Infinite mesh of resistors.
K.
A not so easy problem for you - but I have the difficult problem of
moving house at the moment(which must be worse than having a baby-for a
man anyway).
The following is an infinite mesh of 1 ohm resistors (each,well some,
labelled R).....
. . . .
| | | |
R R R R
| | | |
-R-------|------R--------X-------R-------|-------R-------|-----R-
| | | |
R R R R
| | | |
-R-------|------R--------|-------R-------Y-------R-------|-----R-
| | | |
What is the equivelant resistance of the network between the two points
labelled X and Y on the diagram?
--
K.
R E Sheskey
In article <RExq6AAM...@litak.demon.co.uk>, postm...@litak.demon.co.uk
says...
For people (like myself) who barely know how to screw in a lightbulb,
much less analyze a circuit, would you (or someone) mind explaining
the relevant electrical principles? Assuming that can be done briefly
without giving away the answer...
Robert Sheskey
Richard Walter
>
> For people (like myself) who barely know how to screw in a lightbulb,
> much less analyze a circuit, would you (or someone) mind explaining
> the relevant electrical principles? Assuming that can be done briefly
> without giving away the answer...
>
> Robert Sheskey
Sure.
Ohms law is V=IR.
Where V is the voltage across a resistor, in volts.
I is the current through the resistor, in amps.
R is the resistance of the resistor, in ohms.
So, if you hook up a 1 volt battery across a 5 ohm resistor, then
I = (V/R) = (1/5) = .2 amps will flow though that resistor. Current
is defined as flowing from a higher voltage to a lower voltage.
Another way of looking at that equation is if you force .3 amps of
current through a 10 ohm resistor, then you will get V = IR = (.3)(10)
= 3 volts drop across the resistor.
The second law that is relavent is that all of the current flow
into a node must equal all of the current flow out of that node.
This means that current cannot accumulate at one place.
Also, for the purposes of most puzzles, voltage sources are assumed
to be ideal, in that they can supply an infinite amount of current,
if required.
So, for resistor network problems, if the problem asks for what the
equivalent resistence between two nodes X & Y is, one method to solve
the problem is to imagine that a 1 volt battery is connected such that
node X is +1 V and node Y is ground. Then determine how much total
current, I, flows out of node X. The equivalent resistence of the
network is then R = V/I = 1/I (since voltage is 1 volt).
-Richard Walter
rwa...@auspex.com
--
===============================================================================
"Disclaimer: This posting represents the poster's views, not those of
Auspex"
===============================================================================
Doug McKean
R E Sheskey wrote:
>
> In article <RExq6AAM...@litak.demon.co.uk>, postm...@litak.demon.co.uk
> says...
> >
> >A not so easy problem for you - but I have the difficult problem of
> >moving house at the moment(which must be worse than having a baby-for a
> >man anyway).
> >
> >The following is an infinite mesh of 1 ohm resistors (each,well some,
> >labelled R).....
> > . . . .
> > | | | |
> > R R R R
> > | | | |
> > -R-------|------R--------X-------R-------|-------R-------|-----R-
> > | | | |
> > R R R R
> > | | | |
> > -R-------|------R--------|-------R-------Y-------R-------|-----R-
> > | | | |
> >What is the equivelant resistance of the network between the two points
> >labelled X and Y on the diagram?
> >--
> >K.
>
> much less analyze a circuit, would you (or someone) mind explaining
> the relevant electrical principles? Assuming that can be done briefly
> without giving away the answer...
>
> Robert Sheskey
Absolutely. Maybe an example that would hit you right in the
money roll. I worked at a military facility that needed a pure ground.
The reasons I cannot go into. First, they drilled a well down 360 feet
into the earth. Then, they took a cable or three, and lowered them all
the way down to the bottom of the well. Finally, they filled it up
with concrete to seal it. The ground plane of the facility had to be
attached to this cable. But, the ground plane in the facility had to
maintain a certain, let's say, ground level. In order to get that,
you might have to use pure copper plates underneath the entire facility.
