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Finding Equivalent Resistance

Discussion in 'Electronic Design' started by [email protected], Jun 1, 2005.

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  1. Guest

    This is basic Electrical Engineering puzzle - hounting me since 1993. I
    hope this is a correct place to ask this and such questions. Simple
    mathematicle induction fails in this 2D arrangement.

    If '+' denotes a join, find equivalent resistances between any two
    adjacent and diagonally opposite "+" in the figure below. The mesh
    extends infinitely in all the directions.

    | | |
    --+--R--+--R--+----
    | | |
    R R R
    | | |
    --+--R--+--R--+----
    | | |


    Extending the question, what could we say about an infinite triangular
    or hexagonal mesh like this?

    Thanks in advance,
    -Bhushit
     
  2. Hint






















    Inject a 1A current into one node and out of the other. Use symmetry and
    superposition to calculate the voltage drop between the nodes.

    martin
     
  3. John Bailey

    John Bailey Guest

    The paper: Infinite Resistive Lattices by Atkinson and van Steenwijk
    provides somewhat more details on the Fourier Transform method of
    solving periodic networks. It develops solutions for multi-
    dimensional, trangular and hexagonal lattices.
    http://www-th.phys.rug.nl/~atkinson/resist.pdf

    Three things:
    1) The solution seems to be missing something. It is a methodology
    lacking an explaination of (a key detail in) why it works.

    2) When I tried to open the page just now, it came up with an error
    message.

    3) I have a hard copy and probably the pdf file on my system. If you
    cannot find the web version, message me for a copy.



    John Bailey
    http://home.rochester.rr.com/jbxroads/mailto.html
     
  4. Fred Bloggs

    Fred Bloggs Guest

    damned little chance of that ever becoming a topic of discussion amongst
    these pathetic old trolls...
     
  5. FWIW, there was some discussion 13 years ago in sci.physics, here:
    http://tinyurl.com/cvh8k

    Quoting one extract from one post:
    This is problem 2.27 of
    " -Introduction to Modern Network Synthesis-, by M. E. Van Valkenburg
    (Wiley, New York, 1960). Although it is intended that much of the
    book be covered in a one-semester undergraduate course, "[we] assume
    that the reader is familiar with the elementary methods of network
    analysis."

    The solution approach looks like the one Martin hinted at.
     
  6. r.e.s.

    r.e.s. Guest

    http://atkinson.fmns.rug.nl/public_html/resist.pdf
    seems to work.
     
  7. Joel Kolstad

    Joel Kolstad Guest

    This is almost never true of today's undergraduates in EE programs and nowhere
    not ubiquitous among graduate students. I've looked at those old network
    theory books and it's amazing how 'deep' the area goes!

    (Granted, when you look at the types of problems engineers typically work on
    today, it's hard to argue they should be reading Van Valkenburg rather than
    whoever writes a good Java or Linux book or something...)
     
  8. r.e.s.

    r.e.s. Guest

    WLOG, take R = 1 and suppose a square lattice with unit spacing.
    The paper cited by John Bailey doesn't say so (afaics), but
    inspection of the table of values there suggests that between
    opposite corners of a square of side n in this network, the
    resistance is exactly R_nn = (2/PI)*sum(1/(2*k-1),k=1..n)
    for n=0,1,2,... (Between adjacent lattice points it is 1/2.)

    --r.e.s.
     
  9. That seems right for R_nn. But what about R_nm? As you say, at R_01
    it is 1/2. Also, at R_12 for example, the exact answer is 4/PI-1/2.

    An iterative algorithm I devised (not the first person to invent the
    idea, by any stretch, but it is what I came up with when thinking
    about the problem before) is to arbitrarily place a voltage of 1V at
    the node under consideration (n,m) and 0V at the node of (0,0). I
    then arbitrarily set all other nodes of some arbitrarily sized finite
    grid (extending from -S to S in both dimensions) to an assumed guess
    of 0.5V to start. I then go to alternate nodes on the finite grid,
    first those spanning from 1-S to S-1, stepped by 2, and compute the
    node voltage as the mean of the adjacent node voltages (up, down,
    left, and right, in the case of a square grid.) Once that is done,
    the other alternate nodes on the finite grid are then visited, those
    spanning from S to S, stepped by 2, and compute the node voltages
    there in the same fashion. (Just be sure not to do any such averaging
    for the node at (0,0) or at (n,m).) Repeat the process over and over
    until the voltages settle. At that point, the resistance can be
    computed by assuming that current flows accurately based on the
    potential differences between (0,0) and its adjacent squares. (Which,
    of course, determines the current that the node at (n,m) must
    inevitably supply.) Since your starting potential is 1V between them
    and since you know the required current to be supplied by examining
    the nodes next to (0,0), the resistance is 1/sum.

    Jon
     
  10. Guest

    In alt.engineering.electrical wrote:

    | This is basic Electrical Engineering puzzle - hounting me since 1993. I
    | hope this is a correct place to ask this and such questions. Simple
    | mathematicle induction fails in this 2D arrangement.
    |
    | If '+' denotes a join, find equivalent resistances between any two
    | adjacent and diagonally opposite "+" in the figure below. The mesh
    | extends infinitely in all the directions.
    |
    | | | |
    | --+--R--+--R--+----
    | | | |
    | R R R
    | | | |
    | --+--R--+--R--+----
    | | | |
    |
    |
    | Extending the question, what could we say about an infinite triangular
    | or hexagonal mesh like this?

    For AC, this could be even more interesting based on the lengths between
    nodes and the AC frequency involved.
     
  11. r.e.s.

    r.e.s. Guest

    [r.e.s. wrote ...
    [>> R_nn = (2/PI)*sum(1/(2*k-1),k=1..n) for n=0,1,2,...
    The OP asked only about points that were "adjacent and diagonally
    opposite" (which would reasonably be interpreted as referring only
    to R_11, or possibly to both R_11 and R_01) -- I mentioned the
    apparent formula for R_nn (n=0,1,2,...) because of its simplicity.

    Are you aware of any formulas of comparable simplicity in the more
    general cases?

    --r.e.s.
     
  12. Yes, but I don't recollect them, now. I only recall what I wrote.

    Jon
     
  13. Mark P

    Mark P Guest

    Resistance (ideally or "ohmically") is independent of frequency.
     
  14. operator jay

    operator jay Guest

    Sure and why don't you include self and mutual inductances, and the
    capacitances, for spice.
     
  15. Don Kelly

    Don Kelly Guest

    Martin Demello has the right approach. However, you have to do it in two
    steps. Inject current into a node with the return connection a "circle" at
    infinity. The current splits equally in 4 directions. Now disconnect and
    draw current from the other node with injection at infinity. Superimpose.
     
  16. The Phantom

    The Phantom Guest

    This simple technique doesn't work for the case of a pair of adjacent *diagonal* nodes.
    Work out the details and you'll see why.

    It does work for a pair of adjacent "orthogonal" nodes, however.
     
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