Resistors in a cube (one on each edge)

[eLiu]

Hey all,
I've been reviewing physics (e&m) for a placement test I have to take this fall--I want to get out of that class. Anyway, I stumbled across a rather interesting problem...

It's 12 resistors laid out in a cube--1 resistor on each edge. All have resistance R. If you orient the cube so that you look directly into one of its faces, the current enters through the front, upper left corner, and it exits through the back, bottom right corner.

I was wondering how one would figure out the equivalent resistance the "regular" way. That is, how to add up the reistances in series/parallel?

I found a different way to do it...but I took advantage of the fact that all the resistances are equal. Here's what I did: entering through that corner, the current splits evenly (3 parts)--I/3. Then travel down or across one edge, and the current will have to split 2 ways--I/6. Then if you come down the last edge to reach the exit point...notice that two sections of I/6 must recombine...giving you I/3 for that last edge.

So, the idea is that the potential difference between entrance/exit must equal the sum of the edges we traversed...which results in R(total) = R/3 + R/6 + R/3 = 5R/6.

That's all nice and happy...but I haven't been able to get it to work out properly by using conventional methods of combining resistances until you turn the whole cube into 1 resistor... >.<

Thanks in advance,
-Eric

 

[CTho9305]

Are you familiar with nodal or mesh analysis? Either of those techniques would probably work, but I didn't learn them until freshman year (nodal) and sophomore year (mesh), and I was given that problem in AP Physics.

 

[JSSheridan]

My first thought is to try reducing the topography of the problem from 3-D to 2-D. Then you can tell from inspection if the circuit can be reduced using series and parallel combinations. One unpleasant fact you will have to deal with is that some circuits can't be reduced with this method. In such a case, Thevenin's Theorem or Norton's Theorem comes in handy.

 

[eLiu]

Never heard of mesh/nodal analysis or Thenevin's Theorem...the only physics class I've taken is AP Physics C. I don't really like E&amp;M a whole lot, so I'm trying to get out of it.

I tried to reduce it to a 2D drawing, but at least for me, it becomes even harder to draw... The problem I'm having is that along essentially 3 paths, the resistors are kind of "interdependent." So I've tried breaking them into parallel/series relationships in various ways...but I'm not getting the right answer, so that tells me I'm combining incorrectly

However...I did find that if only consider 2 faces of the cube (in 2D), the total resistance is half that of the entire cube...in a way that makes sense, but I don't really know how to justify that.

I'll give those 2 theorems a look...thanks.

 

[JSSheridan]

You are not combining resistors incorrectly, you are incorrectly combining resistors. In other words, it can't be done for this circuit, I'm sorry to say. Two resistances can be combined in parallel if, and only if, both nodes are common (shared). And two resistances can only be combined in series if, and only if, two (no more) share a single node. To find the right answer, you need to use a different method. Good luck. Peace.

 

[jmcoreymv]

Mesh analysis would be a good way to solve it.

 

[eLiu]

Originally posted by: jmcoreymv
Mesh analysis would be a good way to solve it.

Looks like I'll be googling this...it isn't in the course I'm preparing for, but I'm now very curious...

JSSheridan, so you're saying that the 3D structure of this system makes it so that the combining method becomes insufficient? Hm...that's comforting...at least now I know I wasn't missing something obvious...more like I don't know what I'm doing >.< D'oh.

I guess the potential difference thing I came up with in the original post works out fine (I can't think of any reason why it's wrong...and it matches the correct answer), but I'll look up this mesh thing tomorrow.

Thanks everyone I'll post back if the mesh analysis makes no sense to me..lol

-Eric

Edit: Wait...by mesh &amp; nodal analysis, do you mean Khirkov's (horrible mispelling) Laws? Like the sum of the currents flowing into/out of a node = 0 and the sum of the potentials in a closed loop = 0...?

If so, didn't I basically use nodal analysis in the solution that I did find? And...would you use mesh analysis by like...linking the exiting wire back into the entering wire with some kind of power source in between...? (To make a closed loop that is)

 

[jmcoreymv]

Basically mesh analysis is using kirchoffs law that says the sum of all voltages around a closed loop is 0, so it would let you find the current in each loop. Although now Im sort of remembering that one of the two methods (mesh or nodal) will not work if the circuit cant be drawn in a planar fashion (with no lines crossing over each other. I think it may be mesh that wont work, in which case youd have to use nodal.

 

[CTho9305]

Originally posted by: eLiu

Originally posted by: jmcoreymv
Mesh analysis would be a good way to solve it.

