At home, we often use portable extension cords - sockets for temporary ( usually remaining permanently) turning on household appliances: electric heater, air conditioner, iron with high current consumption.
The cable for this extension cord is usually selected according to the principle of whatever comes to hand, and this does not always correspond to the required electrical parameters.
Depending on the diameter (or cross-section of the wire in mm2), the wire has a certain electrical resistance for the passage of electric current.
The larger the cross-section of the conductor, the lower its electrical resistance, the lower the voltage drop across it. Accordingly, there is less power loss in the wire due to its heating.
Let us carry out a comparative analysis of the power loss for heating in the wire depending on its transverse 
sections. Let's take the most common cables in everyday life with a cross-section: 0.75; 1.5; 2.5 mm.sq. for two extension cords with cable length: L = 5 m and L = 10 m.
Let's take as an example a load in the form of a standard electric heater with electrical parameters:
- supply voltageU = 220 Vol T ;
— electric heater power P = 2.2 kW = 2200 W ;
— current consumption I = P/U = 2200 W / 220 V = 10 A.
From reference literature, let's take resistance data for 1 meter of wire of different cross sections.
A table of resistances of 1 meter of wire made of copper and aluminum is given.
Let's calculate the loss of power spent on heating for the cross-section of the wire S = 0.75 mm.sq. The wire is made of copper.
Resistance of 1 meter wire (from the table) R 1 = 0.023 Ohm.
Length of cable L=5 meters.
Length of wire in cable (round trip)2 L =2
·
5 = 10 meters.
Electrical resistance wires in the cable R = 2 · L · R 1 = 2 · 5 · 0.023 = 0.23 Ohm.
Voltage drop in the cable when current passes I = 10 A will: U = I R = 10 A 0.23 Ohm = 2.3 V.
The power loss due to heating in the cable itself will be: P = U I = 2.3 V 10 A = 23 W.
If the cable length L = 10 m. (same cross-section S = 0.75 mm2), the power loss in the cable will be 46 W. This is approximately 2% of the power consumed by the electric heater from the network.
For cables with aluminum conductors of the same section S = 0.75 mm.sq.. the readings increase and amount to L = 5 m-34.5 W. For L = 10 m - 69 W.
All calculation data for cables with a cross section of 0.75; 1.5; 2.5 mm.sq. for cable length L = 5 and L = 10 meters are given in the table.
Where: S – wire cross-section in mm2;
R 1– resistance of 1 meter of wire in Ohm;
R - cable resistance in Ohms;
U – voltage drop in the cable in Volts;
P – power loss in the cable in watts or as a percentage.
What conclusions should be drawn from these calculations?
- — With the same cross-section, copper cable has a greater margin of safety and less losses electrical power to heat the wire R.
- — As the cable length increases, losses P increase. To compensate for losses, it is necessary to increase the cross-section of the cable wires S.
- — It is advisable to choose a cable with a rubber sheath, and the cable cores should be multi-core.
For the extension cord, it is advisable to use a Euro socket and Euro plug. The pins of the Euro plug have a diameter of 5 mm. A simple electric plug has a pin diameter of 4 mm. Euro plugs are designed to carry more current than a simple socket and plug. The larger the diameter of the plug pins, the larger the contact area at the junction of the plug and socket,hence lower contact resistance. This contributes to less heating at the junction of the plug and socket.
The effect of the conductor material is taken into account using resistivity, which is usually denoted by the letter of the Greek alphabet ρ and represents conductor resistance with a cross section of 1 mm 2 and a length of 1 m. Silver has the lowest resistivity ρ = 0.016 Ohm.mm 2 /m. Below are the values resistivity for multiple conductors:
- Wire resistance for silver - 0.016,
- Wire resistance for lead - 0.21,
- Wire resistance for copper - 0.017,
- Wire resistance for nickel - 0.42,
- Wire resistance for aluminum - 0.026,
- Wire resistance for manganin - 0.42,
- Wire resistance for tungsten - 0.055,
- Wire resistance for constantan - 0.5,
- Wire resistance for zinc - 0.06,
- Wire resistance for mercury - 0.96,
- Wire resistance for brass - 0.07,
- Wire resistance for nichrome - 1.05,
- Wire resistance for steel - 0.1,
- Wire resistance for fechral -1.2,
- Wire resistance for phosphor bronze - 0.11,
- Wire resistance for chromal - 1.45
Since alloys contain different amounts of impurities, the resistivity may change.
