To all, My son is trying to do a science experiment trying to determine
which American coin would conduct electricity the best. I'm the father,
and while I am handy, I am not electrically educated. Here's what we
have done so far:
1. Battery in batter holder to get a good connection
2. Holder to digital meter to test amperes
3. tester to coin
4. Coin to christmas light (resistance)
5. christmas light to battery holder completing circuit
We've tested this thing with AAA zinc-oxide batteries thru D-cell
Alkaline and we always get the ampere reading for each type of battery.
The conclusion would be that all the coins conduct equally well. Did we
miss anything? According to what I'm reading, resistance is accumulative
when done in a series circuit. So while the christmas light provides a
constant resistance, the each of the coins must be the same resistance
for the numbers to come out exact.
Any help / opinions would be greatly appreciated.
The resistance will be so low it would be hard to measure. You will probably
see more resistance in your connection to the coin than in the coin itself. It
will certainly be dwarfed by the resistance of your test lead.
How are you measuring amps? You have to measure it in series...Also, I could
be wrong, but I dont think you are going to have much resistance on any of
the coins....and least not where you would see any sort of readings....but i
have never tried it so (not to mention I would have used Canadian coins...)
But you have read right...in series resistance is accumulative....
You can probably answer this question without any testing, most new coins
have a coating on them which will be carrying the current, and I assume all
coins have the same coating. Current is only carried by the outer layers of
any conductors. So from these assumptions the only real effect is the
geometry of the coins. The smaller the coin the less resistance it should
have (also dependent on the kind of metal the coin is made). But if you
really want to test it, increase the resistance by using coins in series
(sugg: lay them flat and solder them together). Note that to make the
comparison fair have all the soldered coin chains should be of the same
length or divide by the length to normalize the result. That way you should
obtain the conductivity of the alloy used in the coins. Also, get ride of
the Christmas light, their resistance will vary with temperature, meaning
that the current reading will fluctuate over time and may not be accurate.
To be able to measure the very small resistances of the coins don't use any
other loads (i.e. resistance) in series but the coins. Also dip the coin in
an acid solution to get ride of any dirt they may have, dirt is sometimes
conductive! (Coke or Pepsi, 20 min, will do the trick, do not drink!!!). If
you got to this point and your not bored, than you are a very good father!!
GOOD LUCK!! Current = Voltage/Resistance, to see a large variation in
current increase the voltage, please only use a DC battery.
As I recall, measuring very small resistances such as a coin would have is
done using the wheatstone bridge. I searched wheatstone bridge at google
and got 18,200 hits. There are some excellent pages on this subject. The
experiment that you have suggested can grow to intersting proportions.
What you have missed is the significance of "resistance is
accumulative ... in a series circuit."
The coins are metal, and are *very* good conductors with *very*
low resistance. The lamp in series is not a particularly good
conductor, hence it has a much higher resistance than a coin...
by several orders of magnitude! The high resistance of the lamp
is what controls current flow. (In fact, the wires used to hook
up everything will also have a significantly higher resistance
than the coins.)
Worse than that, an incandescent lamp does something really odd.
It changes resistance as it heats up. Hence the more voltage it
has across it, the higher it's resistance. If you put something
in series with it that has enough resistance to lower the
voltage to the lamp, the lamp will not get as hot and will have
less resistance. That is going to swamp any changes caused by
different (and miniscule) coin resistances, making it even more
difficult to measure.
To measure the resistance of something like a coin requires
a very sensitive meter and a very high current. If you
really want to do this, you might think in terms of a "battery
eliminator" type power supply that can deliver 10-20 Amps at an
adjustable 0-15 VDC. It actually *is* an interesting
experiment, because it can't be done without considering such
things as how to make a high current connection to the coins and
get a good connection each time, since you want the coin to be
what changes, not just the resistance of your connection.
The real merit of the experiment would be to show the relationships
between resistances of various parts of the circuit and the currents
that flow, with a comparison to measuring higher resistances (say
something like 100 or 1000 Ohms) where a typical multimeter works
fine. (Do a web search on "low resistance" as a string, plus the
single word "measurement". Also do a web search on the string
"wire kelvin", which should turn up some discussion of the actual
technique to use.)
What coating is that, and why would it carry the current?
Bullshit. Where did you get that idea?
Another strange idea.
Which will no doubt result in some interesting variations, because the
measurement would not indicate what the coin resistance is, but rather
what the quality of the connection between each coin is.
What has more resistance, solder or the coin? And by how much?
It does vary, but that is not the significance.
I'm impressed, you got a few things right!
Wrong. What he wants is a constant current source. Then he needs
to measure the voltage across the item being tested. DC is easier
to handle (because of the requirements for measuring very low voltages),
but AC would actually be a better and more accurate measurement if the
meters were equally accurate.
Incorrect, or rather, incorrectly assumed. Whether current flows
along the skin or through a conductor is still one of those points on
which we all agree to disagree. There has been no difinitive proof
either way although both camps make their good points.
