Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

On Feb 11, 2:25*pm, Cross-Slide wrote:
Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

The way a homopolar generator works showed me you can't really trust
your intuition.

Spin copper disk between magnets- current.
Spin copper disk with magnets attached to the disk- current.
Hold disk still and spin the magnets around the disk (exact converse
of first case)- no current.


Dave

Dave__67 fired this volley in news:73a26711-3806-
:

Hold disk still and spin the magnets around the disk (exact converse
of first case)- no current.


I think you entirely missed the point on that one. The pickup (the
brushes) have to remain fixed relative to the magnets in order to see the
potential in the disk. The potential is only developed under the magnet.

So, to remain stationary with respect to the magnets, the pickup would
have to rotate with them.

So your case is NOT the "exact converse of the first case", because you
didn't rotate the pickup.

'sides.... you can't have an "exact converse" of a three-variable
arrangement by changing only two variables.

LLoyd

Lloyd E. Sponenburgh wrote:
Dave__67 fired this volley in
news:73a26711-3806-
:

Hold disk still and spin the magnets around the disk (exact converse
of first case)- no current.


I think you entirely missed the point on that one. The pickup (the
brushes) have to remain fixed relative to the magnets in order to see
the potential in the disk. The potential is only developed under the
magnet.

So, to remain stationary with respect to the magnets, the pickup would
have to rotate with them.

So your case is NOT the "exact converse of the first case", because
you didn't rotate the pickup.

'sides.... you can't have an "exact converse" of a three-variable
arrangement by changing only two variables.

LLoyd

Can you even have the converse of a 3 variable system??

OK, how bout this non-intuition:

The Doppler effect:

You are standing still, an ambulance with sirens blasting approaches, then
recedes. Observe frequency change.
OK, now Ambulance is sitting still with sirens blasting (as they do in
effing Yonkers, NY), and you approach, recede at same speed as the ambulance
previously. Different frequency change!!
goodgawd....

Lots of inneresting counter-intuitive stuff, just can't remember many.
Marketers of course milk the **** out of our now topsy-turvy intuitive-less
world.
Like, Tobaccer companies telling us not to smoke....
WTF??????????????????????????
--
EA

On Feb 11, 3:02*pm, "Lloyd E. Sponenburgh"
lloydspinsidemindspring.com wrote:
Dave__67 fired this volley in news:73a26711-3806-
:

Hold disk still and spin the magnets around the disk (exact converse
of first case)- no current.

I think you entirely missed the point on that one. *The pickup (the
brushes) have to remain fixed relative to the magnets in order to see the
potential in the disk. *The potential is only developed under the magnet.

On 2/11/2010 11:25 AM, Cross-Slide wrote:


Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

The prof is 'way overdue for a vacation.

He changed the resistors from being in series with
the cell to being in parallel with the inductor.

V(d)-V(a) = V(r1)= V(r2) = 1.0 V

BTW
I(r1)= V(r1)/r1=0.01 A
I(r2)= V(r2)/r2=0.001111... A

--Winston

--
Support the blind and deaf. Hire a professor today!

"Cross-Slide" wrote in message
...


Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

Nice one! What the 'prof' fails to acknowledge is that in
a changing mag field the wires connecting the resistors have
a voltage gradient along their length (think armature wires
in a generator). He is playing on the normal assumption
that the voltage along a wire is the same everywhere, given
it's much lower resistance than the resistors.
Art

On Thu, 11 Feb 2010 14:35:20 -0800, "Artemus" wrote:


"Cross-Slide" wrote in message
...


Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

Nice one! What the 'prof' fails to acknowledge is that in
a changing mag field the wires connecting the resistors have
a voltage gradient along their length (think armature wires
in a generator). He is playing on the normal assumption
that the voltage along a wire is the same everywhere, given
it's much lower resistance than the resistors.
Art



There is no net l.dB/dt in the loop formed by the meter leads and the segment
under test. So you only end up measuring the resistive voltage drop, not the
induced emf. He's not stating/ignoring that point though :-)

I couldn't see what was so counter intuitive. Maybe EE students are more
easily puzzled now than in days of yore.


Mark Rand
RTFM

On Feb 11, 5:35*pm, "Artemus" wrote:
"Cross-Slide" wrote in message

...



Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

Nice one! *What the 'prof' fails to acknowledge is that in
a changing mag field the wires connecting the resistors have
a voltage gradient along their length (think armature wires
in a generator). *He is playing on the normal assumption
that the voltage along a wire is the same everywhere, given
it's much lower resistance than the resistors.
Art

I finally got a chance to watch it...

I don't think that's it.

In the first case the voltage was the 'driving force', in the second
case the current is forced.

The second case can be properly modeled with an ideal current source
of zero internal resistance, I believe.

Dave

Cross-Slide wrote:

Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

Experiencing that as a video is really frustrating! I want to ask Prof
Lewin questions, I want to see the connections, I want to try changing
them, etc. Is one scope connected across R1 & the other across R2, with
lengths of wire between? I think. And that wire has voltage induced in
it. (Am I able calculate the effect of that? No.)

As someone else said, the scopes cannot be literally connected to the
same points. A mind experiment comes to mind :-) - a scope on the left
connected to *points* A & D and one on the right also connected to A &
D, as P. Lewin *implies*. Now I switch them by leaving them connected &
simply moving them. Do the displays change? Obviously not. What makes
the displays different? That the scopes are not literally connected to
A & D, but across the resistors.

Bob

On Thu, 11 Feb 2010 11:25:09 -0800 (PST), Cross-Slide
wrote:



Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

This professor is trying to confuse, bedazzle and impress rather than
teach or elucidate. Nowhere does he explain that meters connected
across the resistors are not connected to points A and D. Wires are
assumed to be of zero resistance or they would be schematically shown
as resistors, but the changing magnetic field induces a potential
difference along each wire, with total loop induced voltage said to be
1 volt. He deceptively neglected to address the relevant matter of
how this EMF was proportionally distributed in top wire, bottom wire
and the two resistors, a matter which would depend on the physical
geometry in a way left totally unaddressed in his performance. He very
briefly mentioned path of integration and showed some (incorrect)
definite integrals but nowhere did he show or speak of line
integrals.

He should be taken to the academic woodshed. He should be reprimanded,
and required to take and pass 9 credits from graduate student
instructors (for which English is a third language) in each of
political science, women's studies, elementary education for K thru 3
including a student teaching assignment, and boxing at 6 AM MWF. After
completing that penance, he should then assigned to clean the latrines
of all engineering fraternity houses for two years, and then be fired.

On 2/11/2010 9:11 PM, Don Foreman wrote:
On Thu, 11 Feb 2010 11:25:09 -0800 (PST), Cross-Slide
wrote:



Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

This professor is trying to confuse, bedazzle and impress rather than
teach or elucidate.

Yes.

(...)

After completing that penance, he should then assigned to clean the
latrines of all engineering fraternity houses for two years, and then
be fired.

Yes.


--Winston

On Thu, 11 Feb 2010 23:11:24 -0600, the renowned Don Foreman
wrote:

On Thu, 11 Feb 2010 11:25:09 -0800 (PST), Cross-Slide
wrote:



Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

This professor is trying to confuse, bedazzle and impress rather than
teach or elucidate. Nowhere does he explain that meters connected
across the resistors are not connected to points A and D. Wires are
assumed to be of zero resistance or they would be schematically shown
as resistors, but the changing magnetic field induces a potential
difference along each wire, with total loop induced voltage said to be
1 volt. He deceptively neglected to address the relevant matter of
how this EMF was proportionally distributed in top wire, bottom wire
and the two resistors, a matter which would depend on the physical
geometry in a way left totally unaddressed in his performance.

Yes, not to mention the voltages induced in the loops connected to the
meters.

He very
briefly mentioned path of integration and showed some (incorrect)
definite integrals but nowhere did he show or speak of line
integrals.

He should be taken to the academic woodshed. He should be reprimanded,
and required to take and pass 9 credits from graduate student
instructors (for which English is a third language) in each of
political science, women's studies, elementary education for K thru 3
including a student teaching assignment, and boxing at 6 AM MWF. After
completing that penance, he should then assigned to clean the latrines
of all engineering fraternity houses for two years, and then be fired.

What do you really think? ;^)


Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

On Feb 12, 8:26*am, Spehro Pefhany
wrote:
On Thu, 11 Feb 2010 23:11:24 -0600, the renowned Don Foreman


Yes, not to mention the voltages induced in the loops connected to the
meters.


