homework and exercises - What would be the effective resistance of the ladder of resistors having n steps

I'm a tutor. This is a high school level problem. In high school, every one have might have solved a problem of effective resistance of a ladder of resistors having infinite steps. Now the problem is little different. what if it has n steps instead of infinite steps. How to calculate effective resistance in that case?

Answer

I'll take it step by step here. First I'll write the answer for the first few cases with circuit analysis. Then I'll apply a reduction to show the pattern that the problem arrives at.

N=1

$$Z = R+R=2R$$

N=2

$$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$

N=3

$$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + \frac{1}{\frac{1}{R}+\frac{1}{R + R}}}} = R \left( 1+\frac{1}{1+\frac{1}{1 + \frac{1}{1+\frac{1}{1 + 1}}}} \right) =\frac{13}{8} R$$

At this point the continued fraction is clearly identifiable. I do find it a little troubling because it doesn't scale with $N$, but the continued fraction grows as $2N$. This could be remedied by not considering adding 2 resistors on the end each time, but instead add one resistor in parallel, one in series to that one, then a new in parallel between the two new ones, and use a different index for that.

Computational programs should have some easy function to write continued fraction to some $N$ number of fractions. Unfortunately I can not find this for Maple, but I need such a procedure in order to give an answer for your question. I will define such a thing right here. I intentionally use $n$ in the definition and not $N$ to avoid the inevitable confusion.

$$F(1)=1$$ $$F(n+1) = 1+\frac{1}{F(n)}$$

With this, I can answer your question.

$$Z = R F(2 N) $$

And I can give you a sample of the function that I just created.

$$F(1..6) = [1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \frac{13}{8}]$$ $$F(\infty) = \frac{ \sqrt5+1}{2}$$

I think this is the best form with which the problem can be answered. The fraction can not be reduced easily into some concise algebraic form because the entire point of this exercise is to not introduce assumptions, and the large fraction is the algebraic form of the answer. The finite values of the fraction, however, are about as easy to implement procedurally as possible.

Closed Form Solution

This should be my final edit, and this expression pretty much wraps up the problem.

$$F(n) = \frac{\varphi^{n+1}-(1-\varphi)^{n+1}}{\varphi^n-(1-\varphi)^n}$$ $$\varphi = \frac{ \sqrt5+1}{2}$$ $$Z = R \frac{\varphi^{2 N+1}-(1-\varphi)^{2 N+1}}{\varphi^{2 N}-(1-\varphi)^{2 N}}$$