Seven Resistor Puzzles
Chris Cole
Keith Lynch wrote:
>
> Resistors can be regarded as line segments each with some positive
> real number resistance. When resistors are linked in series (so that
> to get from point A to point B requires that you go through all of
> them) the total resistance between points A and B is the sum of the
> individual resistances. When resistors are linked in parallel (so
> that to get from point A to point B you can go through any one of
> them), the *reciprocal* of the total resistance is the sum of the
> *reciprocals* of the individual resistances. If there is no path at
> all between two points, the resistance is infinite. Between any point
> and itself, the resistance is zero.
>
> To keep it simple, all resistors in the following problems will have
> the same resistance, 1 ohm.
>
> Puzzle 1: What is the resistance between the opposite corners of a
> square, if there's a resistor from each corner to each other corner
> (i.e. six resistors, three of which meet at each corner).
>
> Puzzle 2: For each of the five regular solids (tetrahedron, cube,
> etc), what's the resistance between adjacent vertices, if each edge
> is a resistor?
>
> Puzzle 3: What is the resistance between adjacent points in an
> infinite two-dimensional grid of resistors (i.e. like a flyscreen
> made entirely of resistors, or a piece of standard graph paper).
>
> Puzzle 4: Like puzzle 3, only with an infinite *three*-
> dimensional grid (i.e. like a jungle-gym). (Extend this to even
> higher dimensions if you like.)
>
> Puzzle 5: For puzzles 3 and 4, what if the points weren't adjacent?
> What's the resistance between the point 0,0 and the point X,Y? Or
> between 0,0,0 and X,Y,Z?
>
> Puzzle 6: Is there any arrangement of resistors which can give an
> irrational total resistance?
>
> Puzzle 7: Is there any arrangement of resistors which can give a
> transcendental total resistance?
This question is in the rec.puzzles archive:
==> physics/resistors.p <==
What is the resistance between various pairs of vertices on a lattice
of unit resistors in the shape of a
1. Square with all vertices connected,
2. Platonic solid,
3. N dimensional Hypercube,
4. Infinite square lattice,
and
5. between two small terminals on a continuous sheet?
==> physics/resistors.s <==
The key idea is to observe that if you can show that two points in a
circuit must be at the same potential, then you can connect them, and
no current will flow through the connection and the overall properties
of the circuit remain unchanged.
1. Square with all vertices connected
The circuit between two diagonally opposite vertices of a square with
all vertices connected is composed of two circuits in parallel, one of
which is the diagonal itself (call this R0) and the other of which is:
---R1--o--R4---
| | |
A --| R3 |-- B
| | |
---R2--o--R5---
The general formula for 5 resistors connected like that (using the
shorthand that 135=R1*R3*R5 and so forth) is:
123 + 124 + 125 + 135 + 145 + 234 + 245 + 345
R= ---------------------------------------------
13 + 14 + 15 + 23 + 24 + 34 + 35 + 45
Then the resistance between the diagonal vertices is 1 / (R + R0). If
all the resistances are 1 ohm, the answer is 1/2 ohm.
2. Platonic Solids
For the cube, there are three resistors leaving the two "connection
corners". Since the cube is completely symmetrical with respect to the
three resistors, the far sides of the resistors may be connected
together. And so we end up with:
|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
| | |---WWWWWW---| | |
*--+---WWWWWW---+-+---WWWWWW---+-+---WWWWWW---+---*
| | |---WWWWWW---| | |
|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
|---WWWWWW---|
This circuit has resistance 5/6 times the resistance of one resistor.
Same idea for 8, 12 and 20, since you use the symmetry to identify
equi-potential points. The tetrahedron is a hair more subtle:
*---|---WWWWWW---|---*
|\ /|
W W W W
W W W W
W W W W
| \ / |
\ || |
\ | /
\ W /
\ W / <-------
\ W /
\|/
+
By symmetry, the endpoints of the marked resistor are equi-potential.
