Simple theory question on capacity of wiring

[Jon K.]

First post, simple question. Oddly enough, there is great dissention amongst my circle of friends as to just what forces are at work here. Of note, neither myself nor my friends are electrical engineers . . . . . obviously.

I have a power source. . . . . . say 110 V AC.

I have a 12 gauge wire running from said power source to a load.

The 12 gauge wire will comfortably carry 10 amps.

If I run another identical wire in parallel from the same power source to the same load,

Did I just double my capacity? Can I now carry 20 amps?

Thank you for any wisdom!

 

[berkeman]

First post, simple question. Oddly enough, there is great dissention amongst my circle of friends as to just what forces are at work here. Of note, none of my friends are electrical engineers . . . . . obviously.

I have a power source. . . . . . say 110 V AC.

I have a 12 gauge wire running from said power source to a load.

The 12 gauge wire will comfortably carry 10 amps.

If I run another identical wire in parallel from the same power source to the same load,

Did I just double my capacity? Can I now carry 20 amps?

Thank you for any wisdom!

Welcome to the PF.

That is pretty much correct, as long as the two sets of wires are separate enough from each other that their self-heating does not combine.

In general, the current-carrying capacity of a wire is based on its thermal rise from the I^2 * R losses from that current. You don't want the wire to heat up so much that it catches the insulation on fire, or anything else that is nearby on fire (like wood in a house). As long as the two pairs of wires are far enough apart that they cannot heat each other up (thus reducing the overall current rating of the two sets of wires), then you can run 2 * 10A = 20A.

 

[MATLABdude]

I'll add another asterisk to Berkeman's post: if you're using two wires to split current, but one has significantly more resistance than the other (e.g. it's thinner, or the ends have some corrosion, or it's much longer, or it's been kinked somewhere) the current will not split equally (more or less) between the two. This can be problematic if you expect the total current to be higher than the ampacity of a single conductor.

 

[es1]

One way to know how to derate the cable for the effects MATLABdude highlights is to use MIL-STD-975. There are some caveats about temperature, cable insulation, and the ambient environment.

This handy calculator does the math for you.
http://circuitcalculator.com/wordpress/2007/09/20/wire-parameter-calculator/

The awesome part is one of the notes for the main table. It has this equation.
IBW = ISW x (29-N)/28

IBW -> Current capacity of each wire in the bundle
ISW -> Current capacity of one wire
N -> Number of wires in bundle (for 2 <= N < 15)

I've always wondered how they came up with that equation. I am pretty sure it is just empirical. i.e. Some dude measured a bunch of cables, looked at the data then said "this will do".

 

[Jon K.]

Thank you guys for the replies.
es1, that is indeed a handy calculator. I will be sure to bookmark it for future reference.
Jon

 

[jaymin nayak]

i want to know that if we provide more & more insulation on wire than it's current capacity will increase or not? What is the effect of insulation on current carryin cacity?

 

[yungman]

i want to know that if we provide more & more insulation on wire than it's current capacity will increase or not? What is the effect of insulation on current carryin cacity?

no, insulation do not increase current carrying capability. In fact extreme insulation will hurt current carrying capability. Heat generated when current pass through the wire, the better it is for heat to be conducted away from the wire, the more the current handling capability. You put a lot of insulation around the wire, you trap the heat and it will cause the wire to burn sooner. It is not the current that burn the wire, it's the heat that generated from current passing through the resistance of the wire that ultimately burn the wire.

On the side note, why you need 12 gauge big wire to carry say 10A, but in the integrated circuit level or in power transistor, they use very thin wire like 30+ gauge and they spec the device can carry 10A?

The reason is the bonding wire in the integrated circuit is very short, even though the resistance per unit length is much higher than the 12 gauge, but the total resistance of the bonding wind is very very low because it is very short. It does not generate enough heat to burn anything. It is not the size of the wire, it is the resistance of the wire that create heat when current passes through that burn the wire. If you can keep the temperature of the wire low, you can pass a lot more current in a small wire that otherwise need a much bigger wire.