Kirt's Cogitations™ #256
The Resistor Cube Equivalent Resistance Conundrum
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Cog·i·ta·tion [koj-i-tey'-shun] – noun: Concerted
reflection; meditation; contemplation.
Kirt [kert] – proper noun: RF Cafe webmaster.
The Resistor Cube Equivalent Resistance Conundrum
You have probably seen somewhere along the line in your electronics career the resistor cube problem. The 12 edges of the cube each contain a 1 Ω resistor, and the challenge is to calculate what the equivalent resistance is between two opposing corners. It is a daunting problem using straight circuit analysis, since it requires writing and solving multiple mesh equations. There are lots of opportunities for making mistakes.
One option if you had the time and facilities would be to build the model in a circuit simulator and let it determine the result. Usually, though, the cube is thrust upon you in a compromising situation, like in a job interview. If you are an electrical engineer and cannot figure it out on the spot, forget that circuit design job. If you are an electronics technician, you will be forgiven for not solving it, but you had better demonstrate an understanding of the method once it is presented.
As it turns out, there is a relatively simple analysis based on symmetry and a fundamental level of understanding of currents and voltages.
The traditional method used involves recognizing sets of equipotential points within the vertices of the cube, then shorting them together to enable calculation of parallel resistances. Finally, those resistances are added in series to arrive at the resulting equivalent resistance. The process is illustrated below.
After explaining the traditional method, I will present my solution, which is a little more intuitive and direct method for arriving at the same answer. Solving via the traditional method actually requires the same knowledge of how currents are divided at nodes.
Finally, LTSpice is used to arrive at an answer via a Spice-based circuit simulator.
This is the cube structure consisting of 12 resistors electrically connected between the 8 vertices. Each resistor is 1 Ω, but any value can be used so long as they are all the same.
Here is where the intuition comes into play. Color coding is used to help keep track of the resistors and associated nodes (below). Due to symmetry, the potential (voltage) at the three nodes labeled "α" are equal. Since no current flows between nodes with a potential difference of 0 V, they can be shorted together without affecting the circuit's integrity. The same can be done for the nodes labeled "β."
Once you short those nodes, you obtain the equivalent circuit shown below. As you can see, there are two sets of three resistors in parallel, in series with one set of six resistors in parallel. So, you have 1/3 Ω in series with 1/6 Ω in series with 1/3 Ω, which equals 5/6 Ω.
Now I will present my method of solving the resistor cube problem. The structure is repeated again here.
Kirchhoff's current law, which states that the sum of the currents entering and exiting a node is zero, is essential in the analysis.
The first step is to recognize that at a node where equal resistances exist, current entering the node will be distributed equally between the number of output branches - in this case three. For convenience sake, I assigned an input current of 3 amperes at the corner labeled "A," so that 1 amp will flow through each output branch. Note that 1 A flows through each branch.
On the far side of each of those branches is another node with two output branches. Again, due to symmetry, the input current will divide evenly so that ½ A flow into each branch. Looking at the cube's output node labeled "B," it is apparent that the same situation exists as with "A."
Take a moment to sum the currents into and out of each node to verify that they all add up as required.
Now that you know the current through each branch, and you know that each branch has a single 1 Ω resistor in it, Ohms law allows you to calculate the voltage across each resistor.
The next step is to sum the voltage from input node "A" to output node "B." Any path you take travels along three edges, and all total to 2½ volts.
Finally, apply Ohms law, which says that the resistance is equal to the voltage divided by the current. As with the other analysis method, the resulting equivalent resistance is 5/6 Ω.
You can see that in reality, being able to make the assumptions in the traditional solution requires an understanding of the current division principles in my method. So, IMHO it is simpler to add voltages and then plug voltage and current into Ohm's law to arrive at the answer than to risk shorting nodes incorrectly. It's so simple, a caveman could do it.
As a verification of the result, the resistor cube circuit was simulated using the free LTSpice program, by Linear Technology. Resistors are labeled in accord with the labels in the traditional method of analysis. A 3 amp current source is placed at input node N001. The resulting voltage is the predicted 2.5 V. Again, Ohm's law for 3 amps and 2.5 volts yields a resistance of 5/6 Ω.
R§b N001 N003 1 Analysis Report: V(n001): 2.5 voltage
R§d N002 N004 1
R§a N002 N001 1
R§f N004 N003 1
R§h N005 N007 1
R§l N006 0 1
R§i N006 N005 1
R§j 0 N007 1
R§g N003 N007 1
R§k N004 0 1
R§e N002 N006 1
R§c N001 N005 1
I1 0 N001 3
Thanks to RF Cafe visitor Les Carpenter for sending me this solution that rearranges the resistors in delta and star configurations in order to "simplify" the solution. My head is still hurting from looking at it.