john@hp-pcd.UUCP (john) (10/15/85)
<<<<<< The answer to the cubic resistors problem is .833333 ohms. If you draw a 2-D schematic you will find it impossible to reduce using normal series/ parallel formula. It can be solved once you realize that it is very symmetrical. There are eight nodes in the circuit of which two are the end terminals. The remaining six can be broken into two groups of three each. These three nodes have exactly the same resistance to either end point and so will be at exactly the same voltage. If you short the three nodes in each group together then NO current will flow in the shorting bar because there is no voltage drop across it. Since there is no current in the shorts they will not affect the current distribution or total resistance of the circuit. The circuit with shorting bars in can easily be reduced to 1/3 +1/6 +1/3 which is .83333 ohm. John Eaton !hplabs!hp-pcd!john