scruffythewild

Hey, This is just a general question and I'm not sure if this belongs in chat or here but mods please move it if needed.. Anywho, I have had bad experiences using smaller wire then required and almost starting fires. I get very paranoid about this now. But my story is I have 2 general power lines to my battery opposed to one combined line to prevent fire. I have one for my Strobe Lights and Special Horns etc... (First Responder I am allowed to have and operate them.) This is a smaller gauge wire and is fine. (Fused and all). Then I have another for my fog lights with a higher gauge wire and I run a whole other line to prevent any high power electricity from getting to my lower gauge wire. But my question is ultimately the wires end up at the same place and Touch each other at the positive battery terminal. Now is it possible that the electricity from the higher current wire can still transmit to the lower power line when they meet and the battery terminal and completely defeat my method of running different wires? Thanks for any help or even input...Rob

By the way for me even though its the opposite in my mind when I say higher gauge wire I mean it can handle more and vica versa...(to me a 10 is higher then 14 gauge wire etc...)

SoCalxJ4.0

They are going to pull the same amount of power from the same place. As long as you are using adequate gauge wire, like what is recommended from manufactures, say for your lights, everything should be fine.

scruffythewild

To be honest, One day, I am going to replace everything with 16 or 14 gauge wire to be safe...Thanks for your input anyway, Rob.

SoCalxJ4.0

That will work.

Sails

Each cable/wire will carry the amperage that is required individually. Ultimately all your wiring will lead back to the battery and 'touch' but each circuit should be sized to the demand.
The higher capacity, lower the number, really means that the wire has more surface area since the current runs on the outside of the wire. Connecting giant cables to switches is both expensive and in not all that practical hence why relays are used. Here smaller amounts of amperage are used to activate the relay which has larger shorter cables carrying the required amperage, smaller switches and better power delivery.
Example of this is the headlight wiring upgrade, power runs directly from the battery to the relays, close to the headlights, larger gauge wire, smaller OEM wiring runs to the headlight switch reducing load on switch but allowing for higher wattage to be used by the headlights with no fried wires.
Each circuit will try to use what the end device wants, if the wire is too small it will fry. Too small will also lead to a voltage drop and dimmer lights or slower motors, etc.
So the simple answer is No.

DortDizzle

Yep kinda

Current is dependent on load. Each wire will carry the load that is attached to it.

For example,

100 watt lamp. Battery Voltage of 12 VDC. Power = Voltage * Current, so Current = Power/ Voltage. 100 watts/12 volts = 8.3 amps of current.

If you had two wires that each fed 100watt bulb each wire would carry 8.3 amps, and the total draw from the battery would be the sum (16.6 amps)

If you only used one cable to power the lamps it would need to be able to carry the full 16.6 amps.

Cables are sized based off their current carrying capability. This has to do with cross sectional area (diameter) and the resistivity of the material (copper).

Power is losses in cables because they are resistive and convert current to heat (power dissipation)

The resistivity of a cable depends on its length and cross sectional area.

R = p *(l/A), where R is the cable resistance, p is resistivity of the material, l is the cable length, and A is the area cross section.

You don't really need to compute this unless you are designing cable harnesses, but for a material (copper) the resistance will increase if the length of the cable is increased or if the cross section area (wire size) is decreased. Higher resistance = More power dissipation = heat = melted cables and fires. (what was said earlier about voltage drop is correct, just different formulas moved around voltage drop means power dissipation)

Power = I(current)^2*R(resistance)

So the longer the cable run the larger the gauge that will be required to keep the cable within capacity. That is why it is common to see amp wiring with such a large gauge, it has to carry large current from the front to the back of the vehicle (and marketing BS comes into play)

There are some good cable calculators out there on Google that will give you an idea of what gauge wire to use. You can always go overkill to prevent cable melting, but that can be expensive.

One thing for DC circuits current is carried within the cross section of the cable, not just the outer surface (skin effect with AC causes current densitys to move toward the skin, but I am just being a dork now).