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
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.
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.)
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.
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
Sometimes the best solution to a hi-tech problem is a low-tech answer.
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.
Do a web search for the "4 wire kelvin" technique. It does
*not* measure the "connective resistances".
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
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.
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.
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"
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 |
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
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' -
It might be more fun to figure out how coin machines tell the difference
between different coins.
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 -
( 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.
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.
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.
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
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.
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
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.
As you have probably determined, comparing relative conductivity of coins
using an electrical/electronic circuit is prohibitively
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
magnet will induce a current in the coin that will produce a counter magnetic
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
distance will indicate higher conductivity. Weight of the coins shouldn't
an effect on the experimental results.
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
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.
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.
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
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.