A large complex I might add. Many tens of thousands of square feet.
Using the above solution to the problem, pure copper cables were put on
the floor in a square grid pattern with a minimum spacing to guarentee the
necessary grounding. And, saving an enormous amount of money.
Good Enough?
*******************************************************
Doug McKean
doug_...@paragon-networks.com
-------------------------------------------------------
The comments and opinions stated herein are mine alone,
and do not reflect those of my employer.
-------------------------------------------------------
*******************************************************
Tom Maciukenas
In article <322C86...@paragon-networks.com>,
Doug McKean <doug_...@paragon-networks.com> wrote:
: Absolutely. Maybe an example that would hit you right in the
: The reasons I cannot go into. First, they drilled a well down 360 feet
: into the earth. Then, they took a cable or three, and lowered them all
: the way down to the bottom of the well. Finally, they filled it up
: with concrete to seal it. The ground plane of the facility had to be
: attached to this cable. But, the ground plane in the facility had to
: maintain a certain, let's say, ground level. In order to get that,
: you might have to use pure copper plates underneath the entire facility.
: A large complex I might add. Many tens of thousands of square feet.
: Using the above solution to the problem, pure copper cables were put on
: the floor in a square grid pattern with a minimum spacing to guarentee the
: necessary grounding. And, saving an enormous amount of money.
Hmm... Why would someone have to take such precautions, unless they were,
oh I don't know, maybe PERFORMING AN ALIEN AUTOPSY???!?!?!?!?!?!!!!!
C'mon Doug, you know something! Spill your guts!
:^)
-tOmm
Dean Edmonds
In article <50hrsq$n...@dfw-ixnews4.ix.netcom.com>,
R E Sheskey <rshe...@ix.netcom.com> wrote:
>
>For people (like myself) who barely know how to screw in a lightbulb,
>much less analyze a circuit, would you (or someone) mind explaining
>the relevant electrical principles? Assuming that can be done briefly
>without giving away the answer...
Certainly. There are only two rules. The first is for resistors connected
in series (i.e. end to end), as follows:
A ---R1---o---R2---o---R3---o-...-o---Rn--- B
The total resistance between points A and B is simply the sum of the individual
resistances. I.e: R1 + R2 + R3 + ... + Rn
The second case is for resistors connected in parallel (i.e. side by side):
---R1---
| |
o--R2--o
A --| |-- B
o--R3--o
| |
...
| |
---Rn---
In this case, the total resistance between points A and B is the inverse of
the sum of the inverses of the individual resistances. I.e:
1
-----------------------
1 1 1 1
-- + -- + -- + ... + --
R1 R2 R3 Rn
Any resistor network can be solved by combining sets of parallel and series
resistors until you end up with a single resistance. E.g:
---- 3 ----
A ----| |--- 10 ---- B
---- 7 ----
First solve for the two parallel resistors as follows:
1
----- = 2.1
1 1
- + -
3 7
The original resistor network then simplifies to:
A ---- 2.1 ---- 10 ---- B
which can be solved by summing up the resistances, giving a result of 12.1
=============================================================================
- deane | Before I answer your question, define `obsessed'.
Mike Naylor
rshe...@ix.netcom.com (R E Sheskey) wrote:
>For people (like myself) who barely know how to screw in a lightbulb,
>much less analyze a circuit, would you (or someone) mind explaining
>the relevant electrical principles? Assuming that can be done briefly
>without giving away the answer...
Don't worry, I won't give away the answer (because I don't know it!) But I
can tell you that if two one ohm resistors are wired in series (end to
end), then the resistance is additive. The total resistance is two ohms.
If two one ohm resistors are wired in parallel (side by side), then the
total resistance is 1/2 the value of one of the resistors, or 1/2 ohm.
