Chapter 3 Introduction to Basic Electrical Circuit Materials 55
equivalent, compute the square/rectangular conductor’s
area in square mils and then divide that area by 0.7854.
For example, if a 1" by 1/4" bar of copper is to be
used as a conductor, first convert the inch measurements
to mils (1" = 1000 mils and 1/4" = 250 mils). Then
multiply the dimensions in mils to get the area in square
mils (1000 250 = 250,000 sq. mils). Finally, divide the
square mil area by 0.7854 ( 250,000/0.7854 = 318,309).
This results in the original 1" by 1/4" copper bar having a
surface area equivalent to a round conductor of 318,309
cmils.
Wire charts commonly express the electrical values
for conductors solely based on their size in cmils. To deter-
mine the current carrying capacity, or resistance value, of
a rectangular conductor, the rectangular conductor must be
converted to circular mils.
Conductor Insulation
Materials with only a few free electrons do not con-
duct electrons well. These materials are called insulators.
To keep the electron flow contained to the conductor
path, and to prevent contact with other conductors and
people, conductors are coated. This protective coating is
called insulation. As the name implies, insulation is
made of insulating material such as rubber, plastic,
or other synthetic materials. There are many types
of insulation commonly used today. These materials
include thermoplastic, neoprene, TeflonTM, nylon, and
polyethylene.
The type of conductor insulation is determined by
the application of the electrical system. In determining
insulation type, many factors must be taken into consid-
eration. Some questions one might ask would include:
• Will the insulation be exposed to extreme heat gen-
erated by an industrial furnace or the extreme cold
of a freezer in a food processing plant?
• Will the insulation need to withstand exposure to
the acidic vapors in a chemical plant or to wet,
acidic soil?
• Will the cable be exposed to oil from manufactur-
ing equipment?
• Will the insulation give off toxic fumes if burned?
Resistance ohms per
Gauge Diameter Circular 1000 ft. at 25 C (68°F)
no. mils mil area copper wire aluminum wire
4/0 460.0 211,600 0.04901 0.07930
3/0 409.6 167,800 0.06182 0.1000
2/0 364.8 133,100 0.07793 0.1261
1/0 324.9 105,600 0.09825 0.1590
1 289.3 83,690 0.1264 0.2005
2 257.6 66,370 0.1593 0.2529
3 229.4 52,640 0.2009 0.3189
4 204.3 41,740 0.2533 0.4021
5 181.9 33,100 0.3195 0.5072
6 162.0 26,250 0.4028 0.6395
7 144.3 20,820 0.5080 0.8060
8 128.5 16,510 0.6405 1.016
9 114.4 13,090 0.8077 1.282
10 101.9 10,380 1.018 1.616
11 90.74 8234 1.284 2.04
12 80.81 6530 1.619 2.57
13 71.96 5178 2.042 3.24
14 64.08 4107 2.575 4.08
15 57.07 3257 3.247 5.15
16 50.82 2583 4.094 6.50
17 45.26 2048 5.163 8.18
18 40.30 1624 6.510 10.3
19 35.89 1288 8.210 13.0
20 31.96 1022 10.35 16.4
21 28.46 810.1 13.05 20.7
22 25.35 642.4 16.46 26.2
23 22.57 509.5 20.76 32.9
24 20.10 404.0 26.17 41.5
25 17.90 320.4 33.00 52.4
26 15.94 254.1 41.62 66.4
27 14.20 201.1 52.48 83.2
28 12.64 159.8 66.17 106.0
29 11.26 126.7 83.44 131.0
30 10.03 100.5 105.2 168.0
31 8.928 79.70 132.7 212.0
32 7.950 63.21 167.3 262.0
33 7.080 50.13 211.0 333.0
34 6.305 39.75 266.0 422.0
35 5.6 31.52 335.0 536.0
36 5.0 25.00 423.0 671.0
37 4.453 19.83 533.4 ____
38 3.965 15.72 672.6 ____
39 3.531 12.47 848.1 ____
40 3.145 9.88 1069 ____
Figure 3-3. Wire chart.
D = 1 mil
area = 1 cmil
D = 2 mils
area = 4 cmils
D = 15 mils
area = 225 cmils
D D D
Figure 3-4. The area in circular mils is found by squaring
the diameter in mils.