What is the resistance of a cube

What is the resistance of a cube that has a 1 ohm resister on each side measured between opposite corners?

Reply to
Kilowatt
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+AF8- .833333 ohms

ARM

Reply to
Alan McClure

Flatten out the cube. Assume your two corners are 'X' and 'Y', and the corners that are attached to 'X' through one resistors are 'A', 'B', and 'C'. The corners thusly connected to 'Y' are 'D', 'E', and 'F'. You then have three resistors XA, XB, and XC and three symmetrical at the other corner: YD, YE, and YF. Inbetween there are six resistors; AD, AE, BD, BF, CE, and CF. By symmetry we can see that the voltage at A=B=C and D=E=F, so we can consider all the resistors in these three sets in parallel. Thus the total resistance is (XA||XB||XC)+ (AD||AE||BD||BF||CE||CF)+(YD||YE||YF), or 1/3 + 1/6 + 1/3.

Reply to
krw

in article OZskd.877943$ snipped-for-privacy@bgtnsc04-news.ops.worldnet.att.net, Kilowatt at snipped-for-privacy@charter.net wrote on 11/10/04 10:20 AM:

By inspec tionand symmetry it is

(1/3 + 1/6 + 1/3) ohms.

I hope we get you homework problem done in time.

Bill

Reply to
Repeating Rifle

Gawd! Is that one still floating around? That was one of the questions given in the early stages of the basic electronics course in my junior year of HS back in 1963...

Reply to
John McGaw

Seems it never goes away. When I first saw it in the late 60's it was presented to me and a friend of mine by a retired engineer from GE. My friend who was in college at the time took about 2 days and several sheets of paper to work it out. I was a junior in HS and worked it out in my head in about fifteen minutes. Not bragging, I saw the symetry and the easy way to work it. My friend attacked it in a way that did not depend on the symetry of the network.I am sure his better understanding of networks led him in the direction he went and my ignorance in mine.

Reply to
Jimmie

Build the cube and drive with a 3 amp ideal current source. Apply Kirchoff's Current Law in presence of symmetry to see how the current splits evenly along edges of the cube. Then add up the voltage drops along any 1 path, divide by 3 amps, and you have the resistance.

Now a more interesting question is replace the 1 ohm resistors with 470-ohm, silver fourth band (10% tolerance) resistors. How does the "tolerance" of this effective network compare to the 10% tolerance of the individual components? For simplicity, assume that the individual resistors fall within a uniform distribution from 423 through 517 ohm.

Reply to
Ar Fai Ve

Why?

When considering tolerances, one isn't interested in a design that will probably work given the distribution of component values within their tolerance limits. One wants a design that definitely will work even if all the components are at their worst case permitted values.

So the tolerance of the cube made of 10% resistors (of the same nominal value) is still 10% isn't it?

Sylvia.

Reply to
Sylvia Else

But I think his point is more about how 'balanced' it will be. Presumably the cube is used instead of a single resistor for a reason. If some resistors are high, while others low, how will the imbalance affect some of the current distribution. Will the non-uniformity upset the current balance enough to dissipate too much power in some resistors?

Admittedly, it is a 'contrived' problem, but it does illustrate that "things are not always what they seem..."

daestrom

Reply to
daestrom

However... things are more like thay used to be than they are now..

Northstar

Reply to
Northstar

Part of any design is understanding the variability in its characteristics and how that variability is a function of all the individual circuit components' variabilities. There are lots of reasons for doing this. One is to compare the robustness of two designs -- hopefully choosing the one that is less susceptible to individual component variability. Another area is in diagnosing and localizing circuit faults.

No.

Reply to
Ar Fai Ve

No, sorry, I don't get this. If a design might not work properly when some components are at the limit of their tolerance, then the design should specify components with closer tolerances.

Reply to
Sylvia Else

What's to not get? All designs are subject to the robustness of their constituent components. It is this up-front analysis that actually leads to the conclusion of that certain tolerance value resistors are sufficient to get the job done. Imagine if you are presented with a brittle design that works only when all components are within 0.00000001% tolerance, then the design is unrealistic, costly, and plain bad.

Reply to
Ar Fai Ve

I do not dispute that, but I cannot see its relevance.

What I should probably have asked is what you actually mean by the "tolerance" of the effective network.

When we say that a resistor has a tolerance of 10%, we mean that its true value may differ from its nominal value by up to 10% (of its nominal value). The obvious meaning of the tolerance of the network is therefore the extent by which its true value may differ from the value calculated using the nominal value of its resistors. You introduced the issue of the distribution of values, but there's no way of using that concept in calculating the tolerance of the network according to the usual meaning, so you presumably meant something else.

What?

Sylvia.

Reply to
Sylvia Else

Yes, but how do you know, "If a design might not work properly when ...."?

You have to do the analysis with the various components at their limits and understand what happens if a certain one is at its high limit while another is at its low limit. That's all, just saying such analysis is an 'interesting' problem.

If it turns out that it won't work under such circumstances, then you're absolutely right. Time to change the design and/or specs. But, "How do you know?"

daestrom

Reply to
daestrom

I have no disagreement with that, but that's not what the proposed modified question required. There was talk about a distribution of values, not worst case scenarios.

Sylvia.

Reply to
Sylvia Else

Perhaps I was reading it differently. I'm thinking more of 'works' includes getting a certain voltage division between vertices, or the total resistance is 5/6 ohm with a certain confidence when the thing is constructed of individual parts whose tolerances are spread with a certain distribution. Kind of like making them on an assembly line and you want to know how many will pass a test that measures the voltage between two vertices. Using just standard 10% components, if the distribution of values is 'normal', or 'flat', or'chi', or whatever, what would be the variation in performance. (i.e. how many will fail the QC test and have to be re-worked).

After all, if building up a resistor network to make one 'composed resistor' with 10% tolerance parts, given enough individual parts, the variations

*should* cancel out in the overall system and give you a 'composed resistor' with even better tolerance than the parts. And how will the variation of such a resistor network be distributed compared the distribution of variation in the individual components.

daestrom

Reply to
daestrom

Such calculations are certainly possible. I'd question whether it would be a realistic approach to production.

As something of an aside, I think it used to be the case that resistors were marked according to the actual tolerance they achieved. That is, you made bunch of resistors, then measured them. The ones within 1% were marked accordingly, and sold at a high price. Next came the 5% ones, at a lower price, then the 10% ones, and finally, the 20% ones (no tolerance band). So if I bought 10% resistors, the one thing I could be sure about was that they were not within 5% of their nominal value.

I doubt it's still true, though.

Sylvia.

Reply to
Sylvia Else

If they did that, why not just mark the measured value on them and sell them as 1% (or .0001%, just need a better tester) tolerance resistors?

Reply to
Michael Moroney

That's a good question.

No market? I suppose it's a form of yield management.

Those designs that require 1% resistors would mostly be because they needed a consistency in value, wouldn't they, rather than because they need to be able to choose a value? I don't remember seeing many very odd valued 1% resistors. Typically, more like the values in the 5% range, just with 1% tolerance.

I did once come across 1% resistors that had an extra band, and the next lower multiplier. I don't know about others, but over time I've tended to have the colour combination for common resistor values in my head, so I recognise them directly, without having to decode them. The extra band and lower multiplier was a damned nuisance.

Sylvia.

Reply to
Sylvia Else

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