An interesting problem was posted on Texas Instruments web site a few years back about an infinite resistor network.  While I got to this post about five years too late to post a timely solution I thought I might post it anyway.

Here is the problem.

What is the resistance of the following infinite resistor network?

And here is a solution.

From inspection the solution is clearly greater than 1.5 Ohms and less than 2 Ohms.  Let’s call the resistance of the network as seen from the left x.

If we add a couple of extra resistors to the left hand side then, because the right hand side extends to infinity, we should expect to see exactly the same resistance.

So now we can simply equate these two networks and solve for x.

x = 1 Ohm in series with (1 Ohm in parallel with x Ohms)

= 1 + x  / (1 + x)

Now rearrange to the standard quadratic equation form, ax² + bx + c = 0

x  - 1 =  x / (1 + x)

(x – 1) (x + 1) = x

x² – 1 = x

x² - 1 – x = 0

x² – x - 1 = 0

and solve using the general solution (or by factors), x = (-b +/- sqrt(b² – 4ac)) / 2a

x = (1 +/- sqrt(1² – 4 * 1 * -1)) / 2

= (1 +/- sqrt(5)) / 2

There are two solutions, but only the positive solution applies because we are using positive resistance values  (but note that negative resistance does exist, as does complex resistance or reactance).

So:

x =  (1 + sqrt(5)) / 2

~ 1.618 Ohms

There are a heap of other solutions that you might try including sums of arithmetic series and iterative methods, but this is the analytical result.

If you have doubts then build the network.  You will achieve 3 decimal digit precision after about 10 resistors (subject to the tolerance values of your resistors).

Now that we have this solution we can extend the network in other dimensions to create a whole lot of interesting solutions to infinite resistance networks.