As always, click the above comic strip to enlarge it; you'll notice that there's a really peculiar question which involves an infinite 2 dimensional array of resistors. Haha. Anyone game for this one? :p It's solvable!

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Well, I was thinking of Kirchoff's Laws as a possible way of solving this, so let me just direct you through my methodology first.

Oh yes, a simple application of Kirchoff's Laws won't work in this question - I'll explain why later.

You start off with a single point in this infinite array and reconstruct the infinite array from scratch - let us assume we have this point (0, 0) that is the origin. We pump in 1A of current into this origin, and thus from this origin, we have four current vectors as such, coming out as shown below.

It should make sense to you that the current is divided equally into four, because all four paths are of equal resistance. We're taking one black line to be 1/4 A of current. Now, we then extend the network as such:

Notice that now each blue line is 1/12 A of current, because at each junction there are 3 paths of equal resistance. We can go on and extend the network and you'll see this:

Each red line is now worth 1/12 A of current as well! Surprise, surprise! This is because you've two '1/12 A' currents going into one junction, and then leaving into two junctions of equal resistance. It should then make sense that the two paths of red carry 1/12 A each.

So let's view our small portion in the infinite array of resistors:

Notice that in order to get from point O to point A, you go from one black to one blue to one red, and you experience a potential drop of:

V = (1/4)(1.0) + (1/12)(1.0) + (1/12)(1.0) = 0.417 V

And this method will only get us the potential drop between these two points, and nothing else.

Why? You might be thinking that as in the cube resistor problem in the previous post, we can simply divide the potential drop by the current to obtain the resistance between these two points. But this problem doesn't allow for that, because the initial current was 1 A, and the final current we dealt with at point A isn't 1 A but 1/12 A. This means we can no longer divide the potential drop by the initial current to obtain the equivalent resistance.

The previous problem has the starting current and end current being the same value, since by Kirchoff's First Law we know that all current going into a junction must equal all current going out of the junction if there is to be no buildup of charge.

The actual mathematics required for this question is beyond me, and it involves Fourier analysis. If you're still interested however, you can have a look at the solution at www.geocities.com/frooha/grid/node2.html.

I guess this post is another fiasco! :(

I guess this post is another fiasco! :(

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