What is the resistance between various pairs of vertices on a lattice of unit resistors in the shape of a 1.
physics/resistors Added material
Problem
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?
Show solution
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:
3(1+4)(2+5) + 12(4+5) + 45(1+2)
R = -------------------------------
3(1+2+4+5) + (1+2)(4+5)
123 + 124 + 125 + 135 + 145 + 234 + 245 + 345
= ---------------------------------------------
13 + 14 + 15 + 23 + 24 + 34 + 35 + 25
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.
Additional material
[ Editor: the infinite-lattice results above are from Noam Elkies
<elkies@math.harvard.edu> (September 1996). The following are from
the same thread and later. ]
dennis@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 equivalent resistance from A to B ?
183/73
-- Chuck Carroll <ccarroll@nyx.cs.du.edu>, who also gave the general
formula above
Make the denominator 63
[ Editor: and of the general formula: ]
Replace 45 in the denominator by 25
-- Xavier Llobet i Sales <llobet@crpp.uhd.epfl.ch>
[ Editor: the general formula above already has 25. With it the
resistance is 183/63 = 61/21 ohms. ]
[ Editor: two notes sent to the archive, November and December 1996. ]
I have derived a different method of determining certain inter-nodal
resistances in very large meshes of single valued resistors (R). It
involves cutting all but the terminal connections between three series
resistors and the mesh and replacing the resistance value of the three
resistor series with the original nodal value (3ER).
This can result in some interesting configurations. One is: one
resistor (R) paralleled by a string of three resistors (each ER).
Because the mesh is essentially infinite the resistance across the
single resistor must be equal to ER. Calculation shows that ER equals
2R/3.
If the one resistor (R) is paralleled by two strings of three
resistors (each ER), it is equivalent to the common puzzle where one is
asked for the inter-nodal resistance of any one resistor in a
two-dimensional mesh of one ohm resistors. Under the ER concept the
one ohm resistor is paralleled by two strings, each being two ohms,
giving the expected answer: one half ohm.
The concept can be extended to any four resistor square in the two
dimensional mesh with the question: what is the resistance between the
diagonal corners of the square? The apparent answer is: two parallel
strings of two ER resistors (each 2R/3) have a resistance of 2R/3. For
a twelve resistor cube in a three-dimensional mesh the diagonal
resistance becomes 5R/9, although the logic of determining the value of
ER is uncomfortable.
The reason I have emailed this to you is not to ask you to validate
the concept, but only to ask whether you have heard of this concept
before. The puzzle to find the resistance values is not new, but, if
it is not in the puzzle archives, I am considering posting it.
-- Ernest Steele <ernsteele@aol.com> (November 1996)
I have developed an approach to the analysis of infinite, or at least
sufficiently large, single valued resistor meshes. It is original as
far as I know and I am wondering whether you have knowledge of its
prior use.
Consider a two-dimension infinite lattice of one ohm resistors (R).
Designating one resistor as Ro, cut the lattice with two parallel lines
orthogonal to and just beyond the nodes of Ro. Revalue the resistors
parallel to the lines of cut to be the resistance that existed between
their nodes before the cuts. Designate this effective resistance as
RE. Now cut the ladder of resistors as shown.
....o-RE-o-RE-o-RE-o-RE-o-RE-o-RE-o-RE-o -RE-o-RE-....
l l l l l l l l l
R Ro R R R R R R R
l l l l l l l l l
....o-RE-o-RE-o-RE-o-RE-o-RE-o-RE-o-RE-o -RE-o-RE-....
Revalue the terminal one ohm resistor to be RE. All nodal resistances
remain the original infinite lattice values.
....o-RE-o-RE-o-RE-o-RE-o-RE-o-RE-a-RE-o
l l l l l l l l
R Ro R R R R R RE
l l l l l l l l
....o-RE-o-RE-o-RE-o-RE-o-RE-o-RE-b-RE-o
-a-RE-o
Evaluate the nodal conductance, A-B, of the end l l
loop. R RE
l l
Cab= 1/R+1/3RE -b-RE-o
....-a
The end of the lattice now has the form l
Cab
l
Since all nodal resistances are those ....-b
of the original infinite lattice, Cab
must equal 1/RE, whence RE= 2R/3.
Repeated reduction of the end loop with the same treatment at the
opposite end of the RE lattice leads to a single square loop of four
resistances of equal value RE. Since RE= 2R/3 and R=1 the resistance
between diagonal nodes of this square is 2/3 ohm.
Stopping end loop reduction one step early gives Ro paralleled by two
series strings of three RE resistances, making the nodal resistance of
Ro equal to 1/2 ohm. This is the consensual value for the nodal
resistance of any resistor in an infinite lattice of one ohm
resistors.
The ladder of resistances above could have been shown shorter, but is
shown longer than necessary to make the point that the original
lattice need not be infinite to give an acceptable accuracy. A
fourteen by fourteen lattice gives better than one percent accuracy
whether the peripheral resistances are zero or infinite. A minor point
is that the usual method of determining inter-nodal resistance by
current injection and extraction is not valid, if the lattice is not
infinite.
Questrel Guest Account, gave me the answer to the four resistor loop
as 2/pi versus my 2/3. I am surprised that pi could come out of a
lattice analysis. My concern is whether my method and my answer are
valid. I submitted the four resistor loop problem to rec-puzzles and
received no answers, so I have submitted it to Alan Gottlieb's Puzzle
Corner in the Tech Review.
I have applied the above approach to the resistance between the
interior diagonal nodes of a cube of twelve one ohm resistors in a
three dimensional mesh and got 5/21 ohm.
-- Ernest R Steele <ernsteele@aol.com>
[ Editor: the diagonal value derived above is 2R/pi = 0.6366R, not
2R/3 = 0.6667R. ]
ernsteele@aol.com (Ernsteele) wrote:
> You have an infinitely long ladder whose rungs are one ohm resistors,
> and whose rails are strings of one ohm resistors. What is the
> resistance between the ends of one of the rungs?
>
> If that is too easy, what is the resistance from the end of one rung
> to the opposite end of the next rung to it?
x
o---o
| |
1| |1
o---o
1
For a semi-infinite ladder of resistors, if its resistance is x, then
adding 3 1-ohm resistors as shown above leaves the total resistance
unchanged, and so
1/x = 1+1/(x+2).
The solution is x = sqrt(3)-1.
x C
o---o
| |
1| |1
o---o
A x B
The infinite ladder of resistors can be obtained from two
semi-infinite ladders joined with two 1-ohm resistors as shown above.
The total resistance R between points A and B is given by
1/R = 1/x+1/(x+2).
Using the above value for x we get R = sqrt(3)/3 = 0.5773502... ohm.
The total resistance between points A and C is given by
1/r = 1/(x+1)+1/(x+1) = 2/(x+1).
Using the above value for x we get r = sqrt(3)/2 = 0.8660254... ohm.
-- Ilan Mayer <ilan@isgtec.com> (February 1999)