This is something my electronics instructor threw out over 30 years ago. There is a cheating method, and the real solution. I figured it out using assumptions in my math, but never really worked it out. Thought some of you guys might enjoy it. He claimed the one correct solution used 8 sheets of paper and network analysis. Some of my fellow students did it with real resistors, which really was cheating. It was a free grade for the course, guess I was close enough because he gave it to me. The Problem: All resistors are 10 ohms. What is the resistance between A and B? BTW, I don't have the answer to this, though I remember my answer. Is that evil or what?
You don't need 8 sheets of paper, just a sharp mind. I can show it in 1/2 page max. And it isn't trickery, just logic.
Looking forward to seeing how you do it. Wouldn't be a puzzle if there wasn't a catch though. When you come up with a solution please PM it to me, and well see if it attracts anyone elses ire.
Is the answer 25/3 Ohms? I might have made mistake in calculation, but check if my approach was correct. I used star-delta transformations. That answer holds true if C,D,E are equi-potential points. At a glance they did seem to me.
Nice method recca. I just did it by brute-force! Actually if you pick the symmetry you can get a relatively simple resistor topology and the answer pops out the same as yours. Dave
I believe I know the shortcut to which Bill is referring. I remember this problem from a the electrical section in a Physics course. hgmjr
Interesting, I'd never seen that approach. If you want to do a check there are actually 2 different sets of points with equal potentials, which is how I approached it. When I worked it the first thing I did was flatten out the schematic, which didn't help nearly so much as I'd hoped it would. I'd also forgotten about computers and SPICE simulators, silly me. We didn't have anything like that 30 years ago.
That is what I did, and I ended up with a series arrangement of 3 parallel sets of resistors. It is only when you look at the cube symmetrically that it becomes clear how to do this. Dave
I wonder is a tessaract cube would be harder? It is possible to diagram a 3d model of a 4D cube. It's distorted, but much in the same way the 2D representation of 3D is. What do you think, even worth drawing up? The other set of equal potential points is exactly symmetrical to the first set. Symmetry is the hook on this problem. I have to wonder if any other instructors will use this for their classes. With SPICE probably not, but it is a lot more fun when you have to do it by hand.
Just had an offline conversation with Caveman, if you really want to make it hard start with 10 ohms and increase each resistor by 1 ohm, so that every resistor is different. Of course, it would leave me out on solving it, I needed the symmetry dagnabit!
You could also use Delta to Y and Y to Delta conversion formulae listed here http://www.allaboutcircuits.com/vol_1/chpt_10/13.html
Actually, if you go through the problem, it is a lot more complex than that. If all the resistors are different it's not even close. Delta to Y and visa versa have to do with conversions, making two different circuits that look identical to each other as seen from the outside world. From a black box point of view (where you can not see how they are wired) there is no telling them apart. This is important for later concepts, namely attenuators. The cube circuit is a network, and if the parts are all the same, it repeats 3 times. Where it gets interesting is they are all intertwined, and hard to separate from each other. -------- Looking at it again you may have a point. I'll have to play with it and see. Sorry about that.
The big problem is that the circuit looks a lot more scary than it actually is. These kind of problems are used to teach students how to visualize different circuit topologies. The trick is to give designators to the resistors. That takes some of the confusion away from the circuit. Then, if you look at it like a mesh, (ie. remove the 3d effect) it becomes easier to understand. Take a look here. They have a Java simulator that demonstrates the circuit you have posted, complete with a voltage source. http://server.oersted.dtu.dk/personal/ldn/javalab/Circuit09.html Best, :Flubbo.
The dangers of sleep posting, what can I say? Recca02 actually posted this way back, post #3 I think, but I missed the reference.
Symmetry of the cube, I'll forward you the PM if I can. He'll post it if he wants. I may get ambitious and try doing the non-repeating resistor scheme while using the Delta conversions. I honestly don't know if that would work or not. After this dead horse stops wiggling I'll probably do a hypercube just for fun. Wonder how many students we'll make miserable in the long run with this?
If you take a 6-sided die, hold it between your fingers at points A and B, and allow it to rotate on that pivot, you will notice that ignoring the dots, there are three positions that look identical in position. In these three positions, the three corners that are only one edge away from A will be switching places, and the three corners that are only one edge away from B will be switching places. Since all of the edges have the same resistance, rotating the cube in this manner generates the exact same schematic. This means that if you apply any generic voltage V between A and B, the first 3 points will have the same voltage as one another, and the second 3 points will have the same voltage as one another. Since this is true, you can connect these equivalent points without changing the characteristics of the circuit. (Similar to the logic of a balanced bridge circuit.) And you result in 3 Resistors in parallel + 6 resistors in parallel + 3 Resistors in parallel. Result = (1/3 + 1/6 + 1/3)*R = 5/6*R This answer does just require simple logic, but it does provide some examples of good circuit theory: 1. If the schematic is the same, it is the same circuit. Therefore, if the circuit acts differently, it has a different schematic. (Usually parasitics or a soldering mistake.) 2. It illustrates the bridge circuit concept. 3. It shows how analogies to other concepts can be used to help understand electronics. The cube shape actually simplifies the analysis by bringing the power of geometry into play.
This circuit has one tricky thing about it that we never think about when starting our analysis. Apparently, mesh analysis doesn't work, because the circuit is not planar. So, after many deadends, I simply went with a nodal analysis. My 1. I calculated the KCL equation for each of the 6 unknown nodes. 2. I drove node A to 1V and node B to ground. 3. Using Excel, I solved the equations for V1 to V6. 4. The currents through R1, R2, and R3 total to the current from the 1V source. 5. Using ohms law, I calculated the equivalent resistance. Included is the Excel spreadsheet. And just in case you don't like opening other peoples spreadsheets, I included a pdf to see the result. Enjoy!
Caveman, you approached the initial question the same as I did (however explained it more elequantly than I would have ). Like the Excel analysis. There are hundreds of different analyses that you can do on this cube: e.g. calculating the resistance between any adjacent node for any arbitrary resistor values (its in the symmetry again). Also try a forum search for the calculate the resistance between two nodes on an infinite resistor network. Dave