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        riddles >> easy >> Sierpinski circuit
        
(Message started by: JocK on Jan 30^th, 2005, 10:55am)

Title: Sierpinski circuit
Post by JocK on Jan 30^th, 2005, 10:55am

Attached network represents an electric circuit. This circuit contains an infinite number of 1 Ohm resistors, each represented by a simple link. What is the resistance between two outer corner contacts?

Title: Re: Sierpinski circuit
Post by SMQ on Feb 16^th, 2005, 11:30am

Your diagram seems to have been eaten by the switch to the new forum software, and wasn't entirely clear to me (was it meant to be a standard Sierpinski gasket composed of 1 Ohm resistors or were there some extra links in there? If there were extra links were they resistors too, or wires?)

In any case, after a couple hours of crunching and double-and-tripple-checking nasty algebra on paper, I don't believe there exists a finite positive solution to any interpretation of the diagram that I considered. The standard Sierpinski has a trivial solution of 0; the other configurations all led to negative solutions.

If anyone wants me to go into detail on my methods (the basic method was always the same: replace each of the three suboccurrances of the gasket in the corners with a triangle of resistors of value Re, apply voltages of +1 and -1 volts at the bottom two corners, solve the resulting diagram using KCL to find the current between the two bottom corners, giving the resulting effective resistance Rs across the side in terms of Re, observe that the suboccurrances of the gasket are symmetrical and, by definition, cungruent to the whole gasket and so Rs = 2Re/3, solve for the value of Re) I can post the complete analyses later this week...

-- SMQ

Title: Re: Sierpinski circuit
Post by JocK on Feb 16^th, 2005, 12:59pm

on 02/16/05 at 11:30:01, SMQ wrote:

Hope William will be able to get the picture back. (I can't add any attachments anymore.)

The circuit is not a 'standard' Sierpinski gasket: extra links indeed have been added. These extra links are essential in rendering the total resistance finite. The resistance of all links (including the 'extra' ones) is 1 ohm each. (So the length of the link is not of any relevance to the resistance of the link.)

Title: Re: Sierpinski circuit
Post by SMQ on Feb 17^th, 2005, 11:45am

Ok, JocK was kind enough to email me the orginal diagram, and since I run my own web server I can host my own images ;D, so here it is:

http://www.dwarfrune.com/~smq/sierpinski/fig0.gif

And here's my analysis which I think shows :: there is no positive finite solution to the circuit as presented. ::
Due to the figures I don't see a good way to hide the rest of this, so
Don't Read any further unless you want to see how I approached this!

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http://www.dwarfrune.com/~smq/sierpinski/fig1.gif
My basic premise, which I present without full proof, is that the entire circuit can be functionally replaced with a simple triangle of equal-valued resistors (fig. 1).

My argument: 1) The circuit is only ever interfaced at the corners, both in the final problem statement (find the resistance between two outer corner contacts) and in its recursive inclusion within itself. 2) Under these conditions, any finite network constructed exclusively of resistors can proveably be replaced by a simple network of resistors joining the interface points. 3) It is permissible to (initially) assume for analysis that the given infinite network also has a finite representation, that is the limit of its resistance as the depth of recursion approaches infinity converges; failure of the analysis would then invalidate this assumption. 4) Since the given circuit is symmetrical with respect to its corners, the values of the equivalent resistances, R_E, must all be equal, and the resistance R_S between any two corners is then ²/₃R_E.

http://www.dwarfrune.com/~smq/sierpinski/fig2.gif http://www.dwarfrune.com/~smq/sierpinski/fig3.gif
At the "top" level the circuit consists of three copies of itself joined at the corners by a network of 1 Ohm resistors (fig. 2). By replacing each of the copies with the equivalent network (fig. 3), it is possible to solve for the resistance across one side of the network in terms of R_E. Note that this solution is also the resistance R_S of fig. 1. Note also that since the circuit in fig. 3 is symmetrical left-to-right, all points along the dotted line are equipotential and may be joined and only half the circuit needs to be solved.

