A more interesting resistance problem.

Suppose there is an infinite network of one ohm resistors on a square grid.

It is relatively easy to determine the resistance across a side of a square (across one resistor in the network).

How do you determine the resistance between two nodes on a small square diagonal?

How do you determine the restance between a node and the second node over in a straight line?

How do you determine the resistance between two nodes arranged as in a chess knight move? That is one nod and another node that is two steps in one direction and one step at a right angle.

Bill

Reply to
Repeating Rifle
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The last one was reported (somewhere.... can't remember) as being one of the questions asked by Google in its search for software developers.

I thought I was making progress, but then realised that I didn't know how to calculate the resistance between two points on an infinite sheet of resistive material either.

It's probably obvious, but it's too close to being pure-math for my liking.

Sylvia.

Reply to
Sylvia Else

This is a theoretical question. Infinite number of parallel resistor paths? Resistance is zero isn't it.

Reply to
Greg

I wouldn't think so, any more than the sum of an infinite number of postive numbers is necessarily infinite.

Sylvia.

Reply to
Sylvia Else

One approach is to join together all the nodes at a very large distance away from one of the measurement node points, and attach a current source from that equipotential point to the measurement node. That makes the problem essentially trivial.

Reply to
bruce varley

in article 4193025d$0$25116$ snipped-for-privacy@news.optusnet.com.au, Sylvia Else at snipped-for-privacy@not.at.this.address wrote on 11/10/04 10:10 PM:

There is good reason. It is infinite. Problems of the type you described are usually solved by conformal transformation. The potential for an infinitesimal wire carrying a charge is logarithmic with radius. That means that your problem reall should measure the distance betweden two circle on a sheet.

It turns out that the conformal transformation needed is readily obtained. See Smythe for example. The pattern of circles is the same as that for a rectangular transmission line chart that is hardly used anymore. The semi-infinite chart has been modified by another conformal transformation into the Smith Chart that has entered wide use.

Bill

Reply to
Repeating Rifle

My first response was :

Try superposition: Apply a 1A current source between one node and the reference which is the "infinite" edge of the grid. Note the current distribution and voltage drop in each element between the nodes of concern. Now use a 1A current sourve out of the second node (to the infinite boundary) and and do the same. Superimpose- you will have 1A in at one node and out the other and 0 at the "infinite" boundary. You can then sum the voltage drops between. In fact you can also use symmetry and simply double the result from the first part.

However, while this works for the adjacent node problem, it may not be right for the other cases, giving a)1/2 ohm, b)3/4 ohm c) 3/4 ohm and d) 5/6 ohm and I question cases b)c) d) .

The problem that I see is that in the diagonal case, for example, 2 out of the 3 nodes adjacent to the "second node" will be at the same potential while the remaining is at an unknown potential so symmetry is lost and with it the assumed current distribution may also be lost.

Reply to
Don Kelly

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