In practice, it is often necessary to calculate the resistance various wires. This can be done using formulas or using the data given in table. 1.
The effect of the conductor material is taken into account using the resistivity, denoted by the Greek letter? and having a length of 1 m and a cross-sectional area of 1 mm2. Lowest resistivity? = 0.016 Ohm mm2/m has silver. Let us give the average value of the resistivity of some conductors:
Silver - 0.016 , Lead - 0.21, Copper - 0.017, Nickelin - 0.42, Aluminum - 0.026, Manganin - 0.42, Tungsten - 0.055, Constantan - 0.5, Zinc - 0.06, Mercury - 0.96, Brass - 0.07, Nichrome - 1.05, Steel - 0.1, Fechral - 1.2, Phosphor bronze - 0.11, Chromal - 1.45.With different amounts of impurities and with different ratios of components included in the composition of rheostatic alloys, resistivity may change somewhat.
Resistance is calculated using the formula:
where R is resistance, Ohm; resistivity, (Ohm mm2)/m; l - wire length, m; s - cross-sectional area of the wire, mm2.
If the wire diameter d is known, then its cross-sectional area is equal to:
It is best to measure the diameter of the wire using a micrometer, but if you don’t have one, you should wind 10 or 20 turns of wire tightly onto a pencil and measure the length of the winding with a ruler. Dividing the length of the winding by the number of turns, we find the diameter of the wire.
To determine the length of a wire of a known diameter made of a given material necessary to obtain the required resistance, use the formula
Table 1.
Note. 1. Data for wires not listed in the table should be taken as some average values. For example, for a nickel wire with a diameter of 0.18 mm, we can approximately assume that the cross-sectional area is 0.025 mm2, the resistance of one meter is 18 Ohms, and permissible current equal to 0.075 A.
2. For a different value of current density, the data in the last column must be changed accordingly; for example, at a current density of 6 A/mm2, they should be doubled.
Example 1. Find the resistance of 30 m copper wire with a diameter of 0.1 mm.
Solution. We determine according to the table. 1 resistance of 1 m of copper wire, it is equal to 2.2 Ohms. Therefore, the resistance of 30 m of wire will be R = 30 2.2 = 66 Ohms.
Calculation using the formulas gives the following results: cross-sectional area of the wire: s = 0.78 0.12 = 0.0078 mm2. Since the resistivity of copper is 0.017 (Ohm mm2)/m, we get R = 0.017 30/0.0078 = 65.50 m.
Example 2. How much nickel wire with a diameter of 0.5 mm is needed to make a rheostat with a resistance of 40 Ohms?
Solution. According to the table 1, we determine the resistance of 1 m of this wire: R = 2.12 Ohm: Therefore, to make a rheostat with a resistance of 40 Ohms, you need a wire whose length is l = 40/2.12 = 18.9 m.
Let's do the same calculation using the formulas. We find the cross-sectional area of the wire s = 0.78 0.52 = 0.195 mm2. And the length of the wire will be l = 0.195 40/0.42 = 18.6 m.
In order to successfully buy a wire, before purchasing it is necessary measure wire diameter, otherwise you may become a victim of deception. You will also have to measure the wire cross-section if you add a new one. electrical point on old wiring, since there may not be letter markings on it. The information below will help you choose the right technique. wire diameter measurements and use it effectively in practice.
At the same time, the question will immediately arise: “What is the point of a company spoiling its reputation?” There may be several explanations for this: But the whole point is that even after committing correct calculations wire gauge, you may still encounter a problem even though you buy wire with suitable diameter. An accident may occur due to the fact that the marking of the wires will indicate a cross-section of conductors that does not correspond to the actual one. This may happen as a result of the fact that the manufacturing plant saved on the material, or the company producing this product did not comply with all the characteristics of the product. You can also find wires on the shelves that have no markings at all, which initially casts doubt on their quality.
