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What's the total resistance of this network?

Discussion in 'Electronic Design' started by Dave Farrance, Mar 1, 2007.

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  1. I'm designing a circuit and found myself unable to figure out the
    resistor values. How embarrassing. After trying to break down the
    problem using equivalent circuits, I found that the problem in its most
    basic form is that I can't figure out the total resistance of this:

    (fixed pitch needed to display this, of course)


    ---------------/\/\/\/-----------
    | d |
    a | b c |
    o----+----/\/\/\/----+----/\/\/\/----+----/\/\/\/----+----o
    | |
    | e |
    ----------/\/\/\/----------------


    Anybody know? Or maybe somebody could point me to a web resource with
    tips on how to get to grips with a tangled network like this?
     
  2. John F

    John F Guest

    Easy :) Rearrange it a bit:

    o-----------o
    | |
    | |
    / /
    \ \
    /e /a
    \ \
    | b |
    +--/\/\/\/--+

    | |
    | |
    / /
    \ \
    /c /d
    \ \
    | |
    o-----------o

    Now calculate :)

    Hint: use Thevenin's theoreme to get the voltage across :)
     
  3. Thanks for the suggestion, but I dunno. It seems to me that Thevenin's
    theorem only works where you can avoid that construct. The interaction
    between the resistances is still overtaxing my brain.

    v1 o-----------o v1
    | |
    | |
    / /
    \ \
    /e /a
    \ \
    | b |
    v2 +--/\/\/\/--+ v3
    | |
    | |
    / /
    \ \
    /c /d
    \ \
    | |
    0v o-----------o 0v


    I think that you're suggesting that I add voltages as above, work out
    the equations at each node, and substitute for the voltages until
    they're eliminated. It doesn't look solvable to me I'm afraid:


    (v1-v2)/e = (v2-v3)/b + v2/c

    (v1-v3)/a + (v2-v3)/b = v3/d
     
  4. Genome

    Genome Guest

    I usually find that not designing in silly resistor networks helps a lot in
    that you don't have to solve them.

    I suppose it's a matter of preference but, if you try my method, I can
    assure you that it's much easier.

    DNA

    DNA
     
  5. Yes, yes, surely I could rearrange the design so that it didn't have
    tangles like that. I know. Trouble is, I've got to fit in with a
    previous design and with cost constraints, and it's hard to avoid.
     
  6. Genome

    Genome Guest

    There's always some damn excuse isn't there :)

    DNA
     
  7. Jon

    Jon Guest

    Thevenin eqiuvalent resistance = (open ckt voltage)/Short Circuit
    Current.


    ---------------/\/\/\/-----------
    | d |
    a | b c |
    o----+----/\/\/\/----+----/\/\/\/----+----/\/\/\/----+----o
    | | |
    | e | |
    Vin ----------/\/\/\/------------- |
    |
    o
    -----------------------------------------------------
    |
    |
    ---
    -
    Write 3 voltage loops, e.g.,
    a, b, e (I1)
    b, d, c (I2)
    e, c, Ground (I3)
    Solve for I3

    Open circuit voltage = Vin
    Req = Vin/I3
     
  8. Jim Thompson

    Jim Thompson Guest

    It's called loop and nodal analysis.

    ...Jim Thompson
     
  9. John F

    John F Guest



    Similar... Transform into
    and now you are able to do thevenin twice:

    e||c v2 b v3 a||d
    v1*c/(e+c)----/\/\/\/----/\/\/\/-----/\/\/\/----v1*d/(a+d)

    now you can easily calculate v2 and v3 as a function of v1.
    then the current across b ...

    Another idea... use superposition of
    1) left v1 and
    2) right v1

    SO many possibilities.
     
  10. krw

    krw Guest

    Thevanize e, c, and v1, and a, d, v1. Then use superposition to
    combine those with b. Stir and serve cold.
     
  11. Thanks Jon and Jim Thompson for pointing out loop analysis.

    This web page gives a technique which I finds works OK if I put in
    actual numbers for the resistor values and I can then get a numerical
    answer after a non-trivial effort.

    http://mathonweb.com/help/backgd4.htm

    If I try to retain the algebraic form while using this technique,
    though, it quickly turns into a monster. Hmm. I'll have to think if I
    can simplify the original form of the puzzle a bit.
     
  12. John Larkin

    John Larkin Guest

    Hell, Spice it!

    John
     
  13. Greg Neill

    Greg Neill Guest

    Well, that's certainly one way to do it.

    If I had to analyze this circuit, I think I'd start
    by performing a Delta-Y transformation for resistors
    a,b,e. Should be clear sailing from there.
     
  14. Aha. Thanks. Problem solved.

    -----------------/\/\/\/----------
    | d |
    | ab c |
    o----- \ ----- -----/\/\/\/----+----o
    | / a+b+e |
    | \ |
    | ae / be |
    | ----- \ ----- |
    | a+b+e | a+b+e |
    ----/\/\/\/----+-----/\/\/\/-----
     
  15. A few months ago, someone posted the formula for total
    resistance of this arrangement, but I can't find it through
    Google.

    The Thevenin's equivalent solution is shown here:
    http://www.broadcast.net/hallikainen/theory6.html
     
  16. The Phantom

    The Phantom Guest

    Just say "Om mane padme hum", and stare at it for a while. Say to
    yourself, "if it weren't for that pesky b resistor, the total resistance
    would be the parallel combination of the two arms; something like:"

    (a+d)(c+e)
    -------------
    a+c+d+e

    Just a plain product over the sum.

    Ah, yes. Now I see it. It's:

    a(de + b(c+e) + c(d+e)) + d(ce + b(c+e)
    ---------------------------------------
    a(b+c+d) + b(c+d+e) + e(c+d)

    I know you're looking for a solution method which I haven't given. :)
    But I see in another post that you have been tipped off to the delta-Y
    transformation. This result provided for checking purposes only.
     
  17. Robert Baer

    Robert Baer Guest

    Here is a rare case when delta-to-wye conversion is useful; take the
    3 points described by b, c, and d (note the delta configuration, and
    convert it to a wye resistor configuration B, C, and D (same points; now
    have an extra point in "middle".
    The derivation of that conversion is fairly simple but i have never
    seen it mentioned in any university of note (over the past 50 years).
    Once you have the wye values, the reduction is simple.
    BTW, this is the classical bridge, so if there is any symmetry, even
    on a ratio basis, take advantage.

    There are at least 4 different ways to solve the "resistor cube"
    problem (what is the resistance across the farmost corners, given all
    sides are of one ohm resistors); many "cheat" based on the symmetry.
     
  18. Robert Baer

    Robert Baer Guest

    I prefer funny resistor networks, as the silly ones keeps the clowns
    agitated.
     
  19. Fred Bloggs

    Fred Bloggs Guest

    It's one thing to come up with a tangled formula and entirely something
    else to make sense of it:
    View in a fixed-width font such as Courier.
     
  20. kell

    kell Guest

    Here's a closely related problem:
    http://groups.google.com/group/sci....894955a1365?lnk=st&q=&rnum=2#1ae15894955a1365
     
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