A wire structure in the form of a cube has a resistance of one ohm in each edge. An electrical circuit is formed by connecting a one volt battery to a pair of diagonally opposed vertices of the cube. What is the current, in amperes, in the wire connecting the battery to the cube structure?
Let's say that the negative pole is connected to vertex A, and the positive pole is connected to vertex H. Also, call the vertices connected via one edge to vertex A, vertices B, C and D, while those connected via one edge with vertex H are vertices E, F and G.
By symmetry, vertices B, C and D are all at the same potential, and therefore could be connected directly or considered as one point. The same is true of vertices E, F and G.
Therefore, A is connected to common point BCD by three parallel 1-ohm resistances, for a resistance of 1/3 ohm.
Common point BCD is connected to common point DEF by six parallel 1-ohm resistances for a resistance of 1/6 ohm.
Common point DEF is connected to vertex H by three parallel 1-ohm resistances for a resistance of 1/3 ohm.
These three resistances are set in series, for a total of 5/6 ohm. Since the current is equal to the voltage divided by the resistance, it is 1/(5/6) = 6/5 = 1.2 amperes.
Posted by Charlie
on 2005-05-17 15:27:58