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Science fair question - Please help - no laughing

Discussion in 'Electrical Engineering' started by eye guy, Jan 18, 2004.

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  1. eye guy

    eye guy Guest

    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.

    Ken
     
  2. Todd Howard

    Todd Howard Guest

    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....
    Good Luck
    TH
     
  3. 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
     
  4. EEng

    EEng Guest

    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.
     
  5. EEng

    EEng Guest

    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.

    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

    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:
     
  6. Brian

    Brian Guest

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

    Coated coins, what world do you live on?!?
     
  7. 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.
     
  8. 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.
     
  9. Bert Hickman

    Bert Hickman Guest

    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 |<----------
    -----------------------
    ^ + - ^
    | V |
    | |
    | |
    | Measure |
    the Voltage

    You will encounter a couple of challenges. For small (ampere level)
    current levels, the voltage developed across the coin will be very small
    - of the order of millionths or thousandths of a volt, so you will need
    to borrow a precision digital voltmeter. Another possibility would be to
    use an analog galvanometer (a bit more difficult but possible). You'll
    also need to come up with a current source that can deliver a constant
    current in the range of perhaps 10-20 amperes while you take the
    measurement. This could be a small DC power supply and a known value
    power resistor or perhaps a battery and a power resistor. In either
    case, you'll want to simultaneously measure the applied current. This
    can be done by either using a DC ammeter or by measuring the voltage
    drop across the power resistor that's connected in the current path.

    See the following for more information on how this test can be performed
    and the theory of operation:
    http://www.allaboutcircuits.com/vol_1/chpt_8/9.html
    http://www.cirris.com/testing/resistance/fourwire.html

    Coins will differ significantly in resistance depending upon the alloys
    used and construction. Copper and aluminum are excellent conductors,
    while steel and certain nickel-copper alloys are relatively poor
    conductors. Some coins are actually "sandwiches" of a combination of
    metals (US dimes and quarters), others are plated (US penny newer than
    1982), and others may have an outer ring of one alloy and an inner disk
    of another (Canadian $2 coin) or 1 Euro and 2 Euro coins. The resistance
    of coins does vary significantly. How do you think I know this? :^)

    Good luck and best regards,

    -- Bert --
    --
     
  10. EEng

    EEng Guest

    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.
     
  11. EEng

    EEng Guest

    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.
     
  12. Brian Reay

    Brian Reay Guest

    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.

    --
    73
    Brian
    G8OSN
    www.g8osn.org.uk
    www.amateurradiotraining.org.uk for FREE training material for all UK
    amateur radio licences
    www.phoenixradioclub.org.uk - a RADIO club specifically for those wishing
    to learn more about amateur radio
     
  13. Chimera

    Chimera Guest

    Very smart. Very smart indeed.

    I spent ages mulling over this and knew I'd missed something.

    Chimera
     
  14. Kilowatt

    Kilowatt Guest

    You need to put many of the same type of coin together and read from one end
    to the other.
     
  15. EEng

    EEng Guest

    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.
     
  16. EEng

    EEng Guest

    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.
     
  17. Guest

    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?
     
  18. Ben Miller

    Ben Miller Guest

    (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
     
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