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Totally newbie question

Discussion in 'Electronic Design' started by SpiralCorp, Jun 28, 2007.

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  1. SpiralCorp

    SpiralCorp Guest

    Greetings duders, I've just recently taken up the study of electronics
    as a hobby and I have some very newbie questions. Since my only source
    of input is some random books, I need someone to go to with these
    queries. What better place than an electronics newsgroup, right?

    Ok, so I'm reading about resistances, cells and whatnot. And I stumble
    upon this line "...a resistor of about 1.000 ohms should always be
    used in series with the galvanometer in experiments of this kind,
    connecting the galvanometer directly will cause too much current to
    flow; possibly damaging the galvanometer and making the acid boil..."

    Clearly the experiment is meant to illustrate chemical energy cells.
    Now here's my question, why would a resistance stop it from boiling?
    Wouldn't it suck just as much energy from the acid as the galvanometer
    alone and dissipate it as heat?

    I'm kind of lost with the concept here... I guess I need someone to
    explain it from another perspective. I'm also interested in this
    because I've been wondering how you would safely draw a specific
    current and voltage from the utility mains. I know about voltage
    divider networks... but the same question remains, wouldn't the
    resistances just dissipate that ridiculous amount o energy that I'm
    not using? I must be missing something...

    Thanks in advance for the explanation.
  2. art of electronics by win hill, loads better than a newsgroup

  3. Tim Williams

    Tim Williams Guest

    "about 1.000 ohms" sounds really silly. The practice of indicating a
    precise number by extending the decimal completely negates the "about"
    calling for generality! Unless you meant ",", in which case "1,000" = 1k
    ohm, which is a more common value than a 0.1% tolerance "about 1 ohm"

    But anyway, it limits current. I = V/R. Increase R (where the galvanometer
    and cell have some finite resistance already) and I falls.

  4. A series resistor limits the current.
    A galvanometer (ammeter) has a very low resistance (ideally zero), so
    if you placed it directly across the battery it would effectively
    "short it out" causing a large current to flow through the
    galvanometer, much higher than it is designed to measure, so it would
    most likely get damaged.

    What book did you get that from?, sounds like some ancient physics
    text book :->
    Beginners cannot *safely* work on mains powered equipment. Stick to
    battery powered equipment or use a mains plugpack or bench power
    supply until you get the required experience to play with mains stuff.
    Pretty much. Putting a resistor and ammeter across a battery doesn't
    accomplish much except make the resistor heat up and the ammeter
    needle move!

  5. Bruce Varley

    Bruce Varley Guest

    Your source of information is almost surreal, it must be from the pre-1950s.
    Throw it away and get a good modern introduction to electronics. For
    galvanometer read a DVM, and for cells read any battery you can buy
    anywhere. BTW, when you're initially messing round with basic electronics,
    using batteries is a good idea. They're guaranteed safe, and guaranteed
    ripple free DC. Some plugpacks can deliver DC with AC ripple on the top when
    they're loaded, that can make for strange results.
  6. Guest

    In some Nordic countries "," is the decimal point and "." is the
    thousands seperator. So you sometimes see numbers like 1.000.000,5 in
    documents. I have no idea why things evolved this way :p
  7. Some calculators like various HP models even allow you to swap between
    the two.

  8. SpiralCorp

    SpiralCorp Guest

    Thanks for the replies everyone. I guess my completely out of context
    example does make the book seem awfully dated. The book is Teach
    Yourself Electricity & Electronics from the McGraw Hill folks. It just
    goes into a fair bit of historical detail in the beginning.

    As for the decimal point, slebetman is right. I'm from latin america
    and here we use the comma as the decimal point like so 1.000.000,001.
    Anyway, back on topic... so I'm right about the resistor just taking
    up all the heat? So it would be completely ridiculous to just put a
    couple of resistances in a voltage dividing network in order to get a
    specific voltage out of the utility mains? (I figured it might) And
    don't freak out, I'm not about to go plug anything into the mains yet.
    I'm just curious as to how it all works.
  9. It's not a completely ridiculous idea, voltage dividers are used all
    the time to generate specific voltages (often ratiometrically to a
    changing input voltage). It's just that they can be quite inefficient
    for any useful amount of current drain on the device you intend to
    power with it, and the output voltage will vary with the load you put
    on it (it therefore has poor "voltage regulation" with varying loads).
    They are mostly used to generate "reference" voltages that do not draw
    any significant current.
    It also gets tricky when you use a resistor divider for AC circuits,
    as capacitive effects come into play.

    Keep it up, you're on your way!

  10. ectoplasm

    ectoplasm Guest

    They're talking about acid, this means they assume you use an acid
    battery in the experiment.

    But OK. A galvanometer is used to measure tiny electrical currents,
    it's a delicate instrument. Any current (=Ampere) meter, should never
    be connected to any voltage source directly, because they have low
    resistance. Especially a galvanometer. Let's say your battery is 10
    volt, and the galvanometer is 1 ohm. It results in a current of I =
    U / R = 10 / 1 = 10 Ampere. The galvanometer will be blown away by
    this huge current; it can be used for small currents of maybe only
    0,001 Ampere !

    So that's why the resistor must be used: you put it *in series* with
    the galvanometer. You should know when you put things in series, the
    battery voltage will be divided over these things. And the biggest
    resistance takes the highest voltage. So in this case, most of the 10
    volts will be over the 1.000 (thousand) ohm resistor. Because its
    resistance is 1000 times higher than that of the galvanometer at 1
    ohm, it will take a thousand times higher voltage. The galvanometer
    will take only about 1/1000 of 10 volts: 0,01 volt. The current
    flowing through it now will be I = U / R = 0,01 / 1 = 0,01 Ampere.
    The voltage divider would bring down the voltage, but as soon as you
    start to draw current from the lowered voltage, you are changing the
    situation again, the output voltage will change.
  11. joseph2k

    joseph2k Guest

    Since it would be more in standing with your knowledge level, please refer
    your posting to "sci.electronics.basics". This newsgroup (not a chat room,
    not a website, not a blog, not a listserver) is targeted at better
    technicians through better engineers.
  12. neon


    Oct 21, 2006
    galvanometers are raly sensitive devices in th eorder of 50uamps su as a prcation a resistor will limit tha current to a small movement.
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