S.O.S. Mathematics CyberBoard

Alright, So i don't know how to post the question but basically it is a cube with 12 equal resistors on each side. A current of 91.9 A is introduced at the point on the lower left hand front face. If anyone can visualize and help i would be most greatful. thanks

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“It is impossible for any number which is a power greater than the second to be written as a sum of two like powers. I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain.â€

er... i am not too sure what you want to do with the current but i'm guessing you want to resolve the cube of resistors into a single resistor. But you didn't say where you were measuring the resistance from too...I'll assume that you are taking the resistance across one vertex to the opposite vertex.

To resolve this resistor cube consider a vertex. Branching out of this vertex are 3 equal resistors, they are in parallel. Each of this 3 resistors branch out to 2 equal resistor(6 in total) and finally they converge back to 3 resistors in parallel.(Kind of hard to explain...easier to show with a cube)

But essentially your cube of resistors is then equivalent to

Starting Opposite
vertex vertex



er... i can't seem to get the thing to stay in place
Its like

.....R
.R..R.R
-R-R-R
.R..R.R
.....R
.....R

The dots are place holders the - are bits of wire. if the R is on top of anotehr R its meant to be in parallel
So that is equivalent to 5R/6.

Hope that helps.