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In Physic, I have to find a way to prove, with equations, that the sum of an infinite network of resistors of 1 $\Omega$ has a limit value. This question is based on the resistor's rules:

In a parallel circuit: $R_E = (R_1^{-1}+R_2^{-1}...)^{-1}$

In a series circuit: $R_E = R_1+R_2...$

Here is my scheme: Click to see the image!

I begin my mathematical reasoning by using Excel to calculate the most precise value. I discovered that Excel does not compute more that 15 digits... but I still got an overview of the possible answer: 2.73205080756888. I also understood the principle of adding the previous number calculated and just add the new loop to the answer, but my teacher told me that using series was difficult...

So, I decided to search online to find a solution and I found that: $$R_{eq} = R \cdot (1+\sqrt{3})$$, but it don't understand how to get up to there. They added a short reasoning: Click to see.

Does anyone could add more steps in the reasoning or give me a hint to use series in this particular case?

Edits


In my excel file, I found that my function to compile the value was : $$(\text{ANS}^{-1}+1)^{-1}+2$$ Where ANS is the result of the calculation of the third or more loop

based on the third loop. The result of my first loop is 3 (sum of the three resistors in serie), the sum of my second loop is calculated based on: $$({3}^{-1}+1)^{-1}+2$$, then you realize when calculating that you are adding the previous numbers, so I have used the previous cell number (ANS)

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3 Answers 3

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Notice that in your ladder

Ladder

the leftmost three resistors can be removed and it is the same ladder again. Therefore, we have

$$ R_{eq} = R + R + R \parallel R_{eq} $$ where $R \parallel T$ means $(R^{-1} + T^{-1})^{-1}$.

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  • $\begingroup$ And I should I be supposed to know the meaning of $R || T$? $\endgroup$
    – Matt
    Oct 29, 2016 at 1:21
  • $\begingroup$ @Matt It simply means the resistance of parallel resistors. $\endgroup$ Oct 29, 2016 at 1:22
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$$R_E=R+R||R_E+R=2R+\frac{RR_E}{R+R_E}$$ Solve it for $R_E$

enter image description here

The reason is when you look through the ladder network (from the arrow in the picture), what you see in the second stage (just behind the $R$ opposite the arrow) is again $R_E$ (since the ladder is infinite) which is parallel to an $R$ and there are two $R$ in series (the two sides of the arrow).

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  • $\begingroup$ And from this point: $2R+\frac{RR_E}{R+R_E}$, how can I obtain: $R_{eq} = R \cdot (1+\sqrt{3})$? How to simplify? $\endgroup$
    – Matt
    Oct 29, 2016 at 3:02
  • $\begingroup$ @Matt Multiply by $R+R_E$ both side of the equation. Group the terms in terms of powers of $R_E$ and you obtain a quadratic equation> Only positive solution is OK $\endgroup$
    – Andrei
    Oct 29, 2016 at 3:07
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    $\begingroup$ $R_E=2R+\frac{RR_E}{R+R_E}=\frac{RR_E+2R(R+R_E)}{R+R_E}$. So $R_E(R+R_E)=RR_E+2R(R+R_E)$, or $$R_E^2-2RR_E-2R^2=0$$ $\endgroup$
    – msm
    Oct 29, 2016 at 3:09
  • $\begingroup$ And concerning the $RR_{E}$, what does it represent? I do understand for the $R +R_E$ which represents the addition of resistors in parallel from the end of the ladder and the opposite of the arrow but I don't understand the multiplication! $\endgroup$
    – Matt
    Oct 29, 2016 at 3:22
  • $\begingroup$ It represents nothing but the numerator of $(R^{-1}+R_E^{-1})^{-1}$. $\endgroup$
    – msm
    Oct 29, 2016 at 3:24
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  1. Three series resistors: $$\text{R}_1=\text{R}+\text{R}+\text{R}=3\text{R}$$
  2. Three series resistors, parallel to the second, first resistor: $$\text{R}_2=\frac{\left(\text{R}+\text{R}+\text{R}\right)\cdot\text{R}}{\text{R}+\text{R}+\text{R}+\text{R}}+\text{R}+\text{R}=\frac{\text{R}_1\cdot\text{R}}{\text{R}_1+\text{R}}+2\text{R}=\frac{11}{4}\cdot\text{R}$$
  3. Three series resistors, parallel to the second, second resistor: $$\text{R}_3=\frac{\left\{\frac{\left(\text{R}+\text{R}+\text{R}\right)\cdot\text{R}}{\text{R}+\text{R}+\text{R}+\text{R}}+\text{R}+\text{R}\right\}\cdot\text{R}}{\frac{\left(\text{R}+\text{R}+\text{R}\right)\cdot\text{R}}{\text{R}+\text{R}+\text{R}+\text{R}}+\text{R}+\text{R}+\text{R}}+\text{R}+\text{R}=\frac{\text{R}_2\cdot\text{R}}{\text{R}_2+\text{R}}+2\text{R}=\frac{51}{15}\cdot\text{R}$$

And so on. So the next one will be:

$$\text{R}_4=\frac{\text{R}_3\cdot\text{R}}{\text{R}_3+\text{R}}+2\text{R}=\frac{153}{56}\cdot\text{R}$$

And the next one:

$$\text{R}_5=\frac{\text{R}_4\cdot\text{R}}{\text{R}_4+\text{R}}+2\text{R}=\frac{571}{209}\cdot\text{R}$$

So, in general:

$$\color{red}{\text{R}_{1+\text{n}}=\frac{\text{R}_\text{n}\cdot\text{R}}{\text{R}_\text{n}+\text{R}}+2\text{R}}$$


In other words:

$$\text{R}_{\text{eq}}=\frac{\text{R}_{\text{eq}}\cdot\text{R}}{\text{R}_{\text{eq}}+\text{R}}+\text{R}+\text{R}\Longleftrightarrow\text{R}_{\text{eq}}=\text{R}\cdot\left(1\pm\sqrt{3}\right)$$

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  • $\begingroup$ I think that the third simplification is wrong, it doesn't correspond to the trending. And how can you calculate those simplifications? I thing I got it for the $1 \pm \sqrt 3$ it is the exact way to express the fraction... $\endgroup$
    – Matt
    Nov 1, 2016 at 0:19

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