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arbitrary resistor network analysis

Discussion in 'Electronic Basics' started by Alex Hunsley, Apr 8, 2004.

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  1. Alex Hunsley

    Alex Hunsley Guest

    I'm aware of the basic rules for calculation resistance networks where you have
    resistors in series, and in parallel combinations, but is this knowledge enough
    to calcuklate the resistance between two points in an arbitrary network (or
    think connected graph!) of resistors?

    For example, take the following graph where every edge represents a resistor:
    (needs to be viewed in a propertional font like courier)

    A
    /|
    / |
    / |
    R0 / | R1
    / |
    / |
    / R2 |
    ---------
    \ |
    \ |
    \ |
    R3 \ | R4
    \ |
    \ |
    \|
    B


    What is the effective resistance betwen A and B in terms of R0, R1 etc?
    Or, to put another way, can I analysis this circuit just thinking in terms of
    parallel and series resistors?

    thanks
    alex
     
  2. CBarn24050

    CBarn24050 Guest

    It's a basic problem covered in any text book. Basicly you mark in the current
    flow around each loop (it doesn't matter that there is no actual flow) and
    produce a voltage equasion for each loop. in this example there are 3 loops
    giving you 3 equasions. The currents are also related ie. ir0=ir1, ir3=ir4,
    ir2=(ir0-ir3). A bit of algebra and your there.
     
  3. Alex Hunsley

    Alex Hunsley Guest

    This makes some sense to me, but I don't understand completely. Do you have any
    resources on the web you could point me at to explain this?
    Or could you explain a little more?
    How do I start marking current in? Do I just assume there is a current, call it
    I, flowing from A to either connections there, and find out the corresponding
    voltage (knowing that V = IR)?
    Suppose I do this for the three loops that exist, and come up with three
    voltages across them, V0, V1, V2, what do I do then with these three voltages?

    thanks
    alex
     
  4. Not for the general case, you can't. When there is symmetry that
    forces zero drop across the center resistor, you can. Do a search for
    "loop equations" and "node equations". They are the general methods.
     
  5. Alex Hunsley

    Alex Hunsley Guest

    Oh yes, I found this one out by putting a few resistors together - when you
    have symettry like that, it makes sense that the resistor crossing the middle
    might as well not be there as far as the resistance calculation goes.

    Aha, thanks, these were the terms I was looking for!

    alex
     
  6. John Larkin

    John Larkin Guest

    The general solution, as has been noted, is to write the loop
    equations and solve them simultaneously, or use equivalent matrix
    methods.

    You can also do this by simple breakdown:

    Apply 1 volt at A and ground B.

    Ignoring R2 for a bit, compute the Thevenin equivalent voltage and
    resistance at the R0:R3 junction (call that Vx, Rx), and repeat for
    the R1:R4 junction (Vy, Ry.) Now add in R2. The two Thevenins and R2
    make a simple voltage divider string, and you then get "loaded" values
    for Vy and Vx. Knowing these, the supply current is apparent and the
    effective net resistance is just its reciprocal.

    More fun than solving simultaneous equations.

    Interestingly, it's often easier to design networks than to analyze
    them, especially since you have the option of picking ones that are
    easier to think about. I once designed a duty-cycle DAC that had two
    trimpots to set gain and offset. The interactions were so bizarre that
    *nobody* could set the pots.

    John
     
  7. Another approach, more related to parralel/series calculation is to do a
    delta-Y conversion. Google for delta-Y, then apply the formula. I guess the
    derivation of the formula is beyond the scope of the answer here, but it can
    be done.

    Cirip
     
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