Initial phase. Phase shift

Measurement of phase shift in AC circuits

Devices and accessories: laboratory panel “Alternating current. Ohm's Law" with a resistor, capacitor and coil, alternating current source - generator G3-118, universal voltmeter V7-40.

Introduction. Let's consider an electrical circuit (Fig. 1) containing (in the general case) active resistance R, inductance L and capacity C, which includes an alternating current source with an output voltage

Where u– instantaneous voltage – voltage at a moment in time t,

U m– voltage amplitude,

 – cyclic frequency of voltage fluctuations.

V L. Such a circuit is called a circuit with focused parameters. According to Kirchhoff's second rule, for this circuit we can write Fig. 1

the following equation:

Where i– instantaneous current value in the circuit, changing like voltage with frequency  ,

u C– voltage across the capacitor.

Let's consider a number of circuits with different loads and the corresponding equations.

1. Let only the source be connected active resistance R(Fig. 2, A). Wherein L=0, C . Resistance is called active because the energy of the electric current is converted into the internal energy of the conductor or into mechanical work.

Equation (2) for this particular case takes the form:

from which it follows that the current

Where I Rm– current amplitude in a circuit with an active load, I Rm = U m / R.

T

Thus, fluctuations in current strength in a circuit containing only active

resistance are in phase with voltage fluctuations (Fig. 2, b). The vector diagram for this situation is shown in Fig. 2, V.

2

. Let the load of the AC source be an inductor L. The active and capacitive resistance of this circuit is neglected (Fig. 3, A).

The Kirchhoff equation (2) for such a contour has the form:

The magnitude of the self-inductive emf is numerically equal to the voltage drop across the inductance L, which we will further denote U L .

From equation (4) we can write that

. (5)

Let us integrate equation (5) and obtain the following expression for the current:

Since there is no direct current component in the circuit, then const=0 .

Thus, the current in the circuit only with inductance has the form

, (6)

Where I Lm– current amplitude. . (7)

Comparing expression (7) with the one given earlier (3), we can conclude that the value  L In the case of an inductive load, it plays the role of resistance. It's called inductive resistance and is designated X L .

From a comparison of formulas (6) and (1) it is clear that the current in a circuit containing a purely inductive load is lagging behind from phase voltage to  radians (Fig. 3, b). In the vector diagram, the voltage vector U Lm turned at an angle  from the current vector in the positive direction - counterclockwise, current vector I m lags behind him.

3. Let only a capacitor with a capacity of WITH without dielectric energy losses (Fig. 4, A).

Fig.4

Capacitor voltage WITH equal to the source output voltage

(8)

Since and , then

(9)

Where (10)

Size (11)

called capacitive circuit resistance. (Index WITH when designating current indicates just for that, that it flows in a circuit with a purely capacitive load).

For DC  , so the capacitor presents an infinitely large resistance. As the frequency of alternating current increases, the capacitance decreases.

From a comparison of formulas (9) and (1) it is clear that the current flowing through the capacitor ahead in phase voltage across the capacitor at  . (Fig. 4, b). On the vector diagram (Fig. 4, V) current vector I Cm rotated by an angle  from U Cm to the side positive direction of rotation.

Resistance X C And X L called reactive. On them not happening converting the energy of an electric current into the internal energy of a load despite the presence of resistance (this is the meaning of their name).

4. Consider an electrical circuit with lumped parameters R, L, C(Fig. 5, A). Under AC voltage U VX alternating current will be established in the circuit I, whose value is the same in all elements - resistor, coil and capacitor, since they are connected in series (we consider the current through each of the voltmeters to be negligible compared to I). The flowing current causes a voltage drop across them: – across the active resistance, – on inductance and – on capacitance. Values U R , U L , U C , U VX The voltages shown by the corresponding voltmeters are indicated. The sum of the voltages must be equal to the voltage applied to this circuit U VX. But this sum can be neither arithmetic nor algebraic, but only vector, since between the voltages there is phase shifts.