1. In order to save money. For example, the plant made wire diameter less by only 2 mm. sq. with a 2.5 mm core, which made it possible to win on one linear meter several kilograms of metal, not to mention the profit in mass production.
2. As a result of great competition, the company reduces the price of electrical wiring, trying to lure most of the consumers. Naturally, this happens due to reducing wire diameter, which cannot be determined with the naked eye.
Both the first and second options take place in the sales market, so you better play it safe and make your own accurate calculations, which will be discussed further.
Three main ways to determine wire diameter.
There are several ways, but each of them is based on determination of diameters cores with subsequent calculations of the final results.
Method one. Using instruments. Today there are a number of devices that help measure wire diameter or wire strands. This is a micrometer and caliper, which come in both mechanical and electronic (see below).
This option is primarily suitable for professional electricians who constantly install electrical wiring. The most accurate results can be obtained using a caliper. This technique has the advantage that it is possible measure wire diameter even on a section of a working line, for example, in a socket.
After you have measured wire diameter, it is necessary to carry out calculations using the following formula:
It must be remembered that the number “Pi” is 3.14, so if we divide the number “Pi” by 4, we can simplify the formula and reduce the calculation to multiplying 0.785 by the diameter squared.
Method two. We use a ruler. If you decide not to spend money on a device, which is logical in this situation, then you can use a simple proven method for measuring the cross-section of a wire or wire?. You will need a simple pencil, ruler and wire. Strip the core of insulation, screw it tightly onto a pencil, and then measure with a ruler total length winding (as shown in the picture).
Then divide the length of the wound wire by the number of cores. The resulting value will be wire cross-section diameter.
But the following must be taken into account:
- how lived longer you wind it around a pencil, the more accurate the result will be, the number of turns should be at least 15;
- Press the coils tightly against each other so that there is no space left between them. free space, this will significantly reduce the error;
- take measurements several times (change the measuring side, the direction of the ruler, etc.). Several results obtained will again help you avoid a large error.
Pay attention to the cons too this method measurements:
1. You can only measure the cross-section of thin wires, since it will be difficult for you to wind a thick wire around a pencil.
2. To begin with, you will need to purchase a small piece of the product before making the main purchase.
The formula discussed above is suitable for all measurements.
Method three. We use the table. In order not to carry out calculations using the formula, you can use a special table in which wire diameter is indicated? (in millimeters) and conductor cross-section (in square millimeters). Ready-made tables will give you more accurate results and will significantly save your time, which you will not have to spend on calculations.
|
Conductor diameter, mm |
Conductor cross-section, mm 2 |
Electrical wiring in modern apartments provides a maximum operating current in the network of up to 25 Amperes. The circuit breakers installed in the switchboard apartments. The cross-section of the wire at the entrance to the room must be at least 4 mm2. When installing internal wiring It is permissible to use cables with a cross-section of 2.5 mm2, which are designed for a current of 16 Amps.
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Wire diameter measurement
According to the standard, the diameter of the wire must correspond to the declared parameters, which are described in the marking. But the actual size may differ from the declared one by 10-15 percent. This is especially true for cables that are manufactured by small companies, but large manufacturers can also have problems. Before purchasing an electrical wire for transmitting currents of great importance, it is recommended to measure the diameter of the conductor. For this purpose they can be used various ways
, differing in error. Before performing measurements, it is necessary to remove insulation from the cable cores. Measurements can be made directly in the store if the seller allows you to remove the insulation from small area
wires. Otherwise, you will have to purchase a small piece of cable and take measurements on it.
Micrometer Maximum accuracy can be obtained using micrometers that have mechanical and electronic circuit
. On the shaft of the instrument there is a scale with a division value of 0.5 mm, and on the circle of the drum there are 50 marks with a division value of 0.01 mm. The characteristics are the same for all micrometer models. When working with mechanical device
- the following sequence of actions should be followed:
- By rotating the drum, the gap between the screw and the heel is set close to the measured size.
- Use a ratchet to bring the screw closer to the surface of the part being measured. The eyeliner is performed by rotating the hand without effort until the ratchet is activated. Calculate the transverse diameter of the part according to the readings on the scales located on the stem and drum. Product diameter equal to the sum
values on the rod and drum.
Mechanical micrometer measurement Working with an electronic micrometer does not require rotation of the units; it displays the diameter value on the liquid crystal screen. It is recommended to check the settings before using the device, as measurements are taken in millimeters and inches.
Vernier caliper
The device has reduced accuracy compared to a micrometer, which is quite sufficient for measuring a conductor. Vernier calipers are equipped with a flat scale (vernier), a circular dial or a digital display on a liquid crystal display.
To measure the transverse diameter, you must:
- Clamp the conductor to be measured between the jaws of the caliper.
- Calculate the value on the scale or view it on the display.
An example of calculating size on a vernier
Ruler
Measuring with a ruler gives a rough result. To perform measurements, it is recommended to use tool rulers, which have greater accuracy. Using wooden and plastic school products will give a very approximate diameter value.
To measure with a ruler you need:
- Remove insulation from a piece of wire up to 100 mm long.
- Wind the resulting piece tightly around a cylindrical object. The turns must be complete, that is, the beginning and end of the wire in the winding are directed in one direction.
- Measure the length of the resulting winding and divide by the number of turns.
Measuring the diameter with a ruler by the number of turns
In the example above there are 11 turns of wire which are about 7.5mm in length. By dividing the length by the number of turns, you can determine the approximate value of the diameter, which in in this case equals 0.68 mm.
On the websites of stores selling electric wires, there are online calculators that allow you to calculate the cross section based on the number of turns and the length of the resulting spiral.
Determination of cross-section by diameter
After determining the diameter of the wire, you can begin to calculate the cross-sectional area in squares (mm2). For cables of the VVG type, consisting of three single-core conductors, calculation methods are used using a formula or a ready-made table of diameters and areas. The methods are also applicable to products with other markings.
According to the formula
The main method is to calculate using a formula of the form - S=(n/4)*D2, where π=3.14, and D is the measured diameter. For example, to calculate the area with a diameter of 1 mm, you will need to calculate the value: S=(3.14/4)*1²=0.785 mm2.
Online calculators are available on the Internet that allow you to calculate the area of a circle by diameter. Before purchasing a cable, it is recommended to calculate the values in advance, put them in a table and use it in the store.
The video from user Alexander Kvasha demonstrates checking the cross-section of wire cores.
According to the table with common diameters
To simplify the calculation, it is convenient to use a ready-made table.
The order of using numbers from the table:
- Select the type of wire that you intend to purchase, for example, VVG 3*4.
- Determine the diameter from the table - a section of 4 mm2 corresponds to a diameter of 2.26 mm.
- Check real value wire diameter. If there is a match, the products can be purchased.
Below is a table of the ratio of sections of the main types copper wiring to diameters and current (at a voltage of 220 V).
An additional criterion for matching the cross-section to the diameter is the weight of the wire. The method of determining diameter by weight is used when testing thin wire for winding transformers. The thickness of the product starts from 0.1 mm, and it is difficult to measure it with a micrometer.
Brief table The correspondence of vein diameters by weight is given below. Detailed data is available in stores specializing in the sale of electronic components.
| Diameter, mm | Section, mm2 | Weight, g/km |
| 0,1 | 0,0079 | 70 |
| 0,15 | 0,0177 | 158 |
| 0,2 | 0,0314 | 281 |
| 0,25 | 0,0491 | 438 |
| 0,3 | 0,0707 | 631 |
| 0,35 | 0,0962 | 859 |
| 0,4 | 0,1257 | 1,122 |
When calculating the wire diameter for fuses, the conductor material should be taken into account. A brief table of cable diameters from common types of material and current strength is given below.
