Toshiba TV29C90 problem; Image fades to black...

[JW]

On Thu, 28 Nov 2013 00:08:40 +1100 "Phil Allison" <phil_a@tpg.com.au>
wrote in Message id: <bfm971FsmubU1@mid.individual.net>:

Electrons leave by one wire and return to the source by the other.

If they meet "resistance" then work is done by the source in propelling them
through that resistance.

That's why superconductors make electrons lazy.

 

[William Sommerwerck]

"Tim R" wrote in message
news:14068614-d5d4-4c8a-99a3-dc785da59025@googlegroups.com...

> You can't get power from nothing.

Oh? Consider Reagan or Bush fils.


> If power came out of the wire, something in the wire now has less power.

It doesn't come from the wire -- it comes from the generating device.

 

[William Sommerwerck]

"Tim R" wrote in message
news:2c693612-4f67-4b6b-9cc9-bc0bbff88d6d@googlegroups.com...
On Wednesday, November 27, 2013 10:03:41 AM UTC-5, William Sommerwerck wrote:

"Tim R" wrote in message

If power came out of the wire, something in the wire now has less power.

It doesn't come from the wire -- it comes from the generating device.

Ah. The generator slows down, WITHOUT the wire knowing. Right...
I admit I don't understand it. You appear to not realize you don't
understand it.

You might be right. But the wire is a conduit, not a source.

Think of varying the nozzle opening on a garden sprayer. The hose "knows"
nothing. It just delivers more or less water, based on the water pressure and
how far the nozzle is opened.

 

[David Platt]

In article <7af4589d-4ffa-4990-a554-4c417ff7576f@googlegroups.com>,
Tim R <timothy42b@aol.com> wrote:

Certainly. But you can measure the speed of the water in the hose, and
there will be a difference. Or if you want to be closer to the
electrical load scenario, have the water in the hose run a small
turbine, and measure the energy of the water before and after. You will
find the mass unchanged and the velocity decreased, so kinetic energy of
the water molecules has decreased by exactly the amount that went into
work done by the turbine (and heat and pressure losses).

What is the equivalent change in the electron stream going through the lamp?

I believe it's one of potential energy, created by the "packing
together" of electrons in opposition to their electrostatic repulsive
force.

In a battery, or a capacitor, you charge up the device by segregating
the electrons onto one side of the barrier (packing an excess of them
in) and creating a corresponding deficit of electrons on the other
side. Because the electrons have the same charge, and you're putting
more of them on one side than you have protons, and because like
charges repel, you have to do work to "pull" the electrons out of the
"+" side of the accumulator and "push" them into closer proximity on
the "-" side (overcoming the net repulsive force). It's analogous to
pumping water up-hill, and storing potential energy in the water's
increased altitude (gravitational P.E.).

Allowing electrons to flow through your circuit is analogous to
letting water flow down-hill. The electrons aren't individually
changed by this process, but they end up less tightly spaced, and in
an environment with relatively more positively charged particles to
counteract their tendency to repel one another.

That's how I see it, anyway.

 

[Tim R]

On Wednesday, November 27, 2013 7:52:32 PM UTC-5, Phil Allison wrote:

"Nightcrawlerďż˝"





Electrons travel at one speed. Close to that of light.



** Bollocks.



In a wire, electron flow is very slow. Like inches per hour.

Yes. I know that.

I don't disagree with anything posted in response to my question, including the fact that power was the wrong term.

But none of it is relevant. I'm not being argumentative, I'm asking a question to which I really don't know the answer. How does the energy from the power plant really move through the wire and really get converted into light and heat in the lamp?

Electrons within a molecule exist at discrete energy states, and when dropping from one state to another release energy - that's one method light is produced.

But electrons traveling through a wire, however slowly, are essentially loose from the copper atoms. And yes, they are bumping into each other like loosely coupled train cars, and that's how electricity can move at near the speed of light while electrons move at inches per hour.

My uneducated guess would be that though the drifting speed of the electrons is slow, they are bouncing rapidly within each other, and they shed some of that speed through collisions in the light bulbs filament, basically dumping kinetic energy. But that's just from thinking about it, not an EE class like most of you have had.

If that guess is true or even close, none of you have come close to explaining it.

 

[NightcrawlerŽ]

"Tim R" <timothy42b@aol.com> wrote in message news:2c693612-4f67-4b6b-9cc9-bc0bbff88d6d@googlegroups.com...

On Wednesday, November 27, 2013 10:03:41 AM UTC-5, William Sommerwerck wrote:
"Tim R" wrote in message
If power came out of the wire, something in the wire now has less power.



