Guide

1

Calculate the maximum

**load on the electricity network.To determine the maximum capacity of the users that can be simultaneously connected thereto.Then define the conductor material from which the wiring will be done.To do this, take better copper wire, it has a higher conductivity than aluminum and not as quickly burns at a high load.** 2

Calculate wire size required for proper load distribution.To do this, the total power of the consumers, which can be found in the technical documentation to them, divide it by the nominal voltage of the network.The result is a maximum current value which should flow through it (I = P / U).Household and industrial networks are made so that the init
ial tension was the same on all the connectors to plug (socket).

3

After determining the maximum current flowing through the network, locate the section of the wire, which is made of a network.Please note that the maximum current density for the aluminum wire is 5 A / mm², and for copper - 8 A / mm².Install a fuse nominal value, the minimum exceeds the maximum current in the circuit, to avoid burn-conductors in the network in the event of a short circuit.

4

Example If dacha need to calculate the electrical load

**, add all electrical power, which may be included in the network.Lighting 10 lamps for 100 watts (1 kW), boiler 4 kW, refrigerator 0.5 kW, 2.5 kW microwave, smaller household consumers 2 kW.In total, will receive 10 kW = 10,000 watts.Since home network rms voltage is 220 V, calculate the maximum current in the network I = 10000 / 220≈45,46 A. For network devices using aluminum conductor cross-section of at least 45,46 / 5≈10 mm² or copper 45.46 /8≈6 mm².Install a fuse rated at no less than 46 A.**