1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the current through an icosahedron of resistors

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider an icosahedron all of whose edges are 2 ohm resistors. If a 12 V battery is connected in opposite vertices, what is the current flow through each resistor?


    2. Relevant equations

    v=ir

    3. The attempt at a solution

    I found somewhere that the equivalent resistance of the icosahedron would be R/2. So i use v=ir to find i=v/r =12/1=12amps and since there are 30 edges in the shape, i take 12/30 because of symmetry so the current through each resistor is 3/5amps?

    Im not sure how to actually reduce a 3D shape of resistors, it has been a while since ive taken a class on circuits. Any help would be much appreciated.
     
  2. jcsd
  3. Apr 28, 2012 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Welcome to PF.

    Not quite. Taking 12A/30 would work if all 30 resistors were simply wired in parallel, but that is not the case here. Moreover, it's not even certain that all 30 resistors have the same current.

    To start, think about one of the vertexes that is connected to the battery, and try to answer these questions:
    1. How many resistors are connected to that vertex?
    2. How does the 12A current get divided up among those resistors?
     
  4. Apr 28, 2012 #3

    vk6kro

    User Avatar
    Science Advisor

    There are many of these 3 dimensional figure problems and they all depend on you knowing a simple trick.

    Because the resistors are all the same size and because the figure has symmetry, you can say that certain points will have the same voltage on them.

    If you connect a wire between these points of symmetry, no current can flow in the wire, (because they have the same voltage on them), so it can't affect the operation of the circuit.

    But once you do this, groups of resistors become paralleled and these are in series with other parallel groups.

    Then you have a chance of solving the problem.
     
  5. Apr 29, 2012 #4
    So each of the vertices has 5 resistors connected to it. Which means if a voltage source was connected to one of these vertices, then 5/12 amps would flow through each resistor. Then each of those resistors is connected to 4 more. but of those 4, two are connected to each other so they cant have current flowing through them. Does that mean that of the 30 total resistors because of symmetry 10 of the resistors don't have any current flowing through them?
    If that was true then the remaining 10 would have to have (1/2) of (5/12)?

    so 10 of the resistors have 5/12 amps going through them; 10 have 0 and the other 10 have 5/24 amps?
     
  6. Apr 29, 2012 #5

    vk6kro

    User Avatar
    Science Advisor

    It would be better to work out the total resistance first.

    You don't know how much current is flowing until you know the total resistance.

    First, though, you need a good image of a icosahedron.

    This may be OK:
    240px-Icosahedron.svg.png

    Now, work out how many resistors are in parallel and how many of these parallel combinations are in series.

    Then when you know the total resistance you can say what the total current is.
     
  7. Apr 29, 2012 #6
    This brings me back to my original problem, as i have no idea how to reduce this circuit. I cant tell which are in series and which are in parallel.
     
  8. Apr 29, 2012 #7
    are the 5 resistors touching any 1 vertex in parallel? so the opposite vertex would have the same situation. which would leave 10 resistors in the middle also connected in parallel? then these 3 groups would all be connected in series? so if what i am thinking is true then the total resistance would be

    2/5 +2/10 +2/5=1ohm?

    but this brings my back to the first conclusion that i had, so i must be doing something wrong, because if Reff=1. then the total current would be 12amps, which is what i had before. unless the current through each resistor is just 12/5. which makes sense for the entrance 5, and exit 5. are the rest of them all the same?
     
    Last edited: Apr 29, 2012
  9. Apr 29, 2012 #8

    vk6kro

    User Avatar
    Science Advisor

    In the diagram above, apply 12 volts to the top and bottom points of the figure.

    Start at the top. You have 5 resistors radiating from this point and then you run a piece of wire around the other ends of these resistors.
    Can you see that they are now in parallel?

    So, what is the resistance of 5, two ohm resistors in parallel?

    Then do the same to the bottom of the figure and then to the middle. You can ignore any resistors that are shorted out by a piece of wire.
     
  10. Apr 29, 2012 #9
    You need to slow down and reread vk6kro's first post.

    Have you figured out which nodes are the same voltage by symmetry? Have you then connected a wire spanning nodes of the same voltage? If you do this the problem is simple.
     
  11. Apr 29, 2012 #10

    vk6kro

    User Avatar
    Science Advisor

    Yes, that answer is right, but you can now derive it instead of using a figure you found somewhere.

    But remember the trick of connecting points of equal voltage.
     
  12. Apr 29, 2012 #11
    Thanks a lot for the help :biggrin:
     
  13. Apr 29, 2012 #12

    vk6kro

    User Avatar
    Science Advisor

    We like happy customers. Welcome to Physics Forums.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook