How Is Total Resistance Calculated in a Cubic Resistor Network?

[Martin]

Twelve ideal R-ohm resistors are connected in a cubic configuration (i.e., each one forms an edge of a cube), as shown http://www.his.com/~mhtesler/Cube2.jpg . What is the total resistance R_AB between the opposite vertices A and B?

 

[Galileo]

There are 6 ways for the current to get from A to B. All 6 ways have resistance 3R.
So I guess R_{AB}=\frac{R}{2}.

 

[Gokul43201]

My guess is 5R/6 {or R*(1/3 + 1/6 + 1/3)}

 

[Gokul43201]

It would be harder to find the effective resistance between any other pair of points (for example, 2 neighbouring points) ...

 

[keffi]

Current passing from b to each three immediate resistors are equal.
Let I10, I11, I12 be currents passing through R10, R11, R12, the last three resistors of the cube.
Current passing at A is given by
IA= IB = I10 + I11 + I12
= V10/R10 + V11/R11 + V12/R12
and since the voltages and resistances are equal
IA = It = 3*V10/R10
Vt = It*RAB
= 3V10*Rt/R10
R10*Vt /3*V10 = Rt
For parallel connection V is constant
Vt = V10
thus
Rt = R10/3 or
Rt = RAB = R/3

 

[Rogerio]

The current flows equally through 3 resistors, then 6 resistors and then through the last 3 resistors. So, the total resistance is
R/3+R/6+R/3 = 5R/6 .

 

[Martin]

I see some correct answers here; for the others, here’s how it goes:

Imagine a current I entering one of the vertices (say, node “A”). Because of symmetry (i.e., each of the three paths leaving node A “looks” the same to the current I), the current I divides evenly into three equal currents I₀

(1) I₀ = (1/3)I

among the three braches A-1, A-2, and A-3, as shown http://www.his.com/~mhtesler/Cube%20Analysis3.jpg .

As each I₀ enters nodes 1, 2, and 3, it too—again, because of symmetry (i.e., each of the two paths leaving each of the nodes 1, 2, and 3 “looks” the same)—divides evenly into two equal currents I₀₀

(2) I₀₀ = (1/2)I₀

at the branch pairs 1-5 and 1-6, 2-4 and 2-6, and 3-4 and 3-5.

The currents then recombine at nodes 6, 5, and 4 into three equal currents I₀, which, in turn, recombine into I at node B.

The cube is simply a bunch of flexible wires connecting resistances. Let’s simplify our visualization by stretching out the cube and laying it flat http://www.his.com/~mhtesler/Cube%20Analysis-Planar.a.jpg (note that the wires are actually connected only at the points indicated by the solid dots “•”).

Now, calculate the voltage between node A and B along any path:

(3) V_AB = I₀R + I₀₀R + I₀R.

Substituting (1) and (2) into (3), we get

(4) V_AB = (1/3)IR + (1/6)IR + (1/3)IR = (5/6)IR.

The total resistance between nodes A and B is, therefore

(5) R_AB = V_AB/I = (5/6)R ohms.

Here’s another way of looking at it: Since the voltage drop from A to 1 is the same as the voltage drop from A to 2 which is the same as the voltage drop from A to 3, which is the same as the voltage drops from 6 to B, 5 to B, and 4 to B (because each equals (1/3)IR), then nodes 1, 2, and 3 are all at the same potential, and nodes 4, 5, and 6 are at the same potential. In other words,

V₁₂ = V₁₃ = V₂₃ = 0

and

V₆₅ = V₆₄ = V₅₄ = 0.

We can therefore connect each of these two sets of three nodes together with a short circuit and not change anything. If we do that, the circuit would appear http://www.his.com/~mhtesler/Cube%20Analysis-Planar.b.jpg . It should be obvious that this configuration is just a series connection of three parallel connections of 3, 6, and 3 R-ohm resistances, which works out to be (5/6)R ohms.

 

[check]

LoL. When my dad was teaching years ago, he used to give his class this problem. It would take them an hour or so to solve it and give him an hour or so to catch up on some work. heh