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?enter image description here


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.


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


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


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

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.

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

  • 2
    $\begingroup$ It is worth mentioning explicitly that $F(n)$ is the ratio of the (n+1)th and nth Fibonacci numbers. In other words, you've got $F(n)=\frac{Fib(n+1)}{Fib(n)}$). $\endgroup$ – spencer nelson May 31 '11 at 18:55
  • $\begingroup$ @Spencer Exactly. I was holding off as it was already there in Ted Bunn's answer. $\endgroup$ – Alan Rominger May 31 '11 at 18:57

For any given $n$, you can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all $n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an $n$-step ladder in terms of an $(n-1)$-step ladder. If I'm not mistaken, the $n$-step ladder can be thought of as a single resistor $R$ in series with a parallel combination consisting of another $R$ and an $(n-1)$-step ladder. The rules for series and parallel resistance give $$ R_n=R+{RR_{n-1}\over R+R_{n-1}}. $$ The initial one-step ladder has $R_1=2R$, and you can use the above formula to work your way up the chain from there.

I made Mathematica work out the first steps, and they seem to be related to the Fibonacci numbers, specifically $$ R_n=R{F_{2n+1}\over F_{2n}}. $$ I'm sure it's possible to prove that by induction.

  • $\begingroup$ I obtained this same result, see my final form of $F(n)$. There are all kinds ways to arrive at the result, but formulating the problem in terms of a series (as several answers have done here) will lead you to a sequence that can be mathematically shown to be the ratio of $n+1$ to $n$ Fibonacci numbers as you've found. Good work making this observation. $\endgroup$ – Alan Rominger May 31 '11 at 18:46

Is interesting. I think you want to proceed as follows: (1) Without loss of generality assume the input voltage is 1, and the resistors are single ohm. (2) Define r = resistance of the infinite ladder. (3) Now look at the circuit where we add one more step on the lefthand side. We have to solve for the voltage where its connected, as a function of r. (4) The new current is 1 minus this voltage. And the total resistance of the configuration is r'=1/(1-V(r)). (5) Adjust r until r=r' (6) Scale by the R given in the original problem. I.E. the result is r*R.

A bit of algebra in steps 4 and 5.....


Work backwards to make the math easier and also make the pattern emerge more quickly. I did this, but will show results in forward order

R(1) = 2R = (1+1/1)R

R(2) = (1+(2/3)) R

R(3) = (1+(5/8)) R

R(4) = (1+(13/21)) R

R(5) = (1+(34/55)) R

Call this general form ---> R(N) = (1+((a(N)/b(N))) R

Define a(1) = b(1) = 1.

For N > 1 --->

a(N) = a(N-1) + b(N-1)

b(N) = a(N) + b(N-1) = a(N-1) + 2 b(N-1)

You still need to crank through the math for each R(N), since one must know a(N-1) and b(N-1) to get the next term; but it is more straightforward. Maybe someone up on their mathmatics of series can get that into a closed form for us.

  • $\begingroup$ You've started on the path I would use to teach this problem. Starting at the last one added and then reusing the answer through each successive step of the circuit to get a series is the easiest way to solve the problem. I believe I've obtained a closed form, but it's no more special than the observation from Ted Bunn's answer that the series consists of Fibonacci number ratios. $\endgroup$ – Alan Rominger May 31 '11 at 18:50

Here is a way an electrician solves the problem:

Let $U$ be a voltage at the input of the ladder of the resistors. Let $I_k$ be current in the horizontal arm of k-th section, $I_0$ be current in the horizontal arm of 0-th section (initial section), $I_n$ be current in the horizontal arm of n-th section (last section) So i am working actually with $n+1$ sections. Successively applying to closed circuits Kirchhoff's voltage law, we find:




$$-R(I_{n-1}- I_n)+2RI_n=0$$ After simplification:



$$3I_n-I_{n-1}=0$$ Now, the effective resistance:

$$R_e=\frac{U}{I_0}$$ Next step is to solve the recurrence relation. Surprisingly, Wolfram Alpha does a good job: Link here. There is $f(k)$ instead of $I_k$

So $R_e=R\frac{(7-3\sqrt 5)(3-\sqrt 5)^n+(3+\sqrt 5)^{n+1}}{(1+\sqrt 5)(3+\sqrt 5)^n-2(\sqrt 5-2)(3-\sqrt 5)^n}$

This answer is consistent with that of Zassounotsukushi

  • $\begingroup$ This is equivalent to work with 1/R = conductance in place of resistance, isn't it? $\endgroup$ – Georg Jun 2 '11 at 10:06

take only first 2 resistors and rest as $x$ now the vertical resistor and your $x$ will be in parallel, effective resistance would be $Rx/R+x$ with series in horizontal resistor.

Now equivalent resistance would be

$$Req. = R+ (Rx/R+x). $$

  • take Req. as $x$ again

Form quadratic equation and solve for $x$. This will be the answer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.