It has been supplied, in jest I hope, that this query originated someday in the course of the final ice age, simply after a profitable mastodon hunt. A loop evaluation or a node evaluation would yield the identical reply as proven right here, however with rather more psychological effort. Generally, if we preserve an open eye, there may be a neater approach.
This query is definitely demonstrative of the advantages that symmetry can supply in some circuit evaluation circumstances. To wit, what’s the resistance between the other corners of a dice made up of twelve equal resistors (Determine 1)?
Determine 1 A resistor dice consisting of 12 equal resistors on every edge the place we’re prompted to find out the resistance at every nook. Supply: John Dunn
You possibly can write the node or the loop equations to unravel this drawback, however the course of will probably be tedious. You’ll need to forgive me, however utilizing SPICE can be dishonest, at the very least for my part. Attempting to see what would go in parallel with what may go away you babbling incoherently. Nonetheless, utilizing some easy observations and recognizing the place symmetry applies will lead easily to the proper consequence.
In Determine 1, we select node A and node H as reverse corners of the dice. We inject a present into these reverse corners and simply to make issues simple, we let that present be one ampere.
First, we take a look at present divisions.
As a result of we have now symmetry, we will state that three resistors main away from node A will every carry one-third of the injected present. Node B, node C and node D will every obtain one-third of an ampere from their respective resistors related again to node A.
Once more, attributable to symmetry, we will additional state that the 2 resistors main away from node B, node C and node D will every carry one-half of their respective nodes’ arriving currents. Node B will ship one-sixth of an ampere every to node F and node G, node C will ship one-sixth of an ampere every to node E and node G and node D will ship one-sixth of an ampere every to node E and node F.
Subsequent, we take a look at present summations.
Node E will obtain two currents of one-sixth of an ampere every from node C and node D, node F will obtain two currents of one-sixth of an ampere every from node B and node D whereas node G will obtain two currents of one-sixth of an ampere every from node B and node C.
Node H will obtain three currents of one-third of an ampere every from node E, node F and node G.
To make sure, this entire chain of reasoning is a linguistic mouthful, however a visible examination of Determine 1 will make all of it very simple to see.
If we let every resistor be one ohm, the voltage drops throughout every resistor that carries one-third of an ampere will probably be one-third of a volt whereas the voltage drops throughout every resistor that carries one-sixth of an ampere will probably be one-sixth of a volt.
If we add up the voltage drops from node H to node A, the result’s five-sixths of a volt it doesn’t matter what path we select. Due to this fact, the corner-to-corner resistance of the dice involves five-sixths of 1 ohm.
Quod erat demonstrandum.
John Dunn is an electronics marketing consultant, and a graduate of The Polytechnic Institute of Brooklyn (BSEE) and of New York College (MSEE).
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