Best regards,
Spehro Pefhany
--
"it's the network..." * * * * * * * * * * * * *"The Journey is the reward"
* * * * * * Info for manufacturers:http://www.trexon.com
Embedded software/hardware/analog *Info for designers: *http://www.speff.com

The internal resistance of an oscilloscope is very high, high enough
it will look like an open circuit (and hence no current), so I think
the voltage would be negligible and ignorable.

Dave

On Fri, 12 Feb 2010 05:34:48 -0800 (PST), Dave__67
wrote:

On Feb 12, 8:26*am, Spehro Pefhany
wrote:
On Thu, 11 Feb 2010 23:11:24 -0600, the renowned Don Foreman


Yes, not to mention the voltages induced in the loops connected to the
meters.


Best regards,
Spehro Pefhany
--
"it's the network..." * * * * * * * * * * * * *"The Journey is the reward"
* * * * * * Info for manufacturers:http://www.trexon.com
Embedded software/hardware/analog *Info for designers: *http://www.speff.com

The internal resistance of an oscilloscope is very high, high enough
it will look like an open circuit (and hence no current), so I think
the voltage would be negligible and ignorable.

Dave

No current means that there will be no voltage drop due to wire
resistance.

An open loop of wire in the changing field such as the one that the
good Dr. is discussing will have 1V induced in it, just like the
secondary of a transformer.

So a voltmeter with the leads "shorted" together will read 1V
(polarity depending on the orientation and voltmeter polarity).

On Feb 11, 8:32*pm, Bob Engelhardt wrote:
Cross-Slide wrote:

Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

Experiencing that as a video is really frustrating! *I want to ask Prof
Lewin questions, I want to see the connections, I want to try changing
them, etc. *Is one scope connected across R1 & the other across R2, with
lengths of wire between? *I think. *And that wire has voltage induced in
it. *(Am I able calculate the effect of that? No.)

As someone else said, the scopes cannot be literally connected to the
same points. *A mind experiment comes to mind :-) - a scope on the left
connected to *points* A & D and one on the right also connected to A &
D, as P. Lewin *implies*. *Now I switch them by leaving them connected &
simply moving them. *Do the displays change? *Obviously not. *What makes
the displays different? *That the scopes are not literally connected to
A & D, but across the resistors.

Bob

The scopes are literally connected to points A and D.

Think about This mind experiment...
You have two resistors connected in a ring around a ( tree for the
sake of visualization) When you are looking at the south resistor, on
the south side of the tree, and current is flowing through the
resistor, The East point is Positive and the west point is negative.

Since the current is flowing around the ring..
As you look at the North resistor, on the north side of the tree, the
West point is positive, and the East point is negative. It HAS to be,
since the current is still flowing in the same direction..

The EXACT SAME TWO POINTS are either negative or positive depending on
which side you stand as you measure them.

Your whole life you have analyzed circuits that had a "point
source" ( for lack of a better description) of power, either a voltage
or current source.

What the professor showed was how NON-INTUITIVE it becomes when the
power ( voltage or current source) is distributed along the length of
the circuit.

The professor did an amazing job of getting some people to think, and
others to deny what they are seeing.
Just like he warned at the beginning, and I warned in the title.
The deniers reacted with some hostility, and refuse to consider the
alternate way there might be to analyze a circuit.

Someone suggested that the circuit switched from a parallel to series
circuit.
Well, with only Two elements, ( two resistors) it is either or both
series and parallel.
Since the power source is distributed around the circuit, it is not a
separate circuit element.

On Feb 12, 11:54*am, Cross-Slide wrote:

The scopes are literally connected to points A and D.


No, they're not. They're connected at the topo and bottom of the 2
resistors, with some wire in-between.

Think about This mind experiment...
You have two resistors connected in a ring around a ( tree for the
sake of visualization) When you are looking at the south resistor, on
the south side of the tree, *and current is flowing through the
resistor, The East point is Positive and the west point is negative.

Since the current is flowing around the ring..
As you look at the North resistor, on the north side of the tree, the
West point is positive, and the East point is negative. It HAS to be,
since the current is still flowing in the same direction..

The EXACT SAME TWO POINTS are either negative or positive depending on
which side you stand as you measure them.