Wire resistance calculated using the formula below:
R=(ρ?l)/S
- R - resistance,
- Ohm; ρ - resistivity, (Ohm.mm 2)/m;
- l—wire length, m;
- s is the cross-sectional area of the wire, mm2.
The cross-sectional area is calculated as follows:
S=(π?d^2)/4=0.78?d^2≈0.8?d^2
- where d is the diameter of the wire.
You can measure the diameter of the wire with a micrometer or caliper, but if you don’t have them on hand, you can tightly wrap about 20 turns of wire around a pen (pencil), then measure the length of the wound wire and divide by the number of turns.
To determine the length of wire needed to achieve the required resistance, you can use the formula:
l=(S?R)/ρ
Notes:
1. If the data for the wire is not in the table, then some average value is taken. As an example, a nickel wire with a diameter of 0.18 mm, the cross-sectional area is approximately 0.025 mm2, the resistance of one meter is 18 Ohms, and the permissible current is 0.075 A.
2.The data in the last column, for a different current density, must be changed. For example, with a current density of 6 A/mm2, the value must be doubled.
Example 1. Let's find the resistance of 30 m copper wire with a diameter of 0.1 mm.
Solution. Using the table, we take the resistance of 1 m of copper wire, which is equal to 2.2 Ohms. This means that the resistance of 30 m of wire will be R = 30.2.2 = 66 Ohms.
The calculation using the formulas will look like this: cross-sectional area: s = 0.78.0.12 = 0.0078 mm2. Since the resistivity of copper is ρ = 0.017 (Ohm.mm2)/m, we get R = 0.017.30/0.0078 = 65.50 m.
Example 2. How much manganin wire with a diameter of 0.5 mm is needed to make a rheostat with a resistance of 40 Ohms?
Solution. Using the table, we select the resistance of 1 m of this wire: R = 2.12 Ohm: To make a rheostat with a resistance of 40 Ohms, you need a wire whose length is l = 40/2.12 = 18.9 m.
The calculation using the formulas will look like this. Cross-sectional area of the wire s = 0.78.0.52 = 0.195 mm 2. Wire length l = 0.195.40/0.42 = 18.6 m.
Electrical resistance of the conductor: 1) a quantity characterizing the resistance of a conductor or electrical circuit to electric current;
2) a structural element of an electrical circuit included in a circuit to limit or regulate the current.
Electrical resistance of metals depends on the material of the conductor, its length and cross-section, temperature and condition of the conductor (pressure, mechanical forces of tension and compression, i.e. external factors affecting the crystalline structure of metal conductors).
Dependence of resistance on material, length and cross-sectional area of the conductor:
where is the resistivity of the conductor;
l – conductor length;
S is the cross-sectional area of the conductor.
Dependence of conductor resistance on temperature:
or 
,
where R t – resistance at temperature t 0 C;
R 0 – resistance at 0 0 C;
- temperature coefficient of resistance, which shows how the resistance of the conductor changes in relation to its resistance at 0 0 C if the temperature changes by one degree;
T – thermodynamic temperature.
Resistance connections: serial, parallel, mixed.
a) Series connection of resistances is a system of conductors (resistances) that are connected one after the other, so that the same current flows through each of the resistances:
I = I 1 = I 2 == I n .
Voltage when resistors are connected in series equal to the sum of the voltages at each of the resistances:
.
Voltage across each series-connected resistance proportional to the value of this resistance:
.
Voltage distribution across series-connected circuit elements (voltage divider) :
,
U is the voltage on the section of the circuit with resistance R1;
R – connection impedance;
R 1 – resistance of the circuit section with the selected resistance.
equal to the sum of individual resistances and it is greater than the largest of those included:
.
Total circuit resistance in series connection n identical resistances :
,
where n is the number of resistances connected in series;
R 1 = value of individual resistance.
b) Parallel connection of resistances: a sign of such a connection is the branching of current I into separate currents through the corresponding resistances. In this case, the current I is equal to the sum of the currents through a single resistance:
.
Total voltage at parallel connection equal to the voltage across a single resistance:
U = U 1 = U 2 = = U i .
Relationship between current and resistance in parallel connection: when resistances are connected in parallel, the currents in individual conductors are inversely proportional to their resistances:
.
