This article explores the electrical conductor sizing, focusing on measurement units such as circular mils and square mils, and their applications in determining conductor area and resistance. It includes practical examples, such as calculating wire diameters, areas, and resistances.
The size of electrical conductors varies widely according to their use and the amount of current they carry. Therefore, special units of measurement have been established for conductors.
Circular Mils
Many applications require very small wires. A unit of measurement much smaller than the inch is needed. The unit mil has been selected as the basis for measuring electrical conductors. One mil is equal to 1000th of an inch (1/1000 inch or 0.001 inch). Rather than stating that a conductor’s diameter is 0.050 inch, it is stated that it measures 50 mils.
In the metric system, there are 1,000,000 micrometers μ(μm) in 1 meter (m). Because there is 0.0254 meter in 1 inch, and 1 mil is equal to 0.001 inch, it can be calculated that 1 mil is equal to 25.4 micrometers.
Note: Micrometers are sometimes called microns.
Conductor size is usually expressed as unit area rather than diameter. To measure circular conductors (wires and cables), the unit circular mil is used. To measure rectangular conductors (bus bars), the unit square mil is used.
One circular mil is the area of a circle 1 mil in diameter. One square mil is the area of a square in which each side measures 1 mil. The linear measurement in mils is denoted by the abbreviation “m.” The area in circular mils is abbreviated “cm,” and square mils is abbreviated “sq m.”
NOTE: In the metric system of measurement, the symbol “m” denotes meter, and the symbol “cm” denotes centimeter. When dealing with the size of conductors, the reader is cautioned to recognize the abbreviations “cm” and “sq m” as circular mil and square mil, respectively.
To determine the circular mil area of a wire, the following equation may be used:
$$A\ =\ d^2$$
Where,
d = diameter, in mils (m)
A = area, in circular mils (cm)
Electrical Conductor Size Calculation Example
Calculate the circular mil area of a wire that measures 0.025 in. in diameter.
What values are known?
d = 0.025 in.
What value is not known?
A = ?
What equation can be used?
$$A\ =\ d^2$$
Begin by converting inches to mils. Recall that 1 mil is equal to 1000th of an inch. Therefore,
$$0.025\ in.\ \times\ 1000\ =\ 25\ mil\ (m)$$
Now substitute the known values, and solve for A:
$$A\ =\ d^2$$
$$A\ =\ \left(25m\right)^2=625\ cm$$
Electrical Conductor Diameter Calculation Example
What is the diameter of a wire in inches if it has an area of 5184 cm?
What values are known?
A = 5184 cm
What value is not known?
d=?
What equation can be used?
$$d\ =\ \sqrt{cm}$$
Substitute the values, and solve for d:
$$d\ =\ \sqrt{cm}=\sqrt{5184}=72\ mils$$
Recall that 1 mil is equal to 1000th of an inch. Therefore,
$$72\ mils\ \div\ 1000\ =\ 0.072\ in.$$
Square Mils
Areas of bus bars are more conveniently expressed in square mils. The square mil area of a conductor may be determined by multiplying the thickness in mils by the width in mils. Use the equation:
$$\ A=\ TW$$
Where,
A = area, in square mils (sq m)
T = thickness, in mils (m)
W = width, in mils (m)
Square Mil Area Calculation Example
Calculate the square mil area of a bus bar ¼ in. thick and 2 in. wide.
What values are known?
T = 1/4 in.
W = 2 in.
What value is not known?
A = ?
What equation can be used?
A = TW
Change the thickness (T) in inches (fraction) to inches (decimal).
T = 1/4 in.= 0.25 in.
Next convert inches to mils. Recall that 1 mil is equal to 1000th of an inch.
T = 0.25 in. × 1000 = 250 m
W = 2 in. × 1000 = 2000 m
Now find the area (A).
$$A=\ TW=250\ m\ \times\ 2000\ m\ =\ 500,000\ sq\ m$$
Busbar Size Calculation Example
Determine the number of square mils in a bus bar that measures ½ in. by 1 in.
What values are known?