Hence they can be connected together, and so it becomes a simple:
*---+---WWWWW---+----*
| |
+-WWW WWW-+
| |-| |
|-WWW WWW-|
3. Hypercube
Think of injecting a constant current I into the start vertex. It
splits (by symmetry) into n equal currents in the n arms; the current
of I/n then splits into I/n(n-1), which then splits into
I/[n(n-1)(n-1)] and so on till the halfway point, when these currents
start adding up. What is the voltage difference between the antipodal
points? V = I x R; add up the voltages along any of the paths:
n even:
V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1)^[(n-2)/2])}
n odd:
V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1)^[(n-3)/2])}
+ I/(n(n-1)^[(n-1)/2])
And R = V/I i.e. replace the Is in the above expression by 1s.
For the 3-cube: R = 2{1/3} + 1/(3x2) = 5/6 ohm
For the 4-cube: R = 2{1/4 + 1/(4x3)} = 2/3 ohm
This formula yields the resistance from root to root of two (n-1)-ary
trees of height n/2 with their end nodes identified (when n is even;
something similar when n is odd). Coincidentally, the 4-cube is such
an animal and thus the answer 2/3 ohms is correct in that case.
However, it does not provide the solution for n >= 5, as the hypercube
does not have quite as many edges as were counted in the formula
above.
4. The Infinite Plane
For an infinite lattice: First inject a constant current I at a point;
figure out the current flows (with heavy use of symmetry). Remove that
current. Draw out a current I from the other point of interest (or
inject a negative current) and figure out the flows (identical to
earlier case, but displaced and in the other direction). By the
principle of superposition, if you inject a current I into point a and
take out a current I at point b at the same time, the currents in the
paths are simply the sum of the currents obtained in the earlier two
simpler cases. As in the n-cube, find the voltage between the points of
interest, divide by I.
As an illustration, in the adjacent points case: we have a current of
I/4 in each of the four resistors:
^ |
| v
<--o--> -->o<--
| ^
v |
(inject) (take out)
And adding the currents, we have I/2 in the resistor connecting the two
points. Therefore V=R x I/2 and effective resistance between the
points = R/2 ohm.
The equivalent resistance between two diagonally adjacent vertices is
2R/pi. 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.
. . . .
| | | |
R R R R
| | | |
-R-------|------R--------X-------R-------A-------R-------B-----R-
| | | |
R R R R
| | | |
-R-------|------R--------|-------R-------Y-------R-------C-----R-
| | | |
For instance, if Z is the vertex labeled A in the above diagram
we find that that the X-B resistance is (2-4/pi)R (the resistance
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.
5. Continuous sheet
The resistence is (rho/dz)log(L/r)/pi where rho is the resistivity,
dz is the sheet thickness, L is the separation, r is the terminal
radius.
cf. "Random Walks and Electric Networks", by Doyle and Snell, published
by the Mathematical Association of America.
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Keith Lynch
Noam Elkies
In article <5pcjt6$8...@clarknet.clark.net>, Keith Lynch <k...@clark.net> wrote:
>Resistors can be regarded as line segments each with some positive
>real number resistance. When resistors are linked in series (so that
>to get from point A to point B requires that you go through all of
>them) the total resistance between points A and B is the sum of the
>individual resistances. When resistors are linked in parallel (so
>that to get from point A to point B you can go through any one of
>them), the *reciprocal* of the total resistance is the sum of the
>*reciprocals* of the individual resistances. If there is no path at
>all between two points, the resistance is infinite. Between any point
>and itself, the resistance is zero.
(Note that this is not sufficient to answer any of the puzzles -- for
starters one needs Kirchhoff's laws, though even that isn't really
enough if you want to make sense of the infinite networks of resistors).
>[...]
>Puzzle 3: What is the resistance between adjacent points in an
>infinite two-dimensional grid of resistors (i.e. like a flyscreen
>made entirely of resistors, or a piece of standard graph paper).