If four one ohm resistors are wired in a square, then to find the
resistance from one corner of the square to the diagonally opposite corner,
you first find the equivalent resistances of the two pairs of series
resistors, which would be two ohms each, and then find the resistance of
the two two ohm resistors in parallel, which is one ohm.
In the above example, you might need to find the resistance from one corner
to an adjacent corner, which means that you would have a one ohm resistance
and a three ohm resistance in parallel. If two unequal resistors are wired
in parallel, the total resistance is equal to the product divided by the
sum. In this case, (3 X 1)/(3 + 1) = 3/4 ohm.
In cases where three or more unequal resistors are wired in parallel, the
total resistance is equal to the reciprocal of the sum of the reciprocals.
For example, a two ohm, a three ohm, and a four ohm resistor in parallel
have a total resistance of 1/(1/2 + 1/3 + 1/4) = 1/(13/12) = 12/13 ohms.
Generally, problems of multiple resistors are solved by reducing sets of
resistors to equivalent individual resistors until the entire set has been
reduced to a single resistance. However, in the puzzle under discussion
here, which features an infinite grid of resistors, that method does not
look very promising. There must be an answer though, and I'm working on it.
Mike
--
Play Five by Five Poker at http://www.serve.com/games/
Noam Elkies
In article <RExq6AAM...@litak.demon.co.uk>
"K." <postm...@litak.demon.co.uk> writes:
>A not so easy problem for you [...]
>The following is an infinite mesh of 1 ohm resistors (each,well some,
>labelled R).....
> . . . .
> | | | |
> R R R R
> | | | |
> | | | |
> R R R R
> | | | |
> -R-------|------R--------|-------R-------Y-------R-------C-----R-
> | | | |
>
>What is the equivalent resistance of the network between the two points
>labelled X and Y on the diagram?
Indeed a difficult puzzle; it even takes some care to ensure it's
well-defined. The equivalent resistance between two *adjacent*
vertices is well-known to be R/2. The equivalent resistance between
two diagonally adjacent vertices turns out to be 2R/pi ! In fact
the resistance between two vertices k diagonal units apart is
(2R/pi) (1+1/3+1/5+...+1/(2k-1)).
This, together with symmetry and the known R/2 resistance,
suffices to determine all the equivalent resistances in this
network, once you realize the following fact:
if X,Z are any two different vertices, the equivalent resistance
between X and Z is the average of the equivalents resistances
between X and Z' over the four neighbors of Y.
For instance, if Z is the vertex I labeled A in the above diagram
we find that that the X-B resistance is (2-4/pi)R [NB the eq.res.
from X to itself is zero]; taking Z=Y and noting that the unlabeled
neighbors of Y are equivalent to A,C yields the value (4/pi-1/2)R
for the X-C resistance; "etc."
The proofs are left as exercises to the reader ;-)
--Noam D. Elkies (elk...@math.harvard.edu)
Dept. of Mathematics, Harvard University
Noam Elkies
In article <50ihlr$g...@decaxp.harvard.edu> I wrote:
>[...]
> if X,Z are any two different vertices, the equivalent resistance
> between X and Z' over the four neighbors of Y.
Of course this should be "as Z' varies over the four neighbors of Z",
not "of Y".
Doug McKean
Ok, Tom. I'll tell you the whole story.
A space ship was hovering over Rosewell one night.
They were trying to choose one of three places to land.
It looked like they had picked one. Being on patrol
that night, I signaled to them that the landing
site nearest me was NOT a good place to land.
It took them so long to decide whether to land
where they had decided, or, switch to a new site
that they ran out of fuel and crashed.
After saving the only living alien on board,
the alien kept on saying, "GRY! GRY! GRY!"
I kept on saying, "NO! TWO THIRDS! TWO THIRDS!"
It was horrible. I feel much better now.
The nightmare is over. Thanks.
Vladimir A. Pertsel
In article <50ihsl$i...@decaxp.harvard.edu>, elk...@ramanujan.harvard.edu (Noam Elkies) writes:
The problem is from the old book of Zilberman and Surkov,
which was used as a problems book in the 18-th Moskow school
some 25 years ago. The problem is easily solved when You
find the correct approach.
