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American wire gauge chart is rated in Ampacity. Isn't it wise to be rated in watts?

Let say a wire is rated to conduct a certain amount of current what's going to happen when the voltage goes higher and higher?

edit: My question is concerning the AWG chart that I think should be in watt or VA but after saying that who am I to dare.

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  • \$\begingroup\$ The voltage you apply to the circuit isn't usually dropping across the wires, but across a load. The current that flows through the load however is also running through the wires. The voltage drop across the wires is not a quantity that is part of the actual circuit design. The load current and thus the current through the wire, is. \$\endgroup\$ Dec 5 '20 at 13:42
  • \$\begingroup\$ Also longer cables can dissipate more heat. A power dissipation rating based on the wire diameter doesn't make much sense. \$\endgroup\$ Dec 5 '20 at 13:47
  • \$\begingroup\$ Why are fuses (basically wires) rated in amps. Think about it. \$\endgroup\$
    – Andy aka
    Dec 5 '20 at 13:54
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Look at it like this. The power delivered to the load doesn't matter. Only the power dissipated in the wire.

Here's an extreme example.

I have wire rated for 1 ampere with insulation rated for 1000VDC.

The first circuit uses 1000V into a load of 1000 ohms, for a current of 1 ampere. It's delivering 1000 watts to the load. (More or less.)

schematic

simulate this circuit – Schematic created using CircuitLab

The second circuit uses only 10 volts into a 1 ohm load. That's 10 amperes of current for a load power of 100 watts.

The load power is lower, but that poor 1 ampere wire is going to melt its insulation off if not straight up catch on fire - despite delivering less power to the load.

The load power doesn't matter.

It is the power dissipated in the wire that matters, and that depends only on the current and the wire properties (resistance of the wire itself.)

Calculating the actual power dissipated in the wire requires knowing the resistance per foot for the wire and the length of the wire. You'd need a table for various lengths as well as for the various wire sizes - and it wouldn't tell you anything more useful than the table with only amperes and wire size.


Yes, I ignored the losses in the wires when calculating the load power while depending on the losses in the wire for the heating. It's a simplification.

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Current flowing in the wire heats it via resistive losses (RI^2).

What determines the usual wire gauge current ratings is basically how much resistive heating it can take before the heat damages the insulation. Thus a wire inside a conduit without any airflow can safely conduct much less current than a wire cooled by the airflow of a fan. Likewise an uninsulated wire, or a enameled wire, can withstand a higher temperature therefore higher current, because the insulation (or lack thereof) withstands higher temperatures.

Wattage doesn't matter, since power is volts*amps, and volts don't heat wires, amps do.

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  • \$\begingroup\$ "Likewise an uninsulated wire, or a enameled wire, can withstand a higher temperature" so let say a enameled wire rated for 10 amps can handel 10 amps from 10volts up to 10kvolts? \$\endgroup\$
    – autodidact
    Dec 5 '20 at 13:23
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    \$\begingroup\$ @autodidact Voltage is a differential measurement, the reference point is important. It doesn't matter whether your supply voltage is 10V or 10kV relative to some other point (e.g. another cable), the voltage drop across the length of the cable will be the same as it's proportional to current (\$V_{drop} = I_{cable}R_{copper}\$), and not supply voltage. Where supply voltage matters is whether the insulation can withstand the voltage differential to another point (e.g. ground, adjacent contductor, air, etc.) without breaking down which is a seperate rating unrelated to AWG. \$\endgroup\$ Dec 5 '20 at 13:29
  • \$\begingroup\$ @bobflux - "Wattage doesn't matter, since power is volts*amps, and volts don't heat wires, amps do." - erm, it is power dissipation, in watts, that causes heating - \$P=I^2R\$. This power is equivalent to \$P=IV_{drop}\$, where \$V_{drop}=I_{cable}R_{copper}\$ is the voltage drop along a length of cable caused by a current flowing through it. The voltage is irrelevant from a rating point of view as it can be directly calculated from the current rating which is specificed based on an allowed temperature rise (power dissipation) for a given resistance of cable, which is proportional to area. \$\endgroup\$ Dec 5 '20 at 13:36
  • \$\begingroup\$ @TomCarpenter we all agree with you but the point is does a awg chart has to be rated in amps or in watts? \$\endgroup\$
    – autodidact
    Dec 5 '20 at 13:40
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    \$\begingroup\$ @autodidact: Amperes. All you really care about is "will this wire carry X amount of current?" The power can be calculated from the current and the wire resistance if you need it, but nobody is interested in the power. The usual question is "will this 18 AWG wire safely carry 20 amperes?" I don't care how much power the cable dissipates, only how much current I can push through it with out it catching fire. \$\endgroup\$
    – JRE
    Dec 5 '20 at 13:55
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The wires have some resistance. They dissipate some energy to heat and the heating power of the wires is R*I^2 where R is the resistance of the wires and I is the current. Of course, if you have a certain prepared piece of cable you can find how much dissipation power in the wires you can allow before the cable becomes too hot (=the air cannot cool it as fast as the heat is generated). But that wattage gets doubled as soon as you double the cable length because doubling the length also doubles the cooling surface.

Instead of watts quantity watts/foot or watts per meter would be scientifically ok rating how much dissipation in wires can be allowed. But making practical calculations is complex by using it. You must know the resistance per foot and the current to calculate the dissipated power in the wires per foot with formula P=R*I^2.

One should see that telling maximum allowed amperes for a cable type is an equivalent rating because it holds inside it the allowed dissipation power per foot and the resistance per foot automatically. In math you find that fact by dividing both sides of the equation P=R*/I^2 by the length of the cable.

About the wattage and voltage: It's possible that you have not kept things clear in your thoughts. The voltage (=U) where a device is connected to via its cable causes current (=I) and the wattage (=U * I) is a sum of how much is dissipated in wires and how many watts goes to the device. With formula P=R*I^2 you can keep the dissipation power in the wires separated assuming you know the current and total resistance of the wires.

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