>
>Puzzle 4: Like puzzle 3, only with an infinite *three*-
>dimensional grid (i.e. like a jungle-gym). (Extend this to even
>higher dimensions if you like.)
>
>Puzzle 5: For puzzles 3 and 4, what if the points weren't adjacent?
>What's the resistance between the point 0,0 and the point X,Y? Or
>between 0,0,0 and X,Y,Z?
>
>Puzzle 6: Is there any arrangement of resistors which can give an
>irrational total resistance?
>
>Puzzle 7: Is there any arrangement of resistors which can give a
>transcendental total resistance?
I see that Chris Cole already posted the appropriate segment of
the FAQ which answers puzzles 1-3 and the 2-dim. part of puzzle 5.
Since the resistance between diagonally adjacent points in a 2-dim
grid already involves pi, this answers 6 and 7 as well. Two years
ago Chris Thompson (ce...@cus.cam.ac.uk) posted to sci.math.research
some information on the higher-dimensional case, citing references:
[1] E.W. Montroll: Random Walks in Multidimensional Spaces,
Especially in Periodic Lattices, J. Soc. Indust. Appl. Math.
4 (1956) 241-260;
[2] G.N. Watson: Three Triple Integrals, Quart. J. Math.
Oxford Ser. 19 (1939) 266-276.
Already for the 3-dim. grid, closed forms are only barely available
and involve complete elliptic integrals! At any rate the equivalent
resistance can be expressed as definite integrals, and so evaluated
numerically (with some care needed since the integrals are improper).
Puzzle 8: Show that *any* positive real number can be obtained
as the total resistance of an arrangement of unit resistors, and
that this arrangement may be realized in the plane (without
crossings).
--Noam D. Elkies (elk...@math.harvard:edu)
Dept. of Mathematics, Harvard University
Randy Gardner
> Puzzle 1: What is the resistance between the opposite corners of a
> square, if there's a resistor from each corner to each other corner
> (i.e. six resistors, three of which meet at each corner).
I belive you mean a cube.
A square is 2d, a cube is 3d, so it may be misleading.
--
--ran...@slip.net
--Randy Gardner
Randy Gardner
> > Puzzle 1: What is the resistance between the opposite corners of a
> > square, if there's a resistor from each corner to each other corner
> > (i.e. six resistors, three of which meet at each corner).
> I belive you mean a cube.
Sorry, i seem to have read it wrong, it's a square. :-(
--
--ran...@slip.net
--Randy Gardner
Patrick Hamlyn
Chris Cole <ch...@questrel.com> parrotted:
]Keith Lynch wrote:
]>
[cut]
]>
]> To keep it simple, all resistors in the following problems will have
]> the same resistance, 1 ohm.
]>
[hack]
]>
]> Puzzle 2: For each of the five regular solids (tetrahedron, cube,
]> etc), what's the resistance between adjacent vertices, if each edge
]> is a resistor?
]>
[chop]
]This question is in the rec.puzzles archive:
]
[shred]
]
]2. Platonic Solids
]
]For the cube, there are three resistors leaving the two "connection
]corners". Since the cube is completely symmetrical with respect to the
]three resistors, the far sides of the resistors may be connected
]together. And so we end up with:
]
] |---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
] | | |---WWWWWW---| | |
] *--+---WWWWWW---+-+---WWWWWW---+-+---WWWWWW---+---*
] | | |---WWWWWW---| | |
] |---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
] |---WWWWWW---|
]
]This circuit has resistance 5/6 times the resistance of one resistor.
]
[maim]
This doesn't answer the asked question, which was *adjacent* vertices, not
diagonally opposite vertices.
For adjacent vertices on a cube I get 3/5.