To make the life easier consider the following variant:
Change all the resistors for the capacitances and use
the definition of the capacitance instead of formulae.
Just put some charge at one node; think; replace it for
the opposite charge at the second node; consider the
interferencion of the fields and charges, calculate the
capacitance according to its definition.
The case with the resistances is, in fact, just dual
for the case with the capacitances, and may be solved
either by the dual idea (less evident) or by reduction
to the solved problem
--
From the Holy Land, with respect
/\ /\ Vladimir A. Pertsel S/W engineer
((ovo)) E-mail: \|/ Sagantec Israel
():::() vold...@sagantec.co.il (o o) tel.972-4-8572781
--PVA------------------------ooO-(_)-Ooo------------------
an ancestor of mine by the name of Noah was once the
commanding admiral of the combined fleets of my planet.
Dennis Yelle
In article <50inec$9...@excalibur.net5c.io.org> de...@io.org writes:
>In article <50hrsq$n...@dfw-ixnews4.ix.netcom.com>,
> >
> >For people (like myself) who barely know how to screw in a lightbulb,
> >much less analyze a circuit, would you (or someone) mind explaining
> >the relevant electrical principles? Assuming that can be done briefly
> >without giving away the answer...
>
>in series (i.e. end to end), as follows:
>
> A ---R1---o---R2---o---R3---o-...-o---Rn--- B
>
>The total resistance between points A and B is simply the sum of the individual
>resistances. I.e: R1 + R2 + R3 + ... + Rn
>
>
>The second case is for resistors connected in parallel (i.e. side by side):
>
> ---R1---
> | |
> o--R2--o
> A --| |-- B
> o--R3--o
> | |
> ...
> | |
> ---Rn---
>
>In this case, the total resistance between points A and B is the inverse of
>the sum of the inverses of the individual resistances. I.e:
>
> 1
> -----------------------
> 1 1 1 1
> -- + -- + -- + ... + --
> R1 R2 R3 Rn
>
>
>Any resistor network can be solved by combining sets of parallel and series
>resistors until you end up with a single resistance.
This is not true. There is at least one more case.
The case of a diamond, or bridge. For example:
---R1--o--R4---
| | |
A --| R3 |-- B
| | |
---R2--o--R5---
Ob Puzzle: Given the network of 5 resistors above,
and R1 = 1 ohm, R2 = 2 ohms, R3 = 3 ohms, R4 = 4 ohms, and R5 = 5 ohms,
what is the equivelent resistance from A to B ?
What is the general formula for 5 resistors connected like that?
--
den...@netcom.com (Dennis Yelle)
"You must do the thing you think you cannot do." -- Eleanor Roosevelt
James R Yingst
In article <50inec$9...@excalibur.net5c.io.org>,
Dean Edmonds <de...@io.org> wrote:
>In article <50hrsq$n...@dfw-ixnews4.ix.netcom.com>,
>R E Sheskey <rshe...@ix.netcom.com> wrote:
> >
> >For people (like myself) who barely know how to screw in a lightbulb,
> >much less analyze a circuit, would you (or someone) mind explaining
> >the relevant electrical principles? Assuming that can be done briefly
> >without giving away the answer...
I don't know how to do it briefly, and there's no danger that I'll give
away the solution. :-/
>Certainly. There are only two rules.
Bzzt. Not sufficient in this problem.
[Equivalent resistance formulas for series and parallel deleted.]
If someone wants to use these, they probably will need more rigorous
definitions of "series" and "parallel" as well as where to draw the
equivalent resistors in the simplified circuit, but I'm not going to
bother as I don't think they'll be all that useful anyway in this
problem. Best to grab an introductory physics or circuits book if you
want the details.
>Any resistor network can be solved by combining sets of parallel and series
>resistors until you end up with a single resistance.