This is always the case. Voltage is always relative. From the
standpoint of the current direction, one scope has its "+" side hooked
up on the opposite side of the resistor compared to the other one ( +
of one scope is hooked up where the current goes in to the R, the
other scope has its + side where the current comes out of that R)

Your whole life you have analyzed circuits that had a "point
source" ( for lack of a better description) of power, either a voltage
or current source.

What the professor showed was how NON-INTUITIVE it becomes when the
power ( voltage or current source) is distributed along the length of
the circuit.

The professor did an amazing job of getting some people to think, and
others to deny what they are seeing.
Just like he warned at the beginning, and I warned in the title.
The deniers reacted with some hostility, and refuse to consider the
alternate way there might be to analyze a circuit.


I don't see any deniers, just folks who don't find it that amazing.

Someone suggested that the circuit switched from a parallel to series
circuit.
Well, with only Two elements, ( two resistors) it is either or both
series and parallel.
Since the power source is distributed around the circuit, it is not a
separate circuit element.

I don't think any of those actually watched the whole video where he
says in an inset that he boogered the equation, but he didn't booger
the math.


Dave

On Fri, 12 Feb 2010 08:54:58 -0800 (PST), Cross-Slide
wrote:

On Feb 11, 8:32*pm, Bob Engelhardt wrote:
Cross-Slide wrote:

Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

Experiencing that as a video is really frustrating! *I want to ask Prof
Lewin questions, I want to see the connections, I want to try changing
them, etc. *Is one scope connected across R1 & the other across R2, with
lengths of wire between? *I think. *And that wire has voltage induced in
it. *(Am I able calculate the effect of that? No.)

As someone else said, the scopes cannot be literally connected to the
same points. *A mind experiment comes to mind :-) - a scope on the left
connected to *points* A & D and one on the right also connected to A &
D, as P. Lewin *implies*. *Now I switch them by leaving them connected &
simply moving them. *Do the displays change? *Obviously not. *What makes
the displays different? *That the scopes are not literally connected to
A & D, but across the resistors.

Bob

The scopes are literally connected to points A and D.

Think about This mind experiment...
You have two resistors connected in a ring around a ( tree for the
sake of visualization) When you are looking at the south resistor, on
the south side of the tree, and current is flowing through the
resistor, The East point is Positive and the west point is negative.

Since the current is flowing around the ring..
As you look at the North resistor, on the north side of the tree, the
West point is positive, and the East point is negative. It HAS to be,
since the current is still flowing in the same direction..

The EXACT SAME TWO POINTS are either negative or positive depending on
which side you stand as you measure them.

If the only current present is circulating around the tree, then the
voltage measured from east to west is zero regardless of where you
stand when measuring it. In order for there to be circulating
current, there must be induced EMF from north to south, with polarity
in the west leg opposite that of polarity in the east leg, and the sum
of these potentials must be equal to the sum of the voltage drops on
the resistors to satisfy Kirchoff's law.


Your whole life you have analyzed circuits that had a "point
source" ( for lack of a better description) of power, either a voltage
or current source.

Good luck with transformers, motors, antennae and transmission lines.

What the professor showed was how NON-INTUITIVE it becomes when the
power ( voltage or current source) is distributed along the length of
the circuit.

Nothing about electricity is intuitive until it is understood. There
is nothing intrinsically more non-intuitive about induction than there
is about conduction. It merely needs to be explained.

The professor did an amazing job of getting some people to think, and
others to deny what they are seeing.

Legerdemain. Mystify rather than illucidate. Brilliant!

Cross-Slide wrote:
The scopes are literally connected to points A and D.

No, no, NO. Two instruments connected to the same 2 points will not
give different readings just because you say that one of them is reading
the right-side voltage & the other is reading the left. Absurd on the
face of it!

Think about This mind experiment...
You have two resistors connected in a ring around a ( tree for the
sake of visualization) When you are looking at the south resistor, on
the south side of the tree, and current is flowing through the
resistor, The East point is Positive and the west point is negative.

Since the current is flowing around the ring..
As you look at the North resistor, on the north side of the tree, the
West point is positive, and the East point is negative. It HAS to be,
since the current is still flowing in the same direction..

We're not talking about polarity, but magnitude.

Bob

On 2/12/2010 8:54 AM, Someone Else wrote:

(...)

I built and tested the prof's circuit just now.

The voltage across R1 and R2 were equal in
voltage and polarity.

How could they not be?