For vertices 'diagonally opposite on the same face', I get 3/4
Sorry about the funny quote character, despite liberal chopping I still need to
con the usenet server into thinking I have more original than quoted text
--
Patrick Hamlyn, Multiprogramming Pty Ltd
Posting from Perth, Windsurfing capital of Australia
Ilias Kastanas
In article <5pdrfd$u20$1...@news.fas.harvard.edu>,
Noam Elkies <elk...@ramanujan.harvard.edu> wrote:
...
>Puzzle 8: Show that *any* positive real number can be obtained
>as the total resistance of an arrangement of unit resistors, and
>that this arrangement may be realized in the plane (without
>crossings).
Represent x as a continued fraction:
x = a_0 + 1/(a_1 + 1/(a_2 + ... ...)))
(a_0 >= 0 and the other a_i > 0, if x is irrational; for rational x the
fraction terminates at some a_k).
So we form a series arrangement of: a_0 unit resistors and the parallel
arrangement of: a_1 unit resistors and the series arrangement of: a_2 unit
resistors and the parallel... etc.
If the basic unit resistor is a piece of wire of length 1, we can
make our planar arrangement's endpoints A and B be arbitrarily close, or
have any distance < a_0 + 1. We can achieve any distance, however large,
by implementing resistance 1 as a series of a pair of two unit resistors in
parallel, and iterating.
We can realize resistance x, then, between any two points. Can we
also short-circuit any two points? Unlike the positive rationals, 0 cannot
be implemented with finitely many resistors; but we can certainly place
infinitely many resistance 1 paths between A and B.
Ilias
Alexander
Alexander.
Seth Breidbart
>Puzzle 6: Is there any arrangement of resistors which can give an
>irrational total resistance?
>
>Puzzle 7: Is there any arrangement of resistors which can give a
>transcendental total resistance?
For infinite arrangements, yes. For finite, obviously not.
Seth
Joseph W. DeVincentis
In article <5pdrfd$u20$1...@news.fas.harvard.edu>,
Noam Elkies <elk...@ramanujan.harvard.edu> wrote:
>Puzzle 8: Show that *any* positive real number can be obtained
>as the total resistance of an arrangement of unit resistors, and
>that this arrangement may be realized in the plane (without
>crossings).
(This works equally well in binary, but people are more familiar with
decimal, so that's how I've written this.)
Let R be our number.
Write R as a0 + a1/10 + a2/100 + a3/1000 + ... where all the a's are
integers 0 through 9.
First, put a0 unit resistors in series.
Add, in series with this, a group of 10 paths in parallel, where each
path consists of a1 unit resistors in series.
Add, in series with this, a group of 100 paths in parallel, where each
path consists of a2 unit resistors in series.
Etc.
/dev/joe
Keith Lynch
In article <5pdrfd$u20$1...@news.fas.harvard.edu>,
Noam Elkies <elk...@ramanujan.harvard.edu> wrote:
> Puzzle 8: Show that *any* positive real number can be obtained
> as the total resistance of an arrangement of unit resistors, and
> that this arrangement may be realized in the plane (without
> crossings).
In article <5pgg52$bg4$1...@geraldo.cc.utexas.edu>,
Joseph W. DeVincentis <dev...@wilma.che.utexas.edu> wrote:
> Write R as a0 + a1/10 + a2/100 + a3/1000 + ... where all the a's are
> integers 0 through 9.
> First, put a0 unit resistors in series.
> Add, in series with this, a group of 10 paths in parallel, where each
> path consists of a1 unit resistors in series.
> Add, in series with this, a group of 100 paths in parallel, where each
> path consists of a2 unit resistors in series.
> Etc.
Sorry, this won't work except for numbers whose decimal expansion
terminates. Otherwise, what would be connected to the second node?
For instance, to get a resistance of pi, how many resistors in
parallel would be attached to the second node? It would have to be
10^N where pi terminates after N decimal digits. I don't think so.
