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AC Voltage - How is it transfered?

  1. Nov 24, 2011 #1
    Sorry but have only a basic understanding of electric circuits? My understanding of DC circuits is that the p.d between either terminal of the power supply is the electrical energy used by the electron in moving through the circuit. How does the voltage change in an AC circuit if the voltage is periodically changing? and if the same elctrons are moving through the same component i.e. light bulb would, these same electrons not eventually lose all their energy? Thanks
     
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  3. Nov 24, 2011 #2

    sophiecentaur

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    To understand electrical conduction better you need to realise that the electrons drift through a metal very slowly as the current 'flows'. There are huge numbers of them (billions of billions) and they all move at high speed in random directions because of thermal agitation but their average speed along the wire will only be a mm per second. That is contrary to the picture most people have in their heads of electrons zapping around a circuit. It is far more like a bicycle chain moving slowly round and transferring power for your foot to the wheel very fast.

    At this low average speed, you can appreciate that, if you apply an AC voltage, changing direction 100 times per second, the electrons won't ever get very far at all (say 1/100mm back and forward) . This does not stop the 'bicycle chain effect' though and power is transferred through the wire just the same.
     
  4. Nov 24, 2011 #3
    Mentor comment (D H):
    This post is nonsense. Normally we delete such nonsense; this is our #1 job as mentors. I have left this here as an object lesson. Look for follow-on comments, also in red, in the posts that follow.


    Actually, the electrons are moving very close to the speed of light, even with a low voltage pulling at them. Concerning your question as to the electrons losing energy as they slosh back and forth, this is true if the conductor is not a super-conductor. But there is a source of electrical potential (voltage) driving the AC circuit, either a generator or an oscillator. The generator or oscillator actually separates valence electrons from atoms, thus creating a electromagnetic force (positive and negative potentials) which drive the electrons in the circuit -- gives the electrons momentum and energy to overcome the loss from sloshing. Because we are talking about AC, the generator / oscillator creates a voltage potential +/- then -/+ then +/- ...etc.

    One thing I didn't mention in the paragraph above is that due to the separation of charges (moving the valence electrons) by the generator or oscillator this causes a chain reaction in the conductor (the wiring of the circuit). The valence electrons in the wiring that are close to the connection of the generator / oscillator are attracted or repulsed. This leaves a surplus or deficient of electrons in that area of the wire which then attracts or repulses valence electrons further down the wire. All this happens close to the speed of light, thus a continuous flow of electrons through the wiring of the circuit.
     
    Last edited by a moderator: Nov 25, 2011
  5. Nov 24, 2011 #4

    sophiecentaur

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    Where did you get hold of that bit of 'information'?
    The only occasion that electrons flow that fast in a circuit is when they are in a CRT. You had better read your text book again about electron drift velocity. The mean velocity is in the order of mm per second. Also, what has superconductivity got to do with this?
     
  6. Nov 24, 2011 #5
    Totally agree. Electrons in a wire are drifting at a few mm per second.
     
  7. Nov 24, 2011 #6
    Mentor comment (D H):
    Emphasis mine.


    Thanks, the explanations really helped. Sophie the bicycle chain analogy was really good and thanks taterz for such detail. If I can use the bicycle analogy, is AC comparable to moving the pedal back and forth periodically but if the voltage is appiled by the pedal, is it as if the voltage is passed through the links(electrons) of the bicycle chain from one to the next. Or like what I am taking from you taterz is that the electrons alternate direction but some move directly around the circuit. Sorry for being nuisance. Thanks


    Mentor comment (D H):
    ZB08: taterz's details are detailed nonsense.

    That ZB08 was led astray by taterz's nonsense is the number one reason we why we moderators moderate at this site. Uninformed users don't know the difference between nonsense and valid answers. ZB08, please do not take that I am calling you an "uninformed user" the wrong way. All of us were at one point uninformed. This applies to even the greatest of scientists. It is not a slight.
     
    Last edited by a moderator: Nov 25, 2011
  8. Nov 25, 2011 #7

    sophiecentaur

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    Mentor comment (D H):
    I deleted sophiecentaur's response to taterz. What's left is good solid information.


    I could recommend this link about Fermi levels if you want an indication of the electron energies involved. The "rest energy" that you quote is hardly relevant as it refers to how much energy is available if an electron is annihilated. The Fermi energy for copper is about 7eV and it is only the few electrons at the top of the 'well' that are encouraged to move about (mixing my classical and QM, I'm afraid).

    I am really anxious that ZB08 should not be further confused by your red herring about speed. I would just ask you where the electric field would come from on a piece of copper wire, across which there might be a PD of a few microvolts, suddenly to accelerate an electron to "relativistic speeds"? The field (in V/m) is very very low, so the potential across the space between atoms would be immeasurably small.

    Perhaps you could give a reference to where you reckon you got your idea from so we could put it to bed.
     
    Last edited by a moderator: Nov 25, 2011
  9. Nov 25, 2011 #8

    D H

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    [strike]Thread locked pending moderation.[/strike]

    My cleanup of this thread involved a good amount of work, and that's a reason we mentors typically delete nonsense and replies to it rather than correct.

    To ZB08:
    The speed at which electrons move down a wire and the speed at which signals propagate down the wire are very different. Those valence electrons move but slowly while the signal moves very quickly.

    Also, please do not take my calling you an "uninformed user" as a slight. It was not intended as a slight. We all have been uninformed at some point in our lives. In fact, the more I learn, the more I learn just how uninformed I still remain.


    To our informed users:
    When you see nonsense, don't just reply to the posted nonsense. Please, please, please hit the report button. This site gets a lot of traffic. We mentors don't have time to look at each and every post. We rely on our informed users telling us about post that violate our rules by reporting the post in question.


    I've re-opened the thread, so feel free to carry on with asking questions and providing answers.
     
    Last edited: Nov 25, 2011
  10. Nov 25, 2011 #9

    D H

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    Here are a few hyperphysics pages that are relevant to the topic at hand:

    Microscopic view of electric current: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html
    Microscopic view of Ohm's law: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html
    Fermi energies for metals: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi2.html

    The latter starts addressing the quantum mechanical view of how current and signals travel down a wire. The valence electrons in a metal form what is called a Fermi gas. "Valence electron" and "Fermi gas" are a couple of good terms to use in a search engine. Another is "free electron model".
     
  11. Nov 25, 2011 #10

    I think you have in mind the thermal phenomena where indeed only the electrons within an energy of KBT from the top of the conduction band "participate".
    In the case of electric conduction in metals all (or most) of the electrons in the conduction band participate. Their momenta are all shifted in the direction of the electric field by the same amount (at equilibrium), of the order mvDrift.
    If we look at the measured density of conduction electrons for copper we can see that is very close to 1 electron per atom.
     
    Last edited: Nov 25, 2011
  12. Nov 25, 2011 #11

    sophiecentaur

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    Absolutely.

    Also:
    I will use the 'report button' more often. I did wonder whether a spike might come up out of my seat if I used it too freely!
     
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