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The first one, the effective resistance is $2R$, then $5R/3$ then $13R/8$ etc.... My job is to find the pattern/equation so I can find the total resistance when $20$ resistors are connected. Of course I can do the long way by using this formula I created $R_n = (R_{n-1} \times R)/(R_{n-1} +R) +R$. Can you guys help me find an equation where I can just input the value of $n$ and simply get an answer. Thanks!

The second question is: What would I get if I had an infinite number of resistance.

Help is very much appreciated :)

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The values of effective resistance reminds me of Fibonacci series en.wikipedia.org/wiki/… –  lab bhattacharjee May 19 '13 at 14:40
See question math.stackexchange.com/q/353261/752. –  Américo Tavares May 19 '13 at 15:03
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marked as duplicate by Américo Tavares, Micah, Joe, Amzoti, TMM May 19 '13 at 16:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers

up vote 2 down vote accepted

Let $R_n$ be the resistance of a circuit with $2n$ resistors, so $R_1 = 2R$. The circuit with $2(n+1)$ resistors is obtained from the circuit with $2n$ resistors by placing one resistor in parallel and another one in series. So

$$ R_{n+1} = R + \frac{1}{\frac{1}{R} + \frac{1}{R_n}} = R + \frac{R R_n}{R+R_n}. $$

Substituting $R_n = \frac{p_n}{q_n}R$ gives

$$ R_{n+1} = R + \frac{R R_n}{R+R_n} = \frac{2p_n + q_n}{p_n + q_n}R $$

and so we can take $p_{n+1} = 2p_n + q_n$ and $q_{n+1} = p_n+q_n$. Placing these numbers in a sequence like

$$ q_1, p_1, q_2, p_2, q_3, p_3, \dotsc $$

results in the Fibonacci sequence


for which explicit expressions can be found on the wp page (with a shifted index $n$). In the limit the quotient $\frac{p_n}{q_n}$ of two consecutive Fibonacci numbers tends to the golden ratio $\varphi = \frac{1+\sqrt{5}}{2}$.

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The second question is easier. Suppose that the resistance for infinite number of elementary parts is equal to $R_{\infty}$. Then it should not change if you add one more elementary part. This gives the equation $$R_{\infty}=\frac{RR_{\infty}}{R+R_{\infty}}+R,$$ with the solution $\displaystyle R_{\infty}=\frac{1+\sqrt{5}}{2}R$.

Now concerning the first question, you should first check that the initial recurrence relation $$R_{n+1}=\frac{RR_n}{R+R_n}+R,$$ being rewritten in terms of the variable $$u_n=\frac{1}{R_n-R_{\infty}}+\frac{R+R_{\infty}}{R_{\infty}(2R+R_{\infty})},\tag{1}$$ is equivalent to $$u_{n+1}=\frac{(R+R_{\infty})^2}{R^2}u_n.$$ (this is just a rephrasing of approach of Did in simpler terms). This has the obvious solution $$u_{N}=\left(\frac{R+R_{\infty}}{R}\right)^{2(N-1)}u_1.\tag{2}$$ Now from (1), (2) and $R_1=2R$ one obtains a general formula for $R_N$.

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It is a fact of life that homographic transforms like $$ u:x\mapsto R+\frac{Rx}{R+x}, $$ are conjugate to affine transforms. The first step to make use of this fact for the iterates of $u$ is to determine a fixed point of $u$, here $$ x^*=R\frac{1+\sqrt5}2, $$ and to rewrite everything in terms of $1/(x-x^*)$, here $$ \frac1{u(x)-x^*}=b+\frac{a}{x-x^*}, $$ for some $(a,b)$ you will compute, hence $$ \frac1{u(x)-x^*}-c=a\left(\frac1{x-x^*}-c\right), $$ for some $c$ you will compute. In particular, $$ \frac1{u^n(x)-x^*}-c=a^n\left(\frac1{x-x^*}-c\right). $$ This yields an explicit formula for $R_n=u^{n-1}(2R)=u^n(\infty)$, namely $$ R_n=x^*+\frac1{c(1-a^n)}. $$ Since $x^*$ is known, note that one can skip everything above and identify $a$ and $c$ through the values of $R_1=2R$ and $R_2=\frac53R$, using $$ c(1-a)=\frac1{R_1-x^*}=\frac1R\frac{3+\sqrt5}2,\qquad c(1-a^2)=\frac1{R_2-x^*}=\frac3R\frac{7+3\sqrt5}2. $$ Better still, use $a=1/u'(x^*)$ and the expression of $c(1-a)$ above, to get, for every $n\geqslant1$, $$ R_n=(s-1)R+\frac1s\frac{s^2-1}{s^{2n}-1}R,\qquad s=\frac{3+\sqrt5}2. $$

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