Infinite network of resistors: an alternative solution (let's discuss about it!)

Question

I was trying to solve analytically the infinite network of resistor problem, but I did not make it. So I looked up for the solution: Infinite network of resistors

But I did not like it. It might be that I am misunderstanding something but I would like to hear your opinion about it.

Here there is a picture of how I think about it. sketch

I think to have simply stretched the wires. For this reason I don't like when in the previous solution they simply add two resistances in series, I don't think they are simply in series.

I tried to compute the value of this resistance by recursion (until the solution seem to converge, Python):

import sys
sys.setrecursionlimit(1500)

def deep(R_base, number, limit):

    if number<limit:
        number = number+1
        R_final = 2.0*R_base + (1.0/R_base + 1.0/deep(R_base, number,limit))**(-1.0)
    else:
        R_final = 2.0*R_base + R_base/2.0


    return(R_final)

R_basic = 1.0

n = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 500, 750, 1000]    

for nn in n:

    R_eq = (1.0/R_basic + 2.0/deep(R_basic, 1, nn))**(-1) 

    print nn, R_eq, R_basic*(1.0+3.0**(0.5))

and I got this result:

1 0.555555555556 2.73205080757
2 0.575757575758 2.73205080757
3 0.577235772358 2.73205080757
4 0.57734204793 2.73205080757
5 0.577349678926 2.73205080757
6 0.577350226811 2.73205080757
7 0.577350266147 2.73205080757
8 0.577350268971 2.73205080757
9 0.577350269174 2.73205080757
10 0.577350269188 2.73205080757
100 0.57735026919 2.73205080757
500 0.57735026919 2.73205080757
750 0.57735026919 2.73205080757
1000 0.57735026919 2.73205080757

The second column value is < than the expected value (third column) and the second column value seem to have converged.

I think this is the right value. What do you think about it?

Answer 1

Your approach is OK, but you solve a different problem. In the first solution you calculate the equivalent resistance of an infinite ladder, but in fact is semi-infinite. It starts at a given point, and you keep adding resistors only on the right side.

Your sketch is different. You start from the middle of the ladder, and keep adding resistors on both sides. To get how your problem is related to the previous one, consider the resistor $R$ between the brown points, and it is in parallel with two semi-infinite ladders as in the previous problem. If $R_E$ for the semi-infinite ladder is 2.732, then the equivalent resistor for your problem is $$\frac{1}{\frac{1}{1}+\frac{1}{2.732}+\frac{1}{2.732}}=0.5773$$