To calculate AC circuits, two methods are used: 1) the so-called symbolic is an analytical method using complex variables and 2) graphic– vector diagram method. Let's use the second one.

The construction of a vector diagram for a sequential circuit is carried out in the following order.

1. In an arbitrary direction, for example horizontally, draw current axis and the current vector is plotted on it on a certain scale I m. Instead of the amplitude value, the effective value can be set aside, i.e. instrument reading. This is equivalent to reducing the scale of the diagram by a factor.

2. The vector is laid in the same direction U R, is the voltage drop across the active resistance, which in-phase current The scale for voltage must, of course, be chosen.

3.At an angle  construct a vector to the current vector U C, since the voltage across the capacitor lags current in phase by this amount.

4.At an angle  a vector is drawn to the current axis U L, since the voltage across the inductance leads the current by phase.

Fig.5

5. Find the vector sum of all voltages, a vector is obtained U VX. It can be seen that the current in the circuit I out of phase with the voltage applied to it U VX(Fig. 5, b).  – phase difference current and voltage (in other words, phase shift between current and voltage).

6. Measuring the length of the sum vector, taking into account the voltage scale, gives the input voltage in volts, and the phase angle is measured on the diagram with a protractor or calculated trigonometrically. This is graphic circuit calculation method.

So, if the voltage at the input of the circuit changes according to the law, then current flows in the circuit, and the phase difference  can be either positive or negative.

Analytically quantities I m And  are defined as follows.

From the vector diagram it follows that

(12)

(13)

Instead of amplitude values ​​in formula (13) can be written effective(or effective) values ​​of current and voltage, which are several times less than the amplitude

Equation (14) expresses Ohm's law for alternating current circuits. The role of resistance here is played by the expression in the denominator,

(15)

which is called impedance.

Thus, the alternating current in a section of a circuit is directly proportional to the alternating voltage in that section and inversely proportional to its impedance. This is how Ohm's law can be formulated.

The difference between the current phase and the voltage phase (phase shift) depends on the active and reactance resistance. From Fig. 5, b follows that

Purpose work is to determine the phase shift between current and voltage using phasor diagrams in the most common AC circuits.

Exercise 1

Capacitive Load Circuit (RC Circuit)

To construct a vector diagram, you need to know the voltage drop across all elements of the circuit in question. This is what the upcoming actions boil down to.

AND

measurements. 1.Assemble the electrical circuit ( R.C.-circuit) according to the diagram (Fig. 6), where LP– laboratory panel. Ask your teacher or lab assistant to check it. There is no ammeter in the circuit. Therefore, it is proposed to determine the current strength in the circuit from Ohm’s law by the voltage drop across a resistor with a known resistance R p .

2.Turn on the volt

Fig.6 meter. Press the “ key U~” – AC voltage measurements and the “ key WUA” – automatic selection of the measurement limit.

3. Set the frequency of the G3-118 generator using decade switches and a decimal multiplier 1.10 kHz.

Severe warning! You cannot set zeros on all decade frequency switches of the generator! According to the instructions, the device generates oscillations from 10 Hz to 200 kHz. He should not be forced to do the impossible. Violation of the instructions is accompanied by burnout of the transistors of the generator output stage.

Connect the generator to the network, set the input voltage U VX=3...4 V (it is also the voltage at the generator output.

4.Alternately connecting a voltmeter in parallel with the capacitor WITH and resistor R p, write down his testimony U C And U R accuracy of three significant figures in the corresponding columns of Table 1.

Table 1

5. Carry out similar measurements at generator frequencies of 2.10; 3.10; 4.10; 5.10 kHz, each time checking and maintaining the same input voltage.

6.Turn off the generator. Using the same universal voltmeter, measure the resistance of the resistor R P and write it down in Table 1.

Processing of measurement results. 1.For all frequencies, construct vector voltage diagrams on graph paper. This must be done as stated above (see p. 92). Draw the current axis, plot the current vector on it I (to scale). On the same axis, plot the vector U R(at its own scale). From the end of the vector U R under angle –  /2 build a vector U C(The voltage drop across the active resistance of the capacitor in in this case can be neglected. The reasons for this are briefly written in work No. 325).