Ken, it's refreshing to see an experiment being reported as concisely
and accurately as you have. You son is in excellent hands. Let's
discuss the problems of the experiment first, followed by some answers
and I'll try to keep it 3rd Grade Coloring Book Simple :)
A standard battery is not going to be powerful enough for your
experiment; you're going to need a 12Vdc Power Supply capable of
sourcing at least 1 amp.
Your Christmas light has an odditiy about it. It's resistance will
change as it heats up, thus changing the current and voltage drops
around your circuit. Remember that voltage is a measurement ACROSS a
component (the bulb or coin) while current is a measurement THROUGH
the component. These are related via Ohms Law.
Since one of your "resistances" can change, you can not make a
reliable reading even with the best of meters, so that variable must
The solid resistance that replaces the bulb is going to have to be
high because of the incredibly low resistance of any coin which is
quite a bit less than even the wires and test leads themselves. We
want to keep the current as low as possible so that large changes in
the fixed resistance become evident by the changing voltage drops.
The measurements you make, more importantly the differences in the
measurements, are going to be very small so you're going to have to go
with an extremely high solid resistance in the 10meg or greater range
and measure the voltage across it to at least 3 decimal points, 5
would be better. If you're using a 12Vdc source, most of that will be
across the resistor....probably somewhere in the 11.99998 range! On a
standard meter it will show up as 11.999V. If you begin with a large
silver dollar and end with a dime, you MIGHT see a tiny difference in
voltage across the fixed value resistor but its going to be
infintesimal. It would literally take many hundreds of volts to get a
reading that indicates a different sized coin, so here's a
1. Get a reliable 12V source. Make sure to measure it to as many
decimal points as possible and write that down.
2. Replace the bulb in your circuit with a 10meg standard resistor.
At this high of resistance, the current will be so tiny that a normal
1/4 watt carbon composition resistor like you buy at Radio Shack will
work just fine with no danger of overheating.
3. Knowing the exact voltage being supplied is critical. Turn on the
circuit and measure voltage across the resistor. Write it down.
4. Change the resistor for 1meg and repeat. Write it down.
5. Repeat this using 100K, 10K, 100, and finally 1 ohm resistors.
Remember that as the resistance goes down, the amperage goes up and so
the power factor increases, causing the resistor to heat up
significantly, so you'll need higher wattage resistors for the lower
values. Allow me to suggest the following:
Using Ohms Law: Voltage = V, Current = I, Resistance = R, Power = P
and ignoring the infintesimally small resistance of the coin for now:
V = R x I so R = V / I and I = V / R and for wattage, P = V x I
If V = 12.000 and R = 10meg (10,000,000 ohms) then:
V / R = 12 / 10meg = 1.2uA (microamps) and ....
P = V x I = 12 x 1.2uA = 14.4uW (microwatts) which is well below the
resistors .25W rating, so we know that's safe. Safety first.
Use this chart:
Using a 12V / 1A supply
10meg 1/4W resistor
20 ohms 7.2W THIS IS GOING TO GET HOT!
You're going to want to use two 5W power resistors at 10ohms each from
Radio Shack connected in series to build a 20 ohm resistor. By using
two instead of one big one, you reduce the amount of heat generated
however keep in mind that even a couple watts can cook an egg. The
point is use resistors properlly power rated so you don't produce
By measuring across the fixed value resistors you build a list of
voltages at different fixed resistances. Then simply plug those
values into Ohms Law and calculate the differing currents. From that,
you'll be able to determine the coins resistance to a more accurate
degree, also using Ohms Law:
That is exactly backwards. What is needed is a constant current
source (which ironically, the resistance lamp tends to provide)
and a precision voltmeter that is very sensitive. To increase
the voltage drop across the coin requires a high *current*.
Otherwise the voltage drop across the unknown but very low
resistance will be too low to be measured by the voltmeter.
Which indeed means that what you are suggesting simply is not going to
work (short of going out an buying a rather expensive voltmeter).
Yes, but you are not describing something that will work.
You are assuming that the 12V source will remain at exactly the same
voltage regardless of the current. That isn't going to happen.
Nothing you suggested is going to tell anyone what the resistance of
the coin is, or even come close.
The way to measure the resistance of the coin is to use a
constant current source (which in fact needs to be only that in
name, not in fact) and then measure the voltage drop across the
coin with sufficient current to give a significant reading on
whatever voltmeter is available.
In fact, a real "constant current" source is not needed because
the circuit is static, which means that a power supply with an
adjustable voltage is just fine if a way to set the current
accurately can be devised (and that is pretty easy to do). The
current measurement doesn't have to be calibrated either, it
just needs be something that can precisely set the current to
the same value every time.
Another way to do that is with a bridge circuit and a calibrated
standard resistor. But lacking a standard resistor, and lacking a
method of measuring one... that method isn't very helpful either.
The simpliest way is to use a "long" lead (long enough to have
about the same resistance as one of these coins to be measured)
to one side of the coin, and for each measurement adjust the
supply voltage to the same "standard value" of voltage drop
across that lead between the power supply and the coin. That
will mean (to the degree of precision available from the
voltmeter used) the exact same current will be going through the
different coins for each different measurement. Then, with a
"known" current through the coin, the voltage across the coin is
measured. The accuracy of this setup depends on how accurate
the amperage from the power supply is measured. The precision
depends on the accuracy of the voltmeter used to set the current
each time and to measure the drop across the coins.
