In the case of the resistance between adjacent nodes it is actually
easier than the finite grid case.
Consider 1A injected into one node and drawn out at infinity. At the
injection node 1/4A flows in each of the 4 branches.
Now remove this source and provide on drawing out 1A at the adjacent
node (and current going in at infinity).
Again we have current flowing in at 1/4 ohm in each branch. Now
superimpose to get no current in or out at infinity but 1A going in at
the first node and out at the second node. In the adjoining link the
current will be 1/4 +1/4 =1/2A or half the current injected. So between
the terminals there is 1/2 volt and 1A corresponding to an effective
total resistance of 1/2 ohm.
For a purely resistive grid of any dimension transients can be neglected.
Don
On 08/01/2013 12:45 PM, Raul Miller wrote:
On Mon, Jan 7, 2013 at 9:54 PM, Don & Cathy Kelly <[email protected]> wrote:
The problem that you mention
http://xkcd.com/356/
is not trivial and even the case with the resistance between adjacent points
in an infinite grid is not trivial by use of expansion of series parallel
or star delta conversions.
Yes... I find the infinite case difficult to reason about: With a
finite grid, we can reasonably ignore transient effects. But with an
infinite grid, the transient effects become infinite so ignoring them
seems unreasonable.
Which is why I was looking for treatments of finite cases. Anyways,
I've a lot to think about now (thanks Aai!).
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