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Does charges remain in an extension cord when unplugged?

  1. Sep 20, 2011 #1
    I am referring to this: PSE_JET_EXTENSION_CORD.jpg

    A brainwave hit me and I wondered the following:
    If the moment when i unplug or switch it off (In my country they have a switch for added safety reason), the live wire happens to be at the peak of the AC curve, will my extension cord effectively become similar to a charged capacitor, with the live and neutral rails at different potentials?

    That means to say, if someone opens up the thing, he may get a nasty shock?
     
  2. jcsd
  3. Sep 20, 2011 #2

    clem

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    Whether it is connected or not, there is never any excess charge in an extension cord.
     
  4. Sep 20, 2011 #3
    why issit so?
     
  5. Sep 20, 2011 #4

    Dale

    Staff: Mentor

    Power cords carry current, they do not normally become appreciably charged.
     
  6. Sep 20, 2011 #5

    HallsofIvy

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    There is NEVER "excess charge" in an electric cord. If the cord is attached to a battery (direct current) we have a potential difference causing electric charge (electrons) to move into one end of the cord and out the other. If the cord is "plugged" into and electric circuit (alternating current) we have elctric charge (electrons) moving back and forth very rapidly. In either case, it is the motion of the charges that causes the current and does "work", not "excess charge".
     
  7. Sep 20, 2011 #6
    The answer to your first question is yes. If you disconnect an unloaded extension chord from an outlet, the two wires in the extension chord will form a capacitor, storing a voltage equal to the voltage across the outlet terminals when the chord is disconnected.

    The answer to the second question is no, for two reasons. The first is that the value of the extension chord capacitance is very small due to the significant distance between the wires. This means the chord can only store a very small amount of energy, probably too small to be noticed if someone were to touch both terminals and discharge the capacitor. The second reason is that the charge stored on the capacitor will quickly leak away due to the finite resistivity of the wire insulation.
     
    Last edited: Sep 20, 2011
  8. Sep 20, 2011 #7
    U need to understand what an electric current is first.
    Imagine current being people on a marathon in a circle filling the whole circle. If the oraganizer whistles (when voltage is applied) then all the people start running at the same rate and there is current but when you ask them to stop running there would not be more/less people remaning than when the race was started ( so why would excess charge get stored ) ....
    hope u find my analogy helpful...
     
  9. Sep 20, 2011 #8
    "...live wire happens to be at the peak of the AC curve..."
    Its just that the voltage (w.r.t something) in the live wire is at the peak not the current. if u unplug it it will not disturb its natural curve (of voltage-time)....
     
  10. Sep 20, 2011 #9
    oh man, who is right? some say yes, some say no.

    isnt it through that the build up of charges across two rails that give rise to a difference in potential?

    technically, if it is disconnected when the potentials are at a different voltage, then shouldnt the live and neutral rails be at 2 difference voltage?
    It will effectively become a capacitor, ready to deliver a current if connected. (how big or small the capacitance is, is another thing. I am only arguing if there even IS any infinitesimal capacitance at all.)
     
  11. Sep 20, 2011 #10
    Ok i get where u were confused if u unplug the wires then u have ask is there any voltage source present (after unplugging) that can make the two wires at different potential? No there is not so in absence of potential source the two wires immediately comes to the same potential.
     
  12. Sep 20, 2011 #11
    Another very imp thing u have to understand about current is that when u make current flow in wires u are not pushing extra charges to the wire u are simply making the charges already present in the wires move in certain direction according to the potential...
     
  13. Sep 21, 2011 #12
    sorry, but i think freeze your concept is kind of incorrect.

    1) are u implying that when u open a circuit, the voltage on the live and neutral rail become the same potential? I disagree with this, if you are implying so.

    2) Yes I understand that, but what causes a build up of potential difference then? Isn't it moving excess positive/negative charge onto the conductor?

    Hope you don't take any offence. Just academic argument.
     
  14. Sep 21, 2011 #13
    No no i am not Saying that the potential of live and neutral wire of power sockets comes to at same potential. the wires of for e.g tv, radio when disconnected comes to at same potential!!!
     
