Melted wiring


Hi, all,

Apologies if this is an old topic, but I couldn't find anything relavent anywhere...

I've just moved my three 100watt Hella 220 jumbo's off the front and out of the way of all the rubbish and stuff and mounted them on the roofrack.

They operate via a 30 amp relay and worked fine on the front of the truck. Moving them up onto the roof, however, is a different matter!

After finishing all the mounting, I simply extended the feeder connections from the relay to the lights as it was before, but with a longer run of cable.

OK, that's my first mistake! The second mistake was checking that they worked!

The longer run of wire was not happy and promptly melted.

What cable should I be using for a distance of approx 3 metres?

I knew all that Electrical Eng stuff I did would be useful ....

The magic numbers .....

1 metre of copper wire of 1 square millimetre cross section has a resistance of 17 milliohms ( 0.017 ohms).

If you follow the recommendations for current loading of wires, the suggestion for mains wiring is 10 Amps per square millimetre max. Unfortunately the voltage drop of the wire for 12 Volt systems is much too great at that loading except for very short runs.

Take the example of 10 amps per square millimetre, 30 Amps load and a run of 4 metres. Assume the earth return through the bodywork is very low resistance in comparison.

The wire will be 3 square millimetres to take 30A.
Each metre of wire will therefore be 17 divided by 3, approx 6 milliohm.
4 metres of wire will give a resistance of 24 milliohms.
The voltage drop is the current times the resistance, ie 30x24=720 millivolts.
That's 0.72 Volts out of 12 available.

I would suggest that the max allowable voltage drop should be something like 0.25 volts, which would mean you going up to 10 square millimetres. This is a bit thick to handle, but a compromise of 6 square millimetres, giving a drop of about 0.35 volts would be acceptable.

From this example, you can work out the voltage drops for any other current loadings and lengths of run. Just remember to double the length if you use a separate wire of the same cross-section for earth return instead of the bodywork.

Oh, BTW, I'm sorry if I'm teaching granny to suck eggs, but Volts times Amps equals Watts (in a unity power factor system, which for dc, we don't need to worry about). Your 300 Watts of load at 12 V will therefore take 300 divided by 12 which equals 25 A. .... 6 square millimetres should be OK.

So, it was you who stayed awake in physics...

I've uprated the wiring to something I'd be happy putting 240 volts through, and sorted out a VERY dodgy earth, so now everything works fine. If I do get stuck in the mud as per usual, all I need to do is make sure it's nose up so I've got something to boil water on for a brew!!!

Thanks for your help!

Bob Hewitt