**How to correctly size wires and why it is important**

The cross-sectional size of a wire is a very important factor which needs to be considered when designing electrical circuits, especially when said wire needs to conduct large quantities of current. This is due to the fact that any physical wire will have some measure of resistance and will thus dissipate energy in the form of heat. It is thus the job of any circuit designer that this dissipated energy is within safe levels and that the load being supplied will operate efficiently.

**Selecting an Appropriate Gauge for your Current Rating**

As stated before the cross sectional area of your chosen wire will determine how much current a cable can safely handle. This is due to the following relationship between resistance (R), length of cable (l), resistivity (ρ) and cross-sectional area (A).

From the equation it is easy to determine that at larger cross-sectional areas, a cable will have a smaller resistance and thus dissipate less energy. It is thus a generally accepted consensus that larger diameter wires can conduct larger values of current. This brings us to wire gauge tables. These tables are used to determine the size of a cable needed for a certain application.

Referring to the below table we can see that if we have a current rating of 27A or smaller we can use the 2.5mm^{2} wire gauge.

[Data obtained from South Ocean’s Bare Copper Earth Wire]

**Sizing the cable for Voltage Drop**

Just as in any resistor, the wire will also have a voltage drop developed across it. This voltage drop can thus hinder certain loads if it is too large. The general rule of thumb is to size a wire such that it will have a voltage drop of less than 2.5% of the source voltage. The size of the voltage drop depends on the length of cable as well as the magnitude of the current carried by the cable. To calculate the voltage drop developed, we simply take the voltage drop constant given by the table, multiply it by the current and length of the cable.

For example if we have a cable which is 10m long and conducts 20A and has a cross-section of 2.5mm^{2}. We have a voltage drop of 3.6V.

(Voltage Drop = 20A x 10m x 0.018V

**Working Example:**

We want our cabling system to supply a load of 10kW at a voltage of 230V and a cable length of 10m. This gives us a current of 43.5A when we divide the power by the voltage.

We should however design the cable to carry an extra 20% of current in the case of an emergency. We thus design the cable to carry a current of 52.5A.

The choice of a 10mm^{2} wire should be sufficient for our needs.

We now need to see if the voltage drop is within sufficient limits.

Let’s calculate the voltage drop across the wire:

The voltage developed across the cable is thus within acceptable limits as it is less than 2.5% of the source voltage.

For further information contact Joshb@switchman.com

Article by: Jannes Smit, 3^{rd }year Electrical Engineering student at the University of the Witwatersrand.

Jannes is completing a 6 week learnership at Switchboard Group.

jannes9000@gmail.com