  15. Sep 21, 2011 #14
    Secondly there are 2 aspects of electricity one static like the potential across capacitor which is developed because of one side being excess in charge and other side equally deficient. we do not get static potential (like capacitors) in our home.

    The voltage/power that comes in our home is the result of Faraday's law of electromagnetic induction which is very much different from what u are thinking of how potential develops in two live and neutral wires...
     
  16. Sep 21, 2011 #15

    Vanadium 50

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    Hey everyone...please take a look at the PF Rules about text-speak.
     
  17. Sep 21, 2011 #16

    Dale

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    In circuit theory no component ever gains a net charge. A build up of charge is not necessary to get a potential difference. All that is necessary is for the two rails to be connected to something with a potential difference.

    The potential of an isolated conductor is not really even defined in circuit theory.

    It is possible to model a pair of conductors as a parallel RC circuit to get the stray capacitance and leakage resistance. If you do that then because of the very small capacitance you will find that the time constant is also small, meaning that it quickly discharges whatever charge it might have had.
     
    Last edited: Sep 21, 2011
  18. Sep 21, 2011 #17

    Dale

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    The reason that you are getting different answers is because different people are answering your question with different theories and/or models. That is always a danger if you don't specify the theory you want to use in the question.

    There are basically three ways that you can look at a pair of conductors:
    1) Circuit theory as simple conductors
    2) Circuit theory as a RC circuit with a small stray capacitance and large leakage resistance
    3) Maxwell's equations with complete material properties

    Personally, I would answer the question using 1) which would be a "no". If you prefer to use a more complicated theory then you are welcome to, but I think that you would need a really carefully designed experiment with very sensitive equipment to detect any errors from the approximations in 1).
     
  19. Sep 21, 2011 #18

    cmb

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    A mains cable is a capacitor and it will charge and discharge in each cycle of current, just as any other capacitor will.

    Post #6 is therefore correct.

    The capacitance is around 30~50pF/m for 13A mains cable (like kettle cord), higher for smaller lower current cable.

    Therefore, if you were to [theoretically] disconnect* a 1m worth of 3A cable at the instant of the peak of a 230V RMS cycle, you'd have 325V on, say, a 100pF capacitor = ½.100x10-¹².325² = 5μJ.

    *(both disconnecting the load on the other end of the wire, and then isolate the lines from the mains)

    5μJ wouldn't hurt a fly. You likely generate more static electrical energy pulling your shirt off!

    But the capacitance of cables does become a real problem for long distance grid distribution, and also control cables. HVDC is a solution to the former, and there is guidance on the latter; if of interest you might like to take a look at a few links on this:

    http://www.alstom.com/grid/solutions/high-voltage-power-products/hvdc-transmission-systems/ [they used to have a presentation on this on their website, but appear to have moved it]
    http://www.moeller.net/binary/ver_techpapers/ver949en.pdf
     
    Last edited: Sep 21, 2011
  20. Sep 26, 2011 #19
    Can someone explain more on this. Is this an engineering theory? A simplification of things? I don't see how the live Rail won't get a net charge as compared to the neutral, if i use fundamental physics to analyse the problem.
     
  21. Sep 26, 2011 #20

    Dale

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    Yes, circuit theory is a simplification of Maxwell's equations. It is the primary theory used for building most modern electronics. Circuit theory relies on 3 assumptions:
    A) The system is small enough that the finite speed of light is not important
    B) The net charge on every component in the system is always 0
    C) There is no magnetic coubling between components

    To a very, very good approximation the power cord follows circuit theory with the simple model (1) that I described above in post 17. Nobody will "get a nasty shock" by opening the case and shorting the wire.

    With rather sensitive equipment and carefully controlled experimental set up you may be able to detect the difference between the simple circuit theory model (1) and the more complicated circuit theory model (2). I don't know that you could ever detect the difference between (2) and the full Maxwell's equations (3).
     
    Last edited: Sep 26, 2011
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