--Winston


--
Support the blind and deaf. Hire a delusional today!

On Feb 12, 2:25*pm, Winston wrote:
On 2/12/2010 8:54 AM, Someone Else wrote:

(...)

I built and tested the prof's circuit just now.

The voltage across R1 and R2 were equal in
voltage and polarity.

How could they not be?

--Winston

--
Support the blind and deaf. Hire a delusional today!

Experimental setup details?
Voltmeters or scopes?
Did you connect the 'meters' at the same physical place?
How much wire (and gauge) between the resistors?

Details on your solenoid?


Dave

On Feb 12, 12:54*pm, Bob Engelhardt wrote:
Cross-Slide wrote:
The scopes are literally connected to points A and D.

No, no, NO. *Two instruments connected to the same 2 points will not
give different readings just because you say that one of them is reading
the right-side voltage & the other is reading the left. *Absurd on the
face of it!


And that is what is interesting.
Any answer is absurd.
Two resistors, connected in a Ring, can you measure the voltage drop
across either of them?
If you cannot measure a voltage drop across a resistor, is it passing
current?
If you can measure a voltage across a resistor, How do you account for
the other resistor connected in series with it?
The two resistors are both connected to the same two points. A and D.
If there is no voltage measured from A to D, are the resistors passing
current?
If the Polarity from A-D is the same for either resistor, that too is
impossible.

Can you tell me WHAT the correct answer will have to be?
Any answer can shown to be absurd, so which one do you think it is?

On 2/12/2010 11:38 AM, Dave__67 wrote:
On Feb 12, 2:25 pm, wrote:
On 2/12/2010 8:54 AM, Someone Else wrote:

(...)

I built and tested the prof's circuit just now.

The voltage across R1 and R2 were equal in
voltage and polarity.

How could they not be?

--Winston

--
Support the blind and deaf. Hire a delusional today!

Experimental setup details?
Voltmeters or scopes?

One four - channel scope. Hitachi V-1100A

Did you connect the 'meters' at the same physical place?

Physically, no. Electrically, yes (for this 60 Hz testing
purpose). One channel across R1 and one across R2.
'Return' wires are clipped to the opposite end of each
resistor respectively.

How much wire (and gauge) between the resistors?

Length about 1.5" on one side and 1" on the other side.
Component leads are about 0.018" diameter.
Wire is about 0.022" diameter.

Details on your solenoid?

Solenoid is an unshielded "new-old stock" power transformer
about the same size as a conventional 'doorbell' transformer.
The secondary is not connected to anything but is center
tapped and labeled for 3 V on each side of the tap.
It measures about 7.1 V RMS open circuit across the entire
winding.
The resistor component leads are routed between the outside
of the transformer winding and the inside of the EI core.


--Winston

On Feb 11, 3:16*pm, "Existential Angst"
wrote:
[snip]
Lots of inneresting counter-intuitive stuff, just can't remember many.
Marketers of course milk the **** out of our now topsy-turvy intuitive-less
world.
Like, Tobaccer companies telling us not to smoke....
WTF??????????????????????????
--
EA

Oh, that's not even close to the most ridiculous marketing bull****.
How about Viagra commercials giving dire warnings of impending doom if
you have a hard-on for more than FOUR HOURS !!!! Talk about taking
lemons and making lemonade. And yes, I know that priapism can cause
permanent damage, but it gives the viagra pushers a chance to slip in
the notion that you can get a never ending boner with this stuff.
Maybe Gunner should give it a try.

On Fri, 12 Feb 2010 12:35:07 -0800 (PST), Cross-Slide
wrote:

On Feb 12, 12:54*pm, Bob Engelhardt wrote:
Cross-Slide wrote:
The scopes are literally connected to points A and D.

No, no, NO. *Two instruments connected to the same 2 points will not
give different readings just because you say that one of them is reading
the right-side voltage & the other is reading the left. *Absurd on the
face of it!


And that is what is interesting.
Any answer is absurd.
Two resistors, connected in a Ring, can you measure the voltage drop
across either of them?
If you cannot measure a voltage drop across a resistor, is it passing
current?
If you can measure a voltage across a resistor, How do you account for
the other resistor connected in series with it?
The two resistors are both connected to the same two points. A and D.
If there is no voltage measured from A to D, are the resistors passing
current?
If the Polarity from A-D is the same for either resistor, that too is
impossible.

