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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


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    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?
  22. Jun 27, 2016 #21
    Well technically from my point of view, the bulb wont light up except when the charges arrive there. Because they have the potential energy inside them.

    1)First, Electrons kinetic energy is not significant and it is roughly the same all around the circuit ( that explains why we have a steady current) #GotThat

    2) How exactly the mechanism of energy transfer happens in the bulb? How does the bulb uses its potential energy? I would have assumed that it changed a bit of that energy into kinetic and because of collision and the high amount of resistance it becomes hotter and light up and loses the energy it gained. But you pointed out before that PE to KE is plain wrong. I know you said before you can't explain in to level B but maybe you could use some analogy or something #Didntgetthat

    3) So the reason why the voltage law holds is I guess what Jbriggs said:
    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.
    So they exactly produce the required amount of energy?
  23. Jun 27, 2016 #22


    Staff: Mentor

    Which is wrong. This point of view is not a valid alternative way of looking at things, it is incompatible with the facts.

    Let's look at this idea carefully. A charge could, in principle, have some sort of storage mechanism that it could use to carry energy, with the energy thus always residing in the charge. The charge would then pick up a packet of energy from the source, travel to the load, and then drop the energy off at the load.

    The single biggest problem with this model is the time which we have already discussed. It would take minutes or hours for a DC electron to travel from the source to the load.

    You seem to think that the electric field could simply serve as a signal to the charges, so that a queue of already energized charges could be told to stop or resume delivering energy. That would allow the energy delivery to start and stop quickly while still having the energy carried by the charges.

    However, that idea doesn't work for the first time that a circuit is connected. That first time would still require minutes or hours, which is contrary to experience.

    Also, if you switched to a higher voltage source it would take minutes or hours for the brightness to change. Which is again contrary to experience.

    Finally, a light bulb would never light up on an AC circuit. In an AC circuit the charges go back and forth the same distance every cycle, so there is no net movement of the charge carriers. So the energy would never go more than a mm away from the source, which is also contrary to experience.
  24. Jun 27, 2016 #23
    There is no first time on connected the circuit

    these electrons are simply supplied by the atoms of copper (or whatever material the wire is made of) within the metal wire.
    "The electrons that light the bulb in a flashlight do not have to first travel from the switch through 10 cm of wire to the filament. Rather, the electrons that light the bulb immediately after the switch is turned to onare the electrons that are present in the filament itself. As the switch is flipped, all mobile electrons everywhere begin marching; and it is the mobile electrons present in the filament whose motion are immediately responsible for the lighting of its bulb. As those electrons leave the filament, new electrons enter and become the ones that are responsible for lighting the bulb. The electrons are moving together much like the water in the pipes of a home move. When a faucet is turned on, it is the water in the faucet that emerges from the spigot. One does not have to wait a noticeable time for water from the entry point to your home to travel through the pipes to the spigot. The pipes are already filled with water and water everywhere within the water circuit is set in motion at the same time." - Physicsclassroom http://www.physicsclassroom.com/class/circuits/Lesson-2/Electric-Current

    The analogy that they used is a bit flawed as you have first to open the faucet for the first time to have water in the faucet in the first place.

    However in circuits, You dont need that. Because electrons are already there. Once the electric field is established, it tries to push the electrons. Lets imagine that the light bulb lights when a charge passes a certain point. Electrons that makes the copper wire are already there. Once we have EF, It takes electrons from the atoms that makes the copper wire, Thus moving it while it consumes its PE

    What they didnt show in the website is how the PE is consumed? That why I came to conclude the KE idea
  25. Jun 27, 2016 #24


    Staff: Mentor

    Energy is transfered through the EM field. The key to understanding energy transfer in EM is the Poynting vector and Poynting's theorem. It describes the work done on matter, the energy density of the fields, and energy transfered from one part of the field to another.

    The nice thing about this is that, under a few simplifying assumptions, a component in a circuit can be regarded as a black box with the power flow equal to IV. So even though the fields describe the details, if we just want the overall balance then we can calculate it much more easily. This is the connection between Maxwell's equations and circuit theory.
  26. Jun 27, 2016 #25


    Staff: Mentor

    The presence of the charge carriers is not disputed, it isn't the problem. The problem is the energy. If the energy is carried by the field then it can move as fast as the field changes. If the energy is carried by the charges then it can only move as fast as the charges.
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