T = 1/2 in.
W = 1 in.
What value is not known?
A = ?
What equation can be used?
A = TW
Change the thickness (T) in inches (fraction) to inches (decimal).
T = 1/2 in.= 0.5 in.
Next convert inches to mils. Recall that 1 mil is equal to 1000th of an inch.
T = 0.5 in. × 1000 = 500 m
W = 1 in. × 1000 = 1000 m
Now find the area (A).
$$A=\ TW=500\ m\ \times\ 1000\ m\ =\ 500,000\ sq\ m$$
The square mil area of a conductor may be changed to circular mils by the following equation:
$$cm\ =\ \frac{sq\ m}{0.7854}$$
Calculate Circular Mil Area of Conductor Example
What is the circular mil area of the conductor in the previous example?
What value is known?
A = 500,000 sq m
What value is not known?
cm = ?
What equation can be used?
$$cm\ =\ \frac{sq\ m}{0.7854}$$
Substitute the known value, and solve for cm
$$cm\ =\ \frac{sq\ m}{0.7854}=\frac{500,000}{0.7854}=636,618\ cm$$
Calculate Square Mil Area of Bus Bar Example
Calculate the square mil area of a bus bar 1/8 in. thick and 1 in. wide.
What values are known?
T = 1/8 in.
W = 1 in.
What value is not known?
A = ?
What equation can be used?
A = TW
Change the thickness (T) in inches (fraction) to inches (decimal).
T = 1/8 in.= 0.125 in.
Next convert inches to mils. Recall that 1 mil is equal to 1000th of an inch.
T = 0.125 in. × 1000 = 125 m
W = 1 in. × 1000 = 1000 m
Now find the area (A).
$$A=\ TW=125\ m\ \times\ 1000\ m\ =\ 125,000\ sq\ m$$
Now find the cm.
$$cm\ =\ \frac{sq\ m}{0.7854}=\frac{125,000}{0.7854}=159,155\ cm$$
American Wire Gauge
A special scale has been established for the more common wires used in the electrical industry. The common name for this scale is the American wire gauge (AWG). The wire sizes range from No. 50 AWG, which is the smallest standard size, to No. 4/0 AWG (pronounced “four-aught”), which is the largest standard size for this scale. Figure 1 illustrates the instrument used to measure the gauge of a wire. All wire sizes larger and smaller than the AWG scale are expressed in circular mils.
Figure 1. American wire gauge (AWG) tool.
Chapter 9, Table 8—Conductor Properties—of the National Electric Code (NEC) lists standard wire sizes from No. 18 AWG to 4/0 (0000) AWG. Also listed are the standard sizes from 250,000 circular mils (250 kcmil) to 2,000,000 circular mils (2000 kcmil.) (The unit kcmil means thousands of circular mils). As an example, number 18 wire has an area of 1620 circular mils and 4/0 has an area of 211,600 circular mils.
Circular and rectangular conductors are measured with a caliper or micrometer (Figure 2 and Figure 3). Calipers and micrometers are very accurate measuring instruments. Determining conductor size is relatively easy. Note: Do not confuse the instrument “micrometer” with the unit of measurement “micrometer.”
Figure 2. Micrometer (left) and caliper (right).
Figure 3. Caliper with digital display.
To measure a rectangular conductor, measure the width and thickness and then multiply the values to obtain the square mil area. To measure circular conductors, measure the diameter and then square the number to obtain the circular mil area.
Mil-Foot
A wire 1 mil in diameter and 1-foot long is called a mil-foot of wire (Figure 4). One foot of any stranded wire may be considered to be composed of a number of mil-foot wires in parallel. In other words, a foot of wire with a total cross-sectional area of 5 circular mils equals five individual 1 mil-foot wires.
Figure 4. One mil-foot of wire.
The resistance of any number of identical wires in parallel is equal to the resistance of one wire divided by the total number of wires. For example, the resistance of 1 mil-foot of copper building wire is 10.4 ohms. A stranded wire of the same material and length, but 10 circular mils in cross section, is 10.4 ohms ÷ 10 circular mils, or 1.04 ohms. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. If a 10-circular mils wire is 1000 feet long, its resistance is 1.04 ohms × 1000 feet, or 1040 ohms.
