What is happening in those wires?

Discussion in 'General Electronics Chat' started by Gerry Rzeppa, Sep 9, 2019.

  1. Wolframore

    Well-Known Member

    Jan 21, 2019
    1,160
    356
    upload_2019-9-14_13-7-4.jpeg

    Is equivalent to this at discharge

    upload_2019-9-14_13-17-21.jpeg
     
  2. Wolframore

    Well-Known Member

    Jan 21, 2019
    1,160
    356
    This is more physics than electronics but many of us are interested in both.

    Free electrons doesn’t mean that the conductor is getting flooded with electrons because the net charge has to remain balanced overall. I understand this condition to be mobile electrons and open holes in the copper valence shell. So If you consider this a physical change it has to do with electron and holes caused by the difference in potential from rest to charge or discharge condition. There is also a directional movement of the electrons from the negative to the positive terminal through the conductor.

    Since most conductor are not perfect there is a loss of energy which produces a voltage when the electrons give off energy and end back in the valance shells. Hence the heat which depends on various factors like material, thickness, and current. You can calculate for this. With heat comes changes in dimension. Also a property of metals. Changes to temp can also affect conductivity.

    Others have mentioned the changes to magnetic field as current flows. This is also a characteristic of current.

    I’m not sure if you’re looking for a certain answer but it looks like most it has been covered.

    If there is an imbalance in the electrons due to the push pull effect of your voltage source it’s very small:

    The invariant mass of an electron is approximately 9.109×10−31 kilograms,[66] or 5.489×10−4atomic mass units. On the basis of Einstein's principle of mass–energy equivalence, this mass corresponds to a rest energy of 0.511 MeV. The ratio between the mass of a proton and that of an electron is about 1836.[10][67] Astronomical measurements show that the proton-to-electron mass ratio has held the same value, as is predicted by the Standard Model, for at least half the age of the universe.[68]
     
  3. BobaMosfet

    AAC Fanatic!

    Jul 1, 2009
    734
    181
    I'm 100% in agreement with you on this one. I'm sorry to the TP, but they have so little actual comprehension of the subject that they have no business trying to teach anyone anything about it.
     
  4. SamR

    Well-Known Member

    Mar 19, 2019
    1,015
    399
    No, he understood and pulled the same dumb act on other forums trolling along playing stupid for page after page with no results. He was just looking for attention and kept changing his story as he went along.
     
  5. shortbus

    Expert

    Sep 30, 2009
    6,838
    4,023
    From where I sit it's only "unresolved" to one person. Guess who?
     
    Wolframore and jpanhalt like this.
  6. kubeek

    Expert

    Sep 20, 2005
    5,582
    1,093
    I have a feeling that there is a lot of "this is such basic stuff" attitude in this thread, without looking deep enough. (I have to admit I am after broken arm surgery and might have missed some bits)
    So lets say there is voltage source (not sure if capacitor vs battery changes anything, but should not), and it has two physical terminals in the real world. Does one terminal have slightly more electrons than the other?


    My approach is: intuitively there will be electric field between the two terminals - there should be some electric field between parts of different voltage. For electric field to exist, there has to be more charges at one side than the other. The two terminals have capacitance (is mutual capacitance the term?), therefore one terminal will have a bit more charge on the surface than the other, that induces the electric field. So the "inside" of the wires remains the same, but the outer layer exhibits some minor change in amount of electrons.

    Can you guys tell me if you agree with this, and such explanation suits to the further question, what happens to two wires connected to an AC source? That is, the capacitive charge moving back and forth, and leaving excess electrons on one terminal or the other?
     
  7. nsaspook

    Expert

    Aug 27, 2009
    6,165
    6,953
    I would start another thread for questions on the subject that's absent from the IMO trolling of the OP.
     
    shortbus and jpanhalt like this.
  8. Gerry Rzeppa

    Thread Starter Member

    Jun 17, 2015
    170
    2
    That's great. Thanks. Now please put a resistor in that circuit and show the close-up view of the wire on each side of the resistor.

    MOD: Please use black text, reserve coloured text to highlight a point.
     
    Last edited by a moderator: Sep 15, 2019
  9. Wolframore

    Well-Known Member

    Jan 21, 2019
    1,160
    356
    Let me ask you a question,

    What is a resistor and what is a wire?

    What are the differences between the two. If you have a wire and a resistor of same resistance. What are their electrical differences?
     
    AnalogKid likes this.
  10. KeepItSimpleStupid

    AAC Fanatic!

    Mar 4, 2014
    3,535
    664
    There is nothing fundamentally different.

    When running power wires in a home, it's customary to do voltage drop calculations. <3% drop is typical at 80% full resistive load. I've already dumbed it down a bit.

    The fundamental equation is R=ρL/A where R is resistance. ρ is Resistivity, a fundamental property of a material in (ohm-distance units e.g. ohm-cm), A is the cross sectional area and L is the length.

    If you use the fundamental eqn, L is 2 * distance.