Here's my solution: Express the reciprocal of R in binary. (Unless
it's zero, in which case you're already done.) Attach a long busbar
to the first node, and another long busbar to point B. Between these
busbars place M resistors, where M is the integer part of 1/R. Then,
for each binary digit of 1/R after the radix, place 2, 4, 8, 16, etc,
resistors in series between the busbars, depending on which digit it
is, if the digit is 1. Do nothing if the digit is 0.
So to get 3 ohms, we would have 4 in series, in parallel with 16 in
series, in parallel with 64 in series, etc. (Of course there's a much
easier way to get 3 ohms, but it doesn't generalize.) To get e ohms,
since 1/e = 0.3678794... = 0.01011110001011... one would have 4 in
series, in parallel with 16 in series, in parallel with 32 in series,
in parallel with 64 in series, in parallel with 128 in series, in
parallel with 2048 in series, etc.
I also liked Ilias Kastanas' solution, which involved continued
fractions. That would allow a much more regular arrangement of
resistors than my solution to get a resistance of e, or of sqrt(2),
among many other numbers.
(Posted, and mailed to everyone mentioned.)
Monwhea Jeng
elk...@ramanujan.harvard.edu (Noam Elkies) writes:
>In article <5pcjt6$8...@clarknet.clark.net>, Keith Lynch <k...@clark.net> wrote:
>>Puzzle 7: Is there any arrangement of resistors which can give a
>>transcendental total resistance?
>Puzzle 8: Show that *any* positive real number can be obtained
>as the total resistance of an arrangement of unit resistors, and
>that this arrangement may be realized in the plane (without
>crossings).
SPOILER SPACE :
Hey, neat puzzles.
Given unit resistors, I can make circuits with resistance
k and 1/k for all integer k>1, by hooking the resistors up in series
or parallel, respectively. So for the rest of the problem assume
that I have resistors of this type as well. A resistance r>0
can be written in continued fraction form :
1 1 1 1
r= k0 + --- + --- + --- + --- + . . .
k1+ k2+ k3+ k4+
where the ki's are all integers, and it is understood that each
fraction is nested in the one before it. i.e.
r=k0+1/(k1+(1/(k2+1/(k3+1/(k4+. . .)))))
To see that we can do this, write r=k0+r0, with k0 the integer part of
r, and r0 the fractional part. r0=1/(1/r0), and (1/r0)>1,
so we can write (1/r0)=k1+r1, where k1>1 and is an integer, and
0<=r1<1. If r1=0 we stop writing the fraction. Otherwirse
0<r1<1, and we write r1=1/(1/r1)=1/(k2+r2). Next r2=1/(k3+r3),
r3=1/(k4+r4), . . . We keep on going, either eventually reaching
rn=0, and stopping, or going on forever. Convergence of this
process is left as an exercise to the reader.
We can make r by putting k0 and
Now we see that we can make r by putting 1/k1 and 1/r1 in parallel.
We can make 1/r1 by putting k2 and r2 in series.
We can make r2 by putting 1/k3 and 1/r3 in parallel.
We can make 1/r3 by putting k4 and r4 in series.
We can make r4 by putting 1/k4 and 1/r4 in parallel.
And so on, forever. The arrangement in the plane is
.---k0---.---k2---.---k4---.---k6---.---k8--- . . . .
| | | |
1/k1 1/k3 1/k5 1/k7
| | | |
.--------.--------.--------.--------.--------
where 1/k1 is short for k1 resistors in parallel.
I'm still working on the infinite plane problem. It seems
hard -- am I going to need to pull out the heavy mathematics
artillery or just be clever?
Momo
Monwhea Jeng
dev...@wilma.che.utexas.edu (Joseph W. DeVincentis) writes:
>In article <5pdrfd$u20$1...@news.fas.harvard.edu>,
>Noam Elkies <elk...@ramanujan.harvard.edu> wrote:
>>Puzzle 8: Show that *any* positive real number can be obtained
>>as the total resistance of an arrangement of unit resistors, and
>>that this arrangement may be realized in the plane (without
>>crossings).
