Brain Teasers
The Cube Resistors
Science
Science brain teasers require understanding of the physical or biological world and the laws that govern it.Science
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?
Hint
There are two ways to solve this problem, the easy way and the hard way. Try drawing a circuit map of what's going on.Answer
3r/4Hide Hint Show Hint Hide Answer Show Answer
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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?
alright, your probably in a higher grade than sixth cuz this is way over my head.
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.
I totally agree with stardust
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.
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.
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
BTW the answer is 3r/2 and not 3r/4
Sorry I missed a risistor in parallel, the answer is 3r/4
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.
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
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
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
------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
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.
My Physics II instructor attempted to do a similar problem involving finding equivalent resistance between diametrically opposite vertices. He attempted to do it the hard way and wound up resorting to the solution manual for the sneaky way.
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