2. Construct the sum vector of the two above. Check that the vector sum of the voltage drops across the capacitor and resistor, obtained by you, is equal to the input voltage.

3.On the resulting diagrams, measure the angle between the input voltage vector and the current vector with a protractor I and write it in the column  gr Table 1. This is the required phase difference found graphically.

The phase shift of current and voltage can be found analytically from formula (16), [see Introduction]. Let's denote it  en .

.

4.Compare the angle values ​​obtained graphically and analytically. Their coincidence or similar values ​​confirm the correspondence of the theoretical provisions contained in the Introduction to the experimental results. If the angles differ by more than 5% from each other, then there is most likely an error in the measurements or calculations.

5.Calculate the current in the circuit I and capacitor resistance Z 1 = X C at all frequencies.

6.From formula (11) find the capacity WITH at all frequencies  .

Calculate the average capacitance value over all measurements, as well as the half-width of the confidence interval  WITH.

Exercise 2

Circuit with inductive load (RL circuit)

A coil containing several thousand turns of copper wire and not containing an iron core is used as an inductive load. If there is a ferromagnetic core, the inductance of the coil depends on the current flowing through it. It is desirable for us to have it constant despite the change in current during the experiment.

Measurements. 1.Without collecting the chain , plug in the voltmeter, press the “ R” and the “ key AVP”, measure the DC resistance of the resistor R P and coils R L, write them down in Table 2.

2.Assemble the electrical circuit according to the diagram (Fig. 7).

3. Turn on the voltmeter in parallel with the generator output. Press the “ keys U~" and " WUA”.

Fig.7 4.Set frequency

generator 1.10 kHz, turn on the generator to the network. Set the output voltage using a voltmeter (it is also the input voltage for the load) U BX = 3…4 V.

5.Connecting a voltmeter alternately to the coil terminals L and resistor R R, measure U L And U R accurate to three significant figures.

6. Repeat similar measurements at frequencies 2.10; 3.10; 4.10; 5.10KHz, maintaining the same voltage U BX .

table 2

1. Construct vector diagrams on graph paper according to the data in Table 2. But unlike the ideal inductance discussed in the Introduction, a real coil has some active resistance R L, which you measured in step 1. Therefore its total resistance

(17)

And U L there is a voltage drop across it. In this case, the vector U L not perpendicular to the current vector I. To build U L, it must be represented as the sum of two terms

The first term is in phase with the current and therefore co-directed with the current vector, the second is perpendicular to the current vector and is ahead of it in phase.

To construct a vector diagram based on the results obtained, plot the vector on the selected scale along the current axis U R, add to it a vector of the same direction IR L, then from its end at an angle +/2, construct a vector I L. If you connect the beginning of the first vector with the end of the last, you get a total vector that should be equal to U BX .

Really

.

2.Measure the angle between the current vector and the input voltage vector with a protractor. Let's call it the phase shift angle defined graphic way -  gr .

3. Determine the current in the circuit from Ohm’s law

4. Find the total resistance of the coil using the formula

5.From formula (17) find L at every frequency. Calculate the average inductance value and the half-width of the confidence interval  L.

6.From formula (16) find the phase angle  en between current in a circuit and voltage. Let's call it the angle defined analytically.

Compare angle values  gr And  en at all frequencies. Is there a difference between them and what is it equal to?

Exercise 3

Combined load chain (RCL-chain)

Consider an AC electrical circuit containing all the elements: active resistance R, capacity C and inductance L.

Measurements. 1.Assemble the circuit according to the diagram (Fig. 8).

2.Measure at input voltage 3-4V and frequencies 1.10; 2.10; 3.10; 4.10; 5.10 kHz voltage drop across the resistor, coil and capacitor and write them down in the appropriate columns of Table 3.

Processing of measurement results. 1. Determine the current in the circuit from Ohm’s law if the voltage drop is known U R at a known resistance R P .

I=U R /R P .