>(This works equally well in binary, but people are more familiar with
>decimal, so that's how I've written this.)
>Let R be our number.
>Write R as a0 + a1/10 + a2/100 + a3/1000 + ... where all the a's are
>integers 0 through 9.
>First, put a0 unit resistors in series.
>Add, in series with this, a group of 10 paths in parallel, where each
>path consists of a1 unit resistors in series.
>Add, in series with this, a group of 100 paths in parallel, where each
>path consists of a2 unit resistors in series.
>Etc.
My feeling is that this is not a valid answer (although it was my first
solution). In the circuit design, each resistor and each node should
be at a specific place on the plane, and the in and out lines
should actually be connected to something. In your example one end
gets connected to the first of the a0 resistors, but where does
the other end go?
Momo
David A. Karr
Joseph W. DeVincentis (dev...@wilma.che.utexas.edu) wrote:
>Noam Elkies <elk...@ramanujan.harvard.edu> wrote:
>>Puzzle 8: Show that *any* positive real number can be obtained
>
>First, put a0 unit resistors in series.
>Add, in series with this, a group of 10 paths in parallel, where each
>path consists of a1 unit resistors in series.
>Add, in series with this, a group of 100 paths in parallel, where each
>path consists of a2 unit resistors in series.
>Etc.
I see an obvious point on this arrangement where I should place one
probe of my ohmmeter. Where should I place the other probe?
For the resistance r, I would write:
1/r = g = a0 + a1/10 + a2/100 + a3/1000 + ...
Now place a0 unit resistors in _parallel_ with each other and in
parallel with a1 strings of 10 unit resistors in series,
a2 strings of 100 unit resistors in series, etc.
This results in an infinite number of wires coming into each of
my probe points, but if I am allowed to use some zero-resistance
segments of wire, I can turn this mess into an infinite "ladder"
with finite fanout at each point.
If I am required to use only resistive wires and finite fanout at
any point, I prefer a continued-fraction solution.
David A. Karr
D.H. Kelly
If each side of square and the diagonals are 1 ohm. then the total
resistance is 1/2 ohm.
(inject a unit current into corner- all others "grounded"- determine
distribution of currents. Now draw out unit current from opposite corner
with all the rest "grounded" Superimpose. Result is current of 4/3 amps
into corner and 2/3 in diagonal 1 ohm element. (i.e. half the current in 1
ohm resistor so net res is 1/2 ohm.
Don
Ian Gay
In article <momo.86...@sbphy.physics.ucsb.edu>,
mo...@sbphy.physics.ucsb.edu (Monwhea Jeng) wrote:
>
>>In article <5pcjt6$8...@clarknet.clark.net>, Keith Lynch <k...@clark.net> wrote:
>>>Puzzle 7: Is there any arrangement of resistors which can give a
>>>transcendental total resistance?
No finite arrangement. Any finite arrangement can be solved by Kirchoff's laws,
which yield algebraic equations.
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Monwhea Jeng
elk...@ramanujan.harvard.edu (Noam Elkies) writes:
>>Puzzle 7: Is there any arrangement of resistors which can give a
>>transcendental total resistance?
>Puzzle 8: Show that *any* positive real number can be obtained
>as the total resistance of an arrangement of unit resistors, and
>that this arrangement may be realized in the plane (without
>crossings).
Puzzle 9: Show that any positive real number can be obtained as the
total resistance of an arrangement of rigid resistors of unit
resistance AND unit length, by an arrangement in the plane
without crossings.
Resistors are to be considered perfect 1D segments of unit length.
The rule against crossing means that two resistors may intersect
only at their endpoints, which are 0-dimensional points.
ObPuzzle : What is Puzzle 10?
Momo
Earl Kiosterud
A dry cell is connected to the two terminals of a box. Current I flows. When
an additional dry cell is connected in series, same current, I. When the two
cells and the box are connected in parallel, current 2I flows.
