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Homework Help: Problem understanding circuit diagrams

  1. Mar 16, 2008 #1
    1. The problem statement, all variables and given/known data
    I know how to deal with resistors/capacitors etc in series and parallell circuits, but when it comes to more complex circuits I have no idea how to treat them, and I cant find any guides or info on it and my physics book doesnt tell anything. All solutions to this kind of problems never explain how to think.
    Heres a picture on a "normal" parallell circuit that is easy to understand:

    What I dont understand, is how to treat more advanced circuits when its not trivial if its parallell or a serie. I give some examples of these kind of problems that I dont understand:
    1. (see picture namnlös4)
    You have 12 1-ohm resistors set up in a cube. What resistance will you get if you measure at the ends over the cubes diagonal?
    2. I just cant see how they get the total resistance in this figure (pic circ.jpg)
    2. Relevant equations
    Serie circuits:
    parallell circuits:
    3. Given solutions to the problems
    1. (see picture sol1)
    well I just cant see how they get to that calculation...
    2.also see picture circ.jpg
    the problem that I dont understand in this one is how the deal with the resistors R3 and R5.

    I hope it didnt get too messy. Basically my problem is tips or "formulas" on how to deal with more complex electric circuits.

    Attached Files:

    Last edited: Mar 16, 2008
  2. jcsd
  3. Mar 16, 2008 #2
    None of the 12 1 ohm resistances is in series or parallel with any other. If the resistances were different there would be nothing to do but set up a bunch of simultaneous equations for the currents through the resistances or the potential at the corners of the cube. This is far too much work to do by hand

    Because the resistances are equal however, the network is symmetric. Align the cube with edges of length 1 along the edges of the x, y and z coordinate axes, and the connections to the cube are at the points (0,0,0) and (1,1,1). The network won't change if you rotate it through 120 degrees around the line through (0,0,0) and (1,1,1).
    This rotation will rotate (0,0,1) to (0,1,0) and then to (1,0,0). Since the resistances do not change with the rotation, the potentials cannot change either, so these three points have the same potential,
    If two or more points have the same potential, you can connect them with wires and no current will flow through these wires, so the rest of the circuit will remain unaffected.
    Now suddenly the 3 resistances between the origin and these 3 points ARE parallel and you can replace them with their equivalent resistance. This will work with other resistances in the circuit as well.
  4. Mar 17, 2008 #3
    Do you meant that virtually adding wires, will make them parallell, and therefor they are parallell since these wires doesnt affect the circuit?

    I still dont get how I should proceed, should I just sum all resistances as parallell? How do I know that some of these resistancec are not parallelles inside other parallell? The circuit dont make any sense to me, I cant see any similarity to parallell- or serie circuits...
  5. Mar 17, 2008 #4
    The point of parallel resistances is that two ends of both resistances are at the same potential V_1, and the two other ands are both at another potential V_2. This allows you to conlude that the currents through them will be (V_2-V_1)/R_1 for R_1 and (V_2-V_1)/R_2 for R_2, and that you can replace them with a resistance of R_1*R_2 * (R_1+R_2).
    This is easiest to see if the ends of the resistances are wired together, but in this case symmetry can show that some points must have the same potential, and that those points can be connected by wires without changing any currents, or potentials in the rest of the circuit.
    If you rotate this circuit clockwise through 120 degrees around the diagonal that goes through the 2 corners where it's connected to the voltage source, you can see that
    resistance A ends up in the same place as resistance C, wich ends up in the same place as resistance B. All of the resistances will end up in the same place as another resistance and since all resistances have the same value, the circuit doesn't change at all with the rotation. This means that all voltages at points that get sent into each other with the rotation must be equal. One such set of points is the 3 points at the end of resistances A, B and C (the points at (1,0,0), (0,1,0) and (0,0,1) from my previous reply). There is one more set of points at the same potential.
    It will help to use an actual cube to see this. If you look at a cube along one of the internal diagonals, it's easy to see the 3-fold symmetry.

    1 - find out all the points that are at the same voltage, because they will end up in the same places with the rotation. There only 6 points to consider in total.
    2 - Connect them with wires.
    3 - draw the circuit that results
    4 - replace parallel resistances with their equivalent. This will be an easy step
    5 - the circuit that results from this will be easy.

    this won't make too much sense until you do step 3.

    If you have an idle week or 2 give a formula to compute the equivalent resistance as a funtion of the twelve resistances if they can be different. Bring a lot of paper :)
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