Wednesday, July 16, 2008

2nd Round


This is the second question of the 1st IPhO - try it for kicks! (Click on it to view the enlarged picture - same for all other pictures in this post!)

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I actually remember doing this in Physics ‘S’ paper! Haha. So let’s give it a try, using my well, noob method. Let’s consider a simple system at the start first:


Obviously the resistance here is just the simple sum of two resistors in series:



Let’s continue to add two more resistors; notice that I’ve renamed one of the resistors x (you’ll know why in a second!):



This is a little trickier; the resistance here must take into account the parallel circuitry, which I’ve circled out in red. If you can’t see this, well, let me redraw it in a more accessible manner:


And of course, if you still remember the formula for parallel resistors, this new set-up has a fairly easy to calculate expression for its resistance:



And then let’s do one more addition of a branch to see how everything adds up, to understand how I’m going to simplify the situation:


Well, without redrawing it, I’d expect you to say that the set-up now consists of two parallel set-ups. For those who can’t see yet, I’ve tried to point out using circles again. Look above at the blue circle; I can represent this as a resistor x, and then the diagram is simplified into:


And what do you notice? Haha, we’ve already obtained the resistance for this, which was determined earlier to be:



So what’s x? Well, let’s find out; referring to the blue circle, I see that x’s resistance is simply equivalent to a set of parallel resistors:



Okay, so that you’ve roughly got my trick of simplifying, let’s do a long set so that you’ll be seeing what I’m seeing:

Let’s do the red circle first, and let’s call this new resistor a:


Now, treating the red circle as a resistor a, we can simplify the diagram into:


Alright, so now let’s do the blue circle and let’s call this new resistor b:



So now we can simplify again:


And now let’s do the green circle and let’s call this resistor c:


And now, let’s simplify again:


This couldn’t be simpler, and let’s do the orange circle this time and call it a resistor d, and find its resistance:


With this, we can then simplify our diagram as:


And the overall resistance is therefore simply:


Now, if you haven’t seen or noticed the pattern yet, let me just write out everything in full:


What a monster! You may want to view it in full size, haha. Notice that there is a repeating unit! And if the series goes on and on indefinitely, let me just re-write it in a more digestible way:


Can you spot the repeating unit yet? Haha. If you haven’t spotted the recurring unit yet, it’s simply this monster right here:


But what is this creature? Notice that the overall resistance R is just the repeating unit:


So how do we solve this thing? Well, algebra gives us a good method:


And then now we manipulate this expression into a quadratic equation, as such, to obtain R:


Which can then be solved using the standard quadratic equation (notice I’ve rejected the negative answer since no resistance is negative):


And as some of you might have already noticed, the Golden Ratio (phi) appears in the answer:


Such that:

Well, what is the Golden Ratio? Let’s leave that for another entry, because I’m absolutely tired of typing out equations and expressions for the day. I’ll give a hint though; recall the Fibonacci sequence as:


If you look closely, each succeeding term is obtained via addition of its two preceding terms. Well, if you take any term and divide it by its preceding term, you’ll obtain a number close to the Golden Ratio (phi). Let’s say we take 89 and 55:


And this ratio gets increasingly closer to the Golden Ratio as one continues on into the series.

And wow! What a lengthy post! And it's rather strange isn't it? Then adding more and more resistors forces your resistance into a fixed, finite value instead of an infinite resistance, and the fact that the Golden Ratio pops up in such an instance! Simply amazing! :p

7 comments:

quirK said...
This comment has been removed by the author.
quirK said...

Looks like the trick to using Kirchoff is not adding to the right end, but to the left one instead...

quirK said...

Oh my, this is a brilliant question.

It's Fibonacci.

I didn't realise it.

The answer, ladies and gentlemen, is the Golden Ratio.

quirK said...

No wonder there was a root 5 inside, and no wonder it looked so familiar.

(1 + root 5)/2 (r)

Andrea Liang will love this question, if she can get past the physics. :p

quirK said...

wa I can't stand it sia... you can literally count the sequence off alternating resistors... this is simply wicked!

quirK said...

eh and the time zone of your blog... change it to UTC +8 !!!

yyknosekai said...

Yup, (1 + sqrt5)/2 is indeed phi, the golden ratio of 1.618..... which goes on and on and on. Lol.