2. Calculate all other quantities included in Table 3.

Table 3

3. Construct vector diagrams for this circuit. It is rational to perform this work in the following order.

A) Place the vector along the current axis U R .

b)From the end of the vector U R draw the vector in the same direction IR L .

V)From the end of the vector IR L at an angle +/2, construct a vector I L .

G)From the end of the vector I L draw the opposite vector to it U C .

d)Draw a vector from the beginning of the first to the end of the last vector. This is the total vector of all the vectors listed above. It is clear that in absolute value it must be equal to the input voltage. The direction of the input voltage vector in relation to the current vector in a given circuit gives the phase angle between them.

4.Measure the angle formed by the vector with a protractor U BX and current axis. This is the phase angle between current and voltage, which is determined graphic method. Label it as before  gr .

The active resistance of the capacitor and wires can be neglected due to their smallness compared to the capacitance and active resistance of the resistor and coil.

Of course, instead of calculating voltage drops on inductive I L, capacitive I/  C and active resistance I(R P + R L ) one could limit ourselves to defining the indicated resistance. But this has not been done. Thus, we would like to draw students’ attention to the coincidence of the voltmeter readings U C With I/  C, in contrast U L from I L and highlight the reason for this.

5.Calculate the phase shift angle from formula (16). Let's call it analytical -  en .

.

Compare it with the angle that was obtained graphically.

6.Plot a graph of the phase shift angle  gr from frequency  AC according to tables 1, 2 and 3.

7.Conclusion(for the entire work as a whole) write down in your workbook.

Control questions

1.What is active resistance in an alternating current circuit? Which circuit elements have active resistance? Will they have it in a DC circuit?

2.What is inductive reactance? What does it depend on? By what formula is it calculated in the work? What does inductance depend on?

3.Prove that the voltage across the inductance is ahead of the current in phase. Draw a vector diagram for this case.

4.Prove that the voltage fluctuations across the capacitor are out of phase with the current. Draw a vector diagram in this case.

5.What is capacitance? What does it depend on? How is this work? Are there any of your results on the basis of which it can be stated that the active resistance of the capacitor is small compared to the capacitive one?

6.What is the vector diagram method and how to use it in a specific situation?

1. Kalashnikov S.G. Electricity. M.: Nauka, 1977. §220.

2. Laboratory classes in physics / Ed. L.L. Goldina. M.: Nauka, 1983. P.312.

3. Savelyev I.V. General physics course. M.: Nauka, 1973. T.2. §92-95.

Chains variable current with series connection of active resistance, inductance and capacitance Laboratory work >> Physics

Work "Research" chains variable current with serial connection... measurements Accuracy class Limit measurements ... shift phases between the voltage on the coil and electric shock in it; φ – angle shift phases between source voltage and electric shock chains; ƒ – frequency current ...

Let's do the following experiment. Let's take the oscilloscope with two loops described in § 153 and connect it to the circuit so (Fig. 305, a) that loop 1 is connected to the circuit in series with the capacitor, and loop 2 is parallel to this capacitor. Obviously, the curve obtained from loop 1 depicts the shape of the current passing through the capacitor, and from loop 2 gives the shape of the voltage between the plates of the capacitor (points and ), because in this oscilloscope loop the current at each moment of time is proportional to the voltage. Experience shows that in this case the current and voltage curves are shifted in phase, with the current leading the voltage in phase by a quarter of a period (by ). If we were to replace the capacitor with a coil with high inductance (Fig. 305, b), it would turn out that the current is out of phase with the voltage by a quarter of a period (by ). Finally, in the same way it could be shown that in the case of active resistance, voltage and current are in phase (Fig. 305, c).

Rice. 305. Experience in detecting phase shifts between current and voltage: on the left - experimental diagram, on the right - results

In the general case, when a section of a circuit contains not only active, but also reactive (capacitive, inductive, or both) resistance, the voltage between the ends of this section is phase shifted relative to the current, and the phase shift lies in the range from to and is determined by the relationship between active and reactive resistance of a given section of the circuit.

