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B Why does Kirchhoff's voltage law hold?

  1. Jun 26, 2016 #1
    I know there were previous threads about this but none of these actually pointed out what I am asking for.

    So if we set the negative terminal in the battery to have 0 potential energy and lets say that Voltage of a battery is 12 volt. So at the positive terminal a one coulomb of charge has 12 joules as the charge moves away from the positive terminal it loses its potential energy and gains it as kinetic energy. Once it encounter load, It takes some of its kinetic energy and change it to another form of energy. What ensures me that this load will take the whole 12 J of energy and the charge will end up gaining or losing nothing at the end of circuit ?
    That is basically the voltage law. Why not for example the load takes 10 J (just for example) and the rest of the 2 J is in the charge as kinetic energy? That would still conserve the law of conservation of energy.
  2. jcsd
  3. Jun 26, 2016 #2
    When you coast down a hill on a bike do you spontaneously do anything except trade potential for kinetic energy?
  4. Jun 26, 2016 #3


    Staff: Mentor

    This is almost never true, particularly not in typical circuits. About the only place where this would ever be true is in a particle accelerator or a cathode ray tube.

    In normal circuits the KE of the charges is irrelevant. The charge carriers are very light and travelling very slowly. It is a worthwhile exercise to calculate the KE in a circuit like the one you describe.
  5. Jun 26, 2016 #4
  6. Jun 26, 2016 #5
    But you haven't actually explained why wouldn't this happen? Are the circuits that we encounter in problems are optimal? meaning it only produce the required voltage?
  7. Jun 26, 2016 #6


    User Avatar
    Science Advisor

    The circuits that we encounter in problems are "ideal". This means that all of the batteries produce exactly the indicated potential difference regardless of how much current is passing. One normally assumes that all wires have zero resistance, all resistors are linear and so on.

    None of this has much to do with why charge carriers do not convert the entire potential difference between the terminals to kinetic energy the moment they move from the negative terminal onto the first little bit of wire. That's because there is no potential difference between the terminal and the first little bit of wire.

    If there were a large potential difference across a circuit element with negligible resistance, what would that mean for the current carried?
  8. Jun 26, 2016 #7
    The current would increase.

    "That's because there is no potential difference between the terminal and the first little bit of wire."
    With negligible ressistance, Do you mean there is no difference at all? Or there is a negligible potential difference because as you move a bit away from the positive terminal (Conventional current) you lose potential energy that is just why electrons move?
  9. Jun 26, 2016 #8


    Staff: Mentor

    Because the charge carriers never have a significant amount of KE. The idea that their PE is converted to KE is simply a mistake.

    Furthermore, even if the charge carriers did have a significant amount of KE, it is roughly the same all around the circuit. So the only conversion of PE to KE would be immediately upon closing the switch. Once the current was flowing, the KE of the charge carriers is essentially constant throughout the circuit.

    There are also several other problems with the idea. First and most importantly, the charge carriers don't carry energy, the fields do. Also, different polarity charge carriers can go in opposite directions within the same circuit.
    Last edited: Jun 26, 2016
  10. Jun 26, 2016 #9
    These words pretty much ruins my understanding of how things works. So basically you are saying that there is an electric field through out the circuit from the battery but it is not exerting a force over a distance except at the beginning? How is that even possible?
  11. Jun 26, 2016 #10


    Staff: Mentor

    The forces and distances you are talking about are very small.

    In a typical circuit, energy is not carried by the charge carriers, it is carried by the fields. The purpose of the charge carriers is not to carry the energy, but to set up the charge and current densities to produce and maintain the desired fields.

    That is also why the KE of the charge carriers doesn't matter. What matters is the fields that they produce. That is the mechanism for transporting energy.
  12. Jun 26, 2016 #11
    Energy stored in field? That is quite weird but it is a topic to search about. As you see above, It is ranked B.. This is just a perspective of a high school student.

    Thank you!
  13. Jun 26, 2016 #12
    The weird thing that most references that I read about this topic pretend that charger are energy carrier. I got accustomed to that which seems reasonable. I could link one or more websites and references if you would like.

    I just don't like to accept facts. I would like to deeply understand how the physics work here.
  14. Jun 26, 2016 #13


    Staff: Mentor

    Oh, my apologies. I did not pay attention to the level.

    I don't think that the question can be answered at a B level. I think that all you can say is that the KE of the charges is not significant in a typical circuit and that the mechanism for the energy transfer will be explained in a future course.

    At a B level, what is important to know is simply how to apply Kirchoff's voltage law. The why will have to wait, beyond a basic statement that it is related to the conservation of energy.
  15. Jun 26, 2016 #14


    Staff: Mentor

    It is a common lie, but it is a lie nonetheless.

    Consider, charge carriers move in fractions of a mm/s. So if the charge carriers were the energy carriers then the time between closing a switch and a light bulb becoming bright would be minutes or hours, depending on the length of the wire.
  16. Jun 26, 2016 #15
    I was going to type a summary about the discussion but after this wait a minute..

    The charges don't have to start moving at the beginning of the circuit. When the electric field is established the wave/signal travels at the speed of light (Which is mynext course after this) to inform all other positive charges (conventional current) in the wire so the electrons near the bulb or exactly at the bulb would move instantaneously and light it up. In fact, the electrons in the filament for example start moving directly thus lighting it up without us noticing a delay.
  17. Jun 27, 2016 #16


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    just careful there .... note the hilighted bits

    it's not quite at the speed of light, can vary quite a bit depending on the conductor and insulation
    can be as slow as 0.5 (50%) for particular types of insulated conductors to around 0.95 (95%) speed of light
    for a bare conductor ... this is called the velocity factor, which you are likely to learn about

    This means that it isn't instantaneous, tho appears that way to us for short lengths of cable( wire ) conductors
    because even at say 75% speed of light you are looking at nano to microseconds over the cable distance from the light switch
    to the light globe in the middle of the ceiling of your room

  18. Jun 27, 2016 #17
    Can it be calculated the speed to switch a distant switch on thru a really long coax cable?
  19. Jun 27, 2016 #18
    I mean instantaneous to our eyes. But it has a really really small time interval.

    Also I should have specified about the speed of light sorry.

    So the few question I mentioned in the back still holds, If anyone has another perspective about the topic feel free to reply.
  20. Jun 27, 2016 #19


    Staff: Mentor

    Which demonstrates that it is the fast changing fields which carry the energy and not the slow moving charges.
  21. Jun 27, 2016 #20


    Staff: Mentor

    Which questions still hold?
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