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The admittance model (of power systems) only ever seems to associate one voltage with any given node or busbar (e.g., just V1 at node 1 in the figure below). Is there an implicit assumption here that the busbar is an equipotential surface?

I see in my physics textbook that "the surface of any charged conductor in electrostatic equilibrium is an equipotential surface." It makes no mention, however, of an equipotential surface for a conductor NOT in equilibrium. How workable is the implicit assumption that the busbar is an equipotential surface? I imagine that a busbar at a substation, for example, would basically never be in equilbrium given all the EM waves splashing around everywhere

enter image description here Source: https://en.wikipedia.org/wiki/Nodal_admittance_matrix

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    \$\begingroup\$ Yes, it's in equilibrium at any time scales you are likely to care about with respect to a power substation, for example. There are a few books that discuss this in more intimate detail -- though usually not all in the same place so as to directly answer this question in so many words. (Yes, it was something I was curious about, too.) You've no clue (I'm projecting) the huge numbers of conduction band charges we are talking about in a conductor and their incredible forces they exert on each other at all times. We don't live in that universe. It's a different place. \$\endgroup\$
    – jonk
    Commented Oct 5, 2022 at 22:23
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    \$\begingroup\$ One book that may help and is written for the "educated high schooler" (they stay away from too much calculus, for example, and the book is very accessible) is Matter & Interactions by Chabay and Sherwood. I think they cover this question, in several pieces unfortunately, pretty well and in a way that is easy to follow. The book has been noted for its novel approach to teaching physics... and I agree. It's very good. There are cases where your question may apply in a nuanced way; RF in the context of a Faraday shield with holes in it, for example. And the charges are NOT equally distributed. \$\endgroup\$
    – jonk
    Commented Oct 5, 2022 at 22:24
  • \$\begingroup\$ Instead, they are equally distributed throughout the interior of the conductor. But on the surface (which is obviously different) there will be a very slight excess charge density and, if a voltage is applied, then also a gradient in the slight excess surface charge density. And if there are kinks or bends in the wire? Then new but again very slight differences in the surface charge densities. The whole thing is fun to spend time on. So I recommend it. But... well... it's complicated as you dig into the exact details, including how the electric field itself surrounds various shaped conductors. \$\endgroup\$
    – jonk
    Commented Oct 5, 2022 at 22:31
  • \$\begingroup\$ "Yes, it's in equilibrium at any time scales you are likely to care about" wow that's a really surprising result... thanks for the pointer, I'll have to check out Chabay & Sherwood! \$\endgroup\$
    – artist_and_not_EE_by_training
    Commented Oct 5, 2022 at 23:21
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    \$\begingroup\$ We often pretend that the voltage is the same everywhere along the length of a current-carrying conductor, but in reality, if it's not a superconductor, then it has resistance distributed along its length, and it must therefore have voltage gradient along its length. If the "voltage drop" is great enough to be a factor in our design, then we model the conductor as a resistor that splits the one node into two separate nodes. \$\endgroup\$
    – Solomon Slow
    Commented Oct 6, 2022 at 18:06

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