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Honeycomb Resistor Problem

  1. Jul 11, 2003 #1
    well i hope someone knows how to solve this problem...... :smile:

    well you see in the picture that there is a honeycomd-like circuit......every line that comprises the honeycomb has a resistor with a resistance of R........meaning in a hexagon, there are six resistors......

    the question is to find the total resistance or Req.......and also the current (I) passing through the one i indicated in the drawing

    sorry for the drawing if it is not to accurate......

    i got the answer of 3R for the Req and one-third for the I......is this correct.......actually what i'm looking for is the shortest solution that would solve the problem..........maybe a two liner will be ....

    thanks.......

    hmmmmmmm, i hope the pic would be posted soon......
     

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    Last edited: Jul 11, 2003
  2. jcsd
  3. Jul 11, 2003 #2
    My 1st idea was that currents distribute equally in the parallel lines, i.e., 1/3-1/3-1/3, and 1/4-1/4-1/4-1/4. But that doesn't work with the voltages.
    My 2nd idea was that current would distribute binomially in the parallel lines, i.e. 1/4-1/2-1/4, and 1/8-3/8-3/8-1/8. But that doesn't work with voltages, either.
    Next, tried it with Kirchhoff's rules, but got a very nasty system of equations, couldn't solve.
    Anyone who got a good computer algebra software will get the result easily if he just inputs Kirchhoff's rules right...!
    However, I'm convinced there is an easy solution which I am just too blind to see.
     
  4. Jul 12, 2003 #3
    still waiting...............

    anyone with the software to do the job............
     
  5. Jul 12, 2003 #4
    Solved it

    Let total current = 1, then the currents in the upper branches are 1/2, 5/16, and 3/16. With this information, you can easily find the answer.
     
  6. Jul 12, 2003 #5
    is my answer correct:

    Req = 3R
    I = 1/3:wink:
     
  7. Jul 13, 2003 #6
    No. How did you get this answer?
     
  8. Jul 14, 2003 #7
    hmmmmmmm. got the answer.....

    with the help of MS Excel, the answer is 61/16 or something about 2.8.....

    right????
     
  9. Jul 14, 2003 #8
    Correct.
     
  10. Jul 15, 2003 #9
    hhahahha.......i was correct......excel never makes a mistake.....

    but could you elaborate to me how you got it......the manual way, i mean..........

    any short-cuts for this....
     
  11. Jul 15, 2003 #10

    Doc

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    Always start at the 'outside'. Take resistors that are joined at simple nodes and simplify the circuit. You will find that as you simplify it gets much easier than it looks. If all resistors are the same then you can simplify to reduce the number of resistors starting at the power supply leads also. For instance, if all resistors are the same, you can reduce the left 4 resistors to one resistors since the current HAS TO BE THE SAME IN EACH ONE. Just keep working.
     
  12. Jul 15, 2003 #11
    Doc's method may be easier than mine...

    Well, what I did was, first use the symmetry of the problem. There's only 7 currents J1...J7 you have to find. See here:

    [​IMG]

    Can you use Kirchhoff's laws to write down enough equations so there's a unique solution?
     
    Last edited: Jul 15, 2003
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