| Break current, A | Copper | Aluminum | Nikelin | Iron | Tin | Lead |
| 0,5 | 0,03 | 0,04 | 0,05 | 0,06 | 0,11 | 0,13 |
| 1 | 0,05 | 0,07 | 0,08 | 0,12 | 0,18 | 0,21 |
| 5 | 0,16 | 0,19 | 0,25 | 0,35 | 0,53 | 0,60 |
| 10 | 0,25 | 0,31 | 0,39 | 0,55 | 0,85 | 0,95 |
| 15 | 0,32 | 0,40 | 0,52 | 0,72 | 1,12 | 1,25 |
| 25 | 0,46 | 0,56 | 0,73 | 1,00 | 1,56 | 1,75 |
| 50 | 0,73 | 0,89 | 1,15 | 1,60 | 2,45 | 2,78 |
| 100 | 1,15 | 1,42 | 1,82 | 2,55 | 3,90 | 4,40 |
| 200 | 1,84 | 2,25 | 2,89 | 4,05 | 6,20 | 7,00 |
| 300 | 2,40 | 2,95 | 3,78 | 5,30 | 8,20 | 9,20 |
For multi-core cable
Diameter multi-core cable is determined by the cross-sectional size of one conductor multiplied by their number. The main problem is measuring the diameter of a thin wire.
An example is a cable consisting of 25 cores with a diameter of 0.2 mm. According to the above formula, the cross-section is equal to: S=(3.14/4)*0.2²=0.0314 mm2. With 25 cores it will be: S=0.0314*25=0.8 mm2. Then, using the correspondence tables, it is determined whether it is suitable for transmitting the current of the required strength or not.
Another way to approximately calculate the current strength is to multiply the diameter of a multi-core cable by an adjustment factor of 0.91. The coefficient provides for a non-monolithic wire structure and air gaps between the turns. The outer diameter is measured with little effort, since the surface is easily deformed and the cross-section becomes oval.
When calculating the segment part of the cable, formulas or table values are used. The table shows the standard values for segment width and height.
Photo gallery
Segment cable (far right) 
Cable segment
Table of power consumption of electrical appliances
A common way to determine the required wire cross-section is the peak power calculation method. In order to find out the load, you can use a standard table that summarizes the parameters of power and peak current consumption for household appliances.
| Device type | power, kWt | Peak current, A | Consumption mode |
| Standard incandescent lamp | 0,25 | 1,2 | Constant |
| Kettle with electric heater | 2,0 | 9,0 | Short-term up to 5 minutes |
| Electric stove with 2-4 burners | 6,0 | 60,0 | |
| Microwave | 2,2 | 10,0 | Periodic |
| Electric meat grinder | Likewise | Likewise | Depends on intensity of use |
| Toaster | 1,5 | 7,0 | Constant |
| Electric coffee grinder | 1,5 | 8,0 | Depends on intensity of use |
| Grill | 2,0 | 9,0 | Constant |
| Coffee maker | 1,5 | 8,0 | Constant |
| Separate electric oven | 2,0 | 9,0 | Depends on intensity of use |
| Dishwashing machine | 2,0 | 9,0 | Periodic (for the period of heater operation) |
| Washing machine | 2,0 | 9,0 | Likewise |
| Dryer | 3,0 | 13,0 | Constant |
| Iron | 2,0 | 9,0 | Periodic (for the period of operation of the heating coil) |
| Vacuum cleaner | Likewise | Likewise | Depends on intensity of use |
| Oil heater | 3,0 | 13,0 | Likewise |
| Hairdryer | 1,5 | 8,0 | Likewise |
| Air conditioner | 3,0 | 13,0 | Likewise |
| Computer system unit | 0,8 | 3,0 | Likewise |
| Electric Motor Driven Tools | 2,5 | 13,0 | Likewise |
The current will be consumed by the refrigerator, electrical appliances in standby condition (TVs, radiotelephones), charging device. The total value of power consumption by devices is considered to be within 0.1 kW.