It doesn't come from the wire -- it comes from the generating device.

Ah. The generator slows down, WITHOUT the wire knowing. Right..........

I admit I don't understand it. You appear to not realize you don't understand it.

The frequency control circuit (or a governor) maintains the frequency of the the
generator. The voltage regulator maintains the voltage, at the generator.
As a load is placed on a generator, the controls will increase/decrease the
RPM of the generator to maintain desired frequency. The generator will supply
all of the current that it is capable of. This is dependent on the loads placed
on said generator and the output circuit protection. Each load can only take
what it is designed for. If it draws 15 amps at a specific voltage, that is
what the load will draw. Any loss in the circuit is via radiation. Almost
all electrons return to the generator (yes there is some loss).

Electrons travel at one speed. Close to that of light. The only variables one
will see is via augmentation of the circuit. Adding a capacitor changes path of
electron (not all, mind you) flow until capacitor saturation. Adding a inductive
load (motor) resist the flow of electrons, but does not change electron speed, or
to any appreciable amount, the number of electrons that flows through a circuit
after the motor comes up to speed.

Purely resistive loads only limit the quantity of electrons through a load (similar
to a motor, but a motor actually backfeeds/adds a return current). The quantity
of electrons flowing through a circuit is determined by the circuit's overall
resistance/impedance. Any energy losses are via radiation. Light, EMR, heat.

Examples: Light bulbs emit light, heat, EMR. The type of light bulb (or lamp)
dictates what the ratios are. There is virtually zero electron loss.
Conductors readily accept more electrons from other sources, otherwise,
after being used once, they would hold a charge once the circuit was open.

Motors emit heat, EMR, and (LOL! Hopefully non-visible light). Ratios are
once again determined by the motor.

Resistive loads emit heat, light, and EMR.

My description is for AC only.
Light shall include ultra violet, infrared, and visible.
EMR is electro-magnetic-radiation.
Heat is any derivative of conducted or radiated energy which might include light or
any other form of heat transfer.

This is very basic and is so for a reason. I am not writing a doctoral dissertation.
Meaning, that I am expecting some whom read this to not have a clue. The anal retentive
types need not reply.

What I have written is not complete, nor does it encompass all of the variables. (see above)

 

[Tim R]

On Wednesday, November 27, 2013 8:31:03 PM UTC-5, Phil Allison wrote:

OR the same electrons create a magnetic field ( by simply flowing at that

very slow rate) that produces mechanical force on the rotor of an electric

motor. Electrical energy is transferred to mechanical work.

If this is the case, and I have no reason to doubt it <grin> then the mechanical force goes in both directions. It must act on the moving electron at exactly the same force as it does on the rotor. Surely electrons moving at an inch per hour are not exempt from Newton's second law. And as F=ma, the electron must slow under that force.

Hmm. I said I have no reason to doubt it, but that implies that the kinetic energy of the electron is dependent only on the slow forward motion (KE being mass times velocity squared), and that is the limiting factor on how much energy you can transfer. Is that really true? Or does a high energy electron carry more energy than due to the drift speed, usable energy?

 

[NightcrawlerŽ]

"Wild_Bill" <wb_wildbill@XSPAMyahoo.com> wrote in message news:0Lllu.51713$Lj7.2854@en-nntp-01.dc1.easynews.com...

The difference between the 2 voltage potentials (9V) is what will light the
lamp.

Plus volts (+)-------------- lamp filament
resistance --------------------(-) Zero

In essence, the voltage travels from the negative to the positive. Along the
way a resistance is encountered before flowing through a parallel path to
the positive. This only splits the current flow. To which extent depends
on each respective conductor's length or overall resistance.

The resistive load (the light) emits radiation. Heat, light, and an electro
magnetic field (stable).

 

[NightcrawlerŽ]

"William Sommerwerck" <grizzledgeezer@comcast.net> wrote in message news:l74uh2$tlo$1@dont-email.me...

The simple fucking fact is that it is ELECTRONS that return, not power.

Actually, they just kinda "slosh" back and forth in an AC system.

Oh, like a Metallica concert?

 

[NightcrawlerŽ]

"Tim R" <timothy42b@aol.com> wrote in message news:7af4589d-4ffa-4990-a554-4c417ff7576f@googlegroups.com...

