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!

How to solve the infinete resistor network

  1. May 8, 2007 #1
    find resistance between A & B :bugeye:
    I'm sure it's not 1 ohm
    but i dont know how to solve

    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. May 8, 2007 #2
    No its only 1 ohm if thats the only path, in looks like there are an infininte number of paths such that


    (and thats only in one 1 dim!), but this may be enuf to solve given your choices (hint!). Sometimes an approximation is good.

    1/(1+6/9)=9/15 second term is 1/(15/9+6/9), etc.

    you can try to solve it algebraically as a sum of infinite terms or work out the first 5 or so iterations at which it drops below 1/4. Assuming its monotonic and converging, only one of the given choices makes sense.
    Last edited: May 8, 2007
  4. May 9, 2007 #3
    My apologies, the analysis above is wrong. I was thinking about it more last night before falling asleep, I have figured out one dimensional case I believe, but am still thinkin about 2d case, if i have time today I'll see if I can't find an answer.
  5. May 9, 2007 #4
    Because the Ohm Law is linear, the principle of superposition is applicable.
    Consider two special cases and superpose one on another.
    Apply a votage between A and infinite, assume the current flows into A is I,and V(A)=votage of A, V(∞)=votage of infinite, the votage difference between A and infinite is V(A)-V(∞).
    The current on the resistance between A and B is I/4.
    Apply the same votage between infinite and B, assume the current flows out from B is I,and V(B)=votage of B, V(∞)=votage of infinite, the votage between B and infinite is V(∞)-V(B).
    The current on the resistance between A and B is I/4.

    Then, consider the two cases occuring at the same time.
    That is, V(A) and V(B) are applied at A and B at the same time and current I flows into A out from B:
    V(A) - V(B) = I×R'
    where R' is the equvalent between A and B.
    The principle of superposition tells us that we can superpose Case1 on Case2.
    Consider the resistance between A and B :
    V(A) - V(B) = [ I/4 + I/4 ]×R
    where R = 1 Ω, and
    I/4 and I/4 are the currents in Case1 and Case2 on the resistance respectively.
    With the two equations, one can solve R' = R/2 = 1/2 (Ω)

    My apologies, my English is quite poor and the description is lengthy.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: How to solve the infinete resistor network
  1. Resistor Network (Replies: 2)

  2. Resistor network (Replies: 4)

  3. Network of resistors (Replies: 3)