What's in the box? (no fancy active circuits -- not a trick question).
Earl Kiosterud
ea...@livenet.net
Virginia Beach, VA USA
Quentin Grady
On Sun, 06 Jul 97 23:40:52 GMT, ea...@livenet.net (Earl Kiosterud)
wrote:
>A dry cell is connected to the two terminals of a box. Current I flows. When
>an additional dry cell is connected in series, same current, I. When the two
>cells and the box are connected in parallel, current 2I flows.
>
>What's in the box? (no fancy active circuits -- not a trick question).
>
G'day G'day Earl,
Try a wire.
Dry cells have appreciable internal resistance, r.
They generate a short circuit current E/r
In series the current is 2E/2r = I
In parallel the current is E/0.5r = 2I
Keep the fun questions coming,
Quentin Grady
On 7 Jul 1997 02:34:11 -0600, Charles Carroll <ccar...@nyx.com>
wrote:
>ea...@livenet.net (Earl Kiosterud) writes:
>
>>A dry cell is connected to the two terminals of a box. Current I flows. When
>>an additional dry cell is connected in series, same current, I. When the two
>>cells and the box are connected in parallel, current 2I flows.
>
>>What's in the box? (no fancy active circuits -- not a trick question).
>
>SPOILER:
>
>An insulator. I=0.
>Chuck Carroll http://www.dfw.net/~ccarroll/
G'day G'day Chuck,
Loved that answer. The external resistance, R, is either 0 or oo
Now you have me wondering. Is there any in between answer?
Monwhea Jeng
mo...@dolphin.physics.ucsb.edu (Monwhea Jeng) writes:
>Puzzle 9: Show that any positive real number can be obtained as the
> total resistance of an arrangement of rigid resistors of unit
> resistance AND unit length, by an arrangement in the plane
> without crossings.
>Resistors are to be considered perfect 1D segments of unit length.
>The rule against crossing means that two resistors may intersect
>only at their endpoints, which are 0-dimensional points.
>ObPuzzle : What is Puzzle 10?
>Momo
Oops. I just looked at this again, and I see I wasn't very clear.
I meant an arrangement that uses only rigid resistors of unit
legnth and unit resistance. So you can't use any of that
resistanceless wire that you always use in pictures.
Also, I would say that two resistors intersect if they occupy
the exact same space in the plae -- i.e. the intersect everywhere.
Momo
Keith Lynch
In article <5ppbcd$pvf$2...@paperboy.livenet.net>,
Earl Kiosterud <ea...@livenet.net> wrote:
> A dry cell is connected to the two terminals of a box. Current I
> flows. When an additional dry cell is connected in series, same
> current, I. When the two cells and the box are connected in
> parallel, current 2I flows.
Good one. There are two answers. One is a short answer. The other
is the exact opposite. But both are very simple.
(Posted and mailed.)
David Kastrup
k...@clark.net (Keith Lynch) writes:
> In article <5ppbcd$pvf$2...@paperboy.livenet.net>,
> Earl Kiosterud <ea...@livenet.net> wrote:
> > A dry cell is connected to the two terminals of a box. Current I
> > flows. When an additional dry cell is connected in series, same
> > current, I. When the two cells and the box are connected in
> > parallel, current 2I flows.
>
> Good one. There are two answers. One is a short answer. The other
> is the exact opposite. But both are very simple.
One is an open box (alternatively, empty batteries) and current 0
(stamping, booing, throwing of vegetables).
Another is based on remembering that we're talking dry cells here, not
ideal voltage sources, and thus have a finite source resistance. One
obvious solution (not even needing a linear resistance, but just a
well-defined short-circuit current) is a short-circuited box. The
latter solution sounds somewhat more like an "appropriate" solution of
the puzzle.
BTW, one should state that the current is always to be measured as the
current through the box.