On Wednesday, November 27, 2013 10:59:21 AM UTC-5, William Sommerwerck wrote:


Certainly. But you can measure the speed of the water in the hose, and there will be a difference. Or if you want to be closer
to the electrical load scenario, have the water in the hose run a small turbine, and measure the energy of the water before and
after. You will find the mass unchanged and the velocity decreased, so kinetic energy of the water molecules has decreased by
exactly the amount that went into work done by the turbine (and heat and pressure losses).

What is the equivalent change in the electron stream going through the lamp?

Also, I'm not sure your statement "the wire is a conduit, not a source" is consistent with your earlier statement that the wire
just connects the zero point. To the load, the wire IS source and return, at the point of connection.

Wires contain the electrons that are used. If no action is done to induce current flow, then
the electrons of the wire just stay where they are. The source of voltage provides a path
for electron flow, but is providing extra electrons if by battery/capacitor, or if by
a rectified AC source, just kicking the electrons in the butt and forcing them to move along.

 

[Rheilly Phoull]

On 27/11/13 21:56, dave wrote:

On 11/27/2013 05:08 AM, Phil Allison wrote:


** False concept.

Electrons leave by one wire and return to the source by the other.

If they meet "resistance" then work is done by the source in
propelling them
through that resistance. The work done is heating that resistance
and the
energy released is proportional to the square of the number of
electrons per
second travelling in the loop.

The simple fucking fact is that it is ELECTRONS that return, not power.



... Phil
Do the electrons actually drive around in circles or do they just bump
into each other like a circle of autos?

No, there are special electron traffic lights built into appliances.
If there is no power in flowing the return wire how come a meter will
show it ? There must be a path to and from the source for current to flow.

 

[Phil Allison]

"William Sommerwerck"

The simple fucking fact is that it is ELECTRONS that return, not power.

Actually,

** Read the post for CONTEXT - Bill

 

[Phil Allison]

"Tim R"

On Phil Allison wrote:

The simple fucking fact is that it is ELECTRONS that return, not power.



Way oversimplified.

** Like hell it is.

Where does the power come from?

** The SOURCE !!!


> What changes?

** Not the electrons.

> If I use steam to do work,

** False comparison.

If the same number of electrons goes through the load and returns to that
zero reference point, what is different?

** False requirement.


> Are they going slower?

** No - electrons travel exceedingly slowly all the time at ordinary current
levels in wires.


> You can't get power from nothing.

** Wrong quantity.

Work done and energy are the same.

Power = rate of doing work.


> If power came out of the wire, something in the wire now has less power.

** Insane crap.

Wrong quantity again

Work done and energy are the same thing.

Power = rate of doing work.

Learn some fucking basic physics - imbecile.

 

[Phil Allison]

"NightcrawlerŽ"

Electrons travel at one speed. Close to that of light.

** Bollocks.

In a wire, electron flow is very slow. Like inches per hour.

In a vacuum, the speed of light can be approached IF hundreds of thousands
of volts are used to accelerate them.

Then electron mass increase dramatically stopping you breaking the most
famous law in physics.



.... Phil

 

[William Sommerwerck]

There is something in physics called a Poynting vector (which I never
understood) that might throw some light on this. Or not.

https://en.wikipedia.org/wiki/Poynting_vector

 

[Phil Allison]

"Tim Ratbag"

Phil Allison wrote:

OR the same electrons create a magnetic field ( by simply flowing at that

very slow rate) that produces mechanical force on the rotor of an electric

motor. Electrical energy is transferred to mechanical work.

If this is the case, and I have no reason to doubt it <grin> then the
mechanical force goes in both directions. It must act on the moving
electron at exactly the same force as it does on the rotor. Surely
electrons moving at an inch per hour are not exempt from Newton's second
law. And as F=ma, the electron must slow under that force.


** Inductance reduces current flow as does any back emf.


Hmm. I said I have no reason to doubt it, but that implies that the kinetic
energy of the electron is dependent only on the slow forward motion

** Look - you are simply not interested in facts.

Your constant over snipping and argumentative tone proves that.

No-one wants to read or debate YOUR mad ideas.




..... Phil

 

[Phil Allison]

"Tim R"
Phil Allison wrote:

In a wire, electron flow is very slow. Like inches per hour.

Yes. I know that.

I don't disagree with anything posted in response to my question, including
the fact that power was the wrong term.

But none of it is relevant. I'm not being argumentative, I'm asking a
question to which I really don't know the answer. How does the energy from
the power plant really move through the wire and really get converted into
light and heat in the lamp?

** Because of RESISTANCE in the metallic conductors.
--------------------------------------------------------------

Electrons within a molecule exist at discrete energy states, and when
dropping from one state to another release energy - that's one method light
is produced.


