Electronics > Beginners
Resistance across a 1k resistor cube.
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JacquesBBB:

--- Quote from: IanB on April 22, 2018, 08:17:46 am ---
--- Quote from: hamster_nz on April 22, 2018, 07:47:56 am ---I would be interested in hearing who solved this by which means, from LTspice, to changing to an equivalent structure that is obvious to solve..
--- End quote ---

My solution arose from observing the symmetry of the system. If all resistors have the same value then all parallel interior nodes of the network have the same voltage, which means they can be treated as if they were connected, which allows the network to be viewed as groups of parallel resistors. Following this observation the overall resistance is (1/3 + 1/6 + 1/3) · 1000 = 833⅓

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This is a great solution Ian and Brumby !

I  worked out nearly the same way, considering the symmetries of the system, and  thus three
different voltages,  V, V1, V2, 0 ,  with by symmetry also I/3 or I/6 current, and got  the solution,
but  although this was easy to solve, I  did not use your brilliant shortcut.

which is to say that all summit with same voltage can be considered as being connected, which
then simplifies the solution as one can use the already know rules.

This equipotential trick  is powerful !


 
Wimberleytech:
Here it is.
Wimberleytech:
This was my method to solve years ago...still works!

A couple of months ago I decided that I wanted to build one of these cubes as a piece of desk art.  I toyed with some 3D printer corner designs and such, ordered some 1/2 watt precision resistors...ADD as I am, the project is still incomplete along with the other hundred that are incomplete...lol...but I am always busy!!
Brumby:
A lot of us know that song only too well.
Wimberleytech:

--- Quote from: RoGeorge on April 22, 2018, 09:28:58 am ---OK, then, the cube was easy, try this one: What is the resistance between 2 nodes A and B, in an infinite 2 dimensional square grid of resistors, each resistor having the same value, R?  >:D

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How about this?
Using superpostion, go out to infinity and ground the last (LOL) nodes.
Then insert a current of 1 amp somewhere in the center of infinity (LOL²).
That current will split into fourths (1/4, 1/4, 1/4, 1/4).
Now remove that current source and connected a different current source of 1 amp LEAVING an adjacent node.
The current leaving will split into fourths.
Now, for a linear system, we can add these result yielding 1/2 amp flowing in the resistor between the adjacent points.
1/2 amp flowing through R gives V = R/2
That voltage (R/2) resulted from a 1 amp source, so the equivalent resistance is V/I = R/2/1 = R/2

Edit:
After pondering further, I do not have to ground the array at infinity...just insert the first test current and then the second.
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