The Resistor Cube Problem  RF Cafe Forums
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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
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