Looks like I'll be googling this...it isn't in the course I'm preparing for, but I'm now very curious...

I hate doing it. It's tedious. I prefer it over nodal (you add up currents in nodal analysis, vs voltages in mesh analysis), but I still hate it.

JSSheridan, so you're saying that the 3D structure of this system makes it so that the combining method becomes insufficient? Hm...that's comforting...at least now I know I wasn't missing something obvious...more like I don't know what I'm doing >.< D'oh.

You can map it to a 2d surface trivially - the problem is the fact that you can't pull out two subpieces that share the same start and end nodes to do thevenin/norton swaps. There's some theorem on swapping out a "Y" shape for a triangle, but I can't remember its name, and IIRC, it doesn't help.

I guess the potential difference thing I came up with in the original post works out fine (I can't think of any reason why it's wrong...and it matches the correct answer), but I'll look up this mesh thing tomorrow.

Thanks everyone I'll post back if the mesh analysis makes no sense to me..lol

-Eric

Edit: Wait...by mesh &amp; nodal analysis, do you mean Khirkov's (horrible mispelling) Laws? Like the sum of the currents flowing into/out of a node = 0 and the sum of the potentials in a closed loop = 0...?

Exactly, mesh &amp; nodal analysis are a set of steps based on Kirchoff's laws.

 

[jmcoreymv]

Alright I solved it and got Req= .8333*R. First I drew out the circuit in 3d. Then extrapolated it to a 2d surface. After that I labled each node with a node voltage (V1,V2,V3,etc....V8). Then I wrote out the nodal equations and set them equal to zero. Plugged them into a matrix....etc, used PSpice to verify that Req=.8333*R

 

[uart]

The easiest way to solve that problem is definitely NODAL analysis. You just place a 1 Amp cuurent source between the two diagonally opposite node, take the node at the current source negative as the reference and calculate the voltage a the other node using nodal analysis.

There are7 nodes invovled (apart from the reference node), so this produces a set of seven linear simultaneous equations in 7 unknowns. The only unknown you need to find however is the voltage at the positive current source node, which is numerically equal to the diagonal resistance of the cube.

 

[f95toli]

I have seen this problem several times in various books (and in courses), it is sort of a "standars" problem used to teach you to look for symmetries.
So I don't hink you are "suppose" to solve it using either mesh- or nodal analysis, in the solutions I have seen you use that fact that all resistors have the same value (=symmetires).

Basically, it is a "constructed" problem used for teaching, I doubt you would ever come across such a circuits IRL.

 

[Mark R]

This is a simple problem and needs only the ability to see the symmetry.

There is no need for techniques such as nodal analysis or matrix analysis. I don't even know what they are. There is no need for simultaneous equations or even detailed knowledge of algebra.

The trick is to see that although there are 8 circuit nodes (corners of the cube), there are only 4 voltages. The 2 supply voltages, and 2 intermediate voltages (found on 3 nodes each). Because several nodes all share the same voltage, the analysis can be simplified by connecting them together and simplifying the circuit.

Once you appreciate that there are only 4 circuit nodes, you can reduce the network into a more easily analysed one. In fact, the network reduces nicely into 3 resistances in series.

BTW - I get an answer of 5R/6

 

[eLiu]

Originally posted by: Mark R
This is a simple problem and needs only the ability to see the symmetry.

There is no need for techniques such as nodal analysis or matrix analysis. I don't even know what they are. There is no need for simultaneous equations or even detailed knowledge of algebra.

The trick is to see that although there are 8 circuit nodes (corners of the cube), there are only 4 voltages. The 2 supply voltages, and 2 intermediate voltages (found on 3 nodes each). Because several nodes all share the same voltage, the analysis can be simplified by connecting them together and simplifying the circuit.

Once you appreciate that there are only 4 circuit nodes, you can reduce the network into a more easily analysed one. In fact, the network reduces nicely into 3 resistances in series.

BTW - I get an answer of 5R/6

That's basically what I did...from OP:

I found a different way to do it...but I took advantage of the fact that all the resistances are equal. Here's what I did: entering through that corner, the current splits evenly (3 parts)--I/3. Then travel down or across one edge, and the current will have to split 2 ways--I/6. Then if you come down the last edge to reach the exit point...notice that two sections of I/6 must recombine...giving you I/3 for that last edge.

So, the idea is that the potential difference between entrance/exit must equal the sum of the edges we traversed...which results in R(total) = R/3 + R/6 + R/3 = 5R/6.

