Q.256. To find the equivalent resistance of a circuit of resistors.

Question 256.
Consider an electrical circuit consisting of a cube of 12 identical resistors such that each edge of the cube is a 1 ohm resistor, and each group of three resistors meeting at a vertex are soldered together. Calculate the resistances between nearest neighbor, second-nearest neighbor, and third-nearest neighbor pairs of vertices.

Answer 256.
( I ) Suppose an equivalent resistance is required between adjacent points A and B. (This is the case of the first nearest neighbours) Refer to the figure in the link as under.

Starting from A, the points D and C' are equipotential as they have equivalent paths to B and starting from B, the points D' and C are equipotential as they have equivalent paths to A.
So, sandwiching points D and C' and also D' and C,

(1) A is connected to C' and D by two parallel resistors whose equivalent is 1/2 Ω

(2) At the other end, B is connected to C and D' by two parallel resistors whose equivalent is 1/2 Ω

(3) Between the above two and in series with them is a circuit of resistors as shown in the figure whose equivalent resistance is equal to three parallel resistors, 1 Ω, 2Ω and 1Ω = 2/5 Ω

(4) The above three in series add upto 1/2 + 2/5 + 1/2 = 7/5 Ω

(5) This 7/5 Ω and 1 Ω across A and B are in parallel which equals the final equivalent resistance between
A and B = (7/5) * 1 / (7/5 + 1) = 7/12 Ω.

( II ) Suppose an equivalent resistance is required between points A and D' diametrically opposite on a face of the cube. (This is the case of the second nearest neighbours.)
Refer to the figure in the link as under.

      
(1) C' and B are equipotential points and so are B' and C. Sandwiching them, the circuit will be as shown in figure 2.

(2) Working out equivalent of 5 parallel combinations of resistors, we get the circuit as in figure 3.

(3) In fig. 3, AD and DB' are in series and so are D'A' and A'B' which results in fig. 4.

(4) Fig. 4 is a balanced Wheatstone Bridge. Hence no current flows through the middle 1/2 Ω resistor.

(5) The final circuit is of 1 Ω and 3 Ω resistors in parallel whose equivalent is 3/4 Ω.

(III) Suppose an equivalent resistance is required between diametrically opposite points A and A'.
(This is the case of third nearest neighbour)

AB, AD and AC' are resistors having common point A => B, D and C' are equipotential
A'B', A'D' and A'C are resistors having common point A' => B' D' and C are equipotential
B, D and C' are connected to B', D' and C using six resistors.

Thus, the circuit from A to A' is equivalent to
3 resistors (AB, AD and AC') in parallel connected in series to
6 resistors (BC, BD', DC, DB', C'B', C'D') in parallel connected in series to
3 resistors (B'A', D'A', CA') in parallel
=> equivalent resistance between A and A' is
(1/3) + (1/6) + (1/3) ohm
= 5/6 ohm.

Link to YA!