

The Resistor Cube Problem  RF Cafe Forums

Kirt Blattenberger 
Post subject: The Resistor Cube Problem
Posted: Sun Jun 13, 2010 1:32 pm


Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308 Location: Erie, PA 
Greetings: Most of us have run into the
Resistor Cube problem where it is required to
determine the equivalent resistance between
opposing corners. In this month's edition of
" Kirt's
Cogitations," I show the traditional
solution, as well as my own solution which I
believe is easier and more intuitive. A circuit
simulator proof of the solution is also presented.
This often shows up in job interviews, so
it's good to know. Anybody have a better
solution?
_________________  Kirt Blattenberger
RF Cafe Progenitor & Webmaster





hkroeze 
Post subject: Re: The Resistor Cube Problem
Posted: Wed Jun 16, 2010 9:36 am


Joined: Wed Jun 16, 2010 9:12 am
Posts: 1 
Hi Kirt, Good point to revisit the old resistor
cube network. Your appoach shows that there
are more ways to tackle a problem. However,
I think that your method is not essentially
different than the classic one. The classic
method uses the equal voltage at several nodes,
your method uses the equal currents in several
branches. Both methods are right, because the
network is symmetrical.
This reminds
me of the very elegant solution to the infinite
mesh network, which is the ultimate symmetrical
network. The problem is to find the resistance
between two adjacent nodes, with all resistors
equal.
(by the way, how do you insert
a picture?)
I'll be glad to show the
solution in a next post
Hugo Kroeze
The Netherlands.
PS. I recently visited
the ISMRM 2010 conference in Stockholm. In MRI
there is a lot of RF work done, so the conference
had a large educational session on this subject.
Several lectures mentioned the RFcafe as a good
source of RF information.





johnlawvere

Post subject: Re: The Resistor Cube Problem
Posted: Mon Jun 28, 2010 4:35 pm


Joined: Wed Jun 02, 2010 2:56 pm
Posts: 2 Location: Arizona 
Hello Kirt: When teaching Intro. E&M,
I often assign this cube problem. My assignment
guides the students to write formulas like "I=(VnVm)/R",
then to write KCL at 6 nodes, then to use a
program called "maxima" to invert the resulting
6by6 matrix to find the node voltages, then
find the total current drawn from the source
to find Req. I think it makes the students become
aware of how circuit theory works. Maxima
was developed years ago with the name "MACSYMA"
and has most of the functionality as Mathematica,
but it is available as a free download from
www.sourceforge.net.





Kirt Blattenberger 
Post subject: Re: The Resistor Cube Problem
Posted: Tue Jun 29, 2010 7:55 am


Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308 Location: Erie, PA 
hkroeze wrote:
Hi Kirt,
(by the way, how do you
insert a picture?)
Hugo Kroeze
The Netherlands.
PS. I recently
visited the ISMRM 2010 conference in Stockholm.
In MRI there is a lot of RF work done, so
the conference had a large educational session
on this subject. Several lectures mentioned
the RFcafe as a good source of RF information.
Greetings Hugo:
Sorry for the delayed response. A couple
people have sent me emails with alternate solutions.
One of them used delta/star transformations
to do it. That one has been added to the Resistor
Cube page at the bottom.
Thanks for letting
me know about RF Cafe being mentioned in the
lectures. I'll take all the promotion I can
get!
Thanks for writing.
_________________  Kirt Blattenberger
RF Cafe Progenitor & Webmaster





Kirt Blattenberger 
Post subject: Re: The Resistor Cube Problem
Posted: Tue Jun 29, 2010 8:02 am



Site Admin 

Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308 Location: Erie, PA 
johnlawvere wrote:
Hello Kirt: When teaching Intro. E&M,
I often assign this cube problem. Maxima
was developed years ago with the name "MACSYMA"
and has most of the functionality as Mathematica,
but it is available as a free download from
http://www.sourceforge.net.
Greetings johnlawvere: Yeah,
trying to solve a 6x6 matrix without the help
of a computer can be daunting. That's the purpose
of assigning equal resistances so that the poor
sap who sees it during an interview has an opportunity
to demonstrate knowledge of fundamental principles.
I posted a link to MACSYMA a while back,
but never did download and try it. There is
a lot of high quality open source software that
replicates most or all of the functionality
of the commercial stuff. OpenOffice.org is a
good example that has a version of Excel (called
Calc) that will import excel files, including
macros. Thanks for writing.
_________________  Kirt Blattenberger
RF Cafe Progenitor & Webmaster





kmorgan 
Post subject: Re: The Resistor Cube Problem
Posted: Sat Dec 18, 2010 4:55 pm



