# How to calculate the cross section of the wire?

The question of how to calculate the cross-section of the wire, is relevant for those who are planning repairs in the apartment or those who want to carry out work on the electrification of the country house. There are two ways here, following which you can achieve an acceptable result: the traditional "by eye" method used by most electricians or based on knowledge of the fundamentals of electrical engineering.

Why do we need such calculations if there are special tables in which the manufacturer enters information about their products? The fact is that each company considers the amount of current in its own way: somewhere it is only 120% excess, and somewhere, all 200%. It goes without saying that long work with double overload cannot be a question, since the heated conductor will melt the insulation and in the best case automatic shutdown will work.

## Calculation of wire section: instruction

For accurate calculation, you need to know what material the wire is made of. If it is copper,then the current density in it will be from 6 to 10 amperes with a cross-sectional area of 1 square millimeter; if aluminum, which occurs infrequently, then this parameter does not exceed 4-6 units.

Such a large scatter is caused by the stock of material: 6 amps are the conditions of normal operation, and exceeding up to 10 is already an overload, which is permissible for a short time.

It turns out that the wire, whose area is 1 square millimeter in cross section, can withstand current up to 10A, 2 millimeters to 20A, etc. Knowing the voltage in the network, the average value of which reaches 220V, we obtain an output power of 4.5 kW for 2 mm of conductor.

How to calculate the wire cross section according to the traditional method? It's still easier here: you need to remember that copper wires with a cross-sectional area of 1.5 square meters. mm is used to connect the lighting, and 2.5 square meters. mm already go to the layout of the network in the apartment.

Here you can immediately take into account the applied load. If a computer acts as the main consumer, then 600 W / 220 V will yield 2.7 A, i.e. There is still about 17A in stock, but this does not mean that it is worth testing the network with simultaneous connections of TV irons, etc. Even if the wiring transfers such an attitude, it may not like the Chinese extension cord, as a result of which all short circuit will also occur.

Another point worth noting is the use of copper-based alloys by manufacturers. Such materials have a smaller value of the permissible current density, so you should not save on wiring.