What is the physical reason for the observed phase shift between current and voltage?

If the circuit does not include capacitors and coils, i.e., the capacitive and inductive resistance of the circuit can be neglected in comparison with the active one, then the current follows the voltage, passing simultaneously with it through maxima and zero values, as shown in Fig. 305, v.

If a circuit has a noticeable inductance, then when an alternating current passes through it, an emission occurs in the circuit. d.s. self-induction. This e. d.s. according to Lenz's rule, it is directed in such a way that it tends to prevent those changes in the magnetic field (and, consequently, changes in the current that creates this field) that cause e. d.s. induction. As the current increases, e. d.s. self-induction prevents this increase, and therefore the current reaches its maximum later than in the absence of self-induction. As the current decreases, e. d.s. self-induction tends to maintain the current and zero current values ​​will be reached at a later point than in the absence of self-induction. Thus, in the presence of inductance, the current is out of phase with the current in the absence of inductance, and therefore out of phase with its voltage.

If the active resistance of the circuit can be neglected in comparison with its inductive resistance, then the time lag of the current from the voltage is equal (the phase shift is equal to), i.e., the maximum coincides with, as shown in Fig. 305, b. Indeed, in this case the voltage across the active resistance is , for , and, therefore, all external voltage is balanced by e. d.s. induction, which is opposite to it in direction: . Thus, the maximum coincides with the maximum, i.e., it occurs at the moment when it changes the fastest, and this happens when . On the contrary, at the moment when it passes through the maximum value, the current change is smallest, i.e. at this moment.

If the active resistance of the circuit is not so small that it can be neglected, then part of the external voltage drops across the resistance, and the rest is balanced by e. d.s. self-induction: . In this case, the maximum is separated from the maximum in time by less than (the phase shift is less), as shown in Fig. 306. Calculation shows that in this case the phase lag can be calculated using the formula

. (162.1)

When we have and , as explained above.

Rice. 306. Phase shift between current and voltage in a circuit containing active and inductive resistance

If the circuit consists of a capacitor and the active resistance can be neglected, then the plates of the capacitor connected to a current source with a voltage are charged and a voltage arises between them. The voltage on the capacitor follows the voltage of the current source almost instantly, that is, it reaches a maximum simultaneously with and goes to zero when.

The relationship between current and voltage in this case is shown in Fig. 307, a. In Fig. 307,b conventionally depicts the process of recharging a capacitor associated with the appearance of alternating current in the circuit.

Rice. 307. a) Phase shift between voltage and current in a circuit with capacitance in the absence of active resistance. b) The process of recharging a capacitor in an alternating current circuit

When the capacitor is charged to the maximum (i.e., and therefore have a maximum value), the current and all the energy of the circuit is the electrical energy of the charged capacitor (point in Fig. 307, a). As the voltage decreases, the capacitor begins to discharge and current appears in the circuit; it is directed from plate 1 to plate 2, i.e. towards the voltage. Therefore, in Fig. 307, and it is depicted as negative (the points lie below the time axis). By the moment of time, the capacitor is completely discharged (and), and the current reaches its maximum value (point); electrical energy is zero, and all energy is reduced to the energy of the magnetic field created by the current. Further, the voltage changes sign, and the current begins to weaken, maintaining the same direction. When (and) reaches its maximum, all energy will again become electrical, and the current (point). Subsequently (and) begins to decrease, the capacitor is discharged, the current increases, now having a direction from plate 2 to plate 1, i.e. positive; the current reaches its maximum at the moment when (point), etc. From Fig. 307, but it is clear that the current reaches a maximum earlier than the voltage and passes through zero, i.e. the current is ahead of the voltage in phase, as explained above.

Rice. 308. Phase shift between current and voltage in a circuit containing active and capacitive resistance

But because the turns are shifted in space, then the EMF induced in them will not reach amplitude and zero values ​​at the same time.

At the initial moment of time, the EMF of the turn will be:

In these expressions the angles are called phase , or phase . The angles are called initial phase . The phase angle determines the value of the emf at any time, and the initial phase determines the value of the emf at the initial time.