Really? Can you apply either of these rules anywhere in the given problem?
A common error is to assume that any two resistors near each other must
be either in series or in parallel. Not true - it's entirely possible
for a pair to be neither in parallel nor in series. As an extreme
example, in this problem, *none* of the resistors are in parallel or in
series.
Note that it is possible to find an equivalent circuit in which some of
the resistors are in parallel &/or series, but not the original circuit.
You need other rules to get there though, such as Kirchoff's Voltage and
Current Laws. I don't have the time & inclination to do these justice -
Richard Walter's post is a good place to start.
Note that this problem is in the rec.puzzles archive, but I haven't read
the answer as I'm trying (unsuccessfully) to work it out myself.
Jim Yingst
yin...@u.arizona.edu
Tom Maciukenas
In article <322D74...@paragon-networks.com>,
Doug McKean <doug_...@paragon-networks.com> wrote:
>Tom Maciukenas wrote:
>>
>> C'mon Doug, you know something! Spill your guts!
>
>A space ship was hovering over Rosewell one night.
>They were trying to choose one of three places to land.
>It looked like they had picked one. Being on patrol
>that night, I signaled to them that the landing
>site nearest me was NOT a good place to land.
>It took them so long to decide whether to land
>where they had decided, or, switch to a new site
>that they ran out of fuel and crashed.
>After saving the only living alien on board,
>the alien kept on saying, "GRY! GRY! GRY!"
>I kept on saying, "NO! TWO THIRDS! TWO THIRDS!"
>It was horrible. I feel much better now.
>The nightmare is over. Thanks.
I knew it! The government has been lying to us!
The TRUE answer to the GRY question *IS* gry! When
it says "... the English language", it simply never
mentions that it means "as spoken on Alpha Centauri".
Apparently, on that planet it is a VERY common word,
and the alien certainly seemed convinced that every
one (including Doug) knew what it meant.
Unfortunately, because of Doug's lack of diplomatic
skills, the aliens are now convinced that we are a
gry-less world (such a tragedy, but what are ya'
gonna' do?), and they'll be landing next Tuesday to
take over. I suggest we all party pretty heavily
this weekend. It may be the last chance we'll get
for a while.
-tOmm
Joseph Peterson
Doug McKean (doug_...@paragon-networks.com) wrote:
>The following is an infinite mesh of 1 ohm resistors (each,well some,
>labelled R).....
>What is the equivelant resistance of the network between the two points
>labelled X and Y on the diagram?
A friend of mne had a very similar problem when he was applying for
an electrical Engineering position. If you want the solution, I can
look it up and post it, but I suggest that if you really want to learn
how... look it up in a good Electromagnetic Fields book...
Look up matched loads for a starting point...
-JEEP
***************************************************************************
* Joseph E.E. Peterson * This line intentionally not left blank. *
* Associate Systems Engineer * Remember the name and vote in 2016. *
* Derby Associates Internat'l * Sm:)e *
* je...@derby.com * All these opinions are mine and only mine *
***************************************************************************
Woody Brison
A third rule that might be helpful is that if two nodes
in the circuit are at the same voltage, then they can
be shorted together (in other words, merged).
Woody
Joseph Riel
Has anyone solved the equivalent problem in 3 and higher dimensions?
The method used for the 2 dimensional case is not valid in higher
dimensions (Polya's theorem applies) so there appears to be no easy solution.
Joe Riel
Dr D F Holt
In article <50inec$9...@excalibur.net5c.io.org>,
de...@io.org writes:
>In article <50hrsq$n...@dfw-ixnews4.ix.netcom.com>,
>R E Sheskey <rshe...@ix.netcom.com> wrote:
> >
> >For people (like myself) who barely know how to screw in a lightbulb,
> >much less analyze a circuit, would you (or someone) mind explaining
> >the relevant electrical principles? Assuming that can be done briefly
> >without giving away the answer...