--
David Kastrup Phone: +49-234-700-5570
Email: d...@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut für Neuroinformatik, Universitätsstr. 150, 44780 Bochum, Germany
Wolfgang Flamme
>> Puzzle 1: What is the resistance between the opposite corners of a
>> square, if there's a resistor from each corner to each other corner
>> (i.e. six resistors, three of which meet at each corner).
>
>I belive you mean a cube.
>A square is 2d, a cube is 3d, so it may be misleading.
No, the square is right. 4 edges + 2 dioganal = 6
This will be R||2R||2R = R/2
One diagonal R contributes nothing, so the arangement can even be
short-circuited across the opposite diagonal or be left open.
Wolfgang Flamme
wfl...@mainz-online.de
Wei-Hwa Huang
David Kastrup <d...@mailhost.neuroinformatik.ruhr-uni-bochum.de> writes:
>k...@clark.net (Keith Lynch) writes:
>> In article <5ppbcd$pvf$2...@paperboy.livenet.net>,
>> Earl Kiosterud <ea...@livenet.net> wrote:
>> > A dry cell is connected to the two terminals of a box. Current I
>> > flows. When an additional dry cell is connected in series, same
>> > current, I. When the two cells and the box are connected in
>> > parallel, current 2I flows.
>> Good one. There are two answers. One is a short answer. The other
>> is the exact opposite. But both are very simple.
>One is an open box (alternatively, empty batteries) and current 0
>(stamping, booing, throwing of vegetables).
>Another is based on remembering that we're talking dry cells here, not
>ideal voltage sources, and thus have a finite source resistance. One
>obvious solution (not even needing a linear resistance, but just a
>well-defined short-circuit current) is a short-circuited box. The
>latter solution sounds somewhat more like an "appropriate" solution of
>the puzzle.
(stamping, booing, melting of boxes, chance of electrocution).
Another (safer) answer notes that it was not specified that both
terminals of the CELLS were connected to the box, just that the
two terminals of the BOX were connected to the cells. So,
just connect the chasses of the cells to the box, and you have
the desired result.
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
"Oh, that was when the sand was so hot we were each hopping on one foot."
David Kastrup
whu...@ugcs.caltech.edu (Wei-Hwa Huang) writes:
> David Kastrup <d...@mailhost.neuroinformatik.ruhr-uni-bochum.de> writes:
> >> In article <5ppbcd$pvf$2...@paperboy.livenet.net>,
> >> Earl Kiosterud <ea...@livenet.net> wrote:
> >> > A dry cell is connected to the two terminals of a box. Current I
> >> > flows. When an additional dry cell is connected in series, same
> >> > current, I. When the two cells and the box are connected in
> >> > parallel, current 2I flows.
>
> >One is an open box (alternatively, empty batteries) and current 0
> >(stamping, booing, throwing of vegetables).
>
> >Another is based on remembering that we're talking dry cells here, not
> >ideal voltage sources, and thus have a finite source resistance. One
> >obvious solution (not even needing a linear resistance, but just a
> >well-defined short-circuit current) is a short-circuited box. The
> >latter solution sounds somewhat more like an "appropriate" solution of
> >the puzzle.
>
> (stamping, booing, melting of boxes, chance of electrocution).
Are we talking about the same devices here? Dry cells? In that case
I recommend you try the experience.
Now lead accumulators (and some other fast-charge accumulators) are a
different matter, but dry cells? And electrocution? You can take the
average dry cell into your mouth without anything of the sort
happening. They are just 1.5V. Small chance of electrocution, unless
you happen to "shorten" them on a coil of a transformator.
D.H. Kelly
for a start - puzzle 3 resistance = 1/2 ohm, puzzle 4- R=1/3 ohm puzzle
5-much harder to solve- no standard answer. Puzz. 6,7 No
puzz 1,2 straightforward but taking a bit longer than the 10 sec required
for 3,4. Try current injection and superposition on these(inject at one
node- all others grounded then repeat with neg injection at other node rest
grounded- superimpose currents)
Don kelly.