Knowing how to dimension and choose cables and conductive wires for electrical installation must be part of the mandatory knowledge for building electricians. Choosing the correct electrical cable depends on well-defined standards and calculations. The cables for shower, grounding, plugs, lighting and motors have dimensioning standards with extremely important variations and details.

## How to dimension electrical cables?

Knowing how to calculate how many amperes a cable should support can help with dimensioning using the tables that are provided by the manufacturers. The correct dimensioning of circuit breakers in each phase, circuit and general also has to do with the dimensioning of the conductors.

The NBR-5410 standard stipulates some criteria that must be taken into account when dimensioning an electrical conductor. The first thing that is important to understand is what the minimum cable sessions are stipulated within the facility, and this is what table 47 shows below:

It is important to emphasize that the values presented in the table refer to the minimum criterion, that is, they cannot have cables smaller than these for these certain functions.

Another important point is to know the method of installing the cable, in the case of residential installations, most of these installations are in conduits embedded in masonry. According to table 33 of NBR-5410, the installation method is number 7 and the reference for installation B1, as shown in the table below:

The next step is to find out how much circuit cables are loaded. For this, we will follow the indications in table 46 also of NBR-5410. Observe the cable specifications carefully, as this information will be very important for dimensioning.

Then you should consult the current conduction table. This table can vary according to the type of conductor, the type of insulation, the temperature of the conductor and also the ambient temperature. For residential installations, the cable with PVC insulation and copper conductor is the most used, so we will refer to table 36 of NBR-5410 which is shown below:

In order to continue and to facilitate further understanding, we will create an example and apply it within the tables presented. The design current of the circuit to be dimensioned is 18 amps, the cable used is PVC insulated, aiming to use the tables presented, the number of circuits inside the conduit will be 4 and the reference method is B1, which represents the built-in in masonry.

Follow the column of method B1, observing the number of charged cables that in this case are 2, phase, and neutral. Then you should look for the current value closest to the number of amps in the circuit, in the example, they are 18, that is, look for the closest value, in this case, it was 24. It is very important to use the upper value and never the lower one. If the application of the data was correct, the dimensioning table will show that the correct cable would be 2.5 mm², as shown in the following image:

But the correct sizing doesn’t end here! It is necessary to take into account the number of circuits inside the conduit. Each number of circuits requires a different correction factor and this is what is shown in table 42 of the NBR-5410 below:

As our example used 4 circuits, the correction factor, in this case, is 0.65. You must now use this correction factor in the conducting capacity of the 2.5 mm² circuits within a conduit with 4 circuits. For this you must use the following formula: Iz = Ic x Fc.

- Iz – (the corrected conductor current value), that is, the value we want to find.
- Ic – (the conducting current value of the conductor in the table), that is, 24A.
- Fc – (correction factor), that is, 0.65.

In this case, we have Iz = 24A x 0.65, resulting in Iz = 15.6A, which is the result shown in the image below

It is important to observe if the value you find, in case of 15.6A, is consistent with the necessary conduction of the circuit design current. In the example used, the current is 18A, that is, the 2.5 mm² cables will not be able to conduct the correct current, requiring a thicker cable.

Returning to the design table, the next value within the data presented in the example is 32A. Playing this new value in the formula (Iz = Ic x Fc), of the correction factor you find the following: Iz = 32A x 0.65, resulting in Iz = 20.8A. Therefore, the ideal cable for this example presented is 4.0 mm², which is what the following image shows:

If you want to clear up any doubts that may have remained, check out the video below and see the step by step on how to size correctly following the examples that we present in this article.

## Final considerations

The use of the 2.5 mm² cables, in this case, does not mean that the installation will present an immediate problem. But there will certainly be excessive heating of these cables, considerably increasing the consumption of electricity. In addition, in the long run there may be a melting of the insulating cable, resulting in a series of problems.