The reciprocal of the circuit impedance (total conductance) in a parallel connection, equal to the sum of the conductivities of individual conductors. In this case, the total resistance of the circuit is less than the smallest resistance included:
;
.
Total circuit conductance in parallel connection n conductors:
G pairs = nG 1,
where G pairs is the conductivity of the circuit;
G 1 – conductivity of a single conductor.
Shunting of electrical measuring instruments – expanding the limit of current measurement using an electrical measuring device to which a conductor with low resistance (shunt) is connected in parallel. In this case
,
where I p is the current flowing through the device;
I – current in the circuit;
n = R p /R sh – ratio of the device resistance R p to the shunt resistance R sh.
Additional resistance – a resistance that is connected in series to an electrical measuring device to expand the voltage measurement limit. Wherein
,
where U p is the voltage on the device;
U – voltage in the circuit;
N = R d / R p – the ratio of the value of the additional resistance to the resistance of the device.
Electrical conductivity – physical quantity, reciprocal of the conductor resistance:
.
Superconductivity– a property of many conductors, consisting in the fact that their electrical resistance abruptly drops to zero when cooled below a certain critical temperature Tk, characteristic of a given material.
Relationship between conductivity and resistivity (electrical resistivity) :
;
.
Dependence of conductor resistivity on temperature:
,
where t – resistivity at temperature t 0 C;
0 – resistivity at 0 0 C;
- temperature coefficient of resistance, which shows how the resistivity of a conductor changes in relation to its resistivity at 0 0 C if the temperature changes by one degree.
Tasks: 1. Familiarize yourself with the electrical measuring instruments used in your work. Enter the results in the table. 1.
Table 1.
2. Measure the electrical resistivity.
1. Measure its diameter with a micrometer in several places in the working part of the conductor. Calculate the average diameter.
2. Set the moving contact to 0.5 0.7 from the length of the working part of the conductor. Enter the length value in table 2.
3. Turn on the installation to an alternating current network with a voltage of 220 V. The indicator light should light up.
4. Take current and voltage measurements. Enter the results in Table 2.
Table 2.
5. Disable installation. Set the moving contact to a different value of the working part of the conductor being tested. Turn the unit back on and determine the new current and voltage values.
Note. Changing the length of the working part of the conductor, determining current and voltage are carried out 3-5 times.
6. Since
,
,
(1)
where is the electrical resistivity of the conductor;
ℓ - conductor length;
S is the cross-sectional area.
,
(2)
Where 
- voltmeter error;
- instrument error of the milliammeter;
- set by the teacher;
d, ℓ - are determined by known methods.
10. Write the result as a confidence interval
Electrical resistance is the main characteristic of conductor materials. Depending on the area of application of the conductor, the value of its resistance can play both a positive and negative role in the functioning of the electrical system. Also, the specific application of the conductor may necessitate taking into account additional characteristics, the influence of which in a particular case cannot be neglected.
Conductors are pure metals and their alloys. In a metal, atoms fixed in a single “strong” structure have free electrons (the so-called “electron gas”). It is these particles that in this case are the charge carriers. Electrons are in constant, random motion from one atom to another. When an electric field appears (connecting a voltage source to the ends of the metal), the movement of electrons in the conductor becomes ordered. Moving electrons encounter obstacles on their path caused by the peculiarities of the molecular structure of the conductor. When they collide with a structure, charge carriers lose their energy, giving it to the conductor (heating it). The more obstacles a conductive structure creates to charge carriers, the higher the resistance.
As the cross section of the conducting structure increases for one number of electrons, the “transmission channel” will become wider and the resistance will decrease. Accordingly, as the length of the wire increases, there will be more such obstacles and the resistance will increase.
Thus, in basic formula to calculate the resistance, the length of the wire, the cross-sectional area and a certain coefficient are included that connects these dimensional characteristics with the electrical quantities of voltage and current (1). This coefficient is called resistivity.
R= r*L/S (1)
Resistivity
Resistivity is unchanged and is a property of the substance from which the conductor is made. Units of measurement r - ohm*m. Often the resistivity value is given in ohm*mm sq./m. This is due to the fact that the cross-sectional area of the most commonly used cables is relatively small and is measured in mm2. Let's give a simple example.
Task No. 1. Copper wire length L = 20 m, cross-section S = 1.5 mm. sq. Calculate the wire resistance.
Solution: resistivity of copper wire r = 0.018 ohm*mm. sq./m. Substituting the values into formula (1) we get R=0.24 ohms.
