Science fair question - Please help - no laughing

That is the conventional technique to use but will the necessary equipment, to the required accuracy, be to hand? I think that is more the question in this application.
I'd be tempted toward a bridge type method which compares the various coins 1:1 so that they can be sorted into relative resistance order. There was no requirement to measure the resistance, just find the coin that conducts best so ordering the coins is enough.
Reply to
Brian Reay
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That's true! If all that is needed is to order them, a bridge will do it better than any other method.
Obviously I was assuming that he wanted to pin some numbers on each of those coins. I sure would... :-)
Note too, that it appears he has almost everything required. All that is missing is a 20 Amp 0-15 volt "battery replacement" power supply. (And a bit of study to figure out just exactly how it works.)
Reply to
Floyd Davidson
Very smart. Very smart indeed.
I spent ages mulling over this and knew I'd missed something.
Chimera
Reply to
Chimera
But then you will be measuring, not just the resistance of the coins themselves, but also of the connections between them. Given the extremely low resistance of the coins, it is quite likely that just the /variation/ in the contract resistance will be more than the coin resistance. Which is to say, the results will be quite variable.
Reply to
Floyd Davidson
After all the replies to this OP I think we've all been guilty of acting like engineers and overlooking the most basic parameter. He is NOT looking for any measurements, he is looking for the answer to which coin conducts best and we're all assuming that measurements must be taken, which is a logical procedure EXCEPT for one thing.....
....we're talking about school kids who probably don't have access to labs so the teachers assignment might be taking that into account. Logically, given that all our American coins do conduct, rationally the one that conducts best will be the one with the largest physical dimensions made of the most conductive material. We all know we can refer to a text book to determine which metals conduct better than others and if we assume current flows on the skin, then a larger coin with more surface area will conduct best. If we assume that current flows through a conductor then the largest coin will conduct best.
Answer: The American Silver Dollar conducts the best IF its an older coin in which all the metal is actually silver and not just plated however even if plated, there is still more conductive skin area than any other coin. Quarters today are made largely of zinc, as are nickels and dimes and pennies, which used to be solid copper, are now an alloy of which copper is only one ingredient.
I think a metalurgist could answer this question better. Sometimes we need to stop being so analytical that we overlook the incredibly obvious.
Sometimes the best solution to a hi-tech problem is a low-tech answer.
Reply to
EEng
The issue is, are the techniques being suggested realizable. I've pointed out in the 'serious answer' the kinds of sensitivities any measurement must have when trying to measure something in the micro ohm range. Address just a few of the questions that would challange those of us who claim to be professionals. First, make a connection to a coin the size of a dime or a quarter such that the connective resistances is small compared to the resistance to be measured. Of course, you need a big broad conductive contact, else you're measuring contact resistance. Then consider the currents required to produce voltage drops that are reasonable to measue. Of course, measuring verly large currents is also not the easiest thing to do. This stuff is called propogation of errors, and it's an important aspect of experimental design.
My advice was, and still is, reading the posts, this is more involved than one would first expect. I'd advise, unless someone has come up with a procedure that I've overlooked, to find a different project. Most judges, at least those who've worked at science fairs I've been involved with, would raise many of the objections I've cited.
AJW
Reply to
tony
Do a web search for the "4 wire kelvin" technique. It does *not* measure the "connective resistances".
Not true.
That is true. And that may indeed make this an impossible project. I don't know off hand what the actual resistance of a coin is, and therefore don't know what current would be necessary to measure it. Up to maybe 50 Amps, I can see this being feasible, but if takes more than that it probably is not, unless someone just happens to have access to a nice 100 or 200 Amp power supply!
Whatever, a 25 Amp supply and a 4 digit voltmeter will measure down to 40 microOhms resistance. Make that a 50 Amp supply and a 5 digit meter, and we are down to 2 microOhms.
Which is why it isn't done. Instead, very small voltages are measured.
This again, is true. I pointed out in my first article responding to the OP that the interesting part, which would make it a good science project, is *not* the relative resistances of the coins, but rather the techniques used to measure the resistances and the overall design of a system to do so.
That, for one thing, means that actual measured resistances *are* significant, not just the ordering of coins according to resistance. Clearly one of the side discussions that should be shown with the project would be a theoretical estimation of the resistance for each coin, knowing the approximate composition of metals used to make it. If the theory and the practice don't match, one of them is wrong. If that happens on all coins, clearly the practice is wrong (the measurements aren't correct!). But if it happens only on one coin out of 10... then the theory is wrong, and in fact the coin is probably not made of what it was assumed to be made of.
All of these many different issues of interest that come up in relation to the project are what makes it a good project.
Reply to
Floyd Davidson
A copper disk about .2 cm thinck and a cm in diameter is sub micro ohm.