Can you tell me WHAT the correct answer will have to be?
Any answer can shown to be absurd, so which one do you think it is?

Read my post some way up. The loop formed by the meter leads and the section
of circuit under test has no net induced voltage (no source of MMF in the
loop), therefore, it only reads the resistive voltage drop in the resistor,
not the induced voltage that is generating the current.

This is Field theory 101.

Mark Rand(BSc Hons. E.E)
RTFM

On Fri, 12 Feb 2010 12:46:02 -0800, Winston wrote:

On 2/12/2010 11:38 AM, Dave__67 wrote:
On Feb 12, 2:25 pm, wrote:
On 2/12/2010 8:54 AM, Someone Else wrote:

(...)

Details on your solenoid?

Solenoid is an unshielded "new-old stock" power transformer
about the same size as a conventional 'doorbell' transformer.
The secondary is not connected to anything but is center
tapped and labeled for 3 V on each side of the tap.
It measures about 7.1 V RMS open circuit across the entire
winding.
The resistor component leads are routed between the outside
of the transformer winding and the inside of the EI core.


--Winston


Layout of test leads WRT solenoid and resistor when making measurement?


Mark Rand
RTFM

Ok - now take a hex inverter and drive the input from another output
and so on - until you have six in a ring - hex-ring if you will.

Now power it up and - that happens. Oscillates. Any / all outputs
drive signals that are running at the prop time of a gate. It hits
its own magic frequency. Faster propagation times - faster pulse train.

I was working on one in GaAs that had switches that changed the loop
length for a customer. The project never got off, but a 64 gate version
was something! It had feedback lines from half a ring away back into the
ring.

Martin

On 2/12/2010 4:18 PM, Mark Rand wrote:
On Fri, 12 Feb 2010 12:46:02 -0800, wrote:

On 2/12/2010 11:38 AM, Dave__67 wrote:
On Feb 12, 2:25 pm, wrote:
On 2/12/2010 8:54 AM, Someone Else wrote:

(...)

Details on your solenoid?

Solenoid is an unshielded "new-old stock" power transformer
about the same size as a conventional 'doorbell' transformer.
The secondary is not connected to anything but is center
tapped and labeled for 3 V on each side of the tap.
It measures about 7.1 V RMS open circuit across the entire
winding.
The resistor component leads are routed between the outside
of the transformer winding and the inside of the EI core.


--Winston


Layout of test leads WRT solenoid and resistor when making measurement?

Dangling below the vise that holds the transformer, one on each side.
Why do you ask, Mark?

--Winston

On 2/12/2010 4:18 PM, Mark Rand wrote:

Layout of test leads WRT solenoid and resistor when making measurement?

It doesn't matter.

I did a little experiment. The wire around the transformer is
most of a one turn loop within the core's magnetic field, as I
mentioned earlier. This sports only one tap instead of two.
The scope is set to trigger on positive transitions of Channel 2.



/---------------------\
| |
| |
| |
/-----o------\ |
| | |
| | | /--------------------\
| | | | /----------------\ |
| | | | | | |
AC Hot | | | | | 2 Ch Scope | |
------------------. ,--------------\ | | | | |
| )|( | | | | \----------------/ |
AC Ret | )|( | | \--------o /---o |
------------------' '---------\ | | | |
| | | | | o | o |
| | | | \----|---|---|-------/
| | | | | | |
| | | | | | |
| | | | | | |
| | | | | | |
| CW | | | | | |
| ___ | | \ -----------------/ |
\----|___|---/ | | |
A | | |
| | | |
| | | |
| | | |
| | | |
| | | |
\---------o-------------------o-------/


Here is what I observed.

With the pot fully clockwise, I got a 60 Hz sinewave on
channel 1 which was in phase with the sinewave I saw
on channel 2.

As I rotated the slider, the voltage on channel 1 decreased
until at dead center, the voltage on channel 1 flatlined.

Rotating CCW (anticlockwise), the sinewave reappeared on channel 1
but at 180 degrees phase relationship to channel 2.

Continuing to fully CCW, the voltage on channel 1
increased to it's maximum, still exactly out of phase with
the reference waveform on channel 2.

I think this explains why I wasn't able to reproduce the
results predicted in Doc Lewin's theory.

