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Brainteaser Forum needs your help

  1. Apr 21, 2005 #1
    Nobody seems to be able to solve this brainteaser[What will be the resistance

    Lets see if the electrical engineers can help :smile:

    have fun
  2. jcsd
  3. Apr 21, 2005 #2

    If you're going to let the EEs try, then it's easy!

    Just say how many decimal places you need? :wink:
  4. Apr 21, 2005 #3
    i'll take a stab at it ..since anything farther away than directly in parallel with the resistor in question wont really effect the outcome..
    60 ohms..
    reason being ..100ohms in parallel with 300 in parallel with 300 = 60..
  5. Apr 22, 2005 #4

    In your post #19 of the original brain teaser thread, you said "Congratulations Davorak" for (I assume) his post #18. Are you saying the problem has been solved? Because I don't think Davorak's answer is correct, and even if it is, it took a lot more than 2 lines, which you said were all that was necessary.

    What's going on?
  6. Apr 22, 2005 #5
    In the brainteaser thread it is said:

    Now consider this: Why is it assumed that there are 4 EQUAL resistance paths between A and B? There is a single 100 ohm resistor resistor between nodes A and B. But the other three paths each start out with a single 100 ohm resistor but more resistors are in series with them. All 4 paths cannot offer the same resistance the way I see it.

    I still do not have an answer to the question though.
    Last edited: Apr 22, 2005
  7. Apr 22, 2005 #6
    The way I see it:

    The shortest parallel path has two loops of three series resistors in parallel with resistor a-b(I'm including this though it can easily be removed). As you go further from a-b the rumber of series resistors in the increasing parallel loops increases by two(the number of resistors in parallel w/ a-b are always odd).

    So, this is what I came up with:


    The 1/100 accounts for the resistor between a-b and the 2/(1+2n) accounts for the two loops with an odd number of resistors.

    Plugging the above into maple (I'm way to lazy to actually figure this out for myself) tells me to expect a resistance of about 0.0997 ohms.

    If you remove the 100 ohm resistor a-b (using my crazy logic from above) you get about 15 ohms.
    Last edited: Apr 23, 2005
  8. Apr 23, 2005 #7

    "....Plugging the above into maple (I'm way to lazy to actually figure this out for myself) tells me to expect a resistance of about 0.0997 ohms.

    If you remove the 100 ohm resistor a-b (using my crazy logic from above) you get about 15 ohms...."

    So 100 ohms in parallel with 15 ohms is about .1 ohms? I don't think so!
  9. Apr 23, 2005 #8
    By removing the 100 ohm resistor you have two loops of 300 ohms which the parallel resistance of that would be 150 ohms. Now stack on an infinite number of loops and you would indeed get 15 ohms. I don't know how you came up with 100 in parallel with 15 because you can't simply remove one component and say oh its X thus if I put that component back in it's obvously some f(x)... It's not. If you want you can see it illustrated above in this post by willib who correctly showed that a 100 ohm in parallel with two 300 ohm loops yields 60 ohms ( http://www.google.com/search?num=100&hl=en&lr=&safe=off&q=(1/100+2/300)^-1&btnG=Search ).

    Basically, what maple is saying is that as the number of elements approaches infinity, the resulting resistance approaches zero. It's probably is true for both cases (with or with out the 100 ohm resistor--I didn't use the indefinate sun function) but like I said, I'm too lazy to integrate the summation eqn (which does work).
  10. Apr 23, 2005 #9

    The resistance of the infinite array is not zero. If that were true then if you had a big sheet of material with some given resistivity, then the resistance between two nearby points in the middle of the sheet would have to be nearly zero, and it's not. As the sheet gets bigger, the resistance approaches some limit, but the that limit isn't zero.
  11. Apr 27, 2005 #10

    Just curious. AntonVrba are you watching this thread? Do you have any comments on what several have posted here?
  12. May 9, 2005 #11

    Once again...
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