> -----Original Message----- > Date: Tue, 08 Jan 2013 18:29:12 -0800 > From: Don & Cathy Kelly <[email protected]> > To: Raul Miller <[email protected]> > Cc: [email protected] > Subject: Re: [Jprogramming] xkcd 356 > Message-ID: <[email protected]> > Content-Type: text/plain; charset=UTF-8; format=flowed > > 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
Beautiful! I think that such an amazingly simple argument only exists for two nodes this close together. The resistance between adjacent nodes is also 1/2 ohm. Regards Graham ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
