Electric Theory
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ggunn
PE (Electrical), NABCEP certified
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- Austin, TX, USA
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- Consulting Electrical Engineer - Photovoltaic Systems
Actually, Sir Sydney Fudd's First Law of Opposition is: "If you push something hard enough it will fall over." He discovered it when he pushed his wife down the stairs.
You must be thinking of Teslacle's Deviant to Fudd's First Law: "It goes in, it must come out."
The electrons having a like charge naturally repel each other (why they repel like charges IDK).
Given a conductive path the electrons naturally push each other away to achieve an equal distribution, or said another way, a zero voltage potential between any two points on the conductive path. That is the static case, the electrons try to achieve an equilibrium state and relieve the excess accumulation of electrons at one point relative to another. They are only held in disequilibrium by insulators.
In the dynamic case, the electron is being pushed around by electric and magnetic fields (Faraday, Lenz).
So the electron does not go back to its source (which would be pair production), it goes back to its equilibrium state or equal distribution of equally repelling charges on a conductive path.
The analogy would be water vapor, each water vapor molecule trying to achieve an equal or equilibrium distribution within its domain of movement.
I don't disagree with near instantaneous effect. It doesn't long at all for you to realise you are being electrocuted.........
As you probably already know.
It's the electrical energy created by the electromagnetic wave that is traveling near the speed of light down the wire, not electrons.
The electrical energy moves down the wire in one direction from the source to the load in the form of an electromagnetic wave.
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al hildenbrand
Senior Member
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:thumbsup: ...........
RumRunner
Senior Member
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Whenever there is energy, there is electron. An atom is composed of proton, neutron and electron. We are all surrounded by atoms. . . it is the basic building block of all matter.
Separating the electron from the atom nucleus is what releases the energy. the process called fission. Since electron is negatively charged, applying electric current could cause separation of the electron from its nucleus thus converting that energy that is usable like operating our hair dryer or kitchen toaster. It also can become destructive like "The Fat Man".
In the transformer pri/sec winding scenario, magnetism happens when electric current is applied. Magnetism in itself has no energy. . . it is a force.
Magnetism within the transformer core, forces the electrons from the primary to the secondary windings through the strength of the magnetic field. As you strengthen the magnetic field the more electrons is pushed to the secondary winding.
Magnetism is essential in controlling energy and converting one energy to another.
Always appreciate being educated.......:thumbsup:
K8MHZ
Senior Member
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There are electrons in an atom nucleus?
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- Placerville, CA, USA
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- Retired PV System Designer
Not in a free state, no.
But in a nucleus which is unstable because of an excess of neutrons, a neutron may decay into a proton, an electron, and an electron neutrino. This is known as beta decay (the emitted electron is called a beta particle.)
This does not have any relation to nuclear fission and is not influenced in the slightest degree by any electric field external to the nucleus.
Nuclear fission involves an unstable nucleus splitting into two nuclei of approximately equal mass and giving off significant energy in the process.
ggunn
PE (Electrical), NABCEP certified
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Um, no. Separating an electron from an atom is ionization, not fission. No mass is converted to energy in an ionization process, and simply applying an electric current cannot initiate fission.
This is a good thing; if the movement of electrons between atoms were a nuclear reaction your car battery would be so radioactive it would kill you.
kwired
Electron manager
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Many elements have unstable conditions in outer electron shell which gives them both chemical and electrical properties that you don't find in elements with a stable outer electron shell. Simple loss of an electron(s) isn't fission. Fission would be splitting the entire atom including the nucleus, which is what is done with nuclear fission. When they split a uranium atom what is left is not uranium. When you have a loss or electrons in an element you still have same element it just is left with a positive charge and is looking to react with some other element that has a negative charge.
Elements that are conductive simply have electrons that move more freely then elements that are not considered conductive. All elements will let electrons move if enough voltage is applied, but "insulators" take a lot more voltage before to get the electrons to move. When electrical current flows most do think the theory that particles (electrons) aren't actually moving through the conductor but rather they absorb energy and bump into an adjacent electrons in the adjacent atom - and it is that energy that flows through the conductor not the electrons. The electrons are just the part of the atom that absorb and then pass on that energy.
I disagree on the transformer theory you mentioned. Yes magnetism occurs when current flows through the primary coil, but it doesn't push electrons from primary to secondary. Only energy is exchanged via magnetic forces between primary and secondary. That magnetism then "excites" the electrons in the secondary circuit and they then pass the energy on to adjacent atoms and thus current can then flow in the secondary circuit.
JFletcher
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- Williamsburg, VA
Negative. Fission is splitting the nucleus of an atom, like Uranium 238 or Plutonium 239, into smaller elements. In the process of becoming more stable, smaller elements, some free neutrons are released, which are either fast (unmoderated) or slow (moderated), which strike other atoms' nuclei, continuing the process. If the process is sustained, you have a nuclear reactor. If not, you have a bomb.
The electrons have virtually nothing to do with that reaction.
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"Electricity is a little tweeting bird, chirping in a meadow..., it is a wreath of pretty flowers, that smell BAD..."
FionaZuppa
Senior Member
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it doesnt have to is the answer. it goes back to a place that can provide a home for the charge. look at how lightning works.