You in fact do need good full contacts even with a 4 wire system. Consider the voltage gradients in the coin with a near point contact. There'll be high, not uniform, voltage gradients, and you'llbe measuring sopmething other than what you think you're measuring. The easiest thought experiment to demonstate this is to conside a large, thin disk with point current contacts in the middle of the disk. then, ask yourself what voltage you might monitor as you move your sensing probes along opposite surfaces of teh disk. Toward the edges, the delta V will be much less than near the probes.
AJW
Reply to
tony
Is that from one edge to the other? Or are you talking one side to the other (which is *certainly* going to be to small a resistance to measure)?
Good contact, but you are *not* measuring it, as you stated. If one setup has twice the connection resistance that the next does, both will still give accurate results. That is because either way, the current is adjusted to the same level, and the voltage probe is a *separate* contact, which has no current flowing and is therefore relatively unaffected by the "current" connections.
The current is adjusted to be a constant value, therefore the voltage gradients will be the same, regardless of the relative quality of the probe contact (within reason).
How does that relate to the topic at hand?
Have you actually *used* a low resistance test set? I've measured the resistance of zinc plated copper bus bar 1/4" thick by 1 1/4" inch wide and six inches long, bolted to zinc plated copper terminals that were 1 1/2" x 1 1/2". And all of this is with a few amps flowing through them while they are being measured (straps on a telco -48 V battery plant).
I can't see that measuring quarters is any more complicated.
+----O Constant Current Source O----+ | | | +---------------------+ | +-------->| COIN |
Reply to
Floyd Davidson
Floyd, I reread the OP -- we're both far from the mark. First, I concure that contact resistance isn't important in a 4 wire edge to edge (along the diameter) if you use voltage sensing probes inbound from the current carrying ones.
However, the OP is not an engineer, he's trying to do a rough lash up. It's unlikely he has microvolt capabilities. I'm guessing the resistance as measured between the voltage sensing probes is the order of 10 micro ohms or so (it's probably sub micro ohms face to face, as I assumed he was measuring initially.) So, if he has mV resolutions as one might get from a high end Radio Shack multimetere, he might need a voltage drop between his probes of say 10 mV. OK, 10 mV across 10 micro ohms looks like a thousand amps, probably out of reach of a home hobbiest.
My concern about good contact in the face to face resistance configuration had more to do with obtaining reasonable current distribution in the coin, I wan't concerned about voltage drops across the contacts until the numbers began suggesting huge current needs. I'd suggest in a configuration like that, one would do better by looking at the emission spectra from the vaporized material to differentate among the coins.
BTW, it's a bit of sport to calculate the resistance of a disk across a diameter, say a mm inbound from the outside edges, about where your voltage sensing probes might be. I started to write an equation or two, then decided an equivelent ciruit odf discrete interconnected resistors might be a good approximation. But even doing a quarter disk (That's all that would be needed, symmetry suggests that would be the same as the overall resistance) with a dozen components was more than the task justified.
Interesting science project, but I have no reason to change my opinion it's probably out of the technical reach of a home project. I wouldn't be surprised if one of the posters came up with a test configuration that was 'realizable' - I haven't.
It might be more fun to figure out how coin machines tell the difference between different coins.
AJW
Reply to
tony
FWIW -
Probably the reason anyone has a question regarding where "current flows" is implicit in the way this answer was phrased.
In the strict sense, and by it's very defintion, current does not flow - charge flows. ( Current is defined as charge passing a point per unit time. One rarely uses the idea of the flow of flow, e.g., flow of the flow of the river, flow of the flow of traffic, flow of the flow of charge, etc., but rather flow of water, traffic, charge, etc. - )
The charge field density is equal across the cross-sectional area of a conductor at DC - implicitly if not explicitly defined. The magnetic field density is not - it is not equally distributed in a conductor.
The field effect that redistributes the charge towards the outer portion of the conductor requires a change in field ("AC" as in alternating from steady) to disturb the field symmetry.
No AC, no varying field, no change in distribution.
The formula is old and universally accepted.
Reply to
Hobdbcgv
The differences in the level of conduction in the coins is very small, because of the small differences in metal and the small length of conductor
When faced with this problem, one usually looks to a Wheatstone bridge - easily built, BTW, often built in middle school and high school physics classes. Can't say it will be sensitive enough for the coins, but it is where I would start
In short, the bridge's flow in it's two two-resistor legs is balanced using a bridge resistor between the middle of the resistor pair of the legs, it si balanced, and then the unknown small resistance is added and the change measured. The bridge unablances easily, so that makes it suited for small resistance measurements -
Then you do a quick calculation to find the resistance of the coin do a google on it, and you will get a feel as to whether it is more than he can get a handle on. The other method is to measure a lot of conductor - as in a stack of 20 coins- however, at the 20-stacked level with the metal as the conductor, you will likley get false resistance levels from contact resistance between the coins, since that is a larger than the resistance than the metal in the coin.
