Instruction

1

Domotica to the coil additional coils. This will increase the inductance of

**coils**with constant parameters the remaining structural elements, and the variometer (**coil**with a movable core) - will shift both limit changes in inductance (top and bottom) upwards. When winding additional turns may be that they do not fit on the frame. Do not be tempted to use a finer wire than that used in the original coil, so as not to cause heating of the windings flowing through it current.2

To the coil not having a core, add such. But remember that it needs to be made of such a material, in which the operating frequency

**of the coil**does not occur losses in eddy currents. For an electromagnet operating on direct current, suitable solid steel core, for 50-Hz transformer - core, recruited from the blue plate in the higher frequency coils will have to use cores made of ferrites of different grades.3

Remember that even with the same number of turns and other parameters of the coil of greater diameter will have a higher inductance. It is clear, however, that the wire for its production will need more.

4

The ferrite is manufactured with different magnetic permeability. Replace one ferrite core in the coil to another, which is the value of this parameter above, and its inductance will increase. But it will decrease the marginal rate at which this coil can operate without any appreciable loss in the core.

5

There are

**coils**equipped with special mechanisms to move the core. In order to increase the inductance in this case, slide the core into the frame.6

A closed magnetic circuit, ceteris paribus provides a higher inductance than open. But try not to apply this decision in transformers and chokes operating in the presence of a DC component. It is able to podmanivaya and saturate a closed core, thereby, on the contrary, causing a decrease in inductance

**of the coil**.# Advice 2: How to calculate the inductance of the coil

An inductor can store magnetic energy during the flow of electric current. Its main characteristic is its

**inductance**, which is denoted by the letter L and is measured in Henry (H).**The inductance**depends on its characteristics.*of the coil*You will need

- the coil material and its geometrical parameters

Instruction

1

**The inductance**proportional to the linear dimensions

*of the coil is**of the coil*, magnetic permeability of the core and the square of the number of turns of winding.

**The inductance of a**wound on a toroidal core is: L = ?0*?r*s*(N^2)/l. In this formula ?0 — magnetic constant, which is approximately equal to 1.26*(10^-6) Ng/m ?r — relative magnetic permeability of core material, which depends on the frequency), s is the cross — sectional area of the core, l is the length of the middle line of the core, N is the number of turns of the

*coil**coil*.

Relative magnetic permeability and material, as well as the number of turns N are dimensionless quantities.

2

Thus,

**the inductance***of the coil*is greater, the larger the area of its cross section. This condition increases the magnetic flux through the coil when the same current.**The inductance**inductance in µh can be calculated also by the formula: L = L0*(N^2)*D*(10^-3). Here N is the number of turns, D is the diameter of*of the coil**coil*in inches. The coefficient of L0 depends on the ratio of the length*of the coil*to its diameter. For single-layer*coil*it is equal to: L0 = 1/(0,1*((l/D)+0,45)).3

If circuit

If

*coil*are connected in series, their total**inductance**equals the sum of the inductances of all coils: L = (L1+L2+...+Ln)If

*coils*are connected in parallel, the total**inductance**equal to: L = 1/((1/L1)+(1/L2)+...+(1/Ln))Note

The main parameter characterizing the properties of inductors and chokes is the inductance. The inductance of the coil depends on its size and shape, number of turns and the magnetic permeability of the medium. . Characterizes the energy losses in the coil and is determined by the ratio of its inductive reactance to active resistance

Useful advice

The physical nature of inductance. Inductors have the property to render the reactance to alternating current with little resistance to direct current. Together with the capacitors they are used to create filters performing the frequency selection of electrical signals and for generating delay elements of signals and storage elements...

# Advice 3: How to determine the inductance of the coil

The inductance of the coil can be measured directly or indirectly. In the first case, you will need active or bridge device, and the second will have to use the generator, voltmeter and milliammeter, and then to carry out a series of calculations.

You will need

- - active or bridge meter inductance;
- - generator of sinusoidal voltage;
- - voltmeter and milliammeter AC;
- - frequency;
- - scientific calculator.

Instruction

1

To measure the inductance of the active device, connect the coil and then successively selecting the measurement range switch, select such of them to the result was approximately in the middle of the range. Read the result. If the meter has an analog scale, when reading the results, take into account the price of division, and the coefficient shown next to the corresponding switch position.

2

On the bridge device after each selector turn the knob balancing of the bridge at any of the end positions, and then rotate it all the way in the opposite direction. Find a range in which the handle is possible to balance the bridge. Having achieved the disappearance of the sound in the speaker or headphones or decreasing readings dial indicator to zero, read the indications on the scale of the regulator (but not the gauge). In this case, as in the previous case, consider the division and the factor that should be multiplied by the range statement.

3

To measure inductance indirectly assemble the measuring circuit. The AC voltmeter switched to the limit at which the upper boundary of the range corresponds to a voltage of several volts, connect the parallel output of the generator. There connect and a frequency counter. Also in parallel, they connect the series circuit consisting of the test inductors as well as milliamperemeter AC. Both devices should show the current, not the amplitude value measured value, and can be calculated for sinusoidal oscillations.

4

The generator will enable the generation of the voltage sine wave. Ensure that the voltmeter showed about two volts. Increase the frequency until then, until the readings on the milliammeter begins to decrease. Make sure that they reduce to about half the original value. Select frequency limit corresponding to the measured frequency. Read the testimony of all three devices, and then disconnect the generator and disassemble the measuring circuit.

5

Put readings in SI units. Divide the voltage across the current. Will the inductive reactance of the coil at the frequency at which the measurement was carried out. It will be expressed in ohms.

6

Calculate the inductance according to the formula: L=X/(2πF), where L is the frequency, G (Henry), X is the inductive reactance in Ohms, F is frequency, Hz. If necessary, move the result of the calculation in derived units (e.g., MH, microgenre).

Note

Do not touch the elements of the measurement circuit when it is energized.