Resistance at parallel connection: calculation formula

In practice, it is not uncommon to find the resistance of conductors and resistors for various connection methods. The article considers how the resistance is calculated for parallel connection of conductors and some other technical issues.

Conductor resistance

All conductors have the property of impeding the flow of electric current, it is usually called the electrical resistance R, it is measured in ohms. This is the main property of conductor materials.

To conduct electrical calculations, the resistivity is applied - ρ Ohm · m / mm ² . All metals are good conductors, copper and aluminum are used most, iron is used much less. The best conductor is silver, it is used in the electrical and electronics industries. Alloys with a high resistance value are widely distributed.

When calculating the resistance, the formula known from the school course of physics is used:

R = ρ · l / S, S is the cross-sectional area; L is the length.

If we take two conductors, then their resistance at parallel connection will be less because of the increase in the total cross section.

Current density and conductor heating

For the practical calculations of the operating modes of conductors, the concept of current density - δ A / mm ^{2 is used} , it is calculated by the formula:

Δ = I / S, I - current, S - section.

The current, passing through the conductor, heats it. The greater δ, the more heated the conductor. For wire and cable standards for the permissible density are developed, which are given in the PUE (Rules of the Electrical Installation Device). For conductors of heating devices, there are norms of current density.

If the density δ is higher than the allowable value, the conductor may break, for example, if the cable is overheated, the insulation is destroyed.

The rules regulate the calculation of conductors for heating.

Ways of connecting conductors

Any conductor is much more convenient to depict on circuits as electrical resistance R, then they are easy to read and analyze. There are only three ways to connect resistances. The first method is the simplest one - a serial connection.

The photo shows that the impedance is: R = R ₁ + R ₂ + R ₃ .

The second way is more complicated - parallel connection. Calculation of the resistance for parallel connection is carried out in stages. The total conductivity G = 1 / R is calculated, and then the impedance R = 1 / G.

It is also possible to do it differently, first calculate the total resistance for parallel connection of resistors R1 and R2, then repeat the operation and find R.

The third method of connection is the most complex - a mixed connection, that is, all the options considered are present. The diagram is shown in the photo.

To calculate this circuit, it should be simplified by replacing resistors R2 and R3 with one R2.3. It turns out a simple scheme.

Now you can calculate the resistance for a parallel connection, the formula of which looks like:

R2.3.4 = R2.3 · R4 / (R2.3 + R4).

The circuit becomes even simpler, in it there are resistors having a serial connection. In more complex situations, the same conversion method is used.

Types of conductors

In electronics, in the production of printed circuit boards, conductors are thin strips of copper foil. Because of their short length, the resistance is insignificant, they can in many cases be neglected. For these conductors, the resistance at parallel connection decreases due to the increase in cross-section.

A large section of conductors represents the winding wires. They are produced in different diameters - from 0.02 to 5.6 millimeters. For powerful transformers and electric motors, copper bars of rectangular cross section are produced. Sometimes, during repair, a large diameter wire is replaced by several smaller wires connected in parallel.

A special section of the conductors are wires and cables, the industry provides a wide range of brands for a variety of needs. It is often necessary 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 the track with steep bends. It is replaced by 2 × 120 mm ² , And the problem is solved.

Calculation of wires for heating

The conductor is heated by a flowing current, if its temperature exceeds the permissible value, then the insulation breaks down. The 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 operating cable has greatly increased. There are two ways out - to replace the cable with another one, it is expensive, or parallel to it, lay another one to unload the main cable. In this case, the resistance of the conductor decreases with parallel connection, therefore the heat release falls.

In order to correctly select the cross-section of the second cable, use the PUE tables, it is important not to make a mistake in determining its operating current. In this situation, cooling cables will be even better than one. It is recommended to calculate the resistance when connecting two cables in parallel, in order to more accurately determine their heat release.

Calculation of conductors for voltage loss

When the consumer R _{n is located} at a large distance L from the energy source U ₁ , a rather large voltage drop occurs on the line wires. To the consumer R _n , the voltage U _{2 is} considerably lower than the initial voltage 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 with parallel connection of all equipment, so is the load resistance R _n . Next, determine the resistance of the line wires.

R _l = ρ · 2L / S,

Here S is the section of the line wire, mm ² .

Then the current in the line is determined: I = U ₁ / (R _l + R _n ). Now, knowing the current, determine the voltage drop on the wires of the line: 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.