- Thread starter gahan
- Start date

How long will the run of the wire be?

1. Use the smallest reasonable sized wire for the required current.

Wire is expensive and the larger you go, the more expensive it is.

Wire is heavy and the larger you go, the heavier it gets.

Mechanically, smaller wire is easier to route, easier to protect, easier to fit connectors on and therefore, more reliable mechanically (There are limits - see below).

2. Use a large enough wire so there is no voltage drop. We want whatever it is we are wiring to operate at top efficiency.

3. Maintain an adequate safety margin. We don’t want to melt any wires do we?

The first thing you have to do is determine the current you have to carry. For DC circuits, that’s relatively easy. Some equipment on a car is rated directly in current draw. Auxiliary fans, fuel pumps and things like that are rated in current draw - Amps. Some equipment is rated in Watts - mostly the lighting equipment. The power requirement in Watts will be printed right on the bulb or stamped in the base. To come up with amps use one of the formulas shown here.

Current in Amps = Power in Watts/Voltage in Volts

Current in Amps = Voltage in Volts/Resistance in Ohms

Current in Amps = The Square Root of Power in Watts/Resistance in Ohms

Let’s calculate for a typical 100 Watt Light - the power required is 100 Watts and the voltage is 12 Volts - so the current requirement is 100 Watts/12 Volts = 8.33 Amps. Actually, it's somewhat less than that because the rated output of a lamp is figured at 13.5 volts, not 12 volts - 13.5 volts is typical when your engine in running and the Alternator is working correctly. We'll use the 12 volt figure anyhow - Let’s assume you have to run a wire 6 feet from a relay to the lamp. Using the 10 Amp column you’ll find that you can run 10 Amps on 15 feet of 18 AWG with only 1/2 Volt drop. Go to the next size larger for a safety margin and you’re at 16 AWG. Now in reality, you have to balance the mathematical results with mechanical reliability. If in a high stress area I’d go to 14 AWG as the wire and connectors are physically stronger. If not, use the lower gauge to shave weight.

Note that wire sizes for lighting is more critical than for other applications — The rated output of a lamp is figured at 13.5 volts, not 12. So with a 1/2 volt drop you are at 13.0 volts. And at 95% of the rated voltage, you are only putting out 80% of the rated luminous intensity - for a 100 watt lamp that’s only 80 watts!! Get what you pay for and figure to the high side when you are sizing wire for lighting.

edit; fix table...

Wire is expensive and the larger you go, the more expensive it is.

Wire is heavy and the larger you go, the heavier it gets.

Mechanically, smaller wire is easier to route, easier to protect, easier to fit connectors on and therefore, more reliable mechanically (There are limits - see below).

2. Use a large enough wire so there is no voltage drop. We want whatever it is we are wiring to operate at top efficiency.

3. Maintain an adequate safety margin. We don’t want to melt any wires do we?

The first thing you have to do is determine the current you have to carry. For DC circuits, that’s relatively easy. Some equipment on a car is rated directly in current draw. Auxiliary fans, fuel pumps and things like that are rated in current draw - Amps. Some equipment is rated in Watts - mostly the lighting equipment. The power requirement in Watts will be printed right on the bulb or stamped in the base. To come up with amps use one of the formulas shown here.

Current in Amps = Power in Watts/Voltage in Volts

Current in Amps = Voltage in Volts/Resistance in Ohms

Current in Amps = The Square Root of Power in Watts/Resistance in Ohms

Let’s calculate for a typical 100 Watt Light - the power required is 100 Watts and the voltage is 12 Volts - so the current requirement is 100 Watts/12 Volts = 8.33 Amps. Actually, it's somewhat less than that because the rated output of a lamp is figured at 13.5 volts, not 12 volts - 13.5 volts is typical when your engine in running and the Alternator is working correctly. We'll use the 12 volt figure anyhow - Let’s assume you have to run a wire 6 feet from a relay to the lamp. Using the 10 Amp column you’ll find that you can run 10 Amps on 15 feet of 18 AWG with only 1/2 Volt drop. Go to the next size larger for a safety margin and you’re at 16 AWG. Now in reality, you have to balance the mathematical results with mechanical reliability. If in a high stress area I’d go to 14 AWG as the wire and connectors are physically stronger. If not, use the lower gauge to shave weight.

Note that wire sizes for lighting is more critical than for other applications — The rated output of a lamp is figured at 13.5 volts, not 12. So with a 1/2 volt drop you are at 13.0 volts. And at 95% of the rated voltage, you are only putting out 80% of the rated luminous intensity - for a 100 watt lamp that’s only 80 watts!! Get what you pay for and figure to the high side when you are sizing wire for lighting.

Code:

```
Maximum length in feet for car wiring
Wire Gauge Current Load in Amps @ 12 Volts DC
4 Amp 6 Amp 8 Amp 10 Amp 12 Amp 15 Amp 20 Amp 50 Amp
20 AWG 26' 17' 13'
18 AWG 37' 25' 18' 15' 12'
16 AWG 56' 37' 28' 22' 18' 14'
14 AWG 90' 60' 45' 36' 30' 24' 18'
12 AWG 143' 95' 71' 57' 47' 38' 28'
10 AWG 227' 151' 113' 90' 75' 60' 45'
8 AWG 363' 241' 181' 145' 120' 96' 72' 29'
6 AWG 585' 390' 292' 234' 194' 155' 117' 46'
4 AWG 925' 616' 462' 370' 307' 246' 185' 74'
2 AWG 1515' 1009' 757' 606' 503' 403' 303' 121'
1 AWG 1923' 1280' 961' 769' 638' 511' 384' 153'
0 AWG 2427' 1616' 1213' 970' 805' 645' 485' 194'
```

edit; fix table...

Last edited:

thanks guys looks like standard 20 Gauge trailor wire will work if I can pull it through 14' of poly fuel line.

Wally

Great Info! Thanks Mid and Wally!