高級程度物理(電學)

　　　　Ｄ　　　　　Ｃ
　　　　　┌───┐
　　　　╱│　　╱│
　　　╱　│　╱　│
　Ａ┌──┼┐Ｂ　│
　　│　Ｅ└┼──┛Ｆ
　　│　╱　│　╱
　　│╱　　│╱
　　└───┛
　Ｈ　　　　　Ｇ
For the overall resistance we are interested in this cube is points A & F,
since resistance of AD, AB, AH are the same,
and resistance of BC, BG, DC, DE, HE, HG are the same,
and resistance of CF, EF, GF are the same,
we may assume points B, D, H are same potential and C, E, G are same potential.
So this cube can be divided into 3 sections,
1. AD, AB, AH
2. BC, BG, DC, DE, HE, HG
3. CF, EF, GF
the resistance at each of the 12 wires is 1 ohm
the overall resistance for the first section is
= 1 / (1/1 * 3)
= 1/3
the overall resistance for the second section is
= 1 / (1/1 * 6)
= 1/6
the overall resistance for the third section is
= 1 / (1/1 * 3)
= 1/3
As a result, the overall resistance is 1/3 + 1/6 + 1/3
= 5/6 ohm
PS. This 'cube' cannot be divided into 6 paths of 3 ohm or some of the paths will be counted twice, this circuit may be view as 6 paths of 5 ohm or (2 + 1 + 2)ohm [AD can be seen as two 2 ohm resistors in parallel.]
R= 1 / ((1/5) * 6) = 5/6 ohm
So it will never add mathematically like 1+1+1 = 3.

如下:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/Crazyelectric1.jpg