When calculating the resistance of the power system, the resistance of one wire must be multiplied by the number of wires.
If instead of copper you use aluminum with a higher resistivity (r = 0.028 ohm * mm sq. / m), then the resistance of the wires will increase accordingly. For the example above, the resistance will be R = 0.373 ohms (55% more). Copper and aluminum are the main materials for wires. There are metals with lower resistivity than copper, such as silver. However, its use is limited due to its obvious high cost. The table below shows the resistance and other basic characteristics of conductor materials.
Table - main characteristics of conductors
Heat losses of wires
If, using the cable from the above example, a load of 2.2 kW is connected to a single-phase 220 V network, then current I = P / U or I = 2200/220 = 10 A will flow through the wire. Formula for calculating power losses in the conductor:
Ppr=(I^2)*R (2)
Example No. 2. Calculate active losses when transmitting power of 2.2 kW in a network with a voltage of 220 V for the mentioned wire.
Solution: substituting the values of current and wire resistance into formula (2), we obtain Ppr=(10^2)*(2*0.24)=48 W.
Thus, when transmitting energy from the network to the load, losses in the wires will be slightly more than 2%. This energy is converted into heat released by the conductor into the environment. According to the heating condition of the conductor (according to the current value), its cross-section is selected, guided by special tables.
For example, for the above conductor, the maximum current is 19 A or 4.1 kW in a 220 V network.
To reduce active losses in power lines, increased voltage is used. At the same time, the current in the wires decreases, losses fall.
Effect of temperature
An increase in temperature leads to an increase in vibrations of the metal crystal lattice. Accordingly, electrons encounter more obstacles, which leads to an increase in resistance. The amount of “sensitivity” of metal resistance to temperature rise is called temperature coefficientα. The formula for calculating temperature is as follows
R=Rн*, (3)
where Rн – wire resistance under normal conditions (at temperature t°н); t° is the temperature of the conductor.
Usually t°n = 20° C. The value of α is also indicated for temperature t°n.
Task 4. Calculate the resistance of a copper wire at a temperature t° = 90° C. α copper = 0.0043, Rн = 0.24 Ohm (task 1).
Solution: substituting the values into formula (3) we get R = 0.312 Ohm. The resistance of the heated wire being analyzed is 30% greater than its resistance at room temperature.
Effect of frequency
As the frequency of the current in the conductor increases, the process of displacing charges closer to its surface occurs. As a result of an increase in the concentration of charges in the surface layer, the resistance of the wire also increases. This process is called the “skin effect” or surface effect. Skin coefficient– the effect also depends on the size and shape of the wire. For the above example, at frequency alternating current 20 kHz wire resistance will increase by approximately 10%. Note that high frequency components may have the current signal of many modern industrial and household consumers(energy saving lamps, switching power supplies, frequency converters and so on).
Influence of neighboring conductors
There is a magnetic field around any conductor through which current flows. The interaction of the fields of neighboring conductors also causes energy loss and is called the “proximity effect”. Also note that any metal conductor has inductance created by the conductive core and capacitance created by the insulation. These parameters are also characterized by the proximity effect.
Technologies
High voltage wires with zero resistance
This type of wire is widely used in car ignition systems. The resistance of high-voltage wires is quite low and amounts to several fractions of an ohm per meter of length. Let us remind you that resistance of this magnitude cannot be measured with an ohmmeter. general use. Often, measuring bridges are used for the task of measuring low resistances.
Structurally, such wires have a large number of copper cores with insulation based on silicone, plastics or other dielectrics. The peculiarity of the use of such wires is not only the operation at high voltage, but also the transfer of energy in a short period of time (pulse mode).
Bimetallic cable
The main area of application of the mentioned cables is the transmission of high-frequency signals. The core of the wire is made of one type of metal, the surface of which is coated with another type of metal. Since at high frequencies only the surface layer of the conductor is conductive, it is possible to replace the inside of the wire. This saves expensive material and improves the mechanical characteristics of the wire. Examples of such wires: silver-plated copper, copper-plated steel.
Conclusion
Wire resistance is a value that depends on a group of factors: conductor type, temperature, current frequency, geometric parameters. The significance of the influence of these parameters depends on the operating conditions of the wire. Optimization criteria, depending on the tasks for wires, can be: reducing active losses, improving mechanical characteristics, reducing prices.