Reply to
Hobdbcgv
Like I said, both camps make excellent points, and is the mitigating reason that to this day, we continue to agree to disagree. For every arguement, proof, mathematical formula, physics lesson, etc that proves one side....there is another of like ilk and just as compelling, proving the opposite.
You'll note I'm not taking a stand on which discipline is correct nor will I argue the topic one way or the other. I've often noted that in electronics we are all taught that some things just "are" and that until difinitive proof is realized that all can agree upon, we must by needs, agree to disagree.
Reply to
EEng
The first advice to give the OP is that he cannot expect a reasonable project to be based on reading the resistance of coins face to face. I don't think "sub micro ohm" quite describes it! It would be dependent upon how much of the face is in contact with the probe, and the only way to differentiate coins would be if the *entire* face is in contact (for example, with a copper block larger than the coin and with some form of copper or silver based conductive material used to guarantee uniform electrical contact). Clearly the resistance face to face on a Silver Dollar would be milli-micro-ohms.
There is no point in discussing the techniques involved for that.
The only measurement that makes any sense at all is edge to edge across the diameter of the coin. I have assumed that, and tried to make it clear that is what would lead to a functional project, and several others have too.
After all, the point is to lead the OP towards a useful project, not to waste his time sorting through theory that does not apply.
[huge snip]
He can acquire microvolt capabilities without great expense. Just as he can acquire a power source that will suffice without great expense. This may not be a $25 project, but it isn't like owning a yacht either! ;-)
If he has a 25 A supply, and the coin has 10 micro Ohms of resistance
25 x 0.000010 => 0.00025 volts drop
A microvolt meter will read 250 microvolts. I just took a quick look on eBay and found that in the past month microvolt meters have sold for $100 or less.
We aren't talking $20 multi meters, but even if a meter and power supply had to be purchased, the entire cost could be kept below $200.
I would also suspect that a little bit of begging in the appropriate places might garner free use of someone else's equipment for the duration! It would certainly be worth a try.
It looks fairly easy to me. The fun part would be designing a jig that would grasp various sized coins to make a connection that is uniformly the same each and every time it is used on any given coin.
Size and weight.
Reply to
Floyd Davidson
As you have probably determined, comparing relative conductivity of coins using an electrical/electronic circuit is prohibitively complicated/expensive.
However, there is a series of experiments that your son can do to measure the relative conductivity of ANY round, thin object. Use the same method that is used in vending machines to test conductivity. Roll the coin, slug, washer down an incline and between the poles of a strong
magnet near the end of the ramp. The relative motion between the coin and the magnet will induce a current in the coin that will produce a counter magnetic field that will cause the coin to slow down. How far the coin travels horizontally
after leaving the ramp will depend on the conductivity of the coin. Lower velocity/ distance will indicate higher conductivity. Weight of the coins shouldn't have an effect on the experimental results.
Good luck, ARM
Reply to
Alan McClure
This question is meaningless and should be thrown back at the under-informed person posing it - not you dad. As with many such questions on newsgroups this is tackled by well informed readers who should know better and should always first question the question.
Resistance face to face? Across a diameter? in both cases the tiny or otherwise cross section of contacts dominates and defeats any attempt at measurement.
A meaningful question is; what coin alloy has the least resistivity? and the answer lies in a table of alloys and their resistivity; not in any measurement technique.
Reply to
Arthur Holland
Some truth in that but reading the question also helps- in this case the question was reasonable in the context, see below.
The question was to "determine which American coin would conduct electricity the best"
So, firstly the requirement is to 'rank' the coins and provided the method used is consistant- eg always face to face, or edge to edge etc. and the contact area the same, the result should be valid as a comparison.
You are assuming that the purpose of the question was for the student to get the 'correct' answer. In this context, I suspect the idea was to get the student thinking about a method / process / approach to the problem.
In this context, the question asked was meaningful.
In fact, it is typical of an 'interview' question- the answer is not always the thing that is looked for, it is the approach to solving the problem that the questioner is looking for.
Reply to
Brian Reay
Even assuming you solve the insurmountable problem of eliminating variations of contact resistance between coins, (you coud melt them down, but then they wouldn't be coins any longer) how many thousand dollars worth of coins would this take? Wouldn't the key factor be the measurement technique and instrumentation, rather than the number of coins?
Reply to
ehsjr
(snipped)
This will not work well. There will be little change in this voltage for the small coin resistance changes.
Start with a 12 volt supply (eight "D" cells), and put a 120 Ohm, 2W resistor in series with the coin. This creates a 100 mA constant-current source, given the low coin resistance. Connect the current to the coin with two leads. Now take a voltmeter set to the lowest range (millivolts is better if available), and connect the two voltage test leads across the coin separately from the current leads. You now have a standard Kelvin arrangement for measuring the low coin resistance, which eliminates the effect of any contact resistance to the wires. Multiply the voltage reading by 10 to get Ohms.
Ben Miller
Reply to
Ben Miller

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