How to solve the infinete resistor network

[DAKONG]

find resistance between A & B
I'm sure it's not 1 ohm
but i don't know how to solve

http://www.pantip.com/cafe/wahkor/topic/X5389102/X5389102-1.jpg

 

[denverdoc]

No its only 1 ohm if that's the only path, in looks like there are an infininte number of paths such that

1/Req=(1+(1/3R+1/3R))+(1/3R+1/3R))...

(and that's 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.

 

[denverdoc]

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.

 

[variation]

Because the Ohm Law is linear, the principle of superposition is applicable.
Consider two special cases and superpose one on another.
Case1:
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.
Case2:
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 occurring 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.