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Twelve identical wires of resistance 6 ohm each are arranged to form the edges of cube the effective resistance bw the opp corners of cube is?
The effective resistance between opposite corners of a cube comprised of twelve 6 ohm resistors, one at each edge, is 5 ohms. There are several ways to solve this. One approach is to build a system of 12 equations in 12 unknowns, and solve them. Another approach is this... Consider that there are three resistors leaving the input node, and there are three resistors entering the output node. In between those three resistors, there are six resistors in a criss-cross matrix. (Draw it out, flattened, to see this.)Inspecting the six resistors in the center, you note that they are completely symmetrical. Since they are symmetrical, you can conclude that the voltage at the junction between the three input resistors and the six others is the same voltage. The same goes for the three output resistors. Said another way, the voltage across the three input resistors and the three output resistors is the same. Given two or more nodes in a circuit having the same voltage, you can draw a wire connecting them, i.e. a resistor of zero ohms. This does not change the characteristics of the circuit in any way, because zero voltage across any resistance is still zero amperes. Now that you have made these connections, look at the circuit. It has simplified to three parallel resistors, in series with six parallel resistors, in series with three parallel resistors. Three 6 ohm resistors in parallel is 2 ohms. Six 6 ohm resistors in parallel is 1 ohm. Three more 6 ohm resistors in parallel is 2 ohms. The total resistance is 2 + 1 + 2 ohms, or 5 ohms.
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