13 Aug 06
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.
13 Aug 06
1 edit
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.
13 Aug 06
Originally posted by XanthosNZDon'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.
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.
13 Aug 06
Originally posted by sonhouseA tetraeder has 4 points, we name them A, the incoming point, and D, the outgoing point, and B and C the others.
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.
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.
13 Aug 06
Originally posted by FabianFnasYep, 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.
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.
13 Aug 06
Originally posted by sonhouseThe 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.
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.