Resistivity
The resistivity (K) of a conducting material is the resistance per mil-foot at a temperature of 68° Fahrenheit (F), or 20° Celsius (C). The resistance of the conductor depends upon the material from which it is made, its cross-sectional area, its length, and the operating temperature. (The operating temperature is the ambient temperature plus the increase in temperature caused by the current flow.)
Figure 5 lists the resistivity of various conducting materials. To calculate the resistance of any conductor at 68 °F ( 20 °C), divide the resistance of 1 mil-foot by the cross-sectional area (in circular mils) and multiply by its length in feet. From this information, the following equation can be constructed:
$$R\ =\ \frac{Kl}{A}$$
Where,
R = resistance of the conductor, in ohms (Ω)
K = resistivity
l = length, in feet (ft)
A = area, in circular mils (cm)
Figure 5. Resistivity and temperature coefficients of conducting materials.
Copper Resistance Calculation Example
Determine the resistance of a copper wire 0.15 in. in diameter and 1-mile long.
What values are known?
d = 0.15 in.
K = 10.4
l = 1 mile (5,280 ft)
What value is not known?
R = ?
What equation can be used?
$$R\ =\ \frac{Kl}{A}$$
First convert inches to mils. Recall that 1 mil is equal to 1000th of an inch.
d = 0.15 in. × 1000 = 150 m
Now find the area (A) of the wire.
$$\ A\ =\ d^2=\left(150\ m\right)^2=22,500\ cm$$
Now find the resistance.
$$R\ =\ \frac{Kl}{A}=\frac{10.4\times5,280}{22,500\ cm}=2.44\Omega$$
Electrical Conductor Length Calculation Example
Calculate the length of a copper wire 0.025 in. in diameter if it has a resistance of 5 Ω.
What values are known?
d = 0.025 in.
K = 10.4
R = 5 Ω
What value is not known?
l = ?
What equation can be used?
$$l\ =\ \frac{AR}{K}$$
First convert inches to mils. Recall that 1 mil is equal to 1000th of an inch.
d = 0.025 in. × 1000 = 25 m
Now find the area (A) of the wire.
$$\ A\ =\ d^2=\left(25\ m\right)^2=625\ cm$$
Now find the length of the wire.
$$I\ =\ \frac{AR}{K}=\frac{625\ cm\times5\ \Omega}{10.4}=300.48\ ft$$
Electric Feeder Resistance Calculation Example
What is the resistance of a feeder consisting of two copper bus bars if each bus bar is ¼ in. thick, 2-in. wide, and 250-ft long?
What values are known?
T = ¼ in or 0.25 in.
W = 2 in.
l = 250 ft
K = 10.4
What equation can be used?
$$R\ =\ \frac{Kl}{A}$$
First, convert inches to mils. Recall that 1 mil is equal to 1000th of an inch.
T = 0.025 in. × 1000 = 25 m
W = 2 in. × 1000 = 2000 m
Next, find the area of one bus bar.
A = TW
$$A=\ TW=25\ m\ \times\ 2000\ m\ =\ 50,000\ sq\ m$$
Now find the circular mil area.
$$cm\ =\ \frac{sq\ m}{0.7854}=\frac{50,000}{0.7854}=63,662\ cm$$
Now find the resistance of the two bus bars.
$$R\ =\ \frac{Kl}{A}=\frac{10.4\times\left(2\times250\ ft\right)}{63,662\ cm}=0.082\Omega$$
Electrical Conductor Sizing Key Takeaways
Understanding electrical conductor sizing and its associated calculations, such as circular mils, square mils, and resistance, is critical for designing efficient and reliable electrical systems. These measurements ensure that conductors are appropriately sized to handle current loads safely while minimizing power losses. The principles explored, from converting units to calculating resistance and area, are foundational in applications ranging from residential wiring to industrial power distribution.