==> physics/resistors.s <== 1. Cube 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. In particular, 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. 2. Platonic Solids 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: (n-2)/2 V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )} n odd: (n-3)/2 V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )} (n-1)/2 + I/(n(n-1) ) 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 and voila`! 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=(1 ohm) x I/2 and effective resistance between the points = 1/2 ohm. We do not derive it, but the equivalent resistance between two nodes k diagonal units apart is (2/pi)(1+1/3+1/5+...+1/(2k-1)); that, plus symmetry and the known equivalent resistance between two adjacent nodes, is sufficient to derive all equivalent resistances in the lattice. 5. Continuous sheet I think the answer 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.