    Unfortunately, things always get wierd: Trace thickness is based on Oz copper,

    PCB Copper Thickness
    1/2 oz. 0.7 mils
    1 oz
    . 1.4 mils
    2 oz. 2.8 mils

    Then there is sheet resistance in ohms/square

    http://four-point-probes.com/sheet-...kness-relative-to-semiconductor-applications/
     
  11. Wolframore

    Well-Known Member

    Jan 21, 2019
    1,160
    356
  12. Gerry Rzeppa

    Thread Starter Member

    Jun 17, 2015
    170
    2
    No difference. And no potential difference, either. On the other hand, in this experiment, the wires and the resistor have significantly different resistances, which divides the circuit into the bottom (B) leg which has a measurable negative electrical potential relative to the top (A) leg for the duration of the discharge. The current may be the same in both legs at any moment during the hour it takes to discharge the capacitor, but it seems to me that something must be different in the A and B legs to account for the potential difference.
     
  13. MrChips

    Moderator

    Oct 2, 2009
    19,111
    6,142
    There is no difference.

    Conduct this thought experiment.

    Connect a very long wire between point A and point B. Current X flows through the wire from point A to point B.
    Now take an infinitesimally small portion of the wire and consider what happens between the two ends of this small portion of wire. We will name the two ends of this portion P and Q.

    The potential difference between P and Q is infinitesimally small, almost zero.
    And still, current X flows between P and Q.

    Current flow is the aggregate movement of charges in the medium. There is nothing physically different between point P and Q, or point A and B.
     
    Sensacell likes this.
  14. Gerry Rzeppa

    Thread Starter Member

    Jun 17, 2015
    170
    2
    If there is "nothing physically different between P and Q," then why does the current flow from P toward Q instead of the reverse?
     
  15. Gerry Rzeppa

    Thread Starter Member

    Jun 17, 2015
    170
    2
    My apologies, folks, for taking up so much of your time and energy. This talk of wires as resistors reminded me of a textbook (https://www.amazon.com/Matter-Interactions-Ruth-W-Chabay/dp/0470503475) that I just dug out (after four years) from an almost-forgotten pile in my office. Turns out the answer to my question is on page 766:

    c&s 766a.jpg
    c&s 766d.jpg
    If I read that correctly, the physical difference that accounts for the potential difference between the wires in my sample circuit...

    cap 4.jpg

    ...is the electron "pile up" in or on or around the surface of the wire in the B-leg of the circuit near the bottom of the resistor, and a corresponding dearth of electrons in or on or around the surface of the wire in the A-leg of the circuit near the top of the resistor.

    That makes sense to me. Does it make sense to you folks?
     
    Last edited: Sep 16, 2019
  16. CharlesWMcDonald

    Member

    May 16, 2019
    232
    79
    You are still attempting to get a simple answer to a complex situation. Get yourself an Electrical Engineering degree, study Maxwell's equations and the Lorentz force law among other things. Then you may be able to understand what you are clearly missing now. You will never be get anywhere as long as you are fixated on electrons piling up somewhere.
     
    nsaspook and ericgibbs like this.
  17. Gerry Rzeppa

    Thread Starter Member

    Jun 17, 2015
    170
    2
    Well, I don't have a degree in Electrical Engineering, but I do have a degree in Mathematics (with honors)...

    degree small.png

    ...and I am not aware that anything in Chabay's & Sherwood's explanation that contradicts what we know from Maxwell or Lorentz.

    Here is a video showing experimental evidence of "surface charge pile up" at certain locations in a circuit:



    If you dislike watching videos as much as I, here is a paper by Sherwood & Chabay explaining the rational behind their "unified treatment of electrostatics and circuits" (that includes a description of the experiment in the video):

    https://matterandinteractions.org/wp-content/uploads/2016/07/circuit.pdf

    I find their approach not only sound, but accessible. And it fills the gaps in my mental images, as well.
     
    Last edited: Sep 17, 2019
  18. BobaMosfet

    AAC Fanatic!

    Jul 1, 2009
    734
    181
    Here's what you're missing- there is no 'pile up' of electrons in this specific case. Electrons cannot pile up because voltage is unchanging. There is a specific potential being lost across the resistor, and so electrons don't 'pile up'. Initially, there is no resistance- it takes some incredibly small amount of time for the resistor to resist-- and as it does, the flow slows down throughout the entire circuit simultaneously.

    That's how the physics actually works.
     
    CharlesWMcDonald likes this.
  19. Gerry Rzeppa

    Thread Starter Member

    Jun 17, 2015
    170
    2
    Why should I prefer your model to the contrary [correction: more comprehensive] model presented by Chabay & Sherwood?
     
    Last edited: Sep 18, 2019
  20. BobaMosfet

    AAC Fanatic!

    Jul 1, 2009
    734
    181
    My model is not contrary to their model. The fact that you think it is, shows your lack of comprehension. Not being mean, just stating the obvious. Your problem is you can't wrap your mind around the resistor because the problem is simply too complex for you Instead, you should replace the resistor with a battery and then try to understand how the capacitor actually behaves at the electron level in both DC and AC environments.

    And don't tell me you understand how a capacitor works, because if you truly did, you wouldn't have even posted this thread.

    I leave you with this timeless wisdom: 'Voltage loves an open; current hates it.'

    If you can actually wrap your mind fully around that statement, it will unlock electronics for you in a way that no library of information will, and for the first time in your life, you will actually understand what Kirchhoff and Thevenin were saying. And that is the gateway to fully comprehending Ohm's Law.
     
Loading...