>
>in series (i.e. end to end), as follows:
>
> A ---R1---o---R2---o---R3---o-...-o---Rn--- B
>
>The total resistance between points A and B is simply the sum of the individual
>resistances. I.e: R1 + R2 + R3 + ... + Rn
>
>
>The second case is for resistors connected in parallel (i.e. side by side):
>
> ---R1---
> | |
> o--R2--o
> A --| |-- B
> o--R3--o
> | |
> ...
> | |
> ---Rn---
>
>In this case, the total resistance between points A and B is the inverse of
>the sum of the inverses of the individual resistances. I.e:
>
> 1
> -----------------------
> 1 1 1 1
> -- + -- + -- + ... + --
> R1 R2 R3 Rn
>
>
I do not believe that, even for finite networks. Take the case, where
you have 1 ohm resistors along each of the 12 edges of a cube, and you
are asked to calculate the resistance from one corner to the diagonally
opposite corner. How do you reduce that to sets of parallel and
series resistors?
Derek Holt.
Doug McKean
That's easy. Contact me off line and I'll tell you.
David Karr
ma...@csv.warwick.ac.uk (Dr D F Holt) wrote:
>de...@io.org writes:
>>Any resistor network can be solved by combining sets of parallel and series
>>resistors until you end up with a single resistance. E.g:
>
>I do not believe that, even for finite networks. Take the case, where
>you have 1 ohm resistors along each of the 12 edges of a cube, and you
>are asked to calculate the resistance from one corner to the diagonally
>opposite corner. How do you reduce that to sets of parallel and
>series resistors?
You observe that by symmetry, three of the nodes are at the same
potential, and three others are at a different potential (but equal
to each other). You can wire together any nodes that happen to be
at the same potential without changing how much current flows
through any of the resistors, that is, without changing the resistance
of the network. Wiring together the equal-potential nodes in the
cube gives you three sets of parallel resistors in series.
I believe you're correct, however, that the formuals for resistors
in parallel and in series are insufficient to solve all networks;
you just picked a weak example.
-- David A. Karr (ka...@bbn.com)
Neil Harris
In article <50ud0s$s...@crocus.csv.warwick.ac.uk>, ma...@csv.warwick.ac.uk
(Dr D F Holt) wrote:
>In article <50inec$9...@excalibur.net5c.io.org>,
> de...@io.org writes:
[snip]
>>Any resistor network can be solved by combining sets of parallel and series
>>resistors until you end up with a single resistance. E.g:
>
>I do not believe that, even for finite networks. Take the case, where
>you have 1 ohm resistors along each of the 12 edges of a cube, and you
>are asked to calculate the resistance from one corner to the diagonally
>opposite corner. How do you reduce that to sets of parallel and
>series resistors?
>
>Derek Holt.
Ok, here's how.
Take the two opposite diagonal points X and Y , and make current flow between
them.Observe that the current flow has 3-fold rotational symmetry about the
diagonal axis. Now notice that the three points nearest X are at equipotential,
so we can short them without changing the observed current flow. Likewise
the three points nearest Y. Now we have the equivalent circuit
X -- 3 1 ohm in parallel -- 6 1 ohm parallel -- 3 1 ohm parallel -- Y
which is equivalent to
X -- 1/3 ohm -- 1/6 ohm -- 1/3 ohm -- Y
or a single resistor of 5/6 ohm.
Your point is valid, but the example you gave is amenable to an elegant
(and classic, thanks to Martin Gardner) cheat. Nice puzzle, though!
-- Neil
Duncan Booth
In article <50ud0s$s...@crocus.csv.warwick.ac.uk>, ma...@csv.warwick.ac.uk (Dr D F Holt) wrote:
>In article <50inec$9...@excalibur.net5c.io.org>,
> de...@io.org writes:
>>resistors until you end up with a single resistance. E.g:
>
>I do not believe that, even for finite networks. Take the case, where
>you have 1 ohm resistors along each of the 12 edges of a cube, and you
>are asked to calculate the resistance from one corner to the diagonally
>opposite corner. How do you reduce that to sets of parallel and
>series resistors?
