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Kirchhoff's Voltage Rule

  1. Mar 19, 2013 #1
    Hello,

    I was reading the article given in this link http://www2.ignatius.edu/faculty/decarlo/kirchhoff.htm, of which pertains to Kirchhoff's Rules.

    In this article, these claims are made:

    "1. Voltage Rule. This is based on the conservation of energy: 'the sum of voltages around a closed conducting loop (that is, a circuit) must be zero'

    In other words, since voltage and work are related, we are saying the net work done must be zero."

    I understand that the battery provides the electrons with energy, that they may traverse the circuit; in addition, I understand that when the electrons (which make up the electric current) flow through the resistor collisions between the electrons and the resistor occur, which results in the electrons losing energy, corresponding to the electric potential drop. However, I don't understand why it's necessary that all of the energy is used up as the electrons flow through the circuit, isn't it possible that some (perhaps even all of them) have some remaining energy when they come to the threshold of the positive terminal?

    As I attempted to answer this question, I came across this article: http://farside.ph.utexas.edu/teaching/302l/lectures/node59.html

    They, too, make a claim that I am unsure of:

    "This rule [voltage rule] is also easy to understand...zero net work is done in slowly moving a charge around some closed loop in an electrostatic field."

    Why is zero net work done?

    I'd massively appreciate it if someone could help me resolve these questions.

    Thank you for your time.
     
  2. jcsd
  3. Mar 19, 2013 #2

    tiny-tim

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    Hello Bashyboy! :smile:
    energy is conserved

    where do you think the energy goes? :confused:

    (if, as in your example, it isn't used up, then it's still conserved, isn't it?)

    since energy is conserved, that means the electric force is conservative

    and voltage = electric potential = potential energy per charge

    which by definition of potential energy is:

    minus the work done by a conservative force, per charge ​
     
  4. Mar 20, 2013 #3
    Well, doesn't some of the charge go into heating up the resistor?
     
  5. Mar 20, 2013 #4

    tiny-tim

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    Yes, some of the energy goes into heating up the resistor.

    (the charge of course is constant)

    The potential drop across the resistor (V = IR) is a measure of the heat …

    it tells you where the energy has gone.

    The battery puts energy into the loop, the resistors take energy out of the loop, the whole thing balances.

    All the energy is accounted for by KVL.
     
  6. Mar 20, 2013 #5
    I still don't see how to answer my original problems. For my first question, isn't it possible that an electron is given energy by the battery, it flows through the resistor and loses some energy--but not all of it--, and goes through the entire circuit, coming to the positive terminal with some remaining energy?
     
  7. Mar 20, 2013 #6

    stevendaryl

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    Sure, that's possible. That would be a transient situation, though. When you first connect a battery to a circuit, the current through the battery isn't immediately equal to the steady-state value; instead, the current will increase from zero to the steady-state value. The steady-state value is the value for which the increase in an electron's energy caused by the battery is exactly equal to the decrease in its energy caused by the resistor. Kirkhoff's laws are approximations that ignore those sort of transient changes.
     
  8. Mar 20, 2013 #7

    tiny-tim

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    the battery is like the motor in a car

    the resistances are like the friction in the bearings, the wind resistance, etc

    when the battery, or the motor, has been running for a short time, equilibrium is reached

    the speed of the car adjusts so that the energy lost to friction, wind resistance, etc, exactly equals the energy supplied by the motor

    the current from the battery adjusts so that the energy lost in the resistances exactly equals the energy supplied by the battery :wink:

    EDIT: stevendaryl already said it! :smile:
     
  9. Mar 20, 2013 #8
    Does this somehow explain why the net work is zero?
     
  10. Mar 20, 2013 #9
    The battery isn't throwing the electrons into the wire. It's not the case that the electrons move into the wire at high speed and then gradually slow down on their way to the other end. Their speed is the same everywhere in the wire assuming the diameter of the wire is equal everywhere. On their journey the electrons are constantly giving of energy to the metal atoms. Despite of that their speed doesn't change because they also are constantly receiving energy from the electric field produced by the battery. The electrons move very very slowly, so their kinetic energy is virtually zero. They can't have any energy left when they reach the end of the wire because they never really had any energy to begin with. When you use a rope to pull a heavy object you have to put a lot of energy in to overcome the friction but the rope never really had much energy since it's very light and moves slowly. So when you stop pulling will the rope have any energy left in it?

    This rule does not apply to an electric field that's induced by a changing magnetic field e.g. a transformer. In that case charge that moves around the transformer core in a closed loop will have work done on it. Such a field is called "non conservative". There is a great video about that topic here


    Most electric fields however are conservative fields. That means that in a closed loop no net work is done on a charge. That is because of conservation of energy. Imagine you had two objects that are statically charged and then you take a third charged object and let it move in a circle between the other two. If it was receiving energy on it's path it could continue to circle forever and you would get energy from nothing.
     
    Last edited by a moderator: Sep 25, 2014
  11. Mar 20, 2013 #10
    I honestly think that it is confusing to explain the voltage rule in circuits with the work by charge in electrostatic field. At least one shouldn't think they are literaly the same. Rather, the charge moving in closed loop in the electric field is just analogy to charge moving across the closed cirquit.
    But I might be wrong.
     
