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So I've gotten myself into a healthy, educational Facebook discussion. Link is here if you're a member of that group.

When I thought about it, everyone defines wire capacity by current only. The voltage only comes in when talking about wire insulation.

However, I've always thought about a wire's capacity by total power transmitted, not just the current.

Assuming arcs and corona discharges aren't a problem, could I transmit say 10 kW via a #40 AWG wire at 10mA, 1MV DC? The ampacity isn't violated, but that doesn't look right for wire that's around the width of a hair.

I'm aware of the math behind it (5th year Electronics engineering student) and I can judge what voltage/current I need for X watts versus safety/other constraints, but wires being rated by current only hasn't bothered me up until this point.

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  • \$\begingroup\$ the length of the wire determines the total thermal resistance, thus the temperature rise. \$\endgroup\$ – analogsystemsrf Aug 2 '19 at 5:14
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    \$\begingroup\$ Aare you sure about that @analogsystemsrf . Given that the cooling is a function of length as well, for a straight wire shifting power from A to B. Now if the wire was bundled up like a wire wound resistor, that would have a constant thermal resistance and power and temperature rise proporiotnal to length. \$\endgroup\$ – Neil_UK Aug 2 '19 at 5:24
  • \$\begingroup\$ If the wires are in a strongly insulating jacket, inside bundle of other hot wires, then the only heat-exit path is OUT THE ENDS. The OP stated no mechanical config. \$\endgroup\$ – analogsystemsrf Aug 2 '19 at 15:17
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    \$\begingroup\$ If you make all your 'ideal' assumptions about insulation, corona, safety etc etc, then the answer is - YES IT CAN. The current capacity is based on maximum tolerable heat / temperature rise. Nothing else. In fact this is precisely why long haul transmission is done at such high voltages - to keep the current low for a given power transfer. \$\endgroup\$ – Kripacharya Aug 2 '19 at 15:45
  • \$\begingroup\$ @Kripacharya I see, that makes a lot of sense! \$\endgroup\$ – PNDA Oct 17 at 3:38
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The wire will heat up because of its resistance and current passing through it, and this also makes a voltage drop over wire. The potential of the wire itself is irrelevant, assuming the insulation does not bŕeak down. 10mA flowing in a wire still heats up the same amount whether it comes from 1.5V battery or 1MV generator.

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However, I've always thought about a wire's capacity by total power transmitted, not just the current.

That is indeed your prerogative. You have every right as an independent human being to think how you like.

However, it's not useful to the rest of the engineering community, because the wire thermal limitation in amps is independent of the system voltage, and multiplication is easier to do than division

Of course, people want to shift power. However, voltage for any given system tends to be standardised, at 120, or 240, or 48v, or 33kV. Working in a particular distribution system, if you want twice as much power, you need twice the current rating. Easy peasy.

Wires have a maximum voltage determined by their breakdown, which is a function of insulation thickness and quality, and wire diameter. Once you've bought your reel of wire, there's nothing you can do about any of those terms.

A wire tends to have a minimum insulation thickness given by robustness, regardless of how low the rated voltage is.

Wires have a maximum current determined by their heating and cooling per unit length. If you string a single wire, or run a bundle of wires in an insulated conduit, the thermal resistance per unit length will be radically different. That's why the tables give different entries for single or bundled wires, and often different ratings depending on ambient temperature.

Imagine how it would be if as well as multiple columns for single versus bundled wires, the tables were of power, and had different columns for 12v, 120v, 240v, and then didn't have a column for your voltage.

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  • \$\begingroup\$ The more I read about the matter, the more I think it's just convention to rate by current rather than by voltage, and everyone agreed that "current causes voltage" rather than "voltage causes current". I think my confusion stems from having P=RI^2 and P=(V^2)/R, and having full control over both current and voltage. \$\endgroup\$ – PNDA Aug 2 '19 at 5:53
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    \$\begingroup\$ @Gene I missed the point slightly in my answer, edited. Forget about current causes voltage or vice versa, current and voltage happen at the same time, 'cause' is a loaded word that will lead you astray. It's not convention, it's usefulness that wires are rated for max current and max voltage. Power is simply the product of the actual values you use the wire at, and can be radically less than the theoretical maximum. \$\endgroup\$ – Neil_UK Aug 2 '19 at 7:06
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Sure, with a higher voltage every wire can transmit more power at the rated current than with a lower voltage. But how does this matter?

