Current through Mains Cable

Define "safely".

The power being lost into the wire is determined by the square of the current. So if you double the current, the wire will have to get rid of

A quick bit of arithmetic gives me about 17W per metre at 25A. It probably won't catch fire unless you enclose it in something, but it is going to get pretty warm..

So just run two of the cables in parallel, joining them together at each end.

No, the quoted voltage rating of a cable is the maximum voltage that can be applied between it's cores (AC or DC).

The quoted current rating is the maximum current that each core can carry.

So, your 240V, 13A cable would be safe to use at 24V, but only up to a maximum current of 13A

woodglass...

Cable may not be up to the degree of vibration, chaffing, heat, and oil/fuel contamination that might be expected of cable used in cars.

Power dissipated in the wire: P = I^2R where I is the current and R is the resistance of the wire. Notice how you don't need voltage to arrive at the answer, when you know I.

Therefore, a wire rated at no more than 13 amps cannot carry 25 amps, regardless of voltage.

Ed

Thanks for the replies everyone - It's nice to drop-in on a newsgroup which still values helpfulness and facts. Looks like I either need to double-up the conductors or go out and buy fatter cable.

On Sun, 01 Oct 2006 05:25:42 GMT ehsjr wrote: | David Skinner wrote: |> Hi, quick question if you guys don't mind... |> |> If a piece of mains cable can carry 240VAC at 13A safely, could the same |> cable carry 12VDC at 25A safely? |> |> Just looking to muck around with a couple of bits of kit out of a car - |> it's not for a permanent installation anywhere. I just don't want it to |> catch fire. | | Power dissipated in the wire: P = I^2R | where I is the current and R is the resistance | of the wire. Notice how you don't need voltage | to arrive at the answer, when you know I.

Indeed. The conditions of resistance and current _define_ the voltage that is across the wire over that distance.

On Sun, 1 Oct 2006 10:50:39 +0100 David Skinner wrote: | In article , snipped-for-privacy@bellatlantic.net says... | |> Power dissipated in the wire: P = I^2R |> where I is the current and R is the resistance |> of the wire. Notice how you don't need voltage |> to arrive at the answer, when you know I. |> |> Therefore, a wire rated at no more than 13 amps |> cannot carry 25 amps, regardless of voltage. | | Thanks for the replies everyone - It's nice to drop-in on a newsgroup | which still values helpfulness and facts. Looks like I either need to | double-up the conductors or go out and buy fatter cable.

I don't know about UK, but in the US, doubling up the conductors is not permitted unless the conductor size is already rather large (well beyond the 13 amps level).

If ring circuits are still permitted in the UK, maybe you can set that up if it's not aready so.

This is a 12v application.

Not very relevant to OP's situation, but since you asked... It is permitted in the UK, but the extra protection required to ensure each conductor isn't overloaded (generally a multi-pole ganged breaker at both ends of the cable) means it's rarely economic to use it just because you have lots of undersized cable around.

A ring circuit doesn't count under our Conductors in Parallel regs, as the cable size used is significantly larger than that required just to share the max load in a parallel cable run.