  12. Mar 20, 2013 #11

    sophiecentaur

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    Using the idea that electrons are flowing around a circuit (although there is a mean drift speed, of course) in conjunction with the idea of a "transient situation" can hardly make sense. During any 'switch on' settling time (as little as a few ns duration for some circuits) an electron would hardly have travelled by any significant distance (10-12m, typically). This is just another example of how discussing electric currents in terms of electrons can be nonsensical. Just stick to Current and 'charges' and worry about electron movement when it is relevant - like inside solid state and thermionic devices (black boxes when doing circuit analysis).
     
  13. Mar 20, 2013 #12

    sophiecentaur

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    Yes it does. If all the energy supplied by the battery were not dissipated or transferred within the circuit, where would it go or where would any extra energy come from?
     
  14. Mar 20, 2013 #13

    sophiecentaur

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    Absolutely. A 9V battery, 0.02m long would have a field of 9/0.02V/m across its terminals. Another, physically bigger, battery - say 0.1m long, would have 1/5 of the electric field across its terminals. How could anyone invent a system for circuit analysis that worked on the basis that different batteries of the same voltage would produce different resuts. (Not to mention the lengths of the connecting wires in a DC circuit)
    We use Volts, Potential Difference and Energy transfered etc etc because it works and Volts per Metre wouldn't work.
    Volts per Metre becomes relevant in many instances, such as electron tubes and particle accelerators but it's 'horses for courses' in Science.
     
  15. Mar 20, 2013 #14

    stevendaryl

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    I don't think you're really addressing his question. He's asking about the claim that conservation of energy implies that the changes in voltages around a circuit add up to zero. It's true that for the domain of applicability of Kirkhoff's laws that the voltages add up to zero, but it doesn't follow from conservation of energy.
     
  16. Mar 20, 2013 #15

    sophiecentaur

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    But his question involves an assumption that there is a reasonable answer in terms of electrons buzzing through the wires. I don't think there is one in those terms.
     
  17. Mar 20, 2013 #16

    stevendaryl

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    No, it doesn't.

    Conservation of energy just says that the energy drained from the battery must equal the energy in the circuit minus the energy dissipated away by the resistors. To get that the energy drained from the battery is equal to the energy dissipated away, you have to assume that the energy in the circuit doesn't keep getting greater.

    Mathematically, energy is being supplied by the battery at the rate

    [itex]I V_{battery}[/itex]

    Energy is being dissipated by the resistor at the rate

    [itex]I^2 R[/itex]

    To get these two equal, you need the assumption that the total energy in the circuit doesn't change with time. That will not absolutely true for all times, but it will be true to a very good approximation soon after the battery is connected.
     
  18. Mar 20, 2013 #17

    sophiecentaur

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    I just read this part of your OP. If there were any resistance in the contact with this terminal then the voltage at this point would not be zero. (And, of course, all real batteries do have some internal resistance so, once you start to drain current from them, some energy is, in fact, dissipated internally, so there is some energy 'left' after the charge has 'passed around' a real circuit.
    I wonder if part of your problem with this is that you are thinking in terms of the 'electrons' carrying the energy round the circuit in the form of Kinetic Energy? This is a common misconception which can be resolved once one considers their 1mm/s mean drift speed and the total mass of moving electrons in a wire (1/120,000 of the mass of a copper wire they are moving through).
     
  19. Mar 20, 2013 #18

    sophiecentaur

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    SO the energy in the circuit turns up in the form of magnetic or electric fields and they need to be included in the Kirchoff analysis. But K2 only applies where there is a conservative field. The OP was referring to the simple model of a battery and some resistors. Adding extra factors can't help in explaining the basic concept, surely.
     
  20. Mar 20, 2013 #19

    stevendaryl

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    His question was why does conservation of energy imply that the sum of voltage changes around a circuit must add up to zero? He was trying to understand this by using a picture of electrons moving through the circuit, but his question is perfectly valid, however you try to understand it. How does one prove, from conservation of energy alone, that the sum of voltage changes around a loop add up to zero? You can't prove it from conservation of energy alone. If you let [itex]U_{battery}[/itex] be the internal energy of the battery, and [itex]U_{circuit}[/itex] be the internal energy of the circuit, then conservation of energy tells us that:

    [itex]\dfrac{d}{dt} U_{total} = - I^2 R[/itex]

    where [itex]U_{total} = U_{battery} + U_{circuit}[/itex].

    The additional assumption that's needed is something along the lines that
    [itex]U_{circuit} = 0[/itex] for a perfect conductor. There is no energy associated with having current flowing through a wire. It's not exactly obvious that that's true, and I don't think it is precisely true, but it's approximately true, because as you say, the charges in a conductor don't move very far at all, so there's really no way to build up kinetic or potential energy inside a conductor.
     
  21. Mar 20, 2013 #20

    stevendaryl

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    It's not a "misconception", it's a question. He's essentially asking whether there can be energy in the form of kinetic energy of charges in a circuit that increases with time as the battery is connected. The answer is that the additional kinetic energy due to drift velocity is completely negligible in ordinary circuits, and this energy is therefore ignored in Kirchoff's laws.
     
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