The AWG  American Wire Gauge
 is used as a standard method of denoting wire
diameter,
measuring the diameter of the conductor (the bare wire)
with the insulation removed.
The higher the number  the thinner the wire.
This AWG table is for a single, solid, round
wire. Small gaps between the strands
in a stranded wire make it so it has to be slightly
larger to keep the same currentcarrying
capacity and electrical resistance as a solid wire.
Because a thick wire has less electrical resistance it
will carry more current with less voltage drop than a
thin wire.

AWG 
Diameter
(mm) 
Diameter
(in) 
Circular Mils 
Square
(mm^{2}) 
Resistance
(ohm/1000m) 
14 
1.63 
0.064 
4110 
2.0 
8.54 
12 
2.05 
0.081 
6530 
3.3 
5.4 
10 
2.59 
0.10 
10380 
5.26 
3.4 
8 
3.25 
0.13 
16510 
8.30 
2.2 
6 
4.115 
0.17 
26240 
13.30 
1.5 
4 
5.189 
0.20 
41740 
21.15 
0.8 
2 
6.543 
0.26 
66360 
33.62 
0.5 
1 
7.348 
0.29 
83690 
42.41 
0.4 
0 
8.252 
0.33 
105600 
53.49 
0.31 
00 (2/0) 
9.266 
0.37 
133100 
67.43 
0.25 
000 (3/0) 
10.40 
0.41 
168800 
85.01 
0.2 
0000 (4/0) 
11.684 
0.46 
211600 
107.22 
0.16 
What good is all this data? I put
it here so you can calculate the proper wire size for your
mobile installations. (yes I only included the wire sizes I thought were
important)
The formula for voltage drop is:
VD = 2 x K x I x D/CM.
 K = DirectCurrent Constant. K
represents the dc resistance.
(for a 1,000circular mils conductor
that is 1,000 ft long, at an operating temperature of 75ºC).
K is 12.9 ohms for copper and 21.2 ohms for aluminum.
 I = Amperes: The load in amperes
at 100%
 D = Distance: Where we
specify distances here, we are referring to the conductor
length.

CM =
CircularMils: The circular mils of the circuit conductor as
listed in NEC Chapter 9, Table 8.
Let’s do an
example. An amplifier rated 13.8VDC and 75A is wired
to the Batteries with 10 ft lengths of 4 AWG stranded wire.
Copper
Wire
K = 12.9 ohms,
copper
I =
75A
D = 10 ft
CM = 41740
VD = 2 x
12.9 x 75 x 10 / 41740 =
.46 vdc dropped
(So the voltage at the
amplifier will be 13.34 vdc) 
Aluminum
Wire
K = 21.2
ohms, Aluminum
I = 75A
D = 10 ft
CM = 41740
VD = 2 x
12.9 x 75 x 10 / 41740 =
.76 vdc dropped
(So the voltage at the
amplifier will be 13.04 vdc) 
Just one more to those 16 pillers out there using 15
feet of 1/0 wire....this same formula for
350 amps would leave a voltage drop of 1.32 vdc and would leave 12.48 vdc
at the amplifier.
1.32vdc x 350A x 65% = 300 watts lost
Reference : National Electrical Code (NEC)
