Resistance Cube

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Re: Resistance Cube

Following Rattus' suggestion, I shorted the junction between R4 and R7 with the junction between R8 and R11.

Same thing with the junction between R2 and R6 and the junction between R5 and R9.

I get 5 parallel combinations of R/2 with R1 and R10 still being R.

But I still wind up with a 2 loop circuit. It could be solved using mesh currents, but I'm too lazy for that. Did I miss a couple of points to short?

Steve
 
Re: Resistance Cube

Another way to solve this is to use the cube's symmetry to divide the currents and write the voltage equations. The currents in the 3 legs at the two end corners have to be equal because of the symmetry. The total current into the corner terminal splits into three equal currents because each path is symmetrical. At the opposite corner the three equal currrents come together. The currents in these 6 legs have to be equal. Call that current Ia.

The "Ia" currents split in two at the next corners of the cube and flow in the other six legs. Again due to symmetry all those 6 currents have to be equal. Call that current Ib.

To solve it we write the equations for the currents and the voltage drops across the cube.

Assume we put a voltage "Vt" across the end terminals of the cube. Call the resistance of the total cube Rt and the total current It. All the numbered resistances are the same value, call it R. The voltage drop in any leg will be either IaR or IbR.

Equations are:

Vt = ItRt ohms law for the total cube
It = 3 Ia (since the total current splits into three equal parts at each end)
Ia = 2 Ib (since each Ia splits into two equal parts on the diagonal corners)

Also,
Vt = IaR + IbR + IaR (this is just adding up the voltage drops in each leg between the two end points. Follow any current path and you will get this equation.)
Doing a little algebra:

Vt = 2IaR + IbR = R (2Ia +Ib) since Ia = 2 Ib, Ib = 0.5 Ia, substitute for Ib
Vt = R ( 2 Ia + 0.5 Ia)
Vt = R ( 2.5 Ia)

From the first two equations:
Vt= It Rt and It = 3 Ia, so Vt = 3 Ia Rt and Rt = Vt/(3Ia)

Since Vt = R (2.5 Ia), substitute for Vt in the Rt equation.

Rt = R (2.5 Ia)/ (3 Ia).

The Ia term cancels and we get Rt = R (2.5/3) which is also R= 5R/6.

A long way around to get the same answer you guys grabbed quickly!

As my college EE professor, Dr. Greenfield, used to say,

?So there, the sun is shining, the little birdies are singing, god is in his heaven and all is right with the world.?
 
Re: Resistance Cube

ResistorCube6r.gif

Sam,

I modified your diagram slightly. The points on the red triangle are at equipotential as are those on the green triangle. The other six resistors are simply connected between them, thus electrically parallel.
 
Re: Resistance Cube

By tying the equipotential points together, rbalex has flattened the network which makes the problem even simpler.
 
Re: Resistance Cube

ResistorCube9.jpg


Ok. Using currents and voltages instead of resistances is much better. I agree. (5/6)R

It's a lot clearer when you see that no current flows between the blue resistors.

Edit: There, no red lines.

[ June 12, 2005, 02:50 AM: Message edited by: physis ]
 
Re: Resistance Cube

I used to like the form they used in the link you posted, I still do, but I think I like this one better. And I'm tired of building these images too.

It would be better if I could spin the blue resistors in the depth axis but there's no tool in my software fo it and doing it manually blows.

Edit: Yeah, the red lines don't help alot. :) but I'm keeping the R#'s. :p

[ June 10, 2005, 08:47 PM: Message edited by: physis ]
 
Re: Resistance Cube

Steve, since it looks like you've built one, what does it measure?

Edit: Or, I guess you shorted your schematic. :D

[ June 11, 2005, 01:03 AM: Message edited by: physis ]
 
Re: Resistance Cube

Let me admit that I have been out of touch for a couple weeks (moving pains), and came into this one late. I haven't had time to go through all the pictures and the equations. But here is my reaction to one of the early statements. I am not picking on Bryan.
Originally posted by bphgravity: At a closer look, I have determined there are 3 resistors in parallel that are in series with 6 more resistors in parallel that are in series with the remaining 3 resistors in parallel.
Click to expand...
Not true, I fear. No resistor is in series with any other resistor, and no resistor is in parallel with any other resistor. Here is what I mean:

Let?s look for a series connection. Start at any point, and label it P1. Pretend you are an electron who is about to travel from that starting point (P1), through a resistor. When you get to the other side, and to the next connection point (let us call it P2), do you have an option of which way to go next? That is, are there more than one resistor connected to P2, other than the one that you have already passed through? The answer is ?yes.? Therefore, the resistor you traveled through is not in series with any other resistor. Now look at the whole picture, and you will see that my first statement is true, that no resistor is in series with any other resistor.

Now let?s look for a parallel connection. Pick any resistor. Put your left hand on the left side of that resistor, and your right hand on the right side of that resistor. Next, look to see what other resistor (if any) is connected to the same point as the one your left hand is touching. You see one? Actually, you should see two. Now look to see if either of the two is also connected to the same point that your right hand is touching. I mean connected by wire alone, not connected via another resistor. You will note that no other resistor, other than the first one you started with, is touching both your left hand and your right hand. Therefore, the resistor you put your hands across is not in parallel with any other resistor. Now look at the whole picture, and you will see that my second statement is also true, that no resistor is in parallel with any other resistor.

I recall having seen this problem in college. It is a ?classic.? What I remember about the solution is that the total resistance of the cube is exactly the same as it would be, if you disconnected and removed a specific pair of resistors. That is, the voltage across each of the two resistors is zero. I do not recall which are the two that can be removed. I would have to think that through. Perhaps someone else will get it before I can remember. But I can say that once you disconnect those two specific resistors, what remains is a simple set of series and parallel connections that you can easily resolve into an equivalent resistance.

[ June 11, 2005, 07:49 PM: Message edited by: charlie b ]
 
Re: Resistance Cube

Charlie B.,

bphgravity has implied that there are two sets of three equipotential nodes. By connecting the equipotential points you indeed have R/3, R/6, and R/3. rbalex has actually connected the nodes which flattens the network. There is current in each resistor.

Rattus
 
Re: Resistance Cube

I said that that was what I remembered. I didn't claim to have remembered it correctly. :D It's been a couple of centuries since I last saw this problem.

I still need to work my way through this thread, so I'm not quite ready to give an opinion on the earlier solutions.
 
Re: Resistance Cube

Physis:

I didn't actually build one, I just "shorted the nodes" on paper.

Would anyone like to solve this thing if R4 is half the value of the other resistors :eek:

Steve
 
Re: Resistance Cube

Sam,

This thing is a piece of cake! How about the resistance between opposite corners of a solid cube of copper?

Rattus
 
Re: Resistance Cube

By Steve:

I didn't actually build one, I just "shorted the nodes" on paper.
Click to expand...
That's what I figured, after I posted it looks like you built one.

By Rattus:

How about the resistance between opposite corners of a solid cube of copper?
Click to expand...
Of course I've thought about it. How bout a sphere? :D
 
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