Rules for selecting wire cross-section depending on the estimated current.

Tables of current and cross-section dependence.

When laying power communications, the main issue that arises is the choice of the type and cross-section of the wire that needs to be used. In this case, the type of wire, which determines the material and number of insulating shells (various types of plastic and other materials), as well as the material (copper or aluminum) and type (single- and multi-core) of the conductor, is selected based on the conditions in which the wire will be laid. The cross-section of the wire is determined based on the maximum current that will flow through the wire for a long time. The following tables will help you choose the cross-section of the wire.

Note 1. Current values ​​for 220/380V wires are given according to the standard range of automatic fuses, wire cross-sections are rounded up to the standard cross-sections of wires produced from the corresponding material.

Note 2. The data are given for a temperature of 30°C. For higher temperatures, move to the next (larger) section for every 20°C.

Note 3. When laying several wires in a bundle, the wire cross-section should be increased: for 2-9 wires in a bundle by 80%, for 10-20 wires by 160%.

Note 4: "AWG value" refers to the American Wire Gauge System, and is especially common for speaker cables.

Voltage drop in the wire.

       When organizing the power supply of low-voltage electrical circuits, the choice of cable lengths and cross-sections comes to the fore. This is due to a significant voltage drop in the conductor for low-voltage DC circuits. Simply put, if you decide to build LED lighting for a 40m-long facade and place 3W LED floodlights each, connecting them in parallel at some interval, with the power source placed at one end of the wire, then with a cable cross-section of 2.5 mm2, the last floodlight in the circuit will shine "at half power". This will be due to the fact that the voltage loss at the end of the wire will be about 5V. 

 To calculate voltage losses in a wire there is a simple and clear program "Avral Delta" . Download .

So, if we take one power, then when the voltage decreases, the current should increase, according to the formula:

P = I U. (1)

In this case, losses on the wire during transmission are calculated from Ohm's law:

U = R I. (2)

From these two formulas it is clear that as the supply voltage decreases, the losses on the wire increase. Therefore, the lower the supply voltage, the larger the cross-section of the wire must be used to transmit the same power.

For direct current, where low voltage is used, it is necessary to carefully approach the issue of cross-section and length, since it is these two parameters that determine how many volts will be wasted.

Resistance of copper wire to direct current

The resistance of a wire depends on the specific resistance ρ, which is measured in Ohm mm²/m. The value of the specific resistance determines the resistance of a section of wire 1 m long and 1 mm² in cross-section.

The resistance of the same piece 1 m long is calculated using the formula:

R = (ρ l) / S , where (3)

R — wire resistance, Ohm,

ρ - specific resistance of the wire, Ohm mm²/m,

l — length of wire, m,

S — cross-sectional area, mm².

The resistance of copper is 0.0175 Ohm mm²/m, we will use this value in further calculations.

It is not a fact that manufacturers of copper cables use pure copper, therefore, in practice, the cross-section is always taken with a reserve, and protective machines are used to prevent overload of the wire.

 

From formula (3) it follows that for a piece of wire with a cross-section of 1 mm² and a length of 1 m, the resistance will be 0.0175 Ohm.

Now, using formula (2), we calculate the voltage on the wire:

U = ((ρ l) / S) I ,              (4)

Here are the table data for a length of 1 m and a current of 1 A:

Table 1.
Voltage drop on a 1 m wire of different cross-sections

Calculations for selecting the cross-section of a wire for direct current are carried out according to formula (4).

Table 2.
Voltage drop for different wire cross-sections (top row) and current (left column).
Length = 1 meter

1.  The cases where the wire will overheat, i.e. the current will be higher than the maximum permissible for a given cross-section, are marked in red .

2. Blue color  - when the use of too thick wire is economically and technically impractical and expensive. The threshold was taken as a drop of less than 1 V over a length of 100 m.

   The table is very easy to use. For example, you need to power a certain device with a current of 10A and a constant voltage of 12V. The line length is 5 m. At the output of the power supply, we can set the voltage to 12.5 V, therefore, the maximum drop is 0.5V.

