## Resistor problem

Anything goes, but keep it seemly...

### Resistor problem

Eight points are arranged in an octagon. Each point is connected to each other point with a 1-ohm resistor. (So there are a total of 28 resistors.)

What is the total resistance between two diametrically opposite points on the octagon?

(Recall that two resistors, A ohms and B ohms, have a combined resistance of A+B ohms if they are in series, or 1/(1/A+1/B) ohms if they are in parallel.)

Bill Smythe
Smythe Dakota

Posts: 555
Joined: 11 February 2006

Hmmmm!

Symmetry usually plays a part in questions like this, so to avoid 28 simultaeneous equations in currents, my approach would be (assuming the two contact points are at N and S of the octogon)

a) Take out the N-S resistor for the moment. It can always be reinstated in parallel once the resistance of the rest is known.

b) Eliminate the 3 E-W resistors. Symmetry says that nothing flows through these

I also feel that there should be top/bottom symmetry so that the voltages at the extreme E and W points should be half the difference between N and S.

Further thinking happening!
Bigtone53

Posts: 413
Joined: 19 September 2005

Eliminate the 3 E-W resistors. Symmetry says that nothing flows through these

Every resistor between 2 nodes other than N and S is at the same "distance" from N and S and no current will flow through it. We only need the direct paths between N and S.

The simplified picture would be:
Code: Select all
`          N .--.--.--+--.--.--. |  |  |  |  |  |  | 1  1  1  |  1  1  1 |  |  |  |  |  |  |NE NW  E  1  W  SE SW |  |  |  |  |  |  | 1  1  1  |  1  1  1 |  |  |  |  |  |  | '--'--'--+--'--'--'          S`

Resolve the serial circuits to:
Code: Select all
`         N .--.--.--+--.--.--. |  |  |  |  |  |  | 2  2  2  1  2  2  2 |  |  |  |  |  |  | '--'--'--+--'--'--'          S`

R = (1/2 * 6) + 1/1 = 4
Ruud

Posts: 664
Joined: 28 October 2005

Every resistor between 2 nodes other than N and S is at the same "distance" from N and S and no current will flow through it. We only need the direct paths between N and S.

Ruud,
I am not so sure about this. If you imagine the layout as a flow of water from N to S down an inclined layout (with suitable amendments),there will be flows from L to R and vice versa between the non N-S nodes. Can you really assume that they are simply not there?
Bigtone53

Posts: 413
Joined: 19 September 2005

Bigtone53 wrote:If you imagine the layout as a flow of water from N to S down an inclined layout (with suitable amendments),there will be flows from L to R and vice versa between the non N-S nodes.

From what I've learned (a long time ago), the only thing that matters is the voltage level of the nodes. As you can see in my first diagram, these nodes are all at the same level. The picture would be completely different (and beyond my Math skills) when the resistors had different values.

However, I goofed on the calculation. The result should have been 1/4.
Ruud

Posts: 664
Joined: 28 October 2005

So forgetting my water flow analogy, why no current between NW and SE for instance?
Bigtone53

Posts: 413
Joined: 19 September 2005

An image may help:

Both images have the same connections, but the second shows more clearly how the 6 imtermediate nodes are all at the same level, so no currency flows between them.
Ruud

Posts: 664
Joined: 28 October 2005

Ok, got it. Nice symmetry-based answer Ruud.
Bigtone53

Posts: 413
Joined: 19 September 2005

Nice going, all! You guys nailed it.

Two-dimensionally, think of the 6 intermediate nodes as being on a straight line halfway between the N and S nodes. All the points on this line are at a voltage halfway between N and S, so any resistors connecting them don't matter.

Furthermore, it doesn't matter which two nodes are selected as N and S. Even if you are asked to calculate the resistance between two adjacent nodes, the reasoning (and the answer) is still the same.

Bill Smythe
Smythe Dakota

Posts: 555
Joined: 11 February 2006