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 amps.
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?
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.
Tim
-- Deep Fryer: a very philosophical monk. Website:
--- 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 carry.
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--- 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:
E1 | [R1] | +----E2 | [R2] | GND
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:
120V | [R1] 1R | +----E2 | [R2] 12R | GND
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.
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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