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Wire Gauge ratings

Discussion in 'Electronic Basics' started by Jamie, Mar 7, 2006.

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  1. Jamie

    Jamie Guest

    I have a question about wire gauge ratings. Why do wire gauge charts
    show how many amps that the wire is able to carry, and not the wattage
    that the wire can dissipate safely? Resistors are rated in watts, not

    I would think that if I had a length of wire that was just barely
    capable of conducting a charge at the rate of 10 amps at 12 volts, that if
    I were to hook up 120 volts to the same piece of wire and attempted to run
    THAT charge at the rate of 10 amps, that, since P=I*E, the wire would be
    dissipating heat much more rapidly and could overheat.

    Why isn't voltage taken into account in determining the right diameter
    for a conductor in a circuit?

    - Jamie

    The Moon is Waxing Gibbous (55% of Full)
  2. Tim Williams

    Tim Williams Guest

    You certainly can, although it varies by insulation (plastic insulates
    thermally as well as electrically!).
    When you have several variables appearing in a problem, you can't just take
    two and multiply or divide; you have to pick and choose. You picked the
    wrong ones -- and that's why it doesn't work.

    You may ask that -- well, there are only two numbers in the question!
    True enough, but the others are hiding in the problem. When current flows
    through a wire, its resistance develops a voltage, and *this* voltage
    multiplied by the current equals the wattage.

    The line voltage does not appear across the wire itself. It appears outside
    of it, where there are insulators, so we could care less.

    -- Wires do, in fact, have a voltage rating, but that relates to the voltage
    the *insulation* is rated to handle, under a variety of conditions
    (temperature, solvents, crush, etc.), as rated by the manufacturer.
    Obviously, bare conductor cannot have a voltage rating; it would be
    completely arbitrary.

  3. Ban

    Ban Guest

    Good Lord
  4. John Fields

    John Fields Guest

    OK, lets say that you've got a 100 foot extension cord and it's
    rated to safely dissipate 10 watts. What are you going to do with
    that information? More importantly, how do you know what you can
    plug into it safely?

    On the other hand, if it's rated to safely carry 10 amps all you
    have to do is look at the nameplate of what you're going to plug
    into it to know how much current the extension cord will have to

    You forget that the wire isn't supposed to be a significant portion
    of the load. Using our 100 foot extension cord as an example, if it
    can safely dissipate 10 watts with 10 amps going through it, then
    the voltage dropped across it will be:

    P 10W
    E = --- = ----- = 1V,
    I 10A

    with the balance being dropped across the load, and its
    (the extension cord's) resistance will be:

    E 1V
    R = --- = ---- = 1 ohm,
    I 1A

    and the circuit will look like this:


    Where E1 is the supply voltage, E2 is the voltage appearing across
    the load, R1 is the resistance of the extension cord, and R2 is the
    resistance of the load.

    So, let's take a look at what actually happens with a load hooked up
    to our extension cord.

    First, if we have a 120V supply feeding a load which draws 10 amps,
    the load will look like:

    E 120V
    R --- = ----- = 12 ohms
    I 10A

    and our circuit will look like:

    [R1] 1R
    [R2] 12R

    Now, since the extension cord and the load are in series, that's a
    total of 13 ohms, so the current the extension cord will have to
    carry will be:

    E 120V
    I = --- = ------ ~ 9.2 amperes
    R 13R

    and instead of 120V appearing across the load, there'll be about 119
    there because of the ~ 1V drop in the cord.

    On the other hand, if you wanted to treat the extension cord like a
    resistor, all you'd have to do would be to short out the socket end
    of it and plug it in.

    Since the cord looks about like an ohm, when you first plugged it in
    you'd get:

    E 120V
    I = --- = ------ = 120 amperes
    R 1R

    through it, and the power it would be dissipating would be:

    P = IE = 120A * 120V = 14,400 watts!

    Because it's largely irrelevant since the bulk of the voltage is
    supposed to appear across the load. The diameter is important
    because that, in conjuction with the length of the wire will
    determine its resistance and that, in turn, will determine how hot
    the wire gets with a specified current running through it.
  5. Chris

    Chris Guest

    Good morning, Mr. Fields. If I might ask the master if he'd like a
    second cup of coffee?

    E 1V
    R = --- = ---- = .1 ohm,
    I 10A

    and then through to:

    E 120V
    I = --- = ------ = 1200 amperes
    R .1R


    P = IE = 1200A * 120V = 144,000 watts!!!

    (Extra exclamation points added -- it only makes your point to the OP
    more relevant.)

    I never post before the second cup of coffee anymore. I used to do
    this almost every morning.

    Cream or sugar, sir?

  6. John Fields

    John Fields Guest

  7. ehsjr

    ehsjr Guest

    P = I*E which is also equal to I^2*R

    In the math, voltage is automatically taken into account:
    E = I*R. Substitute I*R for E in your power equation and you get
    P = I*(I*R), which is more commonly written P=I^2*R. That means
    power is equal to current squared times resistance.

    The neat thing about this is that they don't need to
    know the source voltage when making a wire table.
    They know the resistance of the wire and can determine
    how many watts will be produced in it for a given
    current through it. They know how many watts the wire
    can safely dissipate, so it is simple to rate the
    wire in amps.

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