Formula for the effective value of the emf of the primary winding of a transformer. What determines the EMF of the transformer windings and what is their purpose? Magnetic core. Magnetic materials
Let's take a coil with a ferromagnetic core and take out the ohmic resistance of the winding as a separate element, as shown in Figure 1.
Figure 1. Inductor with ferromagnetic core
When an alternating voltage e c is applied to the coil, according to the law of electromagnetic induction, a self-induction emf e L appears.
(1) where ψ
— flux linkage, W- number of turns in the winding, F- main magnetic flux. We neglect the scattering flux. The voltage applied to the coil and the induced emf are balanced. According to Kirchhoff’s second law for the input circuit, we can write:
e c + e L = i × R exchange, (2)Where R obm - active resistance of the winding.
Because the e L >> i × R exchange, then we neglect the voltage drop across the ohmic resistance, then e c ≈ −e L. If the mains voltage is harmonic, e c = E m cosω t, That:
(3)
Let us find the magnetic flux from this formula. To do this, we transfer the number of turns in the winding to the left side, and the magnetic flux Ф to the right:
(4)
Now let's take the indefinite integral of the right and left sides:
(5)Since we consider the magnetic circuit to be linear, only harmonic current flows in the circuit and there is no permanent magnet or constant component of the magnetic flux, then the integration constant c = 0. Then the fraction in front of the sine is the amplitude of the magnetic flux
(6)from where we express the amplitude of the input EMF
E m = F m × W × ω (7)Its effective value is
(8) (9)Expression (9) is called basic formula of transformer EMF, which is valid only for harmonic voltage. With a non-harmonic voltage, it is modified and the so-called form factor is introduced, equal to the ratio of the effective value to the average:
(10)
Let's find the shape factor for a harmonic signal, and find the average value in the interval from 0 to π/2
(11)
Then the form factor is 
and the basic formula of the transformer EMF takes its final form:
If the signal is a sequence of rectangular pulses of the same duration (meander), then the amplitude, effective and average values for half a period are equal to each other and its k f = 1. You can find the shape factor for other signals. The basic formula of transformer EMF will be valid.
Let's construct a vector diagram of a coil with a ferromagnetic core. With a sinusoidal voltage at the coil terminals, its magnetic flux is also sinusoidal and lags in phase from the voltage by an angle π/2 as shown in Figure 2.
Let's take a coil with a ferromagnetic core and take out the ohmic resistance of the winding as a separate element, as shown in Fig. 2.8.
Figure 2.8 – To derive the formula for transformer EMF
When you turn on the alternating voltage e c in the coil, according to the law of electromagnetic induction, a self-induction emf e L appears.
(2.8)
where ψ is flux linkage,
W – number of turns in the winding,
Ф – main magnetic flux.
We neglect the scattering flux. The voltage applied to the coil and the induced emf are balanced. According to Kirchhoff’s second law for the input circuit, we can write:
e c + e L = i * R exchange, (2.9)
where R rev is the active resistance of the winding.
Since e L >> i * R exchange, we neglect the voltage drop across the ohmic resistance, then e c ≈ – . If the network voltage is harmonic e c = E m cos ωt, then E m cos ωt = , whence 
. Let's find the magnetic flux. To do this, we take the indefinite integral of the right and left sides. We get
, (2.10)
but since we consider the magnetic circuit to be linear, only a harmonic current flows in the circuit and there is no permanent magnet or constant component, then the integration constant c = 0. Then the fraction in front of the harmonic factor is the amplitude of the magnetic flux, from which we express E m = Ф m * W * ω. Its effective value is
Or we get
where s is the cross-section of the magnetic circuit (core, steel).
Expression (2.11) is called the basic formula of transformer EMF, which is valid only for harmonic voltage. Usually it is modified and the so-called form factor is introduced, equal to the ratio of the effective value to the average:
. (2.12)
Let's find it for a harmonic signal, but find the average value on the interval
Then the form factor is 
and the basic formula of the transformer EMF takes its final form:
(2.13)
If the signal is a meander, then the amplitude, effective and average values for half the period are equal to each other and its . You can find the shape factor for other signals. The basic formula of transformer EMF will be valid.
