July 13th, 2015, 11:04 PM  #1 
Member Joined: Aug 2014 From: Lithuania Posts: 60 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: $\displaystyle \frac{2}{3}R_{n1}+\frac{\frac{1}{3}R_{n1}*R_n}{\frac{1}{3}R_{n1}+R_n}$ $\displaystyle \frac{2}{3}(6*2^{n1})+\frac{\frac{1}{3}(6*2^{n1})*(6*2^n)}{\frac{1}{3}(6*2^{n1})+(6*2^n)}$ Then it would seem that the total resistance of the circuit is equal to (assuming that R=1Ω): $\displaystyle \lim_{n\rightarrow +\infty} {(\frac{2}{3}(6*2^{n1})+\frac{\frac{1}{3}(6*2^{n1})*(6*2^n)}{\frac{1}{3}(6*2^{n1})+(6*2^n)})}$ 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: 914 Thanks: 331 
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: 60 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: 7,649 Thanks: 2630 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: 914 Thanks: 331 
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: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions  
July 15th, 2015, 05:00 AM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra  Quote:
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: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra 
Following the above method yields a resistance of$${R \over 2}\left(5 + \sqrt{41}\right)=5.70156\,R\;\text{to 5 d.p.}$$

July 24th, 2015, 12:57 AM  #9 
Senior Member Joined: Apr 2014 From: UK Posts: 914 Thanks: 331 
I couldn't follow the method at all 

Tags 
circuit, endless, resistance, total 
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