Cube Resistor Puzzle, October PW
Twelve identical resistors (value R) are arranged one along each side of a cube. At each of the eight corners, three resistors meet in a solder joint. A pair of diametrically opposite corners are (arbitrarily) designated INPUT and OUTPUT terminals. What is the total resistance (Z) of the network between the terminals?
A fixed but unknown voltage is applied across the terminals causing current I to flow. The only known parameter is R.
The circuit has been drawn as a two-dimensional layout (see picture), without crossovers (except where the ohm-meter has been added to see the result of the simulation, the ground also being a requirement of the simulation package’s calculator). A pattern is revealed whereby INPUT current splits into three (R1, R2, R5). Each of these three paths then splits again, this time into two (R1 feeds R4 & R6; R2 feeds R3 & R8; R5 feeds R9 & R10). Pairs of these latter paths then recombine, giving another three resistors (R3 & R4 feed R7; R6 & R9 feed R11; R8 & R10 feed R12). Finally, the final three paths (R7, R11, R12) merge to form the OUTPUT. The pattern is then a series of splits/merges of the INPUT current, first into three, then into six (actually three pairs) and finally back into three that combine to send all the current to the OUTPUT. Current is preserved (Kirchoff’s First Law) so what goes IN also comes OUT.
On inspection, the pattern has symmetry such that R1, R2 & R5 each carry one-third of I (I/3). Each of these resistors feeds a split which is in two equal parts, each part being half of I/3 = I/6. The final merge sees once again three equal parts with I/3 going through R7, R11 & R12. Hence six resistors each carry I/3 and the other six pass I/6.
With only I and R available for calculation, the ‘I-Squared-R’ power equation is applied W = I2R to find the power dissipated by each resistor, summed to give the total power in the network, which is:
W = [6.(I/3)2.R] + [6.(I/6)2.R]
The same power dissipation would be achieved if the same fixed voltage were to be applied to a single component with resistance Z. Then:
W = [(I2.R).5]/6 = I2Z
Divide out I2 to give Z = (5/6)R
The total network resistance is five-sixths of the value of any one of the component resistors. The simulation is made of 1.2kΩ resistors, five-sixths of this is 1kΩ as confirmed by the simulation. Godfrey Manning G4GLM Edgware