Let's construct a vector diagram of a coil with a ferromagnetic core. With a sinusoidal voltage at the coil terminals, its magnetic flux is also sinusoidal and lags in phase from the voltage by an angle π/2 as shown in Fig. 2.9a.
Figure 2.9 – Vector diagram of a coil with ferromagnetic
core a) without losses; b) with losses
In a lossless coil, the magnetizing current is reactive current(I p) is in phase with the magnetic flux Ф m. If there are losses in the core (), then the angle is the angle of losses due to magnetization reversal of the core. The active component of the current Ia characterizes the losses in the magnetic circuit.
Question 7. Ems of transformer windings.
The operating principle of the transformer is based on the phenomenon of electromagnetic induction (mutual induction). Mutual induction consists of inducing an emf in an inductive coil when the current to the other coil changes.
Under the influence of alternating current in the primary winding, an alternating magnetic flux is created in the magnetic circuit
which penetrates the primary and secondary windings and induces an EMF in them
where are the amplitude values of the EMF.
The effective value of the EMF in the windings is equal to
;
.
The ratio of the EMF of the windings is called the transformation ratio
If , then the secondary EMF is less than the primary one and the transformer is called a step-down transformer, while the transformer is called a step-up transformer.
Question 8. Vector diagram of the open circuit of an ideal transformer.
Since we are considering an ideal transformer, i.e. without dissipation and power loss, then the current is x.x. is purely magnetizing – , i.e. it creates a magnetizing force, which creates a flux, where is the magnetic resistance of the core, consisting of the resistance of the steel and the resistance at the joints of the core. Both the amplitude and shape of the current curve depend on the degree of saturation of the magnetic system. If the flow changes sinusoidally, then with unsaturated steel the no-load current curve is almost also sinusoidal. But when the steel is saturated, the current curve becomes more and more different from a sinusoid (Fig. 2.7.) The current curve x.x. can be decomposed into harmonics. Since the curve is symmetrical about the x-axis, the series contains only harmonics of odd order. First harmonic current i ( 01) is in phase with the main flow. Of the higher harmonics, the third harmonic of the current is most pronounced i ( 03) .
Fig 2.7 Current curve X.X
Effective value of no-load current:
.
(2.22)
Here I 1 m , I 3 m , I 5 m– amplitudes of the first, third and fifth harmonics of the no-load current.
Since the no-load current lags behind the voltage by 90 , the active power consumed by an ideal transformer from the network is also zero, i.e. An ideal transformer consumes purely reactive power and magnetizing current from the network.
The vector diagram of an ideal transformer is shown in Fig. 2.8.
Rice. 2.8. Vector diagram of an ideal transformer
Question 9 Vector diagram of the no-load circuit of a real transformer.
In a real transformer, there are scattering and losses in steel and copper. These losses are covered by power R 0 entering the transformer from the network.
Where I 0a – effective value of the active component of the no-load current.
Consequently, the no-load current of a real transformer has two components: magnetizing - , which creates the main flux F and in phase with it, and active:
The vector diagram of a real transformer is shown in Fig. 2.9.
Usually, therefore, this component has little effect on the value of the no-load current, but has a greater effect on the shape of the current curve and its phase. The no-load current curve is clearly non-sinusoidal, and is shifted in time relative to the flux curve by an angle called the magnetic retardation angle
By replacing the actual no-load current curve with an equivalent sinusoid, the voltage equation can be written in complex form, where all quantities vary sinusoidally:
Considering that the leakage emf,
Rice. 2.9. Vector diagram of a real transformer
Rice. 2.11. Vector diagram of transformer voltages, no-load mode
LR 5. Study of operating modes of a single-phase transformer
Name the main design elements of a single-phase transformer.
