Voltage ratings for wire - Page 1
Forum section: electronics / beginners

author: Weez
author_rank: Contributor
timestamp: March 02, 2014, 07:25:17 pm
content: Hey guys,I have another noob question. In my text, there is an AWG chart which shows resistances per 1k ft and CM for many different gauges. It gives some formulas, explains that for a given voltage, a larger wire can carry more current, because it has less resistance than a smaller wire. OK, easy enough, I get that.But it doesn't really get into voltage ratings specifically.Let's say I have a piece of wire, and it's rated at, say, 500v. If I take another piece of the same kind of wire and twist the two together, have I just created a wire capable of handling 1kv? Maybe not quite double (perhaps dependent upon the gauge)?I'll google this later, figured I'd ask here first.

author: mdszy
author_rank: Unknown
timestamp: March 02, 2014, 07:29:12 pm
content: There's no such thing as wire voltage ratings, I don't know where you're getting that from. I can send a million volts at 1nA through a wire and only get 1mW, which any wire can handle. Arcing, of course, is another thing, but that's not dependent on the wire you use...

author: johnk
author_rank: Contributor
timestamp: March 02, 2014, 07:35:38 pm
content: Hi Weez,I think you've got something wrong here.The 500V rating may means that there should be no greater voltage difference between the wire and a surrounding piece of metal than 500V (at least that's what I would guess). I don't think you will be able to drop 500V across a wireTo get the voltage drop over the wire just calculate it by multiplying the current that flows through the wire with its resistance (U=R*I), if that's what you wanted to know.Also you should be very carefull when you handle 1kV. Better ask someone who knows about isolation and so on.

author: Weez
author_rank: Contributor
timestamp: March 02, 2014, 07:45:27 pm
content: Quote from: mdszy on March 02, 2014, 07:29:12 pmI'm getting it from having seen it listed in product descriptions, here's an example:https://www.tubedepot.com/products/18-ga-solid-cloth-wire-600-volt-ratingQuote from: johnk on March 02, 2014, 07:35:38 pmI wasn't referring to voltage drop, just the "rating".I'm not working with anything close to 1kv, was just using that value as an example.

author: olsenn
author_rank: Frequent Contributor
timestamp: March 02, 2014, 07:50:44 pm
content: So your textbook is mentioning the various resistances of different diameter wires, all 1000 feet long. The power that is dissipated in those wires is P = V*I = V^2/R = I^2*R. Since the power loss in a wire is purely resistive/real, it will all be dissipated as heat. All of these wires are the same length (1000ft) but the thicker the wire, the less resistance; thus the more current or voltage (according to the previous equation) is needed to create the same amount of heat (power loss) in those wires.

author: sacherjj
author_rank: Frequent Contributor
timestamp: March 02, 2014, 07:52:03 pm
content: Voltage ratings for wire are ratings on the insulation of the wire. At a certain point, the voltage would arc out of the insulation.

author: Dave Turner
author_rank: Frequent Contributor
timestamp: March 02, 2014, 08:03:49 pm
content: Power rating would be nice though. Any wire melts if it gets hot enough.

author: electronics man
author_rank: Frequent Contributor
timestamp: March 02, 2014, 08:23:39 pm
content: the voltage rating is the insulations breakdown voltage (or just lower than it) the metel its self can carry any voltage its the insulation that cannot whithstand it .

author: Weez
author_rank: Contributor
timestamp: March 02, 2014, 08:59:34 pm
content: Alright cool, tyvm guysThis does beg another question, though, based on your replies...wouldn't a better way of rating be power, instead of volts?It seems that, if the current is low enough, the voltage can be safely exceeded?Also, if you were to take two lengths of wire and strip them of their original insulation, then twist them together and re-cover them with some heat shrink with specs equivalent to the original insulation...you now have a single wire with a greater mass, so its "rating" is now greater...correct?I appreciate all your replies, thanks again...maybe I should stop trying to think outside the box

author: electronics man
author_rank: Frequent Contributor
timestamp: March 02, 2014, 09:29:44 pm
content: no, loss is due to current and not power, that is how electrical transmission works, they up the voltage to decreas the current while having the same power. if you were to strip a wire of isulation then there is no limit to the voltage a wire can carry. you can have a massive voltage and a low current = high power and, low voltage and high current = high power, they will both fail but due to different reasons, the first ones insulation will fail and the second one will get very hot.

author: IanB
author_rank: Super Contributor
timestamp: March 02, 2014, 09:32:25 pm
content: Quote from: Weez on March 02, 2014, 08:59:34 pmNo. You are still not getting it.The wire itself carries current. Current flowing through the wire does two things: it heats up the wire, and it causes a loss of voltage along the length of the wire. Making the conductor thicker causes less heat and less voltage drop. Note that is is purely current that does both of these things. Voltage doesn't come into it at all. Whether the wire is carrying 10 volts or 1000 volts, it makes no difference.Where voltage does come into it is in the insulation of the wire. The insulation is designed to keep the current inside the wire and stop it escaping. This leads to the insulation having a voltage breakdown rating, which is usually the normal voltage expected on the wire plus a large safety margin. If you damage or remove the insulation you will create a weak point and spoil the safety rating. You can never safely exceed the voltage rating on the insulation, whatever the current is inside the wire.

