In reference books on electrical equipment there are tables with sections of wires for application in various conditions. It is possible to measure by a caliper not the section, but diameter. Knowing any of these sizes, it is possible to calculate on a formula and another.

## Instruction

1. A caliper carry out measurement of diameter of a wire in the absence of tension. Any caliper irrespective of whether it is mechanical or electronic, has the metal sponges capable to carry current. If the wire is covered with an insulation layer, carry out measurement of its section without its diameter.

2. Use for expression of diameter and cross-sectional area of conductors of unit of measure, accepted in electrical equipment: respectively, millimeters and square millimeters (electricians call them for short "squares").

3. To transfer the wire section specified in the reference book to its diameter, use the following formula: D=2 √ (S/π) where S is the area of the conductor (mm²), D - diameter of the conductor (mm), π - number "пи", 3.1415926535 (dimensionless size).

4. For back translation (diameter in section) use the same formula transformed as follows: S=π (D/2)², where D - diameter of the conductor (mm), S - the area of the conductor (mm²), π - number "пи", 3.1415926535 (dimensionless size).

5. Accept the section of a multicore wire equal to the sum of sections of the separate conductors entering as its member. It is senseless to summarize their diameters. Calculations can be and multistage. So, for example, to learn the equivalent diameter of a multicore wire, calculate the section of one his vein, increase by their quantity, and then again transfer result to diameter.

6. It is possible to take a wire with a diameter or section exceeding the estimated or specified in the table value, but too thick wires to apply happens inconveniently: they can pull out, for example, the terminal from a klemmnik body weight. To apply wires with a diameter or section it is less settlement or specified in the table it is impossible.

7. Hollow conductors of a cylindrical form (for example, being a part of coaxial cables) have two diameters: external and internal. On them calculate, respectively, two sections: external and internal. Subtract one of another, and then transfer result to equivalent diameter.