How Does Symmetry Solve a Cubic Resistor Network?

[vink]

Dear all,

I'm trying to understand the second figure of the following webpage. The webpage simply solve it by 'symmetry', but I could not figure out its reasoning. Could someone explain how does it work? Thanks in advance.

http://www.schoolphysics.co.uk/age1...ml?PHPSESSID=5b0029c25a5894099c6df916f68d95ac

(b) By symmetry

i1 = 5i2
i3 = 14 i2
i = 24 i2

Therefore:

Total resistance (R) = 7/12 r

 

[haruspex]

The link marked 2i2 can be thought of as shared between two equivalent paths, the 'front' and 'back' paths. So it carries i2 on behalf of each, and since its conductivity is split between them it appears to each to have resistance 2R. A current i1 entering from (say) top left on the front path has two routes to the resistor directly below it. It can go down a single (unlabelled) resistance R or along the sequence R, 2R, R. The former has one quarter the resistance so carries four times the current, 4i2. Now you can add up the currents through different cutsets to find 2i1 = 10i2.

 

[ehild]

I do not understand it either. But you can redraw the circuit by using symmetry. The symmetric points are at the same potential so they can be connected with a wire and considering them a single node. In the second problem, (the third cube in the attachment) the opposite nodes on the upper face of the cube are equivalent, (shown in red) and so are the opposite points on the bottom face (green). Points 4,5 make one node, so resistors (1,4) and (1,5) are parallel. Resistors (2,3) and 2,6) are also parallel. You can redraw the network between A and B and find parallel and series connected resistors, so it is easy the find the resultant resistance. See also https://www.physicsforums.com/showthread.php?t=557461

ehild

 

[vink]

ahhh, thanks for the help, I get it now.

By the way, is anyone aware of any discussion about larger cubic resistor network, such as 3x3, 4x4, nxn etc?

Thanks in advance!