A single-phase transformer consists of a magnetic core (core) and two windings laid on it. The winding connected to the network is called primary, and the winding to which the electricity receiver is connected is called secondary. The magnetic core is made of ferromagnetic material and serves to enhance the magnetic field and the magnetic flux is closed along it.
Features of the design of the transformer magnetic circuit.
The magnetic core of the transformer is in a magnetic field alternating current, and, consequently, during operation, its continuous magnetization reversal occurs and eddy currents are induced in it, which consumes energy that goes to heating the magnetic circuit. To reduce energy losses due to magnetization reversal, the magnetic circuit is made of a soft magnetic ferromagnet, which has a low residual induction and is easily remagnetized, and to reduce eddy currents, and, consequently, the degree of heating of the magnetic circuit, the magnetic circuit is assembled from individual electrical steel plates insulated relative to each other.
3. How are the EMF of the transformer windings determined, what do they depend on?
The EMF of the transformer windings is determined by the formulas: E 1 =4.44*Fm*f*N 1 And E 2 =4.44*Fm*f*N 2
Where fm– maximum value of magnetic flux,
f- AC frequency,
N 1 And N 2– the number of turns of the primary and secondary windings, respectively.
Thus, the EMF of the transformer windings depends on the magnetic flux, the frequency of the alternating current and the number of turns of the windings, and the ratio between the EMF depends on the ratio of the number of turns of the windings.
4. Name the types of energy losses in a transformer, what do they depend on?
When a transformer operates, two types of energy losses occur in it:
1. Magnetic losses are energy losses that occur in the magnetic circuit. These losses are proportional to the network voltage. Energy in this case is spent on magnetization reversal of the magnetic core and on the creation of eddy currents and is converted into thermal energy released in the magnetic core.
2. Electrical losses are energy losses occurring in the windings of a transformer. These losses are caused by currents flowing in the windings and are determined by: Re = I 2 1 R 1 + I 2 2 R 2.
That. electrical losses are proportional to the squares of the currents flowing in the windings of the transformer. In this case, energy is spent on heating the windings.
5. How are magnetic losses in a transformer determined, what do they depend on?
To determine magnetic losses in a transformer, a XX experiment is carried out, in which the current in the secondary winding is zero, and in the primary winding the current does not exceed 10% of I nom. Because When conducting this experiment, the power receiver is turned off, then all the power measured by a wattmeter connected to the circuit of the primary winding of the transformer is the power of electrical and magnetic losses. Magnetic losses are proportional to the voltage applied to the primary winding. Because when conducting the experiment, XX is supplied to the primary winding U nom , then the magnetic losses will be the same as in the nominal mode. Electrical losses depend on the currents in the windings, and since the current in the secondary winding is zero, and in the primary winding the current does not exceed 10% of the rated current, and the electrical losses are insignificant. Thus, neglecting minor electrical losses, we believe that all the power measured during the XX experiment is the power of magnetic losses.
6. How are electrical losses in a transformer determined, what do they depend on?
To determine the electrical losses in the transformer, a short-circuit experiment is carried out. To do this, it is necessary to reduce the voltage on the secondary winding to zero, close the secondary terminals to each other and increase the voltage until the rated currents are established in the windings. The voltage at which rated currents are established in the windings is called short-circuit voltage. As a rule, the short-circuit voltage is insignificant and does not exceed 10% of the nominal value.
Electrical losses in the transformer during the short circuit experiment will be determined :Re= I 2 1nom R 1 + I 2 2nom R 2.
Because When conducting a short-circuit experiment, rated currents are set in the transformer windings, then the electrical losses in them will be the same as in the nominal mode. Magnetic losses are proportional to the voltage on the primary winding, and since In the short-circuit experiment, a small voltage is supplied to the primary winding, then the magnetic losses are insignificant. Thus, neglecting insignificant magnetic losses, we can assume that all the power measured in the short-circuit experiment is the power of electrical losses.