** Fact is, resistance IS a quantum mechanical effect that cause electrons
to emit energy as heat. The atoms making up the conductor get hot and
radiate IR energy and or visible light.

This energy conversion occurs at a rate proportional to the current flow and
the voltage drop.

OR the same electrons create a magnetic field ( by simply flowing at that
very slow rate) that produces mechanical force on the rotor of an electric
motor. Electrical energy is transferred to mechanical work.



..... Phil

 

[Leif Neland]

Wild_Bill udtrykte prćcist:

The previous "no return" statements are the same for AC or DC.

With 2 wires from the battery positve (+) terminal connected to a lamp, there
will be no light or heat.

Without utilizng the battery minus (-) terminal the two wires from the plus
(+) terminal have no difference in voltage potential.

The difference between the 2 voltage potentials (9V) is what will light the
lamp.

Plus volts (+)-------------- lamp filament resistance --------------------(-)
Zero

As the lamp resistance drops the (+) voltage to zero at the lamp's zero
voltage terminal, there is nothing to "return" to the battery.

The current in a circuit with only one path, i.e. no resistors, lamps
etc in parallel, is the same everywhere, both in the positive and the
negative wire.

If the current can't return to the battery, no current can flow.

Saying you don't need a return is like saying you only need the upper
side of the chain on the bicycle, as this is where the pedals pull the
wheel. A chain does not push, so the lower part is not needed.

(And a tyre is only flat at the part touching the road, the rest is
fine)



--
Husk křrelys bagpĺ, hvis din bilfabrikant har taget den idiotiske
beslutning at undlade det.

 

[Tim R]

On Thursday, November 28, 2013 10:22:10 AM UTC-5, Wild_Bill wrote:

What I did say (several times) is that nothng is returned to the power

source.



A (-) zero volt/DC or neutral/AC wire only takes the different voltage

potential closer to the load.. and nothing (depleted electrons, waves,

molecules, atoms etc) is being returned to either power source.

The power is consumed by the load, nothing is left over to return to

anwwhere.

Hmm. My vague memory of storage batteries says that ions migrate from one pole to the other through the electrolyte, then combine with electrons arriving through the return wire to balance the charge. At least, I'm pretty sure it was explained that way in school, which was admittedly 50 plus years ago.

It seems rather unlikely to me that the electron moving down the wire, however slowly, comes to a full stop in the load. Wouldn't the rest back up behind it?

I'm sure Phil will be back shortly to tell us how stupid we are, but to avoid explaining what the correct version is.

Not being a EE, my classes covered the macro level of how wires connect, but not what is really happening at the atomic level, and I'm finding it interesting.

 

[Wild_Bill]

Read what is offered. never suggeted that a circut can exist with only one
wire.

You made that part up in our head, possiblly due to poor reading
comprehension sklls.

What I did say (several times) is that nothng is returned to the power
source.

A (-) zero volt/DC or neutral/AC wire only takes the different voltage
potential closer to the load.. and nothing (depleted electrons, waves,
molecules, atoms etc) is being returned to either power source.
The power is consumed by the load, nothing is left over to return to
anwwhere.

The use of the word "return" for one conductor is just a bad choice of
wording.

The concept and belief that anything is being returned from a load, back to
the power source is fantasy, fiction, foolish, retarded, idiotic etc.

--
Cheers,
WB
..............


"Leif Neland" <leif@neland.dk> wrote in message
news:mn.e2377ddb36023829.130671@neland.dk...

Wild_Bill udtrykte prćcist:
The previous "no return" statements are the same for AC or DC.

With 2 wires from the battery positve (+) terminal connected to a lamp,
there will be no light or heat.

Without utilizng the battery minus (-) terminal the two wires from the
plus (+) terminal have no difference in voltage potential.

The difference between the 2 voltage potentials (9V) is what will light
the lamp.

Plus volts (+)-------------- lamp filament
resistance --------------------(-) Zero

As the lamp resistance drops the (+) voltage to zero at the lamp's zero
voltage terminal, there is nothing to "return" to the battery.

The current in a circuit with only one path, i.e. no resistors, lamps etc
in parallel, is the same everywhere, both in the positive and the negative
wire.

If the current can't return to the battery, no current can flow.

Saying you don't need a return is like saying you only need the upper side
of the chain on the bicycle, as this is where the pedals pull the wheel. A
chain does not push, so the lower part is not needed.

(And a tyre is only flat at the part touching the road, the rest is fine)



--
Husk křrelys bagpĺ, hvis din bilfabrikant har taget den idiotiske
beslutning at undlade det.