How do you calculate the equivalent resistance of a circuit of the shape of a cube and a having a resistor of i kilo ohm in every branch?

Answer 1

Given twelve 1 KOhm resistors, connected in the shape of a cube, in order to determine the net resistance between opposite corners, first draw the cube in two dimensions. (Try this at each step before continuing, so you can understand the lesson as it unfolds.)

There are three resistors leaving the initial vertex, and three resistors entering the final vertex. In between those six resistors, are six more resistors, each pair connected together on one end, and to two other resistors on the other end.

If every resistor has the same value, then (by symmetry), the voltage on the ends of the first three resistors must be the same. Similarly, the voltage on the ends of the last three resistors must be the same.

If two points in a circuit have the same voltage, then (for purposes of analysis) you can consider them to be shorted together. That short does not change the results, as there is no current flowing through that short.

With the bottom ends of the first three resistors shorted, and with the top ends of the last three resistors shorted, the circuit degrades into three resistors in parallel, in series with six more resistors in parallel, in series with three more resistors in parallel.

Three 1 KOhm resistors in parallel have a net resistance of 333 ohms. Six have a net resistance of 167 ohms. Two 333 ohm resistors and one 167 ohm resistor in series have a net resistance of 833 ohms, or 5/6 of 1 KOhms.

Note: This technique does not work if the resistors are not all the same value. In that case, you would need to solve 12 equations in 12 unknowns, looking at the partial currents in each branch.