On 18/01/2004 eye guy opined:-
The variation in each coins resistance will be such a very tiny
proportion of your total resistance that any small variation will be
unmeasurable. The resistance of your test leads plus the
interconnections, will make this method a non starter.
alexandre marsolais wrote on Sunday (18/01/2004) :
Misinformed rubbish! If that were true, power cables would be made
The only skin effect is that found in higher frequencies. At DC there
is no skin effect.
I see, so the higher the current I wish to pass through a cable, the
smaller that cable needs to be. You have just turned the whole
electrical engineering field on its head.
None of which will make a jot of difference....
The resistance of the coins will be so tiny by comparison with the
other resistances in the measurement process, that it will be
undectable by this method.
Measuring the resistance of a coin will be non-trivial because its resistance
is very low. You might want to start out by first guessing what the probable
resistance is, then figuring out how to measure it. Resistance of course (to be
technical, in an isotropic homogenious material -- that just means it's uniform
and there are no surprises) increases as the length of the conductor iucreases,
and decreases as the cross sectional area increases. So, the bigger area - area
of a circle is pi times r squared - the lower the resistance, and the the
longer the path, the greater.
The resistivity of copper (I'll us a copper coin as an example) is 0.0000017
ohm - cms at room temperature. So if we think about a copper coin that's .2 cm
thick and a centimeter across its resistance will be
.0000017 times .2 divided by ( pi times .25) or about .0000004 ohms (.4 micro
Suppose you had a digital voltmeter that could measure say .001 volts as its
smallest voltage increment. Then you have to say, if you wanted a reasonable
chance accuracy - like to measure the resistance within 10%, you'd have to put
enough current through this coin to get a voltage drop of .01 volts -- that's
10 millivolts. OK, the resistance .000004 ohms, so to get .01 volts you'd have
to pump 23,000 amperes through the coin. That, by the way, is 230 watts.
If you could measure microvolts instead of millivolts you'd get to 23 amperes
-- that's still a lot of current but a car pattery could do it for a while.
Other coins will be more resistive than the copper on I've just ran the numbers
on, but you can see it's not an easy problem. Moreover, because contact
resistances are so large you wouldn't be able to meaningfully stack coins to
get greater resistances.
I've judged science fairs, and I'd look very carefully at the experimental set
up if someone tried to do this. If the instrumentation was sensitive enough,
they'd have to be measuring the voltage drop across the coin with voltage leads
that isolated the coin's voltage drop from that of any contact resistance, for
I guess in the end I'm suggesting you find a better project -- this is is
is 1 volt pretty well.
You've received some interesting, and in some cases, incorrect
information. All coins are comparatively good conductors, so the amount
of DC resistance will be very small - at most only thousandths of an
ohm. The resistance is so small that your present measurement approach
cannot measure the small resistance differences between coins. In order
to accurately measure low resistances you will need to use a special
measurement technique known as a "four-wire" or "Kelvin" resistance
measurement. Otherwise the resistance of the test leads and contact
resistance to the coin will be much much larger than the resistance you
are trying to measure.
The technique passes a known current (I) through the coin with one pair
of test leads while the voltage drop (V = I*R) across the coin itself is
measured using a separate pair of test leads. See the ASCII drawing
below (use a non proportional font such as Courier):
| Apply a known current |
| I |
| --> ----------------------- |
--------->| Rcoin |
Um, current flows, not voltage and I'd love to know where you got info
that claims AC is on the outside and DC is on the inside. As far as
I've known all these years nobody has ever proved conclusively exactly
where in a conductor current actually flows. I rather suspsect its
Which is exactly why I said to measure across the various fixed
resistances then use the results, small as they may be, to calculate
the coin resistance. It's the simplest way to do it, but not the only
way. Since its a father/son project, my interests are to also make it
as safe as possible, and while a constant current source will indeed
work, the current necessary to develop a measurable voltage across the
coin would be considerably more than the bulb could withstand.
There are many inexpensive meters that go to 5 decimal points. I paid
$62.50 for mine 2 years ago.
Yes I am. Try it. His experiment was one of those we had to do in
lab while in school in the 70's.
It will if its reliable. My lab power supply is good for 0-36V @ 25A
and remains absolutely stable once set and even if it doesn't, with
each change of resistance, recalibration of the source is usually
necessary. I'm assuming that with each new setup, he recalibrates
the source voltage to as many decimal places as possible. That's what
most people would do.
Siure it will.
You're talking hundreds of amps. Not Safe for a father/son project.
I assume they don't have access to a HV lab.
You seem to have forgotten a few things in the 20-30 years since then.
As I mentioned to the OP in my first post (and as Bert Hickman
has mentioned also in a more recent post), what *you* need to do
is a web search on "low resistance" and the single word
"measurement" and do a search on "wire kelvin". The first will
turn up all sorts of information, the last will provide details
on the 4-wire Kelvin method of measuring low resistances.
That is the only sensible technique for the OP to use.