Turns out the power in the left half of the winding will
appear to exactly phase - cancel the power in the right half
of the winding if R1 and R2 are equal. The power present
across this combination can flip phase *in relation to the
incoming waveform*. Our point of reference was the center
point of the resistors, so we could not see the phase reversal.
We would have just seen decreasing amplitude as resistor
values approached each other and increasing amplitude as a
function of the ratio of the resistance values.

No magic required. No D.C. of different, opposite voltage
produced between ground and taps on the top of the sense
winding. No significant offsets produced by inductive coupling
into the test leads are necessary to explain the earlier
test results.

The original claim is bogus.


--Winston

On Thu, 11 Feb 2010 23:11:24 -0600, the infamous Don Foreman
scrawled the following:

On Thu, 11 Feb 2010 11:25:09 -0800 (PST), Cross-Slide
wrote:



Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

This professor is trying to confuse, bedazzle and impress rather than
teach or elucidate.

If he manages to spark CURIOSITY in a few more kids, isn't this A Good
Thing(tm)? He's teaching. Allow him some poetic license, please.

--
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

On Sat, 13 Feb 2010 06:45:03 -0800, Larry Jaques
wrote:

On Thu, 11 Feb 2010 23:11:24 -0600, the infamous Don Foreman
scrawled the following:

On Thu, 11 Feb 2010 11:25:09 -0800 (PST), Cross-Slide
wrote:



Youtube video, for the electrical types.

http://www.youtube.com/watch?v=eqjl-qRy71w

This professor is trying to confuse, bedazzle and impress rather than
teach or elucidate.

If he manages to spark CURIOSITY in a few more kids, isn't this A Good
Thing(tm)? He's teaching. Allow him some poetic license, please.

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.

Don Foreman wrote:
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.

Very well put! Succinct & dead on.

The razzledazzle is in interpreting the schematic
differently at different parts of his presentation while never
explicitly pointing that out.

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

On 2/13/2010 6:45 AM, Larry Jaques wrote:
On Thu, 11 Feb 2010 23:11:24 -0600, the infamous Don Foreman
scrawled the following:


(...)

This professor is trying to confuse, bedazzle and impress rather than
teach or elucidate.

If he manages to spark CURIOSITY in a few more kids, isn't this A Good
Thing(tm)? He's teaching. Allow him some poetic license, please.

Tell me you're joking, Larry.

--Winston

On 2/13/2010 7:12 AM, Don Foreman wrote:
The razzledazzle is in interpreting the schematic
differently at different parts of his presentation while never
explicitly pointing that out.

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

On Sat, 13 Feb 2010 19:26:16 -0800, Winston
wrote:

On 2/13/2010 7:12 AM, Don Foreman wrote:
The razzledazzle is in interpreting the schematic
differently at different parts of his presentation while never
explicitly pointing that out.

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

I don't know what that means.

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.

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.

On 2/13/2010 9:54 PM, Don Foreman wrote:
On Sat, 13 Feb 2010 19:26:16 -0800,
wrote:

On 2/13/2010 7:12 AM, Don Foreman wrote:
The razzledazzle is in interpreting the schematic
differently at different parts of his presentation while never
explicitly pointing that out.

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

I don't know what that means.

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.

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.

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.

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:

On 2/13/2010 6:45 AM, Larry Jaques wrote:
On Thu, 11 Feb 2010 23:11:24 -0600, the infamous Don Foreman
scrawled the following:


(...)

This professor is trying to confuse, bedazzle and impress rather than
teach or elucidate.

If he manages to spark CURIOSITY in a few more kids, isn't this A Good
Thing(tm)? He's teaching. Allow him some poetic license, please.

Tell me you're joking, Larry.

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

On 2/14/2010 4:20 PM, Larry Jaques wrote:

(...)

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.

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

On Sun, 14 Feb 2010 22:22:41 -0800, Winston
wrote:

On 2/14/2010 4:20 PM, Larry Jaques wrote:

(...)

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.

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.

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.

On 2/14/2010 11:07 PM, Don Foreman wrote:
On Sun, 14 Feb 2010 22:22:41 -0800,
wrote:

(...)

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.

You demonstrated quite the contrary with your experiment.

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.

Induction can produce a voltage difference from end to end of a
good or perfect conductor regardless of what current it may be
passing.

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

Ohmic IR drop is independent of induction and superposition holds.

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