The difference in the initial phases of two sinusoidal quantities of the same frequency and amplitude is called phase angle

Dividing the phase angle by the angular frequency, we obtain the time elapsed since the beginning of the period:

Graphic representation of sinusoidal quantities

U = (U 2 a + (U L - U c) 2)

Thus, due to the presence of a phase angle, the voltage U is always less than the algebraic sum U a + U L + U C. The difference U L - U C = U p is called reactive voltage component.

Let's consider how current and voltage change in a series alternating current circuit.

Impedance and phase angle. If we substitute the values ​​U a = IR into formula (71); U L = lL and U C =I/(C), then we will have: U = ((IR) 2 + 2), from which we obtain the formula for Ohm’s law for a series alternating current circuit:

I = U / ((R 2 + 2)) = U / Z (72)

Where Z = (R 2 + 2) = (R 2 + (X L - X c) 2)

The Z value is called circuit impedance, it is measured in ohms. The difference L - l/(C) is called circuit reactance and is denoted by the letter X. Therefore, the total resistance of the circuit

Z = (R 2 + X 2)

The relationship between active, reactive and impedance of an alternating current circuit can also be obtained using the Pythagorean theorem from the triangle of resistance (Fig. 193). The resistance triangle A'B'C' can be obtained from the voltage triangle ABC (see Fig. 192,b) if we divide all its sides by the current I.

The phase shift angle is determined by the relationship between the individual resistances included in a given circuit. From triangle A’B’C (see Fig. 193) we have:

sin? = X/Z; cos? = R/Z; tg? = X/R

For example, if the active resistance R is significantly greater than the reactance X, the angle is relatively small. If the circuit has a large inductive or large capacitive reactance, then the phase shift angle increases and approaches 90°. Wherein, if the inductive reactance is greater than the capacitive reactance, the voltage and leads the current i by an angle; if the capacitive reactance is greater than the inductive reactance, then the voltage lags behind the current i by an angle.

An ideal inductor, a real coil and a capacitor in an alternating current circuit.

A real coil, unlike an ideal one, has not only inductance, but also active resistance, therefore, when alternating current flows in it, it is accompanied not only by a change in energy in the magnetic field, but also by the conversion of electrical energy into another form. Specifically, in the coil wire, electrical energy is converted into heat in accordance with the Lenz-Joule law.

It was previously found that in an alternating current circuit the process of converting electrical energy into another form is characterized by active power of the circuit P , and the change in energy in the magnetic field is reactive power Q .

In a real coil, both processes take place, i.e., its active and reactive powers are different from zero. Therefore, one real coil in the equivalent circuit must be represented by active and reactive elements.

The initial phases of electromagnetic sinusoidal oscillations of the primary and secondary voltage, with a frequency of the same value, can differ significantly by a certain phase shift angle (angle φ). Variable quantities can change repeatedly over a certain period of time with a certain frequency. If electrical processes are unchanged and the phase shift is zero, this indicates synchronism of sources of alternating voltage values, for example, transformers. Phase shift is a determining factor of power factor in AC electrical networks.

The phase shift angle is found if necessary, then if one of the signals is a reference signal, and the second signal with a phase at the very beginning coincides with the phase shift angle.

The phase shift angle is measured using a device that has a normalized error.

The phase meter can measure the shift angle within the range from 0 o to 360 o, in some cases from -180 o C to +180 o C, and the range of measured signal frequencies can range from 20 Hz to 20 GHz. The measurement is guaranteed if the input signal voltage is between 1 mV and 100 V, but if the input signal voltage exceeds these limits, the measurement accuracy is not guaranteed.

Methods for measuring phase angle

There are several ways to measure the phase angle, these are:

1. Using a dual-beam or dual-channel oscilloscope.
2. The compensation method is based on comparing the measured phase shift with the phase shift provided by a reference phase shifter.
3. The sum-difference method consists of using harmonic or shaped square-wave signals.
4. Conversion of phase shift in the time domain.