   Available - wire with a cross-section of 1.5 square. What do we see from the table? At 5 meters with a current of 10 A, we will lose 0.1167 V x 5 m = 0.58 V. It seems to be suitable, considering that most consumers tolerate a deviation of +-10%.

   But. We actually have two wires, plus and minus. And the total length is 10 meters, and the drop will actually be 1.16 V.

   In other words, in this situation, the output of the power supply is 12.5 Volts, and the input of the device is 11.34.

   And this is without taking into account the transition resistance of the contacts and the imperfection of the wire (the copper “test” is not right, impurities, etc.)

   Therefore, such a piece of cable most likely will not work, you need a wire with a cross-section of 2.5 squares. It will give a drop of 0.7 V on a 10 m line, which is acceptable.

   And if there is no other wire? There are two ways.

   1. You need to place the 12.5 V power source as close to the load as possible. If we take the example above, 5 meters will do. This is always done to save on the wire.

   2. Increase the output voltage of the power supply. This is fraught with the fact that with a decrease in the load current, the voltage on the load may rise to unacceptable limits.

   
   

   For example, in the private sector, 250-260 volts are installed at the transformer (substation) output, in houses near the substation, light bulbs burn like candles. I mean, not for long. And residents on the outskirts of the area complain that the voltage is unstable, and drops to 150-160 volts. A loss of 100 volts! Multiplying by the current, you can calculate the power that heats the street, and who pays for it? We, the column in the receipt "losses". 

   Conclusion on choosing the wire cross-section for direct voltage:

   The shorter and thicker the wire through which the direct current flows, the better. If you look at table 2, you need to choose the values ​​from the top right, without going into the "blue" zone.

   For alternating current the situation is the same, but the issue is not so acute - there the power is transmitted by increasing the voltage and decreasing the current. See formula (1).

   Finally, here is a table in which the DC voltage drop is set to a limit of 2%, and the supply voltage is 12 V. The parameter we are looking for is the maximum length of the wire.

   Table 3. Maximum wire length for 2% DC voltage drop.

   S, mm² I,A	1	1,5	2,5	4	6	10	16	25	35	50	75	100
   1	7	10,91	17,65	28,57	42,86	70,6	109,1	176,5	244,9	-	-	-
   2	3,53	5,45	8,82	14,29	21,4	35,3	54,5	88,2	122,4	171,4	-	-
   4	1,76	2,73	4,41	7,14	10,7	17,6	27,3	44,1	61,2	85,7	130,4	-
   6	1,18	1,82	2,94	4,76	7,1	11,7	18,2	29,4	40,8	57,1	87	117,6
   8	0,88	1,36	2,2	3,57	5,4	8,8	13,6	22	30,6	42,9	65,25	88,2
   10	0,71	1	1,76	2,86	4,3	7,1	10,9	17,7	24,5	34,3	52,2	70,6
   15	-	0,73	1,18	1,9	2,9	4,7	7,3	11,8	16,3	22,9	34,8	47,1
   20	-	-	0,88	1,43	2,1	3,5	5,5	8,8	12,2	17,1	26,1	35,3
   25	-	-	-	1,14	1,7	2,8	4,4	7,1	9,8	13,7	20,9	28,2
   30	-	-	-	-	1,4	2,4	3,6	5,9	8,2	11,4	17,4	23,5
   40	-	-	-	-	-	1,8	2,7	4,4	6,1	8,5	13	17,6
   50	-	-	-	-	-	-	2,2	3,5	4,9	6,9	10,4	14,1
   100	-	-	-	-	-	-	-	1,7	2,4	3,4	5,2	7,1
   150	-	-	-	-	-	-	-	-	-	2,3	3,5	4,7
   200	-	-	-	-	-	-	-	-	-	-	2,6	3,5

   According to this table, our 1.5-liter battery can only be 1 meter long. The voltage drop on it will be 2%, or 0.24V. We check using formula (4) — everything matches.

   If the voltage is higher (for example, 24 V DC), then the length can be correspondingly longer (2 times).