When connecting all available household appliances, the current can reach 100-120 A. This connection option is unlikely, therefore, when calculating the load, common connection combinations are taken into account.
For example, in the morning the following can be used:
- electric kettle - 9.0 A;
- microwave oven - 10.0 A;
- toaster - 7 A;
- coffee grinder or coffee maker - 8 A;
- other Appliances and lighting - 3 A.
The total consumption of devices can reach: 9+10+7+8+3=37 A. There are also calculators that allow you to calculate the current based on power consumption and voltage.
Cable selection according to tables of maximum current in the network
Two types of data from the table above are used for the calculation:
- by total power;
- by the amount of current consumed by the devices.
There are tables of standard values that allow you to determine the required diameter and cross-section, which are then checked on the purchased wire. The found indicator is rounded to big side until it matches the actual cable diameter.
In residential premises, wires with excessive cross-section should not be used, as they have high resistance, which leads to a voltage drop.
For copper cable
To calculate a copper conductor, a table compiled for a voltage of 230 V is used.
| power, kWt | Current, A | ||||
| 0,1 | 0,43 | 0,09 | 0,33 | 0,11 | 0,37 |
| 0,5 | 2,17 | 0,43 | 0,74 | 0,54 | 0,83 |
| 1,0 | 4,35 | 0,87 | 1,05 | 1,09 | 1,18 |
| 2,0 | 8,70 | 1,74 | 1,49 | 2,17 | 1,66 |
| 3,0 | 13,04 | 2,61 | 1,82 | 3,26 | 2,04 |
| 4,0 | 17,39 | 3,48 | 2,10 | 4,35 | 2,35 |
| 5,0 | 21,74 | 4,35 | 2,35 | 5,43 | 2,63 |
| 8,0 | 34,78 | 6,96 | 3,16 | 9,78 | 3,53 |
| 10,0 | 43,48 | 8,7 | 3,33 | 10,87 | 3,72 |
For aluminum cable
To calculate aluminum wire, the table below can be used (data taken for a voltage of 230 V).
| power, kWt | Current, A | Area (with external wiring), mm2 | Diameter (for external wiring), mm | Area (at hidden wiring), mm2 | Diameter (with hidden wiring), mm |
| 0,1 | 0,43 | 0,12 | 0,40 | 0,14 | 0,43 |
| 0,5 | 2,17 | 0,62 | 0,89 | 0,72 | 0,96 |
| 1,0 | 4,35 | 1,24 | 1,26 | 1,45 | 1,36 |
| 2,0 | 8,70 | 2,48 | 1,78 | 2,90 | 1,92 |
| 3,0 | 13,04 | 3,73 | 2,18 | 4,35 | 2,35 |
| 4,0 | 17,39 | 4,97 | 2,52 | 5,80 | 2,72 |
| 5,0 | 21,74 | 6,21 | 2,81 | 7,25 | 3,04 |
| 8,0 | 34,78 | 9,94 | 3,56 | 11,59 | 3,84 |
| 10,0 | 43,48 | 12,42 | 3,98 | 14,49 | 4,30 |
Cable selection according to PUE and GOST tables
When purchasing a wire, it is recommended to look at the GOST standard or the technical specifications according to which the product is manufactured. GOST requirements are higher than similar parameters technical specifications, therefore, you should prefer products made according to the standard.
Tables from the rules for electrical installations (PUE) represent the dependence of the strength of the current transmitted through the conductor on the cross-section of the conductor and the method of installation in the main pipe. The permissible current decreases as the individual cores increase in size or as multi-core cable insulation is used. The phenomenon is associated with a separate clause in the PUE, which stipulates the parameters of the maximum permissible heating of the wires. The main pipe is understood as a box, including a plastic one or when laying wiring in a bundle on a cable tray.