How to measure phase angle with an oscilloscope

The oscillographic method can be considered the simplest with an error of around 5 o. The shift is determined using oscillograms. There are four oscillographic methods:

1. Application of linear sweep.
2. Ellipse method.
3. Circular scanning method.
4. Using brightness marks.

Determination of the phase shift angle depends on the nature of the load. When determining the phase shift in the primary and secondary circuits of a transformer, the angles can be considered equal and practically do not differ from each other.

The phase angle of the voltages, measured using a reference frequency source and using a measuring element, makes it possible to ensure the accuracy of all subsequent measurements. Phase voltages and phase shift angle depend on the load, so a symmetrical load determines the equality of phase voltage, load currents and phase shift angle, and the load in terms of power consumption in all phases of the electrical installation will also be equal.

The phase angle between current and voltage in asymmetrical three-phase circuits is not equal to each other. In order to calculate the phase shift angle (angle φ), series-connected resistances (resistors), inductances and capacitors (capacitors) are included in the circuit.

From the results of the experiment, it can be determined that the phase shift between voltage and current serves to determine the load and cannot depend on the variable current and voltage in the electrical network.

As a conclusion, we can say that:

1. The constituent elements of complex resistance, such as resistor and capacitance, as well as conductivity, will not be reciprocal quantities.
2. The absence of one of the elements makes the resistive and reactive values, which are part of the complex resistance and conductivity, and makes them reciprocal quantities.
3. Reactive quantities in complex resistance and conductivity are used with the opposite sign.

The phase angle between voltage and current is always expressed as the main reasoned factor in the complex resistance φ.

When solving a number of practical problems, it is often necessary to obtain a certain phase shift, not only in magnitude, but also in a given direction. Such examples are described in the article “Transformer connection groups”.

Shift by 30 and 60°.

By connecting the windings in star and triangle, shifts are obtained that are multiples of 30°, depending on what (ends, beginnings) are connected to what and in what direction (from phase A to phase B or vice versa), a shift is obtained in one direction or another.

When connecting in a zigzag - star (see the article "Zigzag connection diagram"), the end of one section is connected to the end of another section and the angle changes by 30°. If you connect not the end to the end, but the end to the beginning, then the vectors will rotate by 60° (see Figure 4, in the article “Some errors in star, triangle, zigzag connections”) In other words, by reconnecting the windings, you can easily get a shift of 30 and 60°.

The following must be kept in mind. Firstly, when reconnecting the windings, not only the angle can change (which is required), but also the voltage (see Figure 4, V, in the article “Some errors when connecting in a star, triangle, zigzag”). Secondly, connecting the windings in opposite directions - an extreme case - or changing the angle between them can reduce the inductive reactance, and this will lead to an increase in current. An increase in current is dangerous for the winding and, in addition, can lead to saturation of the magnetic circuit. The matter is much more serious than it might seem at first glance, and therefore, without making sure that the current has not exceeded the specified value, reconnections cannot be performed.

Shift by 90°.

Let's look at a common example of obtaining a 90° shift. In Figure 1, A the activation of the reactive energy meter is shown. Note: the current winding (thick line) is in phase A, and the voltage winding is connected to the phases B And C. Referring to the vector diagram in Figure 1, b, it is easy to see that in this simplest way a shift of 90° is obtained, which is what is required in this case.

Figure 1. Obtaining a 90° phase shift.

Shift to any angle from 0 to 90°

easy to get with phase regulator– rotary three-phase transformer. It is an asynchronous machine with a locked rotor. By turning the rotor relative to the stator, the phase of the electromotive force (emf) of the rotor is smoothly changed without changing its value (magnitude).

It is necessary to distinguish a phase regulator from a potential regulator, also called an induction regulator. In a phase regulator, only the phase changes; In the potential regulator both voltage and phase change. In addition, the primary and secondary windings of the phase regulator are mutually isolated, while those of the potential regulator are connected.

Let us note in conclusion that any phase shifts can also be obtained by connecting active and inductive resistances and capacitances. Such converters are widely used and are called static.