Zer0's Conundrum #3 has been solved. No additional prizes will be given out. Stay tuned for future conundrums... Welcome to... Zer0's Conundrum #3 So, in Zer0's conundrum, Zer0 will post a problem (usually math or computer science related) and you will have to solve it! First user who solves the problem will win a prize! If no one gets it within a week, there will be no winner for that week. Post your answer here! Rules One entry per person per contest You MUST provide justification (proofs, etc) along with your answer otherwise it WILL NOT count! No collaboration (I don't understand why you would want to...) No looking up the answer or asking someone for the answer No computer programs unless otherwise specified This Week's Prize... 15 forum cash! - courtesy of some mod! This Week's Conundrum... What is the equivalent resistance across the two terminals in the following circuit? Express your answer in terms of R.
Re: Zer0's Conundrum #3 It seems like Pando is going to be the only one solving these things. I just looked at it, then started crying uncontrollably... good luck pando. :|
Re: Zer0's Conundrum #3 Yep. I understand the question completely, just I can't seem to get a pattern that works or anything, the way I'd normally do something mathematical like that.
Re: Zer0's Conundrum #3 Yes Sorry The infinite grid of resistors is much more difficult to solve than this one (part of the reason I didn't use it). The technique to solving that one isn't necessary here. Plus the answer to the infinite grid case is much more well known so its easier to look up
Re: Zer0's Conundrum #3 I'm pretty sure the answer is phi, which is (1+sqrt(5))/2 or approx. 1.6180339887. work: ehh well I just started calculating the resistance with 2 resistors, then 4, then 6, and I noticed that it was a fibonacci sequence...well not exactly...it went more like this: with 2 resistors the resistance was f(4)/f(3), with 4 it was f(6)/f(5), with 6 it was f(8)/f(7), etc so I just took the limit of f(n+1)/f(n) as n->infinity equation of Fibonacci sequence: {(phi^n)-(-1/phi)^n}/sqrt(5) so at infinity it'd be phi^(n+1)/phi^n, or just phi yeah...I can put the actual equation of the limit I took when I get home (I just typed all this out on my phone =P)
Re: Zer0's Conundrum #3 That seems to be just a clever observation. You'll have to prove it mathematically. And remember, give the answer in terms of R, so don't just let R=1.
Re: Zer0's Conundrum #3 Z=R+RZ/R+3 Z^2+RZ=2RZ+R^2 Z^2-RZ-R^2=0 Z=(1+squareroot(5))/2=1.618 R <----- answer i can do it another way too
Re: Zer0's Conundrum #3 ....well then...1.618R but yeah...no freakin' way I'm proving that on a cell phone it's common sense...like adding 1+2 when you add the resistors together in series and parallel mathematically you are creating a Fibonacci sequence... so unless you want me to proved the equation to the fib seq (which I must say I can't)...there's not much to prove =P
Re: Zer0's Conundrum #3 lol, that's not common sense to me. Make a picture and draw some arrows and symbols and stuff when you get on a computer
Re: Zer0's Conundrum #3 The answer's right but algebraically I'm not sure how you got from step one to step two.
Re: Zer0's Conundrum #3 Correct You had a typo on step 1 though, it should be "Z=R+RZ/(R+Z)" Ima say patty wins this one, sorry pando Some mod give him 15 cash pl0x