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?])}

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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