If you have cable rated, say, 10A, and you were to connect two lengths of that cable between both poles of the load and the source would you then have 20A of maximum rated capacity or still 10A?

10A I think

Yes….

….20A – ignore Captaintomo 😆

Current rating is to do with power dissipation in the cable ie heating.

Power = I^2 x R

Where I is current and R is resistance of the cable.

Voltage rating is do to with insulation breakdown.

So no; 10A limit is independant of voltage.

So four lengths in total? (two on + and two on -)
If so, each will take 10A, so 20A in total.
It’s down to the cross sectional area.

OK, ta.

So dumbass questino part II

4x 55W halogen lamps on a 12v supply. About 3m of cable required.

Will 20A cable suffice?

How much total wattage will the 20A 12v cable take?

EDIT: richmars, missed your post. But that’s what I was wondering, 2 on + and 2 on – does that give me 20A total?

Might have missread question, two 10A cables in parallel will give you a 20A cable.

Power (wattage) is:

P = I x V

P = I^2 x R

P = V^2 / R

EDIT all derived using V=IR substituted into P=VI

Yeah I didn’t understand you either. I now change my answer to 20a

Right – so that probably clears things up.

If I have a distribution panel that has, say:

4x 3W LED lamps
1x 1Amp 12v USB charger (Cigarette socket)
1x 20W incandescent bulb
2x 55W halogen lamps
4x 55W halogen lamps
and 2x 20W speakers driven by a nominal 500W mini amp

means that the peak draw from the panel if everything is on is 12+12+20+110+220+40W = 418W

Which in theory could require 34A of current if they were all on at the same time.

So if I were to wire 2x 20A cables to each pole to the panel and then single 3A, 10A or 20A cable as needed to each circuit that would be OK?

Chances of having all on at once are zero, but best to plan….

AC or DC?
Power in AC isn’t the same as in DC. (phases shift between V and I, I think, but I’m no sparky).
With DC, you need 2 wires ‘each way’, from the supply and back.

this is all 12v dc automotive.

Also put a fuse in, that’s a lot of A’s.

4x 55W halogen lamps on a 12v supply

So…..Watts = Amps x Volts

Amps = Watts/Volts = (4 x 55)/12 = 220/12 = 18.333 Amps.

20 Amps will be enough.

Hmmm you’ve edited your post since I wrote the above. However the principle still applies and I’m sure you can work it out.

aye.

Was going to use 3 or 4 of these.

one each on the two 20A lines from battery to panel, and one each on the lighting circuits.

Fuse looks neat.
Not that it matters, but I think the power of the speakers isn’t the same ‘sort’ as the others.

the amp is rated 500W, but there’s no way the diddy little think is drawing that. Id be surprised if it’s more than 1A so I guess you’re right and it’s more like 10-20W.

12 volts x 20 amps goves you a max load of 240Wats.

The cable will present a small load but the lamps are a total of 220 Watts.

Its tight but should be ok.

Going back to the original question; Generally, if the cables are of the same size (gauge), same construction (stranding), same material, same length, same type of termination, run in parallel path, and are properly physically connected properly (clean terminals, no surface contaminant, same bolt torque, etc.), then there should be minor differences in the current carried by the individual cables, and so running two 10A cables in parallel should in theory give a 20A max rating. However if there is a slight imbalance between the two, e.g. one cable isn’t terminated quite as cleanly as the other or got pinched by accident and has a difference cross-section, then one cable would attempt to draw more than the other.

Can you separate the 4x 55W halogen lamps into two 2x 55w? Or use 20A cable for that circuit?

Spent today working on rewiring my mates comp safari racer, well mostly undoing the mess of spaghetti vaguely disguised as wiring by the previous owner, ready to put the new stuff in next weekend. I tend to work on over spec’ing wiring, especially for lights. For instance the 100W spots are wired with independent supplies with 25A cable, and 30A relays (although the independence is for failure resiliance as much as load). Much better to minimise the resistance and hence losses between the battery source and the lights to maximise the light output. Do I recall you fitting lighting loom kit with relays etc to your transporter to improve the light output? Same princilple here.

As Sum says use one run of the right rated cable, safer in the long run if one of the cables/connections becomes a high resistance point.

Cheers Sum & mark90

yes I did uprate the lamp cabling in the T4.

This exercise is for my Landrover 110. I’m putting together an auxiliary electrics panel to manage all the non-MOT stuff. I want to directly wire it to the battery with a panel power switch and then everything else on toggle switches from there.

Given the low chance of actually drawing more than 20A at any one time, I could just use one 20A wire, but since Im getting a roll if wiring two 20A wires to the panel (with idenitcal runs, connectors etc etc, natch) give more redundancy then it’s cheaper for me than buying some 30A as well.

thanks for the help.

In that case I would use a double pole switch for the panel power switch to keep each 20A feed independent. Then have the lights 2 per feed circuit, and split the other smaller loads across the 2.

good idea.

double pole can have two input and two output cant it?

Yes, and each pair are seperate, like two switches in one.

great…off to ebay…