Science fair question - Please help - no laughing

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.

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.

good luck

"eye guy" wrote in message news:Xns9473E1CF8EBF7eyeguy@24.25.9.41...

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.

Try these links:

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 be eliminated.

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 "work-around".

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.

1. Knowing the exact voltage being supplied is critical. Turn on the circuit and measure voltage across the resistor. Write it down.
2. Change the resistor for 1meg and repeat. Write it down.

1. 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

Resistance Power

10meg 1/4W resistor 100K 1/4W 1K 1/4W 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 smoke.

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:

Voltage flows on the outside only in AC. DC flow takes up the whole cunductor.

Coated coins, what world do you live on?!?

accumulative

Here I thought that was extremely well understood as to exactly where the current flow, as well as why and when...

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 hollow.

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 ohms).

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 example.

I guess in the end I'm suggesting you find a better project -- this is is pretty difficult.

AJW

is 1 volt pretty well.

Hi Ken,

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 both.

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.