http://www.dwarfrune.com/~smq/sierpinski/fig4.gif http://www.dwarfrune.com/~smq/sierpinski/fig5.gif
The two resistors spanning the equipotential line can be split into two resistors each (of half their original value) terminating on the equipotential line (fig. 4a), and then resistors terminating on the equipotential line combined (fig. 4 b,c) to eventually yield a circuit of only four resistors (fig. 5). This circuit is readily solved for R_3,0, the resistance between points V₃ and V₀, using KCL as follows:

Let the voltage V₃ be 1 and V₀ be 0, with V₁ and V₂ unknown. By KCL the current into node V₁ is equal to the current out:

1) (1 - V₁)/R_E = (V₁ - V₂)/R_E + (V₁ - 0)/[(4R_E + 15)/(3R_E + 12)]

and likewise at node V₂

2) (1 - V₂)/R_E + (V₁ - V₂)/R_E = V₂/1

Solving for V₁ and V₂ in terms of R_E gives

3) V₁ = (4R_E + 15)/(3R_E² + 17R_E + 15)

4) V₂ = (3R_E + 15)/(3R_E² + 17R_E + 15)

The total current through the circuit I = I₁ + I₂ (fig. 5) = (1 - V₁)/R_E + (1 - V₂)/R_E which gives

6) I = (6R_E + 27)/(3R_E² + 17R_E + 15)

So R_3,0 = (V₃ - V₀)/I = 1/I which is

7) R_3,0 = (3R_E² + 17R_E + 15)/(6R_E + 27)

This result from KCL is half the resistance of a side of fig. 3, which from fig. 1 gives us R_S = ²/₃R_E = 2R_3,0, finally giving

8 ) R_E² + 8R_E + 15 = 0

9) R_E = {-3, -5}

10) R_S = {-2, -¹⁰/₃}

Which, I believe, shows, together with point 3 of my argument above, that there are no positive finite solutions to the given curcuit.

So, where am I wrong? :D I'll have to try to solve this with infinite series (a bit more rigorous than my approach above) and see if I still get the same result...

-- SMQ

Title: Re: Sierpinski circuit
Post by JocK on Feb 17^th, 2005, 2:36pm

on 02/17/05 at 11:45:20, SMQ wrote:

... there are no positive finite solutions to the given curcuit.

So, where am I wrong? :D I'll have to try to solve this with infinite series (a bit more rigorous than my approach above) and see if I still get the same result...

Ooops! :o You are right. I made an error when constructing this problem... :-[

It is actually easy to sea that Sierpinsky-type fractal circuits consisting of resistive links always lead to infinite resistances, unless 'outer resistors' are added. Maybe I'll post a correct fractal circuit (one with finite resistance) later... In case anyone still is interested in trying these... :-/

Sorry for leading you up the wrong path SMQ. You posted a very precise and concise solution! (Leading to the correct answer R = [Infty].)

Title: Re: Sierpinski circuit
Post by Grimbal on Feb 17^th, 2005, 2:42pm

I was about to say that. To get anywhere from the upper corner, you have to cross an infinity of 1 ohm resistors. So, the total resistance is ~~futile~~ infinite.

http://www.florian.net/pic/resistors.gif

Title: Re: Sierpinski circuit
Post by SMQ on Feb 18^th, 2005, 8:19am

on 02/17/05 at 14:36:04, JocK wrote:

Ooops! :o You are right. [...] Sorry for leading you up the wrong path SMQ. [...]

No need to apologize; around here it's the journey not the destination, right? ;D Besides, it gave me something to do with my EE degree and math minor--which are mostly wasted on my current job as an applications programmer...gotta keep the ol' noggin tuned up. ;)

--SMQ

Title: Re: Sierpinski circuit
Post by JocK on Feb 18^th, 2005, 5:05pm

Actually...

... the problem is well-defined if you assume that the fractal pattern extends indefinitely. So: what is the resistance between the two points indicated assuming the fractal pattern keeps 'growing'...? 8)