The Cube Resistors
Science brain teasers require understanding of the physical or biological world and the laws that govern it.
Consider a cube, each edge of which has a resistor of resistance r on it. What is the resistance between two points on the same side of the cube but on opposite corners?
HintThere are two ways to solve this problem, the easy way and the hard way. Try drawing a circuit map of what's going on.
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Comments
cathalmccabe
Mar 08, 2002
| I'm lost on this, I have a circuit map (I think) but I can't see how to begin reducing it down. Any help? |
starlust
Jun 05, 2002
| alright, your probably in a higher grade than sixth cuz this is way over my head. |
MrBean
Jun 11, 2002
| I got it, but I had to bust out the ol' EE book from college. I think I had something like that on a quiz once. I got it wrong then. |
CCat20   
Oct 06, 2002
| I totally agree with stardust |
jimbo   
Nov 14, 2003
| Good puzzle. I am not entirely sure about the answer though. I have an answer that is much smaller than this. In gerneral, the more paths you have from A to B, the less resistance. There are multiple paths to get from one corner of a cube to another so it would seem a much lower resistance. I am not an electronics expert however and there may be some interraction between these multiple paths that I am failing to understand. |
krishnan   
Nov 16, 2003
| I just solved this and got the answer stated but I couldn't explain it without circuit diagrams . Good teaser but requires a lot of knowledge to solve. |
NomadShadow   
Oct 25, 2005
| It was fun, you can perform a full circuit analysis, or you can use the famous delta-star transformations.
BTW the answer is 3r/2 and not 3r/4 |
NomadShadow   
Oct 26, 2005
| Sorry I missed a risistor in parallel, the answer is 3r/4 |
khedron   
Aug 06, 2007
| Some more explanation in the answer would be nice. Even when I don't know the answer, it's nice to feel like I have learned something. |
bfriedfischtom
Mar 28, 2008
| If you do a clever drawing it would look something like that:
r
r r r
r r r
rrrrrr rrrrr
r r r
r r r
r
A dimonond contained within another dimond, the edges connected.
Now if you imagine current and potential you may notice that
the horizontal resistors do not carry any current, therefore can be skipped.
Not it look like this:
r
r r r
r r r
r r r r
r r r
r r r
r
No star-transformation needed to solve this one.
The middle part evaluates to 3r, the left and right branch to 2r each.
2r//2r = r
3r//r = 3r/4
here we go |
bfriedfischtom
Mar 28, 2008
| well the drawing is a problem, 2nd try
------r
----r-r-r
--r---r---r
rrrrrr-rrrrrr
--r---r---r
----r-r-r
------r
------r
----r-r-r
--r---r---r
r---r---r---r
--r---r---r
----r-r-r
------r |
kunk  
Jun 27, 2010
| Nice. I knew the answer had to be less or equal to 3/4 of r. Calculating the resistance without the 2 useless wires is fairly easy. Then I thought that adding a connection can only decrease the total resistance or do nothing. If only had I known the 2 connections really were useless I would have gotten to the final answer. Oh how I wish I were smarter. |
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