3 plastic or metal discs in space, two are very positively charged, one is very negatively charged. the neg disc full of extra electrons will jump over to either of the other two pos discs depending on which one is closer to allow such to happen.
and to boot, the electrons on the neg disc do not have to come from either of the two pos discs.
however, in a generator with mag field cutting across loops of wire, its like a water pump circulating in the cooling system of your car. gets pushed one way, must return on the other side of the pump.
in general, no.
3 plastic or metal discs in space, two are very positively charged, one is very negatively charged. the neg disc full of extra electrons will jump over to either of the other two pos discs depending on which one is closer to allow such to happen.
and to boot, the electrons on the neg disc do not have to come from either of the two pos discs.
however, in a generator with mag field cutting across loops of wire, its like a water pump circulating in the cooling system of your car. gets pushed one way, must return on the other side of the pump.
squeeze hard enough you'll get all sorts of stuff in the nucleus area
in general, no.
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Kirchhoff's current law
or a restatement of the theory of conservation of charge
charge seeks equilibrium
if current (charge/time) leaves a node the current entering and leaving must = 0
think of a xfmr as a node
now consider a system of grids
each grid independent of the other (no Cu interconnects) but sharing the same earth reference (obviously)
assume 2 adjacent grids (1 and 2) have L-G earth faults simultaneously
the Grid 1 source is 'closer' to the Grid 2 fault, ie, lower Z rtn path than to its own source, Grid 2 source
and vice versa
will the fault current 'motivated' or generated by Grid 2, return to the 'closer' (lower Z rtn earth path) Grid 1 source?
and vice versa, Grid 1 i fault rtn to Grid 2 soure?
more likely to share: source 1 will get i from 1 and 2, the same for source 2?
does the source care where the rtn equalization charge comes from?
or a restatement of the theory of conservation of charge
charge seeks equilibrium
if current (charge/time) leaves a node the current entering and leaving must = 0
think of a xfmr as a node
now consider a system of grids
each grid independent of the other (no Cu interconnects) but sharing the same earth reference (obviously)
assume 2 adjacent grids (1 and 2) have L-G earth faults simultaneously
the Grid 1 source is 'closer' to the Grid 2 fault, ie, lower Z rtn path than to its own source, Grid 2 source
and vice versa
will the fault current 'motivated' or generated by Grid 2, return to the 'closer' (lower Z rtn earth path) Grid 1 source?
and vice versa, Grid 1 i fault rtn to Grid 2 soure?
more likely to share: source 1 will get i from 1 and 2, the same for source 2?
does the source care where the rtn equalization charge comes from?
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- Retired PV System Designer
The only complication to that current flow calculation comes for the special case of a capacitor. A net positive charge builds up on one plate, balanced exactly by a net negative charge on the other plate. And a displacement of charge as dipoles within the dielectric line up with the resulting electric field.
If you want to be able to assert that the total current through any cross section of the circuit will be identical you "fake" the calculation by defining "displacement current" which is a function of the dielectric constant and the rate of change of the electric field integrated over the cross section.
If you want to be able to assert that the total current through any cross section of the circuit will be identical you "fake" the calculation by defining "displacement current" which is a function of the dielectric constant and the rate of change of the electric field integrated over the cross section.
for general engineering calcs that has no real bearing
now f you are into high freq electronics or small order tc fault calcs it may
but in general if you account for Xc (1/(jwC) in the calcs it is a non-issue
especially once charged and looking at steady state values >5 tc
on a quantum level, yes
on a practical level, moot
now f you are into high freq electronics or small order tc fault calcs it may
but in general if you account for Xc (1/(jwC) in the calcs it is a non-issue
especially once charged and looking at steady state values >5 tc
on a quantum level, yes
on a practical level, moot
RumRunner
Senior Member
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- Retired EE
You can disagree with anything that I said, but it doesn't make you right. This is social media.
Anything that is said by members is either hypothesis or opinion. . .nothing is binding.
Unlike testifying under oath or audience, with authority, like a group of scientists known for their work, anyone should take someone's word with a grain of salt.
Social media is exactly what this site is. Readers can either take it or leave it. If they find it beneficial then more power to them.
All these things are for entertainment purposes...and to say that you are an authority of a subject matter will only put you in a position where if someone gets hurt or dies because of what you said, you should be prepared to answer and stand by your word. Someone forced to answer has happened before.
If you don't believe me which is likely because of what you've said earlier, check with an attorney specializing in "The Internet Law".
We live in a litigious society. . .and no one seem to take responsibility for their action. People always try hard to find someone else to blame.
I have said what I needed to say regarding the topic.
Thank you for listening.
LarryFine
Master Electrician Electric Contractor Richmond VA
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Happy that you asked yet?
Carultch
Senior Member
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- Massachusetts
You cannot really know exactly where any electron is, within the atom. All that we really see in diagrams of the atom, are probability clouds of where the electron is statistically likely to be. The miniature "solar system" model of the atom is oversimplified.
On rare occasion, the electron could be inside the nucleus, but in general, it isn't. Energy would have to be destroyed for the atom to simply collapse and have the electron coalesce with the nucleus. The electron probability clouds are what give the atom its size. Atoms are so empty that if it were scaled to a stadium, the nucleus would be the size of a marble. All that space is maintained by the electric fields associated with the electrons, as they surround the nucleus.
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