>
>Derek Holt.
>
>
than the rules for resistors in series and parallel and some common sense.
However you are right to dispute that this holds for every grid of resistors,
as it only holds true for these special cases because the resistors are all
the same value.
(By common sense I mean things like the fact that shorting together two points
with a common potential makes no difference to the problem, or that you can
replace an infinite subset of the resistors with a simpler infinite set of
resistors (albeit with an unknown resistance). FWIW, once you see the trick
involved in the infinite grid, you can work out the resistance in your head.
Nice puzzle, I enjoyed it.
--
Duncan Booth dun...@rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?
http://ourworld.compuserve.com/homepages/D_Booth
-----BEGIN GEEK CODE BLOCK-----
Version: 3.1
GCS !d-@ s:- a C++++$ US++++ P+ L !E W++ N++ !o K- w+++$>+++++ O$ M? V
PS+ PE- Y+ PGP t+ 5++ X+ !R tv+ b+++ DI !D G e+++ h--- r+++ y+
------END GEEK CODE BLOCK------
Rick Castello
In article <50mtor$a...@amdint2.amd.com>,
Aw, c'mon, you expect us to believe that?
EVERYONE knows that the world ends on a Thursday!
-Rick
--
Rick Castello Lightspeed Internet Technologies
ri...@lstech.com
Do you like it hot? Check out http://www.pepperplant.com
Need something to chew on? Check out http://www.goldeneagleherbchew.com
Joseph Riel
Joseph Riel (j...@sparc.SanDiegoCA.NCR.COM) wrote:
: Has anyone solved the equivalent problem in 3 and higher dimensions?
: The method used for the 2 dimensional case is not valid in higher
: dimensions (Polya's theorem applies) so there appears to be no easy solution.
After thinking about this some more I believe that I was mistaken, the
method used to solve the infinite mesh in 2 dimensions is applicable to
all higher dimensions. Polya's theorem (about a random walk being nonrecurring
for dimensions greater than 2) implies that the resistance to infinity is
finite for dimensions greater than 2, but this does not make the method
invalid---on the contrary, it would seem that the case that is questionable
is the 2 dimensional case (because it involves infinite voltages).
Any thoughts on this?
Joe Riel
Feico Nater
On 4-09-96 3:53p, in message <50inec$9...@excalibur.net5c.io.org>, Dean Edmonds
<de...@excalibur.net5c.io.org> wrote:
> Any resistor network can be solved by combining sets of parallel and series
> resistors until you end up with a single resistance. E.g:
Alas, it's not so simple. The network below is quite hard to solve.
(Use a non-proportional font to view this)
|----- R1 -------|-------- R2 ------------|
| | |
-----| R3 |-----
| | |
|----- R4 -------|--------- R5 -----------|
I havn't solved the original puzzle either, I only can say the answer must be
less than one and more than a quarter.
Feico
Joseph Riel
Dr D F Holt (ma...@csv.warwick.ac.uk) wrote:
: I do not believe that [any resistor network can be solved by combining
series and parallel resistors], even for finite networks. Take the case, where
: you have 1 ohm resistors along each of the 12 edges of a cube, and you
: are asked to calculate the resistance from one corner to the diagonally
: opposite corner. How do you reduce that to sets of parallel and
: series resistors?
You are right that, in general, a resistor network cannot be solved by
combining series and parallel resistors. In the case of a cube with identical
resistors, however, it is possible to do so by using the trick of shorting
nodes with the same voltage. This is possible due to the symmetry of the
structure and the resistances.
Joe Riel
Charles Carroll
den...@netcom.com (Dennis Yelle) writes:
> ---R1--o--R4---
> | | |
> A --| R3 |-- B
> | | |
> ---R2--o--R5---
> Ob Puzzle: Given the network of 5 resistors above,
> and R1 = 1 ohm, R2 = 2 ohms, R3 = 3 ohms, R4 = 4 ohms, and R5 = 5 ohms,
> what is the equivelent resistance from A to B ?
