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Defining a Unique Voltage

  1. Jul 28, 2015 #1
    The book "Foundations of Analog and Digital Electronic Circuits" by Anant Agarwal says the following things regarding the defining of an unique voltage between two points (a voltage that doesnt depend on the path taken).

    First, it defines the voltage as being the line integral going from x to y of the electric field.
    Then it presents the Faraday's law of induction: the line integral over a closed path of the electric field is minus the rate of change of the magnetic flux through a surface delimited by the closed path.

    Until now all is clear.

    Then the author says a thing which I put in the picture attached. This I dont understand.
    (ps: he says that it doesnt have useful meaning because in circuit theory the voltage must be uniquely defined between the two terminals of an electric element).
    What does he try to say by choosing x and y to be the same? Like x=y? And what would be the surface in this case?

    Then the author tries to define a unique voltage and I understand that to have a unique voltage between two points (so that the voltage doesnt depend on the path taken from x to y), the magnetic field must be constant so that the closed loop line integral is zero and not equal to minus the rate of change of the magnetic flux. I understand this. Because if it's zero then I can say that the line integral going from x to y + the line integral going from y to x is 0.

    But I really dont understand what he tries to say in the picture attached.
     

    Attached Files:

    Last edited: Jul 28, 2015
  2. jcsd
  3. Jul 28, 2015 #2
    You should show more of the context.
    What definition of potential is he talking about?

    I thought that the potential difference between two points should be independent of path, by definition. If it is not, the field cannot be associated with a potential.
    But who knows what is this about?
     
  4. Jul 28, 2015 #3

    Dale

    Staff: Mentor

    He is saying to evaluate the integral along a closed loop.
     
  5. Jul 28, 2015 #4
    Evaluating the integral from one point to the same point is a closed loop? And what is the surface in this case?

    If, for example, I evaluate:

    from 1 to 1 of xdx = 0

    Any such simple integral evaluated from one point to the same point on the x axis will be 0. Then, why evaluating a line integral of Edl from x to x will give a nonzero value?
     
  6. Jul 28, 2015 #5
    The author models the circuit elements. He makes the transition from physics to engineering.
    And he says 3 conditions that the models must respect in order to be valid models. One such condition is that the rate of change of magnetic flux linked with any closed loop outside an element (conductor) must be zero for all time. This is the lumped matter discipline.

    And he starts to derive this condition using Maxwell's equation. He starts by definining the voltage between x and y as being the line integral of Edl from one point to the other. The definition that we all know. Nothing special.

    Then he says that this definition indicates that the voltage depends on the path taken. He uses the word "indicates". So at first glance we might look at it and say it depends on the path taken, I think.
    Then he says "put another way, we know from Maxwell’s Equations that:" and he puts Faraday's law of induction.
    Then he says what I attached in the picture.
    Then he says "In the absence of a time-varying magnetic flux, we can write that the closed loop integral is 0"

    The book can be found at: http://siva.bgk.uni-obuda.hu/jegyzetek/Mechatronikai_alapismeretek/English_Mechatr/Anal&Dig_Circuits/literature/Foundations%20of%20AD%20Circuits.pdf [Broken]
    Or search Foundations of Analog and Digital Electronic Circuits on google if u dont trust the link. The first link should be a pdf.

    In Appendix A.


    In a changing magnetic field, the voltage depends on the path taken. So that condition must hold in order to had a conservative electric field where a voltage can be uniquely defined. But I dont understand that with choosing the points to be the same.
     
    Last edited by a moderator: May 7, 2017
  7. Jul 29, 2015 #6

    Dale

    Staff: Mentor

    Yes, clearly.

    There is no surface, it is a path integral, not a surface integral.

    Yes, that is the trivial loop, and it is trivially always zero.

    Because the trivial loop is not the only loop closed loop. For example there is also the loop ##\int_{-\pi}^{\pi} f(\theta) d\theta## where due to periodicity ##\theta=-\pi## is the same point as ##\theta=\pi##. For the voltage to be well defined the integral along ALL closed paths must be 0, not just along the trivial closed path.
     
  8. Jul 29, 2015 #7
    Thanks for the answer.
    I know that it is a path integral and not a surface integral. But in a path integral, you can have a loop, and you can attach a surface to this loop. Like in Faraday's law of induction.

    Ok, I think I got it. First he defines the voltage as a line integral over an open path. Then he makes the two points the same and by doing so he obtains a closed path.
    From the net: "A path C is called closed if its initial and final points are the same point. For example a circle is a closed path."

    So, when he says "if we choose the two points x and y to be the same" he is not saying x = y in the sense that the length of the path is 0, but in the sense of a closed path and of length different than 0.

    My confusion arose from the fact that I considered this path having length 0.
     
  9. Jul 29, 2015 #8
    Now I am curious of this path of length 0. Or the trivial loop, as it is called. The surface attached to this trivial loop has area 0, I think.

    For example, if I consider Faraday's law of induction.
    The left-hand side of Faraday's law of induction is, of course,0.
    Thus the right-hand side must also be 0, independently of what the magnetic field is in that point (constant or changing).

    Because if its changing, then the magnetic flux is also 0 because the surface attached to this trivial loop has area 0.
    Right?
     
  10. Jul 29, 2015 #9

    Dale

    Staff: Mentor

    Yes, but that is a separate theorem, not what is being described by the author here.

    Yes.

    In the sense of all possible closed paths starting and ending at x=y, including the trivial path with length 0, but also including all other possible paths.

    One path does have length 0, but there are an infinite number of other paths with length greater than zero. For the voltage to be unique that integral must evaluate to 0 for all such paths.
     
  11. Jul 29, 2015 #10

    Dale

    Staff: Mentor

    Yes. The fact that the trivial loop is 0 and the fact that all loops must be the same tells us that all loops must be 0. It is only that the 0 result is obvious for the trivial loop.
     
  12. Jul 29, 2015 #11
    Ok. Thanks.
     
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