My Math Forum Total resistance of an endless circuit

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 July 13th, 2015, 11:04 PM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 62 Thanks: 3 Total resistance of an endless circuit I need to find the total resistance of this endless circuit when R=100Ω. I want to solve this problem mathematically. Here's my try. This circuit is made out of loops: First loop: R, 2R, 3R; Second loop: 2R, 4R, 6R; Third loop: 4R, 8R, 12R; ... The first loop has a total resistance of 6R, the second- 12R, the third- 24R. So I see a geometrical progression here, which can be written as $\displaystyle R_{loop}=6*2^n$ where n is the number of loops. I can express the total resistance of two last loops like this: $\displaystyle \frac{2}{3}R_{n-1}+\frac{\frac{1}{3}R_{n-1}*R_n}{\frac{1}{3}R_{n-1}+R_n}$ $\displaystyle \frac{2}{3}(6*2^{n-1})+\frac{\frac{1}{3}(6*2^{n-1})*(6*2^n)}{\frac{1}{3}(6*2^{n-1})+(6*2^n)}$ Then it would seem that the total resistance of the circuit is equal to (assuming that R=1Ω): $\displaystyle \lim_{n\rightarrow +\infty} {(\frac{2}{3}(6*2^{n-1})+\frac{\frac{1}{3}(6*2^{n-1})*(6*2^n)}{\frac{1}{3}(6*2^{n-1})+(6*2^n)})}$ I don't know if this is correct. Please show me how to solve this problem and where my mistakes are.
 July 13th, 2015, 11:23 PM #2 Senior Member   Joined: Apr 2014 From: UK Posts: 954 Thanks: 342 The resistance of the first loop is 4R + whatever the rest is (the 2R in parallel with the other loops) using // to denote parallel: the total resistance is 4R + (2R//(8R + (4R//(16R + (8R//(16R + etc... ))))))
 July 14th, 2015, 07:32 AM #3 Member   Joined: Aug 2014 From: Lithuania Posts: 62 Thanks: 3 I see, but how should I make an equation out of it?
 July 14th, 2015, 10:15 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra We have two terminals on the left, yes? So the right hand side is closed. Let r be the resistance of the whole circuit. We have a resistance of R+3R that is in series. Then we have the left branch with resistance 2R. The rest has resistance 2r. These last two parte are in parallel.
 July 15th, 2015, 01:37 AM #5 Senior Member   Joined: Apr 2014 From: UK Posts: 954 Thanks: 342 I have no idea how to build a formulae for this, but, you can find an adequate answer by limiting the number of loops. This is how I solve these types of problem in the real world. I get 5.701562119R with 9 loops, more loops just increases the accuracy. Clearly, 9 loops is way over the top since even the best resistors available won't be that accurate. Excel makes it very easy to do....
 July 15th, 2015, 03:20 AM #6 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions Thanks from weirddave
July 15th, 2015, 05:00 AM   #7
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Quote:
 Originally Posted by v8archie We have two terminals on the left, yes? So the right hand side is closed. Let r be the resistance of the whole circuit. We have a resistance of R+3R that is in series. Then we have the left branch with resistance 2R. The rest has resistance 2r. These last two parts are in parallel.
It seems that the end of my post got lost.

The parallel part of the circuit has resistance $r_p$ given by$${1 \over r_p} = {1 \over 2R} + {1 \over 2r}$$

Then the total resistance is $$r = 4R + r_p$$

 July 17th, 2015, 06:44 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra Following the above method yields a resistance of$${R \over 2}\left(5 + \sqrt{41}\right)=5.70156\,R\;\text{to 5 d.p.}$$ Thanks from kaspis245
 July 24th, 2015, 12:57 AM #9 Senior Member   Joined: Apr 2014 From: UK Posts: 954 Thanks: 342 I couldn't follow the method at all

 Tags circuit, endless, resistance, total

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