183/73
> What is the general formula for 5 resistors connected like that?
Using the shorthand that 135=R1*R3*R5 and so forth,
R= 123 + 124 + 125 + 135 + 145 + 234 + 245 + 345
---------------------------------------------
13 + 14 + 15 + 23 + 24 + 34 + 35 + 45
Chuck
--
Chuck Carroll http://www.dfw.net/~ccarroll/
Calvin & Hobbes
Nov. 18 1985 - Dec. 31 1995
Thanks for ten great years, Bill!
Xavier Llobet i Sales EPFL-CRPP 1015 Lausanne CH
In article <515l24$5...@nyx.cs.du.edu>, ccar...@nyx.cs.du.edu (Charles Carroll)
writes:
=den...@netcom.com (Dennis Yelle) writes:
=
=> ---R1--o--R4---
=> | | |
=> A --| R3 |-- B
=> | | |
=> ---R2--o--R5---
=
=> Ob Puzzle: Given the network of 5 resistors above,
=> and R1 = 1 ohm, R2 = 2 ohms, R3 = 3 ohms, R4 = 4 ohms, and R5 = 5 ohms,
=> what is the equivelent resistance from A to B ?
=
=183/73
Make the denominator 63
=
=> What is the general formula for 5 resistors connected like that?
=
=Using the shorthand that 135=R1*R3*R5 and so forth,
=
=R= 123 + 124 + 125 + 135 + 145 + 234 + 245 + 345
= ---------------------------------------------
= 13 + 14 + 15 + 23 + 24 + 34 + 35 + 45
Replace 45 in the denominator by 25
Duncan Booth
In article <N.091096....@hgl2-12.worldaccess.nl>, eff...@worldaccess.nl (Feico Nater)
<50hrsq$n...@dfw-ixnews4.ix.netcom.com> <50inec$9...@excalibur.net5c.io.org> wrote:
>I havn't solved the original puzzle either, I only can say the answer must be
>less than one and more than a quarter.
>
.. might be a spoiler ...
FWIW I make the answer to the original problem (sqrt(3)-1)/2.
(More of a teaser than a spoiler actually, if I got it right that is, since I
haven't bothered to tell you how I got there).
Feico Nater
On 9-09-96 9:33p, in message <neil-09099...@nharris.demon.co.uk>, Neil
Harris <ne...@nharris.demon.co.uk> wrote:
> >>Any resistor network can be solved by combining sets of parallel and series
> >>resistors until you end up with a single resistance. E.g:
> >
> >I do not believe that, even for finite networks. Take the case, where
> >you have 1 ohm resistors along each of the 12 edges of a cube, and you
> >are asked to calculate the resistance from one corner to the diagonally
> >opposite corner. How do you reduce that to sets of parallel and
> >series resistors?
>
> X -- 3 1 ohm in parallel -- 6 1 ohm parallel -- 3 1 ohm parallel -- Y
> which is equivalent to
> X -- 1/3 ohm -- 1/6 ohm -- 1/3 ohm -- Y
> or a single resistor of 5/6 ohm.
This one is easy because all resistors are 1 ohm, so you can easilt figure that
many nodes are equivalent, have the same potential and may be connected. But
that's cheating: Try a cube with different resistors.
I have tried a bridge of only five different resistors. Really hard!
Feico Nater
mark edward hardwidge
Sorry to bring up an old thread, but...how is the resister puzzle solved?
I looked and looked, and I never saw a solution.
I thought I understand how to connect the equal potential nodes, but
that seemed to give me an answer of 1/4 ohm (1||1||1||1), and I thought
I read that the correct answer was 1/2.
Could someone tell me what I'm doing wrong?
--
Mark E. Hardwidge
hard...@uiuc.edu