Captain 

Joined: Sat Dec 18, 2010 8:56 am
Posts: 5 Location: Long Island, NY

I have a question concerning this problem. I
teach amateur radio license classes here on
Long Island and lately I've been doing some
more advanced stuff with some of the already
licensed hams.
Regarding this cube, which
I've seen before, when one does the math with
the "equivalent" circuits described by most
websites, i.e., 3 parallel resistors into 6
parallel resistors and those into another 3
parallel resistors, as long as all the resistors
are the same value all the math works out well.
Simulations also work as expected.
You
can also work the cube as 3 parallel branches
of a series resistor feeding 2 parallel resistors
and those into another series resistor. Again,
all the math works out fine and so do simulations.
However...
If just one of the resistors
is changed, say doubling the value, then an
imbalance occurs and things get weird.
If the above mentioned "equivalent" circuits
are used the mathematically calculated answer
does not jibe with simulations of the cube.
If you simulate the equivalent circuit it does
match the math (the simulation that is).
Running a cube simulation you notice this
immediately. All three branches have different
currents running through them whereas in the
equivalent circuits 2 of the branches have the
same current and the third branch with the changed
resistor is different.
Can anyone help
enlighten me on what's going on? I'm confused
enough normally, this problem is making me more
so.
Thanks.
_________________
Kevin AB2ZI





Kirt Blattenberger 
Post subject: Re: The Resistor Cube Problem
Posted: Wed Dec 22, 2010 12:06 am


Joined: Sun Aug 03, 2003 2:02 pm
Posts: 308 Location: Erie, PA 
Greetings Kevin:
I think you've answered
your own question. Altering any resistor value
so that it does not equal all the others disturbs
the symmetry of the cube, so currents in the
branch circuits at each node are not equal.
You might be tempted to think that by
doubling one resistor value it will cut the
current in that branch in half and therefore
the voltage across it will still be what it
was when at the original value, but the node
is not an ideal voltage (or current) source
that maintains the same value regardless of
the load; its value is dependent on the resistances
in all the other branches.
Symmetry alone
allows the assumptions that permit the simplified
solutions like the series/parallel equivalent
circuits, or my method that depends on Kirchhoff's
current law.
Does that answer your question,
or did I miss the point?
_________________  Kirt Blattenberger
RF Cafe Progenitor & Webmaster





wb9jtk 
Post subject: Re: The Resistor Cube Problem
Posted: Wed Dec 22, 2010 9:14 am



Captain 

Joined: Tue Dec 26, 2006 5:39 pm
Posts: 10 
It's Gordian's knot.
I solved the
problem by BUILDING it and measuring it with
an ohm meter.
And I got a different
answer.





kmorgan 
Post subject: Re: The Resistor Cube Problem
Posted: Wed Dec 22, 2010 10:12 am



Captain 

Joined: Sat Dec 18, 2010 8:56 am
Posts: 5 Location: Long Island, NY

Kirt Blattenberger wrote:
Greetings Kevin:
I think you've answered
your own question. Altering any resistor
value so that it does not equal all the
others disturbs the symmetry of the cube,
so currents in the branch circuits at each
node are not equal.
You might be
tempted to think that by doubling one resistor
value it will cut the current in that branch
in half and therefore the voltage across
it will still be what it was when at the
original value, but the node is not an ideal
voltage (or current) source that maintains
the same value regardless of the load; its
value is dependent on the resistances in
all the other branches.
Symmetry
alone allows the assumptions that permit
the simplified solutions like the series/parallel
equivalent circuits, or my method that depends
on Kirchhoff's current law.
Does
that answer your question, or did I miss
the point?
I'm still a bit confused. Mostly I think
it's the "equivalent circuits" that mess up
the math. Here is an example using 100 ohm resistors
and a 5 v supply. When the cube is modeled you
get 82.7 ohms with 60 ma of total current. 20
ma through each of the 3 branches and 10 ma
distributing nicely among the middle 6 resistors.
When you draw it out as an equivalent circuit
of 3 parallel feeding 6 parallel outputting
to 3 parallel it looks like this:
The circuit on the left behaves like
a normal seriesparallel circuit and the math
and simulations come up with the same answer
as the cube simulation. However, if one of the
middle resistors is changed to 200 ohms, as
is the case on the right, then the math and
a simulation of this
equivalent
circuit work as expected, yielding total resistance
of 84.85 ohms with 58.93 ma of current. 19.64
ma in each input and output set of 3 resistors,
10.71 ma in the five 100 ohm parallel resistors
and 5.36 ma through the 200 ohm resistor.
But... When you model the cube and change
one of these middle resistors this is what you
see: Total resistance is now
85.3 ohms with 58.62 ma total current. Each
of the 3 branch input/output resistors (for
lack of a better description) has a different
total current. One branch has 19.83 ma, the
second 20.69 ma and the last, the one with the
200 ohm resistor, has 18.1 ma. In the
middle resistors, the first branch divides to
10.34 and 9.48 ma respectively, the second branch
divides up to 9.48 and 11.21 ma, and the third
to 11.21 and 6.9 ma (the 6.9 ma flowing through
the 200 ohm resistor). The circuit follows
Kirchoff's law without a problem. The equivalent
circuits are just not equivalent unless the
resistors are all the same value. My
head hurts! Kevin
_________________
Kevin AB2ZI


Posted 11/12/2012



