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Energy/Potential Loss for current through a 90 degree bend wire/resisitor

  1. Mar 8, 2010 #1
    Hi All,

    I know that when water pass through a 90 degree bend pipe, there will be a pressure lose. However, how to calculate the energy/potential loss for electrical current pass through a 90 degree bend wire/resistor.

    Academic paper appreciated!

    THanks
     
  2. jcsd
  3. Mar 8, 2010 #2

    sophiecentaur

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    You want an academic paper to discuss a very flawed analogy?
    Just as soon as the water analogy for electricity starts to fail you need to drop it and move on.
    Just study elementary circuit theory and you won't go wrong.
     
  4. Mar 8, 2010 #3

    Come on dude, at least you can somewhat prove it doesn't happen. If everybody just drops, I don't see anything will actually move on.
     
  5. Mar 8, 2010 #4

    f95toli

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    The bend won't make any difference whatsoever. A resistor is full of scatterers, meaning the current does not really pass through it in a straight line even without a bend.

    The only time a "bend" can make a difference is if you really have ballistic transport, but that never happens in an ordinary conductor and definitely not a resistor.
     
  6. Mar 8, 2010 #5

    sophiecentaur

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    Au contraire - it's up to you to prove that a resistive effect exists. I'm not prepared to trawl through the Science to try and find a loophole that would make your idea right - and I don't think anyone else would be either.
    Until you have looked thoroughly at EM theory and measurements and have found such an effect then you are not really in a position to insist that your analogy applies here.
    I can only ask "Why should it?". Things can only "move on" if people don't waste time without learning the present state of things.
     
  7. Mar 8, 2010 #6
  8. Mar 8, 2010 #7

    sophiecentaur

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    I think that ,unless there were some actual radiation from the bend, there could be no 'resistance' in addition to the existing dielectric and copper losses.
    I don't think the OP was considering anything as esoteric as this, though.
     
  9. Mar 8, 2010 #8

    Averagesupernova

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    It's a well known fact that high frequencies (UHF and up) are impeded by corners.
     
  10. Mar 8, 2010 #9

    f95toli

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    Yes, but that is because corners mess up the impedance of the line which is relevant at hight frequencies, it is a geometric effect which has to do with the field distribution in and around the line; it has nothing as such to do with the current having to "bend".

    Also it is -as Bob S has already pointed out- perfectly possible to make sharp corners at high frequencies as long as you use a miter; the change in dimensions basically compensates for the extra coupling which in turn means that the impedance does not change.
     
  11. Mar 8, 2010 #10

    sophiecentaur

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    Mismatch is a different thing from Resistance. It's an entirely separate issue, involving reflected (coherent) power and it is certainly nothing like analogous to water going round corners. Why won't this thing just DIE?
    At least read a book about it.
     
  12. Mar 9, 2010 #11

    Averagesupernova

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    To both of the posters who posted after I did (#8). Notice that nowhere did I say resistance. In fact I said IMPEDE which implies impedance. I may be wrong about this, but with the level of understanding that the OP seems to have (not have) I thought it was worth mentioning.
    -
    I've heard the old mitered corner thing many times. And to me it's like saying: "We can't build this building on this site because it is too low and water will collect there. So to fix that, we will bring in enough fill to raise it 15 feet. Now we can build on this low-laying site." Ummm, duh, it's no longer low. Same thing with the mitered corner, it's no longer a sharp corner.
     
  13. Mar 10, 2010 #12

    sophiecentaur

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    It is true that more or less 'anything' different on an AC transmission line will produce a mismatch and reflect some of the power flow. The word "impede" is not one which is defined in Electrical Engineering (afaik) so using it in an argument doesn't achieve much. If you place the appropriate matching element in front (or even after) a right angle bend in a microstrip, you can eliminate the mismatch perfectly at a particular frequency. I don't know of an equivalent in plumbing. . . . . . ?
    I don't see what the remotest connection this has with the resistance to a continuous flow of water around a bend.
    I think we should have a special Forum for people who want to invent / sustain analogies rather than try to get to grips with the real thing. What's the point in this one?
     
  14. Mar 10, 2010 #13

    f95toli

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    But the difference is that this corner IS still sharp; the miter just compensates for the fact that the EM field now can couple to a different part of the waveguide; if you don't want to use a miter you can instead change the layout of the ground plane, perhaps use stubs etc...
    Also, a bend at these frequencies will results an impedance mismatch that in turn results in reflections, not resistive losses which is what the OP was referring to.
    The coplanar waveguides I design and use are all superconducting so they are very nearly lossless, but sharp bends etc will of course still mess up the impedance if I do not compensate for them.
     
  15. Apr 28, 2010 #14
    I asked a question, and very grateful for the answers. That's the point.
     
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