# What is the cross section of the wire

The ability to determine the cross-section of the wire is necessary not only for those people whose work is directly related to electrical engineering, but also to ordinary citizens who have decided to independently replace or restore electrical wiring in any room. Moreover, even attracting hired workers to such work, you need to understand the interconnection of the conductor, current and power of the connected load. Although the calculation of the cross-section of wires does not pose any difficulty, it is one of the key points when laying / restoring / repairing power grids.

**From simple to complex**

In order to understand what the term "wire cross section" means and how the corresponding numerical value is used, consider the simplest example. So, suppose that a person is faced with the task of illuminating a room by installing 10 new luminaires in which lamps of 500 W each are installed. It is obvious that a cable with any number of wires or a wire can be used for power supply. The material of the conductors is copper or aluminum.The question arises of how thick the wires should be (let them be two, so as not to complicate the example). Of course, you can, without performing the calculation of the wire section for power, to purchase the cheapest thin wire and, connecting the lamps in series, supply power. This will result in significant cost savings, but this system will soon be out of order. This is because the cross section of the wire was not taken into account. And since the power of all the lamps on the line (10 * 500 W = 5000 W) is known, the diameter of the connected conductors must be selected based on the load. A thinner wire can not stand and burn out, and taken with a reserve means completely unnecessary costs.

**Formula**

What relation does the wire (conductor) cross section have to its diameter? In fact, the connection is direct. It is expressed by the following formula: S = (3.14 * (d * d)) / 4, where S is the area of the circle (cross section) in mm square; d is the measured diameter of the current conducting conductor, in mm. Sometimes, initially moving away from Pi, the formula is simplified by getting: S = 0.8 * (d * d). Such a record is much easier to remember, and the error due to rounding is minimal. In all reference books on electrical engineering, the concept of the cross-sectional area of the cable (wire) conductor, and not the diameter, is used, so recalculation is necessary.

**Calculations**

So, the lamp power is 5 kW.Since the example considers a two-wire network designed for 220 V, then to determine the amount of current consumed by the entire line, you need to use the formula: I = P / U = 5000 W / 220 V = 23 A. After the current is determined, you can begin to select the required wire across the section. It is not difficult to find the tables in which the correspondence of the permissible current for a specific section of an aluminum or copper conductor is indicated. Nevertheless, we recommend using the data of regulatory documents - reference book of the OLC of the current version. There are no restrictions on the choice of the core material: it is possible to lay both copper and aluminum wires. It all depends on the installation features. According to the table we find the nearest current value: for copper it is 19 and 27 A. Obviously, the second option is appropriate. It corresponds to a cross section of 2.5 mm square. For aluminum, the figures are different: 20 and 28 A. Here is a match of 4 mm square. Power is also indicated (6 kW), which is more than the required 5 kW, and this is required.