Here's one about electric current flow:

I was thinking of some of the problems I got in electronics 101 and came up with this one, I suppose its been done before but here goes:
you have a triangular pyramid, triangular base, 6 lines describes it.
So each of those 6 lines outlining the pyramid happens to be a one ohm resistor. So measuring from any apex to another apex will show the same resistance but what is that resistance? Each line is a 1 ohm resistor.

0.5 ohm

You can solve this through repeated use of delta-wye conversions. You could also make a giant mess with mesh analysis.

If anyone can solve this and wants a challenge then find the resistance of the three cases in a similar cube.

The cases being:

1. Between two points that share a side.
2. Between two points that share a face but not a side.
3. Between two points that don't share a face or a side.

Now solve for any regular polyhedron.

Originally posted by XanthosNZ
If anyone can solve this and wants a challenge then find the resistance of the three cases in a similar cube.

The cases being:

1. Between two points that share a side.
2. Between two points that share a face but not a side.
3. Between two points that don't share a face or a side.

Now solve for any regular polyhedron.

Don't forget the cases where the polygon is a solid resistance or the case where its only conductive on the surface but the whole surface is conductive and you measure any two points.

Originally posted by sonhouse
you have a triangular pyramid, triangular base, 6 lines describes it.
So each of those 6 lines outlining the pyramid happens to be a one ohm resistor. So measuring from any apex to another apex will show the same resistance but what is that resistance? Each line is a 1 ohm resistor.

A tetraeder has 4 points, we name them A, the incoming point, and D, the outgoing point, and B and C the others.

Then from A to D the current can tacke this paths:
From A to D directly,
From A to B to D, and
From A to C to D.

What about B to C?
No, those have the same potential so no current is flowing that way, right?.
Just cut that wire and there will be no difference.

Connect resistors in serial is R = R1+R2
Connect resistors in parallel is 1/R = 1/R1+1/R2
Each resistor R is here 1 ohm so you get:
1/Rtot = 1/R + 1/R+R + 1/R+R = 1/1 + 1/2 + 1/2 = 2
Rtot = 1/2 = 0.5 ohms.

Mephisto2 is right and this is the explanation.

Originally posted by FabianFnas
A tetraeder has 4 points, we name them A, the incoming point, and D, the outgoing point, and B and C the others.

Then from A to D the current can tacke this paths:
From A to D directly,
From A to B to D, and
From A to C to D.

What about B to C?
No, those have the same potential so no current is flowing that way, right?.
Just cut that wire and t ...[text shortened]... = 1/1 + 1/2 + 1/2 = 2
Rtot = 1/2 = 0.5 ohms.

Mephisto2 is right and this is the explanation.

Yep, its basically pretty easy, just a one ohmer with 2 sets of one ohmers in series and those sets in parallel with the basic one ohm, so 2 ohms in parallel with another 2 ohms is one ohm and that one ohm is in parallel with a one ohm so the total is 0.5 ohms, you are right, the one resistor left over is in a junction with the same potential at both ends so no current flow so no effect on the whole matrix. You can draw it as a delta with an internal Wye connection. Lot easier if we had graphics here, eh. Hey, Meph, did you know the answer from a previous problem or did you actually work it out? Like Xanth said, I may be just duping the work.

Originally posted by sonhouse
Yep, its basically pretty easy, just a one ohmer with 2 sets of one ohmers in series and those sets in parallel with the basic one ohm, so 2 ohms in parallel with another 2 ohms is one ohm and that one ohm is in parallel with a one ohm so the total is 0.5 ohms, you are right, the one resistor left over is in a junction with the same potential at both ends s ...[text shortened]... revious problem or did you actually work it out? Like Xanth said, I may be just duping the work.

The symmetry makes it easy to see that no current is flowing through the B-C resistor, and from there, it's a simple three parallel scheme of 1 ohm and two times 2 ohms. And that whatever XNZ says.