However, I was looking to see if this problem could be solved the "standard" way.

Hm...uh, if all the resistances were different (arbitrary R1 thru R12), could I still apply the same method? B/c current "splits up" in an evenly describeable manner...like if the first node was connected to a R, 2R, and 3R, then the resistors would receive I/2, I/3, and I/6 respectively (right?). Then after that, would it be the same thing as in this problem--except that the R's may/may not add up so cleanly?

-Eric

Edit: jmcoreymv, doesn't nodal analysis involve adding up currents..? Or did you do voltage just b/c it really doesn't matter for this problem (since R's are equal)?

uart, do you mean to connect the entry/exit nodes with a current source? And would the potential difference be numerically equal to the total resistance only in this case (R's equal) or in any situation? I'm thinking it's the latter...but I just wanted to make sure.

 

[uart]

uart, do you mean to connect the entry/exit nodes with a current source? And would the potential difference be numerically equal to the total resistance only in this case (R's equal) or in any situation? I'm thinking it's the latter...but I just wanted to make sure.

Yep, i's a useful general way of computing resistance, use nodal analysis to find the terminal voltage due to an impressed 1A current source. It works on active circuits with dependant soucres as well, so you can use it to compute the input and output resistance of amplifiers and other active linear circuits.

Mark is correct in pointing out that with all resistors equal that some nodes must have the same voltage by symmetry and since that implies no current will flow between them if shorted they can therefore be "fictitiously" shorted to make the analysis easier without affecting the results.

Remember that with nodal analysis that there is no difficulty in having any values of resistors so the method is the same even when there is no symmetry, it's a powerful technique and well worth learning.

 

[eLiu]

Ah ok...I think I get it then--just write out the current-in = current-out equation for each node...and then row reduce to get solutions?

Thsi is an intro-class...so the 'hardest' circuits I've done are multi-loop, RC circuits. Not really that difficult...find the loops that'll give you X equations for X independent variables and row reduce. Seems like this cube is much the same thing...

But yeah, thanks uart, a general method was basically what I was looking for, since I was afraid the method I used to get the answer only works out (easily) when all the R's are equal.

-Eric

 

[SuperTool]

I remember this problem, but forgot how to solve it

 

[SuperTool]

I am getting resistance of R.
I may be way off. Basically from a vortex, you have 3 edges. I consider an edge followed by square as one path. the first edge resistance R, and I split all shared edges into two resistors of 2R in paralle. So each path has R+(2R+2R)/2 = 3R resistance. I have 3 such paths. So I have total resistance of 3R/3= R.
I didn't give that much thought into it, but it makes sense to me. It may be off a little bit though

 

[SuperTool]

Ok, quick search on google makes me think it's 5/6R, but within order of magnitude is good enough for me
http://rec-puzzles.org/new/sol.pl/physics/resistors

 

[Varun]

I like Nodal, it's not even that hard once you do it a couple times.

However, just use PSpice!! ( I realise that is kind of cheating )

 

[eLiu]

Supertool, good find...thanks

varun, what is PSpice??

 

[SuperTool]

Originally posted by: SuperTool
I am getting resistance of R.
I may be way off. Basically from a vortex, you have 3 edges. I consider an edge followed by square as one path. the first edge resistance R, and I split all shared edges into two resistors of 2R in paralle. So each path has R+(2R+2R)/2 = 3R resistance. I have 3 such paths. So I have total resistance of 3R/3= R.
I didn't give that much thought into it, but it makes sense to me. It may be off a little bit though

Actually this approach works.
I counted some edges as shared that weren't.
It's actually each path is (R+(R+2R)/2)/3=5/6R
So basically you have edge with R, which forks into two R edges, which then continue into shared edges, each with 2R.
and you have 3 such paths. So it's (R+(R+2R)/2)/3 = 5/6 R

 

[Varun]

PSpice is a program used for electrical circuit anaysis. It's super slick, though I prefer using the text method of doing it, there is a GUI as well.

You may have to search the net a bit, the original PSpice program was freeware, but it was a bit of a challenge for me to track down last year when I found it.

It will solve any circuit including phase angles if using AC, by using Nodal.

 

[eLiu]

SuperTool, that's pretty neat...I never thought to do it that way

Varun, if I ever get into a deeper circuits class, I'll definitely go look for that program...but for now, I'm off to the chapter on magnetism, lol.

 

[Mday]

okay, key here is that, FLATTEN THE SUCKER. its very easy to do this problem, btw.