If you want to design a system and need some wiring there are probably two cases:

  • You know your voltage and current already, because they are given parameters: Just pick a wire that can handle the necessary current (or to be more precise: A wire that is staying in the limits for voltage drop / power dissipation).
  • You want to transmit a specific maximal energy: You would want to first choose a proper voltage for your design, that minimzes design costs (taking into account costs for voltage rating and cost for cable diameter). You still would not want to pick a cable based on some calucated max. power, but on the current you calculated with your design voltage.

Imagine having a power rating for a cable. It would give you no information about the current handling capabilites. You would have to also give the voltage, at which that power was calculated.

It is just easier to rate a cable my maximum current and also give a voltage rating for the insulation. That way, everybody can determine by themself what power he is able to transmit via this wire.

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The amount of heat dissipated when you pass a current through a wire is given by I²R, where R is the resistance of the wire. Notice that the heating goes up with the square of the current, and is completely unrelated to the supply voltage.

Suppose I have a piece of wire that the manufacturer has rated to 20A, with insulation good for 600V. 600x20 = 12kW, so the manufacturer could claim that it is 12kW wire.

Now suppose I use that wire on a 220V supply, and try to use it to run a 12kW industrial heater. The current is now 12000/220 = 54.5A. That's about 2.7 times the current that the wire was rated for, and so about 7.4 times the heat dissipated in the wire.

It's quite likely that the wire will overheat, and the insulation will melt.

It's much better to say that it is 20A wire, insulated to 600V, rather than give a wattage figure that is highly misleading.

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However, I've always thought about a wire's capacity by total power transmitted, not just the current.

You should think the other way around:
A wire should not be selected by total power transmitted, but by the total power NOT transmitted.

The total power NOT transmitted is defined by the (undesired) power dissipation in the wire ( \$I^2R\$).

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Assuming arcs and corona discharges aren't a problem, could I transmit say 10 kW via a #40 AWG wire at 10mA, 1MV DC? The ampacity isn't violated, but that doesn't look right for wire that's around the width of a hair.

That's correct, you could transfer a high power through a thin wire and not violate the current rating of that wire.

However, the caveat you make "Assuming arcs and corona discharges aren't a problem" is putting the finger on the weak spot. In order to make that happen, sufficient isolation is needed. One part of that isolation is spacing such that arcing etc. does not happen.

So to transfer high power through a thin wire you'd need very thick and large isolation between the two thin wires.

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Back ground:

So effect(Watt) is only dissipated in ohmic resistance (pure resistance). as the cable heats up / cools down, this resistance changes proportionally - as with "ohms" law, the current is the voltage / resistance.

Thought experiment time!:

Long cable

Think an infinitely long wire - huge resistance, huge voltage drop across said resistance. so back to "ohms" law, something very small over something very large equals something very small, so hardly any current passes the wire.

Small cable

Now think about an infinitesimal small wire, very low resistance, very low voltage drop - so something very large over something very small equals something very large - i.e. huge current -

Now think about it, we put a standard cable in outer space where its super cold - now the ohmic resistance drops to near zero (super conductor!) - We now approach the infinitesimal cable idea - Now we put the cable in Sahara - and the ohmic resistance increases many folds - now we are approaching the infinitely long cable.

The voltage means jack on its own, when you get zapped from walking on a carpet and you touch the door handle you discharge many kV (yes KILOvolts) - you have a fixed resistance around 100kohm - this should kill you! .. Or should it ? .. Obviously the answer is NO - this is because of time - the discharge happens so quickly that you will "barely" feel it.

Now you hook up your DC supply, 20 amps 0.2 volt -- but wait.. 20 amps ? .. I read somewhere that 0.03A (30 mA) is enough to kill you! .. Yes that is correct, but here the voltage is the limiter - not time.

Summary Environment - Length - Voltage - Resistance - Time .. These are the factors that matters. When you get advanced you will realize that the long cable is not possible because of other factors, but that is a story for another time.

Sorry for the wall of text.

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