1. Subscribersonhouse
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    13 Aug '06 06:17
    I was thinking of some of the problems I got in electronics 101 and came up with this one, I suppose its been done before but here goes:
    you have a triangular pyramid, triangular base, 6 lines describes it.
    So each of those 6 lines outlining the pyramid happens to be a one ohm resistor. So measuring from any apex to another apex will show the same resistance but what is that resistance? Each line is a 1 ohm resistor.
  2. Joined
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    13 Aug '06 07:26
    0.5 ohm
  3. Standard memberXanthosNZ
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    13 Aug '06 09:01
    You can solve this through repeated use of delta-wye conversions. You could also make a giant mess with mesh analysis.
  4. Standard memberXanthosNZ
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    13 Aug '06 09:411 edit
    If anyone can solve this and wants a challenge then find the resistance of the three cases in a similar cube.

    The cases being:

    1. Between two points that share a side.
    2. Between two points that share a face but not a side.
    3. Between two points that don't share a face or a side.

    Now solve for any regular polyhedron.
  5. Subscribersonhouse
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    13 Aug '06 13:24
    Originally posted by XanthosNZ
    If anyone can solve this and wants a challenge then find the resistance of the three cases in a similar cube.

    The cases being:

    1. Between two points that share a side.
    2. Between two points that share a face but not a side.
    3. Between two points that don't share a face or a side.

    Now solve for any regular polyhedron.
    Don't forget the cases where the polygon is a solid resistance or the case where its only conductive on the surface but the whole surface is conductive and you measure any two points.
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    13 Aug '06 15:07
    Originally posted by sonhouse
    you have a triangular pyramid, triangular base, 6 lines describes it.
    So each of those 6 lines outlining the pyramid happens to be a one ohm resistor. So measuring from any apex to another apex will show the same resistance but what is that resistance? Each line is a 1 ohm resistor.
    A tetraeder has 4 points, we name them A, the incoming point, and D, the outgoing point, and B and C the others.

    Then from A to D the current can tacke this paths:
    From A to D directly,
    From A to B to D, and
    From A to C to D.

    What about B to C?
    No, those have the same potential so no current is flowing that way, right?.
    Just cut that wire and there will be no difference.

    Connect resistors in serial is R = R1+R2
    Connect resistors in parallel is 1/R = 1/R1+1/R2
    Each resistor R is here 1 ohm so you get:
    1/Rtot = 1/R + 1/R+R + 1/R+R = 1/1 + 1/2 + 1/2 = 2
    Rtot = 1/2 = 0.5 ohms.

    Mephisto2 is right and this is the explanation.
  7. Subscribersonhouse
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    13 Aug '06 16:17
    Originally posted by FabianFnas
    A tetraeder has 4 points, we name them A, the incoming point, and D, the outgoing point, and B and C the others.

    Then from A to D the current can tacke this paths:
    From A to D directly,
    From A to B to D, and
    From A to C to D.

    What about B to C?
    No, those have the same potential so no current is flowing that way, right?.
    Just cut that wire and t ...[text shortened]... = 1/1 + 1/2 + 1/2 = 2
    Rtot = 1/2 = 0.5 ohms.

    Mephisto2 is right and this is the explanation.
    Yep, its basically pretty easy, just a one ohmer with 2 sets of one ohmers in series and those sets in parallel with the basic one ohm, so 2 ohms in parallel with another 2 ohms is one ohm and that one ohm is in parallel with a one ohm so the total is 0.5 ohms, you are right, the one resistor left over is in a junction with the same potential at both ends so no current flow so no effect on the whole matrix. You can draw it as a delta with an internal Wye connection. Lot easier if we had graphics here, eh. Hey, Meph, did you know the answer from a previous problem or did you actually work it out? Like Xanth said, I may be just duping the work.
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    13 Aug '06 17:35
    Originally posted by sonhouse
    Yep, its basically pretty easy, just a one ohmer with 2 sets of one ohmers in series and those sets in parallel with the basic one ohm, so 2 ohms in parallel with another 2 ohms is one ohm and that one ohm is in parallel with a one ohm so the total is 0.5 ohms, you are right, the one resistor left over is in a junction with the same potential at both ends s ...[text shortened]... revious problem or did you actually work it out? Like Xanth said, I may be just duping the work.
    The symmetry makes it easy to see that no current is flowing through the B-C resistor, and from there, it's a simple three parallel scheme of 1 ohm and two times 2 ohms. And that whatever XNZ says.
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