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Equivalent resistance of pentagon of resistors

  1. Aug 14, 2016 #1
    1. The problem statement, all variables and given/known data

    Here: http://imgur.com/a/5rRgu
    Resistor Pentagon.png

    2. Relevant equations


    V=IR
    3. The attempt at a solution
    Basically from reversing the polarity of the cell and looking at a mirror image of the circuit i got the following relationships. First I(9)=0 then I(1)=-I(4) I(5)=-I(6) I(7)=-I(8). I believe i can also get these from a symmetry argument. From there I went ahead and wrote down equations for the currents at points A,B,C and O. I inserted the current through the cell to be I(10) and after a few equations i got I(10)=0 and so the equivalent resistance to be infinite ??????????? I made a mistake somewhere. Is it just a mistake during my calculations or is my approach wrong? I believe the error might be I(9)=0 but i'm not sure
     
    Last edited by a moderator: Aug 14, 2016
  2. jcsd
  3. Aug 14, 2016 #2

    gneill

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    Staff: Mentor

    Hints:
    Use a symmetry argument to determine the potential differences between O and F, and O and A.
    If I(9) is zero, what can you do with the resistor OC? (what's the potential drop across it?)
     
  4. Aug 14, 2016 #3
    So basicaly i can simplify the circuit by removing OC completely from the diagram.
    Now i just have a square of resistors. From the symmetry the pd accross OF and OA should be equal and so that should be E/2
    Since the current through FD and AB should have the same magnitude their pd should be the same. Considering DO and OB these two should also have the same pd. and so the pd accross DB is zero. So i can now remove the segment DB ( from my square). From this the circuit is simplified to 2 identical systems of resistors in series where each 'system' is R and 2R in parrallel. So the total resistance is 4R/3. Is that correct ? ( this is preparation for a contest and the answers are not available online )
    Also what was wrong with my approach before using currents?
     
  5. Aug 14, 2016 #4

    gneill

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    Careful. While the magnitudes of potentials OD and OB may be the same, they will have different signs. Remember, the cell has a polarity and the symmetric currents have opposite directions.
    Hard to say since you gave no details of your equations or your manipulations of them.
     
  6. Aug 14, 2016 #5

    gneill

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    Consider instead replacing the resistor OC with a wire (short circuit). Since no current flows through it you have the choice of removing it entirely or replacing it with a short. Choosing the short circuit option gives you some parallel and series resistor reductions to exploit.
     
  7. Aug 16, 2016 #6
    @rohanlol7, did you ever come up with a solution for this problem. I worked it out the long way (a pain) and confirmed that current i9 is indeed 0, and the other symmetric currents that you showed are also correct. So something definitely went awry in your calculation of I10=0 and infinite resistance. It has been so long since I have done this, I'm not even sure what what the method is called that I used. I think it is called node analysis since I summed currents into nodes. There are basically 4 unknowns: voltages B, C, D, and O. Once those are found, all currents and total effective resistance can be calculated.
     
  8. Aug 16, 2016 #7
    Basically I started by assuming that
    The potential at A is E and F is 0. from there I looked at currents and simplified the circuit by removing unnecessary wires. I'll post a solution tomorrow, going to bed now sorry
     
  9. Aug 16, 2016 #8
    I did the same thing - assumed A=E and F=0. And as @gneill suggested, replacing the i9 resistor with a short allows you to parallel resistors. Continuing on from there, the remaining resistors can be reduced to a single overall resistor.
     
  10. Aug 16, 2016 #9
    I overlooked this before, What you need to prove that I9 = 0 is voltage law and node rule considering you have a lot of time to spend solving 7 or more question.
    You could use mesh currents which is an easy way to prove that you will have 5 equation. You could manipulate four of them and you will get an interesting result

    Tom, How did you make so it gets reduced to single resistor? What I was left with is some sort of two wheatstone bridges connected to each other which seems a good point to stop unless you use some other methods to calculate the equivalent resistance
     
  11. Aug 16, 2016 #10
    @Biker, it is based on the symmetry assumption, which allows resistor OC to be replaced by a short (credit to @gneill). Once that is done, it reduces fairly easily by paralleling and seriesing (my own made-up words maybe) the remaining resistors.

    To solve the long way, I ended up with 4 equations and 4 unknowns by summing currents into nodes B, C, D, and O.
    For example, for node B: i2 = i5 + i7 --> (E-B)/R = (B-O)/R + (B-C)/R
    I quickly realized that, for all 4 equations, R can be ignored because it cancels out.

    EDIT: Sorry that my notation was kind of sloppy. The variable B, for example, really should be VB as it represents the voltage at node B. I was just trying to make it as easy on myself as I could.
     
  12. Aug 16, 2016 #11

    CWatters

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    As others have said it's easy using symmetry.

    The impedance to the left of point O is the same as the impedance to the right so the voltage on point O is half that of the battery. Exactly the same applies to the node at the bottom. If both the nodes are at the same voltage then no current flows.
     
  13. Aug 17, 2016 #12
    Okay so i have no idea if this is correct or not but it should be.
    First We have I(1)=I(4), I(6)=I(5), I(7)=I(8) and I(9)=0 ( all of these by looking at the symmetry of the circuit )
    If we can assume V(A)=E and V(F)=0.
    Since I(1)=I(4), the pd across FO and OA is the same.
    V(O)-0=V(A)-V(O)
    V(O)=E/2
    So I(1)= E/2R
    Since I(9)=0 V(C)=V(O)=E/2 this implies that the pd across BC= pd across BO
    So I(5)=I(2)/2
    Applying kirchoffs rule ( forgot which one xD) in loop AOB :
    I(2)R+I(5)R-I(1)R=0 and solving for I(2)
    We get I(2)=2I(1)/3=E/3R.
    Now the current through the Cell is I(10)
    at A we get :
    I(10)=I(2)+I(1)=E/3R + E/2R=5E/6R
    If the resistance of the circuit id R(eff), we get E=I(10)*R(eff) solving for R(eff) we get R(eff)=6R/5
     
  14. Aug 17, 2016 #13
    I posted a solution ( not sure if correct)
     
  15. Aug 17, 2016 #14

    SammyS

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    That looks perfectly good.
     
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