In practice, it is not uncommon to find the problem of finding the resistance of conductors and resistors for various connection methods. The article discusses how resistance is calculated when conductors are connected in parallel and some other technical issues.
Conductor resistance
All conductors have the ability to prevent the flow of electric current, it is commonly called electrical resistance R, it is measured in ohms. This is the basic property of conductive materials.
Resistivity is used for conducting electrical calculations - ρ Ohm·m/mm2. All metals are good conductors, copper and aluminum are most widely used, and iron is used much less frequently. The best conductor is silver, it is used in the electrical and electronic industries. High resistance alloys are widely used.
When calculating the resistance, the formula known from the school physics course is used:
R=ρ · l/S, S – sectional area; l – length.
If we take two conductors, then their resistance atparallel connection will become smaller due to the increase in the total cross section.
Current density and conductor heating
For practical calculations of the operating modes of conductors, the concept of current density is used - δ A/mm2, 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 (Rules for the Construction of Electrical Installations). For conductors of heating devices, there are current density standards.
If the density δ is higher than the permissible one, the conductor may be destroyed, for example, when the cable overheats, its insulation is destroyed.
The rules regulate the calculation of conductors for heating.
Methods of connecting conductors
Any conductor is much more convenient to depict on the diagrams as an 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=R1 + R2 + R3.
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 totalresistance R=1/G.
You can do it differently, first calculate the total resistance when resistors R1 and R2 are connected in parallel, then repeat the operation and find R.
The third connection method is the most complex - a mixed connection, that is, all the considered options are present. The diagram is shown in the photo.
To calculate this circuit, it should be simplified, to do this, replace the resistors R2 and R3 with one R2, 3. It turns out a simple circuit.
Now you can calculate the resistance in parallel connection, the formula of which is:
R2, 3, 4=R2, 3 R4/(R2, 3 + R4).
The circuit becomes even simpler, it still has resistors connected in series. In more complex situations, the same conversion method is used.
Types of conductors
In electronic engineering, in the production of printed circuit boards, conductors are thin strips of copper foil. Due to their short length, their resistance is negligible, and in many cases it can be neglected. For these conductors, the resistance in parallel connection decreases 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 mm. For powerful transformers and electric motors, rectangular copper bars are produced.sections. Sometimes, during repairs, a large-diameter wire is replaced with several smaller ones connected in parallel.
A special section of conductors are wires and cables, the industry provides the widest choice of grades for a variety of needs. Often you have to replace one cable with several, smaller sections. The reasons for this are very different, for example, a cable with a cross section of 240 mm2 is very difficult to lay along a route with sharp bends. It is replaced with 2x120mm2, and problem solved.
Calculation of wires for heating
The conductor is heated by the flowing current, if its temperature exceeds the permissible value, the insulation is destroyed. PUE provides for the calculation of conductors for heating, the initial data for it are the current strength and the environmental conditions in which the conductor is laid. According to these data, the recommended conductor cross-section (wire or cable) is selected from the tables in the PUE.
In practice, there are situations when the load on the existing cable has greatly increased. There are two ways out - to replace the cable with another one, it can be expensive, or to lay another one parallel to it in order to relieve the main cable. In this case, the resistance of the conductor when connected in parallel decreases, hence the heat generation decreases.
To correctly select the cross section of the second cable, use the tables of the PUE, it is important not to make a mistake with the definition of its operating current. In this situation, the cooling of the cables will be even better than that of one. It is recommended to calculateresistance when two cables are connected in parallel to more accurately determine their heat dissipation.
Calculation of conductors for voltage loss
When the consumer Rn is located at a large distance L from the energy source U1, a rather large voltage drop occurs on the line wires. The consumer Rn receives voltage U2 much lower than the initial U1. In practice, various electrical equipment connected to the line in parallel acts as a load.
To solve the problem, the resistance is calculated when all equipment is connected in parallel, so the load resistance Rn is found. Next, determine the resistance of the line wires.
Rl=ρ 2L/S,
Here S is the section of the line wire, mm2.
Next, the line current is determined: I=U1/(Rl + Rn). Now, knowing the current, determine the voltage drop on the wires of the line: U=I Rl. It is more convenient to find it as a percentage of U1.
U%=(I Rl/U1) 100%
Recommended value of U% - no more than 15%. The above calculations are applicable for any kind of current.
