of A.C. induced across them (240 mV p-p across both of them) when they are not grounded.

If you build the jig (I showed in my 11:10 post yesterday morning,) you will instantly see the effect of the power induced into the resistive inductors formed by R1 and R2.

The prof's claim is bogus in that the circuit he provided cannot be used to show the effect he claims. A slightly different circuit *can* be used to show an effect that is eerily similar to the effect the prof claims, however.

--Winston

On Sat, 13 Feb 2010 14:52:27 -0800, the infamous Winston scrawled the following:

Newp. I wasn't going to argue Faraday with him because I didn't study him in my computer technology course at Coleman. No EE here, sorry. I followed him through Kirchhoff, though. ;)

-- In order that people may be happy in their work, these three things are needed: They must be fit for it. They must not do too much of it. And they must have a sense of success in it. -- John Ruskin, Pre-Raphaelitism, 1850

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Good! That means you grok Ohms Law which states we cannot expect large voltage differences on the ends of a good conductor that is passing Very Little Current.

So you are 'one up' on Prof Lewin!

--Winston

You demonstrated quite the contrary with your experiment. Induction can produce a voltage difference from end to end of a good or perfect conductor regardless of what current it may be passing. Ohmic IR drop is independent of induction and superposition holds.

No, in my first experiment, I was not able to determine any voltage difference across the wire connecting the top of R1 to the top of R2.

When I saw the maximum voltage in my 2nd experiment, it was being dropped across a ~22.8 K ohm resistor. Current was on the order of 5.3 uA, so I did not expect to see measurable voltage drop across ~1.5" of an 0.022" diameter wire (tens of nano Henrys?).

X(L) for say 50 nH at 60 Hz is what, 19 micro ohms? I'd need a meter that could resolve femtovolts while nulling out millivolts. I sure wasn't going to see the 1.0 V difference predicted by the Prof. :)

That wasn't the proper place to look anyway, as the prof had indicated that the two voltages in question were dropped across R1 and R2. If you isolate the top of R1 from the top of R2 you will see that they *do* have different voltage drops and they *are* 180 degrees phase displaced. His circuit shorts the 'meters' together which completely obliterates the appearance of these differences, however.

Yes, and in my experiment, that voltage difference across the entire winding was about 120 mV, open circuit.

Yes! With R1 and R2 connected together as in the Prof's 2nd circuit, the voltage across R1 was summed with the voltage across R2 to yield a single voltage value. When R1 equals R2, the voltage dropped across them both sums to a value that is very nearly zero because of phase cancellation. In my second experiment, I showed that 5.3 uA through a 11.35K ohm resistor can appear to have a voltage drop of very nearly zero (a great deal less than 60 mV) if it is connected with another

11.35K ohm resistor that is also passing 5.3 uA from the same source, but in the opposite direction.

That is what I have been on about.

--Winston

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These two statements sound much less whacky when you recall that EMF induced into a wire falls as the *cube* of the distance from the magnetic field. The loop connecting the top of R1 to the top of R2 was far enough away from the magnetic field to prevent much induction. The wire sections placed on both sides of the core acted as two separate inductors, each of which converted a tiny portion of the field to ~60 mV of EMF.

--Winston

Let me clarify that.

Maximum voltage readings between the center tap and ground occurred when R1 was nearly zero ohms and R2 was at maximum (22.8K). The other maximum occurred with R1 at 22.8K and R2 at nearly zero ohms. The voltage was continuously produced by the inductor sections attached to the top of the resistors but only became visible as the associated resistor *fell* in value.

Our voltage readings came from the inductor - resistor network that had the *lowest* value of series resistance.

--Winston