How does a transformer work?
(b, c) W x. W 2 connects to the load.
U 1 i 1 F. This flow induces an emf e 1 And e 2 in the windings of the transformer:
EMF e 1 U 1, emf e 2 creates tension U 2
· Step-down transformer – a transformer that reduces voltage (K>1).
What is the transformation ratio?
Transformation ratio is the ratio of the effective voltages at the ends of the primary and secondary windings when the secondary windings are open circuit (no-load of the transformer). K=W 1 /W 2 =e 1 /e 2.
For a transformer operating in no-load mode, we can assume with sufficient accuracy for practice that .
What nominal parameters of the transformer do you know and what do they determine?
Rated power is the rated power of each of the transformer windings. Rated current, voltage of windings. The external characteristic is the dependence of the voltage at the terminals of the transformer on the current flowing through the load connected to these terminals, i.e. dependence U2=f(I2) at U1=const. The load is determined by the load factor Kn=I2/I2nom ≈ I1/I1nom, efficiency - η = P2/P1
How to determine the rated currents of the transformer windings if the rated power of the transformer is known?
The rated power of a two-winding transformer is the rated power of each of the transformer windings.
Rated power equation: S H =U1 * I1 ≈ U2 * I2
I1 = S H /U1 ; I2 = S H /U2
What is called the external characteristic of a transformer and how to obtain it?
The external characteristic is the dependence of the voltage at the terminals of the transformer on the current flowing through the load connected to these terminals, i.e. dependence U 2 =f(I 2) at U 1 =const. When the load (current I 2) changes, the secondary voltage of the transformer changes. This is explained by a change in the voltage drop across the resistance of the secondary winding I 2 " z 2 and a change in EMF E 2 "=E 1 due to a change in the voltage drop across the resistance of the primary winding.
The EMF and voltage equilibrium equations take the form:
Ù 1 = –È 1 + Ì 1 " z 1, Ù 2 "=È 2 – Ì 2 " z 2 " (1)
The load value in transformers is determined by the load factor:
K n =I 2 /I 2 nom ≈ I 1 /I 1 nom;
The nature of the load is the phase shift angle of the secondary voltage and current. In practice, the formula is often used
U 2 = U 20 (1 - Δu/100),
Δu=K n (u ka cosφ 2 + u cr sinφ 2)
u ka = 100% I 1nom (R 1 - R 2 ")/U 1nom
u ka = 100% I 1nom (X 1 - X 2 ")/U 1nom
How to find the percentage change in transformer secondary voltage for a given load?
The percentage change in secondary voltage ∆U 2% at variable load is determined as follows: , where are the secondary voltages at no-load and at a given load, respectively.
What transformer equivalent circuits do you know and how are their parameters determined?
T-shaped transformer equivalent circuit:
How does a transformer work?
A transformer is a static electromagnetic device designed to convert, through a magnetic flux, alternating current electrical energy of one voltage into alternating current electrical energy of another voltage at a constant frequency.
Electromagnetic circuit of the transformer (a) and symbolic graphic symbols of the transformer (b, c) are shown in Fig. 1. There are two windings located on a closed magnetic circuit made of sheets of electrical steel. Primary winding with number of turns W x connects to a source of electrical energy with voltage U . Secondary winding with number of turns W 2 connects to the load.
What determines the EMF of the transformer windings and what is their purpose?
Under the influence of supplied alternating voltage U 1 current appears in the primary winding i 1 and a changing magnetic flux appears F. This flow induces an emf e 1 And e 2 in the windings of the transformer:
EMF e 1 balances the bulk of the source voltage U 1, emf e 2 creates tension U 2 at the output terminals of the transformer.
3. In what cases is a transformer called a step-up transformer and in what cases is it called a step-down transformer?
· Step-down transformer – a transformer that reduces voltage (K>1).
Step-up transformer - a transformer that increases the voltage (K<1).