author: w2aew
author_rank: Super Contributor
timestamp: March 02, 2014, 09:33:47 pm
content: Quote from: Weez on March 02, 2014, 08:59:34 pmNo, it doesn't work that way.Wire gauge/diameter determines its resistance. The resistance determines how much power is dissipated (P=I^2 * R). The current flowing in the wire depends on the voltage on the wire, and the load placed on it. Thus, wire gauge determines current capability (independent of voltage).The insulation on the wire determines the voltage rating, to prevent insulation breakdown and arcing. Voltage rating has nothing to with the gauge of the wire.When joining/splicing wire, the series combination will have a current rating of the smaller diameter of the two wires. Voltage rating will be determined by the weaker of the insulation of either wire or the insulation applied over the splice.

author: sacherjj
author_rank: Frequent Contributor
timestamp: March 02, 2014, 09:36:31 pm
content: Quote from: Weez on March 02, 2014, 08:59:34 pmCurrent is low, because your resistance is high with whatever is hooked to the wire. If you exceed the insulation voltage, you can let the high voltage find a shorter path for more current through the air or close by metal. This is why wires are rated by current AND voltage. Because power doesn't mean anything.Voltage is rating the insulation breakdown point. When you exceed this, current arcs out of the wire somewhere.Current is rating the conductor current handling capability within a safe temperature. When you exceed this, things catch fire nearby.QuoteIf your insulation is the same spec as the wire, you still have a voltage rating the same. However, you have doubled your current rating. (Maybe slightly less than double, because they are both heating each other.)

author: Rigby
author_rank: Super Contributor
timestamp: March 02, 2014, 09:48:51 pm
content: Quote from: sacherjj on March 02, 2014, 07:52:03 pmBingo. The current rating is for the metal conducting portion, the voltage rating is the insulation rating.

author: uwezi
author_rank: Unknown
timestamp: March 02, 2014, 09:58:16 pm
content: Quote from: Dave Turner on March 02, 2014, 08:03:49 pmBut this would be the power consumed by the wire and not the power consumed by something connected to the wire - and hence such a rating would not be very practical.Actually the current rating with the specific resistance is a kind of power rating. It tells you the power in W per cm, km or k ft (never seen that before) dissipated in the wire which will lead to self-heating of the wire.

author: Monkeh
author_rank: Super Contributor
timestamp: March 02, 2014, 10:00:09 pm
content: Quote from: electronics man on March 02, 2014, 08:23:39 pmFar, far lower than the breakdown voltage.

author: G7PSK
author_rank: Super Contributor
timestamp: March 02, 2014, 10:01:05 pm
content: Current rating of the wire is not just on the wire size but also on the usage, or positioning of the wire it will be lower if the wire is bunched or in a conduit or enclosed space and if it's run horizontal or vertical voltage rating is purely on the insulation although that would be lowered in many cases if the wire is run at a higher temperature.If you go to a good wire manufacturers website they will give all the ratings and de-ratings as required for various installation types.http://www.usawire-cable.com/index.html

author: Dave Turner
author_rank: Frequent Contributor
timestamp: March 02, 2014, 10:47:19 pm
content: Good Point! It is usually difficult to explain to management that just because power trunking is half empty, it doesn't necessarily mean that more cables can be put in it and that more trunking has to be paid for.

author: Weez
author_rank: Contributor
timestamp: March 02, 2014, 11:45:48 pm
content: Lol, ok thanks guys...I dunno if I'll ever have to trunk my wires, whether horizontally or vertically, but I'll keep that in mind in case I ever doSeriously, I do appreciate the comments.So now I know that voltage rating is referring to the insulation, and it is what it is and can't be exceeded.Btw Alan, I'm subscribed to your youtube channel, it's great!

author: IanB
author_rank: Super Contributor
timestamp: March 03, 2014, 01:42:49 am
content: But do you understand why power doesn't make sense?Let's look at an example to explain that.Suppose we have a device that consumes 1200 watts and we want to power it through a length of cable that has a total resistance of 0.1 ohms (it's quite thin, crap cable...).In the first case, let's suppose our device runs off 120 V, so it needs 10 A. In that case the heat generated in the wire (power lost) will be 102x 0.1 = 10 W. That may not seem too bad, but it could easily make the cable hot if the heat cannot escape.In the second case, let's suppose our device runs off 240 V, so it needs 5 A. In this case the heat generated in the wire will be 52x 0.1 = 2.5 W. It's only a quarter of the heat generated and power lost, even though the cable is the same and the load is consuming the same power.What we see here is that the current through the cable is what matters. If you double the current, you quadruple the heat generated. Clearly, therefore, the cable must be sized for the current expected.Load power doesn't matter, since the power delivered was the same in both examples above.

author: Weez
author_rank: Contributor
timestamp: March 03, 2014, 09:56:56 pm
content: Yes Ian, that example makes it much clearer and easier to understand.