Consider a cube made of resistors

Factual Questions
1

This was going to be a Bricker Challenge question, but I must admit that I am not sure of the right answer. So it’s here rather than in the coming-soon Year-End BC.

Consider a cube made of wire, with each edge a single wire containing a 1-ohm resistor in the middle.

Now consider a connection made with a circuit connected to opposite vertices of the cube (upper left back and loer right front, for instance).

What is the total resistance across the circuit?

I’ve done this two different ways and gotten two different answers, and both make perfect sense to me :slight_smile:

2

The answer is5/6 ohm.I assume you know the trick of the solution; but what other answer are you getting, and how?

3

Oh god.

My high school physics teacher posed this question, and to this day it makes my head hurt just thinking about it.

4

I wish I coudl draw a picture here, but here goes. There are twelve wires (and resistors) in this configuration. Three call (them I1 I2 I3) are connected to the input vertex, three (O1 O2 O3) are connected to the output vertext (diagonally opposite) and the other six are used to to connect each of the I’s to two of the O’s. (For simplicity say I1 is not connected to O1, etc., but it really doesn’t matter).

The simlest thing to do now is to look at the voltage at the three vertices just after I1, I2 and I3. Since the current has passed through 1 ohm in each case (and by symmetry) the voltage will be identical at each point. So if they were shorted together no current would flow. Similarly you can short together the three vertices just one resistor (O1 O2 O3) removed from the output vertex. Again no current woudl flow as they are at equal potential.

Now in your shorted circuit, you have three 1 ohm resistors in parallel from the input to a single point, three 1 ohm resistors in parallel from a single point to teh out put and six 1-ohm resistors connecting the two points.

The resistance rule for adding resistors in series is R = r1 + r2 + … The resistance rule for adding resistors in parallel is R = 1/(1/r1 + 1/r2 + …) so we have

Resistance of whole thing is [1/(1/1 + 1/1 + 1/1) + 1/(1/1 + 1/1 + 1/1 + 1/1 + 1/1 + 1/1) + 1/(1/1 + 1/1 + 1/1)] = 1/3 + 1/6 + 1/3 = 5/6 ohms.

5

Isn’t it just 5/6 ohms? 1/3 to get to the first set of intersections, 6 1 ohm resistors in parallel = 1/6 to get to the second set, and another 1/3? What am I missing?

6

Nothing. We’re all keen to hear what Bricker’s other solution was.

7

This is pretty easy.

Draw your resistor cube. Then pretend 1 amp is going into one corner (the “input” junction), and 1 amp is coming out of the opposite corner (the “output” junction).

You will see that, when one amp flows into the input junction, it is equally split by three 1 Ω resistors. (We’re assuming symmetry, here.) Let’s call these resistors R1, R2, and R3. Thus 1/3 of an amp flows in R1, 1/3 of an amp flows in R2, and 1/3 of an amp flows in R3. And because each resistor is 1 Ω, this means there’s 1/3 V across R1, 1/3 V across R2, and 1/3 V across R3.

Now let’s look at R1. Notice how the current through R1 is split by two 1 Ω resistors? Let’s call these resistors R4 and R5. Because 1/3 of an amp flows in R1, 1/6 of an amp must flow through R4 and 1/6 of an amp must flow through R5. This also means there’s 1/6 V across R4 and 1/6 V across R5.

Same goes for R2. Notice how the current through R2 is split by two 1 Ω resistors? Let’s call these resistors R6 and R7. Because 1/3 of an amp flows in R2, 1/6 of an amp must flow through R6 and 1/6 of an amp must flow through R7. This also means there’s 1/6 V across R6 and 1/6 V across R7.

Same goes for R3. Notice how the current through R3 is split by two 1 Ω resistors? Let’s call these resistors R8 and R9. Because 1/3 of an amp flows in R3, 1/6 of an amp must flow through R8 and 1/6 of an amp must flow through R9. This also means there’s 1/6 V across R8 and 1/6 V across R9.

R4 and R6 are connected to one of the resistors connected to the output junction. Let’s call the latter resistor R10. The current flowing in R4 and the current flowing in R6 must equal the current flowing in R10. Thus the current flowing in R10 is 1/6 + 1/6 = 1/3 amp. This also means there’s 1/3 V across R10.

R5 and R8 are connected to one of the resistors connected to the output junction. Let’s call the latter resistor R11. The current flowing in R5 and the current flowing in R8 must equal the current flowing in R11. Thus the current flowing in R11 is 1/6 + 1/6 = 1/3 amp. This also means there’s 1/3 V across R11.

R7 and R9 are connected to one of the resistors connected to the output junction. Let’s call the latter resistor R12. The current flowing in R7 and the current flowing in R9 must equal the current flowing in R12. Thus the current flowing in R12 is 1/6 + 1/6 = 1/3 amp. This also means there’s 1/3 V across R12.

Now all you have to do is figure out the voltage between the input junction and the output junction. It doesn’t matter what path you take; it will always equal the same value. For example, the voltage across R1 + the voltage across R4 + the voltage across R10 = 1/3 + 1/6 + 1/3 = 5/6 V. And the voltage across R2 + the voltage across R7 + the voltage across R12 = 1/3 + 1/6 + 1/3 = 5/6 V.

The current through the cube is 1 amp. The voltage across the cube is 5/6 V. The resistance is R = V/I = (5/6)/1 = 5/6 Ω.

8

I did it like Crafter_Man. If you draw out the cube and start by concentrating on the top square of resistors. Assume 1 Amp going into the cube it splits evenly, 1/3 A going each way. Follow through a top square resitor out of that junction you come to another junction where the current divides evenly, 1/6 A each way. Continue through the resitor in the top square and at the next junction your 1/6 A is joined by another 1/6 A from a resistor in the top square to make 1/3 A going down through the resistor to the ouput terminal.

So in the 3 1-Ohm resistors in series in that path there is a total of 1/3 + 1/6 +1/3 volts giving a total of 5/6 volts for an equivalent resistance of 5/6 Ohm.

9

Well, suffice to say I’m an idiot. Never you mind what my other solution was. I must have been breathing fumes.

10

Good old Kirchoff.

11

I hear that it was 5/6 Ohm, but only for 20 minutes.