In practice, one often encounters the problem of finding the resistance of conductors and resistors with various connection methods. The article discusses how resistance is calculated in parallel connection of conductors and some other technical issues.
Conductor resistance
All conductors have the property of inhibiting the flow of electric current, it is commonly called the electrical resistance R, it is measured in ohms. This is the main property of conductive materials.
For conducting electrical calculations, the specific resistance is used - ρ Ohm · m / mm 2 . All metals are good conductors, copper and aluminum are most widely used, and iron is much less commonly used. The best conductor is silver, it is used in the electrical and electronic industries. Alloys with a high resistance value are widespread .
When calculating the resistance, the formula known from the school physics course is used:
R = ρ · l / S, S is the cross-sectional area; l is the length.
If you take two conductors, then their resistance with a parallel connection will become less due to an increase in the total cross section.
Current density and conductor heating
For practical calculations of the operating modes of the conductors, the concept of current density is used - δ A / mm 2 , it is calculated by the formula:
δ = I / S, I - current, S - section.
Current passing through the conductor heats it. The larger δ, the more the conductor heats up. For wires and cables, norms of permissible density have been developed, which are given in the PUE (Electrical Installation Rules). For conductors of heating devices, there are standards for current density.
If the density δ is higher than permissible, the conductor can be destroyed, for example, when the cable overheats, its insulation is destroyed.
The rules govern the calculation of conductors for heating.
Conductor Connection Methods
Any conductor is much more convenient to depict on the diagrams as the electrical resistance R, then they are easy to read and analyze. There are only three ways to connect resistances. The first way is the easiest - serial connection.
The photo shows that the impedance is: R = R 1 + R 2 + R 3 .
The second way is more complicated - parallel connection. The calculation of resistance in parallel connection is carried out in stages. The total conductivity G = 1 / R is calculated, and then the impedance R = 1 / G.
You can do it differently, first calculate the total resistance in parallel with the resistors R1 and R2, then repeat the operation and find R.
The third method of connection is the most difficult - mixed connection, that is, all the options considered are present. The scheme is shown in the photo.
To calculate this circuit, it should be simplified, for this, the resistors R2 and R3 are replaced with one R2.3. It turns out a simple scheme.
Now you can calculate the resistance in parallel connection, the formula of which has the form:
R2.3.4 = R2.3 · R4 / (R2.3 + R4).
The circuit becomes even simpler; resistors with a serial connection remain in it. In more complex situations, the same conversion method is used.
Types of conductors
In electronic technology, in the manufacture of printed circuit boards, conductors are thin strips of copper foil. Due to the short length, their resistance is negligible, in many cases they can be neglected. For these conductors, the resistance in parallel connection is reduced due to the increase in cross section.
A large section of conductors is represented by winding wires. They are available in different diameters - from 0.02 to 5.6 millimeters. For high-power transformers and electric motors, copper busbars of rectangular cross section are produced. Sometimes, when repairing, replace a large-diameter wire with several smaller ones connected in parallel.
A special section of conductors is represented by wires and cables, the industry provides a wide selection of brands for a wide variety of needs. Often you have to replace one cable with several, smaller cross-sections. The reasons for this are very different, for example, a cable with a cross section of 240 mm 2 is very difficult to lay along a highway with sharp bends. It is replaced by 2 × 120 mm 2 , and the problem is resolved.
Calculation of wires for heating
The conductor is heated by the flowing current, if its temperature exceeds the permissible limit, insulation breaks down. PUE provides for the calculation of conductors for heating, the initial data for it are the current strength and environmental conditions in which the conductor is laid. Based on these data, the recommended section of the conductor (wire or cable) is selected from the tables in the PUE.
In practice, there are situations when the load on the existing cable has increased significantly. There are two ways out: to replace the cable with another, it can be expensive, or lay another one in parallel to unload the main cable. In this case, the resistance of the conductor with parallel connection decreases, therefore, the heat release decreases.
In order to correctly choose the cross section of the second cable, they use PUE tables, it is important not to make a mistake in determining its operating current. In this situation, cable cooling will be even better than one. It is recommended to calculate the resistance when two cables are connected in parallel in order to more accurately determine their heat dissipation.
Calculation of conductors for voltage loss
When the location of the consumer R n at a large distance L from the energy source U 1 there is a fairly large voltage drop across the wires of the line. To the consumer R n the voltage U 2 is significantly lower than the initial U 1 . In practice, various electrical equipment connected to the line in parallel acts as a load.
To solve the problem, the resistance is calculated in parallel connection of all equipment, so the load resistance R n is found . Next, determine the resistance of the line wires.
R l = ρ · 2L / S,
Here S is the section of the line wire, mm 2 .
Next, the current in the line is determined: I = U 1 / (R l + R n ). Now, knowing the current, determine the voltage drop on the line wires: U = I · R l . It is more convenient to find it as a percentage of U 1 .
U% = (I · R l / U 1 ) · 100%
The recommended value of U% is no more than 15%. The above calculations are applicable for any kind of current.