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ELECTRICAL CONDUCTORS

LEARNING OBJECTIVES

Learning objectives are stated at the beginning of each chapter. These learning objectives serve as a preview of the information you are expected to learn in the chapter. The comprehensive check questions are based on the objectives. By successfully completing the OCC-ECC, you indicate that you have met the objectives and have learned the information. The learning objectives are listed below.

Upon completing this chapter, you should be able to:

  • Recall the definitions of unit size, mil-foot, square mil, and circular mil and the mathematical equations and calculations for each.
  • Define specific resistance and recall the three factors used to calculate it in ohms.
  • Describe the proper use of the American Wire Gauge when making wire measurements.
  • Recall the factors required in selecting proper size wire.
  • State the advantages and disadvantages of copper or aluminum as conductors.
  • Define insulation resistance and dielectric strength including how the dielectric strength of an insulator is determined.
  • Identify the safety precautions to be taken when working with insulating materials.
  • Recall the most common insulators used for extremely high voltages.
  • State the type of conductor protection normally used for shipboard wiring.
  • Recall the design and use of coaxial cable.

ELECTRICAL CONDUCTORS

In the previous modules of this training series, you have learned about various circuit components. These components provide the majority of the operating characteristics of any electrical circuit. They are useless, however, if they are not connected together. Conductors are the means used to tie these components together. Many factors determine the type of electrical conductor used to connect components. Some of these factors are the physical size of the conductor, its composition, and its electrical characteristics. Other factors that can determine the choice of a conductor are the weight, the cost, and the environment where the conductor will be used.

CONDUCTOR SIZES

To compare the resistance and size of one conductor with that of another, we need to establish a standard or unit size. A convenient unit of measurement of the diameter of a conductor is the mil (0.001, or one-thousandth of an inch). A convenient unit of conductor length is the foot. The standard unit of size in most cases is the MIL-FOOT. A wire will have a unit size if it has a diameter of 1 mil and a length of 1 foot.

SQUARE MIL

The square mil is a unit of measurement used to determine the cross-sectional area of a square or rectangular conductor (views A and B of figure 1-1). A square mil is defined as the area of a square, the sides of which are each 1 mil. To obtain the cross-sectional area of a square conductor, multiply the dimension of any side of the square by itself. For example, assume that you have a square conductor with a side dimension of 3 mils. Multiply 3 mils by itself (3 mils X 3 mils). This gives you a cross-sectional area of 9 square mils.

Figure 1-1. - Cross-sectional areas of conductors.

32NE0312.GIF (12021 bytes)

Q.1 State the reason for the establishment of a "unit size" for conductors. answer.gif (214 bytes)
Q.2 Calculate the diameter in MILS of a conductor that has a diameter of 0.375 inch. answer.gif (214 bytes)
Q.3 Define a mil-foot. answer.gif (214 bytes)

To determine the cross-sectional area of a rectangular conductor, multiply the length times the width of the end face of the conductor (side is expressed in mils). For example, assume that one side of the rectangular cross-sectional area is 6 mils and the other side is 3 mils. Multiply 6 mils X 3 mils, which equals 18 square mils. Here is another example. Assume that a conductor is 3/8 inch thick and 4 inches wide. The 3/8 inch can be expressed in decimal form as 0.375 inch. Since 1 mil equals 0.001 inch, the thickness of the conductor will be 0.001 X 0.375, or 375 mils. Since the width is 4 inches and there are 1,000 mils per inch, the width will be 4 X 1,000, or 4,000 mils. To determine the cross-sectional area, multiply the length by the width; or 375 mils X 4,000 mils. The area will be 1,500,000 square mils.

Q.4 Define a square mil as it relates to a square conductor. answer.gif (214 bytes)




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