We know that the electric power coming into our homes is produced very far away from our homes. They are transmitted with the help of cables. But during transmission, energy is lost. The voltage is raised at the power plants before transmission to reduce the loss of energy. The voltage is reduced and distributed to our houses at the power stations. To perform all these functions, we need an electrical component called Transformer. This article will learn about types of transformers, their working principle, and some applications.

What is a Transformer?

A transformer is an electrical device that transfers electrical energy from one circuit to another circuit through electromagnetic induction and mutual induction. It is most commonly used to increase (‘step up’) or decrease (‘step down’) voltage levels between circuits without a change in the frequency of AC between circuits.

Types of Transformers

Depending on the operating voltage, there are mainly two types of transformer. These are as follows:
1. Step-down Transformer: A step-down transformer converts the primary voltage level to a lower voltage across the secondary output. For step-down transformers, the number of windings is higher across the primary side than the secondary side.
Therefore, the overall winding ratio of secondary and primary will always remain less than \(1\).

2. Step-up Transformer: Step-up transformer increases the low primary voltage to a high secondary voltage. In this type of transformer, the number of turns in the primary winding is lower than the secondary winding, so the ratio of the primary winding to the Secondary winding will be more than \(1\).

Working Principle of a Transformer

Transformer works on the principle of Faraday’s Law of Electromagnetic Induction. According to it “the magnitude of induced voltage is directly proportional to the rate of change of magnetic flux through the coil”. A varying current in one coil (primary coil) of the transformer produces a varying magnetic field, which in turn induces a varying electromotive force (e.m.f) or “voltage” in a second coil (secondary coil). Power is transferred between the two coils through the magnetic field; there is no metallic connection between the two circuits.

From the above figure, we can see that a transformer has a magnetic core over which two sets of windings, termed primary and secondary, are suitably placed. When one of the windings is connected to an \(AC\) supply, an emf is induced on the other winding, which is proportional to the number of turns in the primary and secondary coil.

Parts of Transformer

The following are the various transformer parts:

1. Core
2. Winding
3. Insulation
4. Tank
5. Transformer oil
6. Oil Conservator
7. Breather
8. Radiators and fans

1. Core

These are made up of soft iron. It provides a low reluctance path for electromagnetic flux and supports primary and secondary windings. It is made by stacking thin sheets of high-grade soft iron. A thin insulating material separates each sheet to reduce loss due to eddy current.

2. Windings

The transformer carries two sets of winding per phase. These are Primary winding (The set of turns of windings to which supply current is fed) and secondary winding (The set of turns of winding from which output is taken). Out of the primary and secondary winding, the one with higher voltage is known as High Voltage (HV) winding, and the other is known as the Low Voltage (LV) winding.

3. Insulation

Proper insulation is the most important part of transformers. Insulation failures can cause severe damage to transformers. Proper insulation is required for the stability and durability of the transformers. Synthetic materials, paper, cotton, insulating oil, etc., are used as insulation in transformers.

4. Tank

The main tank of a transformer serves two purposes:

1. Protects the core and the windings from the external environment, such as rain and dust.
2. It serves as a container for oil and also support all other transformer accessories.

5. Transformer Oil

Most of the large transformers are oil-immersed. The transformer oil provides added insulation between the conducting parts, better heat dissipation from the coils, and fault detection features. Generally, hydrocarbon mineral oil is used as transformer oil.

6. Oil Conservators

The oil conservator is located above the tank and bushings of the transformer. A rubber bladder is present in some oil conservators of transformers. When a transformer is loaded, then the ambient temperature rises; this causes an increase in the volume of oil inside the transformer. A conservator tank of the transformer provides adequate space for this expanded transformer oil. It also acts as a reservoir for insulating oil.

7. Breather

It is present in all oil-immersed transformers that have a conservator tank. It helps in keeping the oil-from moisture.

8. Radiators and Fans

Most of the power lost in the transformer is dissipated in the form of heat. Radiator and fans help in dissipating heat from the transformer and protection from failure. Dry transformers are mostly natural air-cooled.

Ideal Transformer

In an ideal transformer, there are no losses. There is no magnetic flux leakage, no ohmic resistance in its windings, and no iron loss in the core.

EMF Equation of Ideal Transformer

Let \(N_p\) is the number of turns in the primary winding and \(N_s\) is the number of turns in the secondary winding.
When the \(AC\) voltage is applied to the primary coil of the transformer, the resulting current produces an alternating magnetic flux that links the secondary coil and induces an emf in it. The value of this emf in the secondary coil depends on the number of turns in it. Let us consider an ideal (no losses) transformer in which the primary coil has negligible resistance (No voltage drop across coil) and all the flux in the core links both primary and secondary windings. Let \(\phi\) be the flux linkage in each turn in the core at time \(t\) due to current in the primary coil when the voltage \(V_p\) is applied to it.
Then the induced emf or voltage \((ε_s)\), in the secondary with \(N_s\) turns are
\({\varepsilon _{\text{s}}} = – {N_{\text{s}}}\frac{{{\text{d}}\phi }}{{{\text{d}}t}}\) ……..(1)
And also, the alternating flux \(\phi\) induces an emf, called back emf, in the primary. This is
\({\varepsilon _{{p}}} = – {N_{{p}}}\frac{{{{d}}\phi }}{{{{d}}t}}\) ……..(2)
And for an ideal transformer, \(ε_p = V_p\),
If the secondary is an open circuit or the current taken from it is small, then by approximation, \(ε_s = V_s\).
where \(V_s\) is the voltage across the secondary coil. Therefore, Eqs. (1) and (2) can be written as
\({V_{{s}}} = – {N_{{s}}}\frac{{{{d}}\phi }}{{{{d}}t}}\) ……..(3)
\({V_{{p}}} = – {N_{{p}}}\frac{{{{d}}\phi }}{{{{d}}t}}\) ……..(4)
From Eqs. (3) and (4), we have
\(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}\) ……..(5)
The above relation is obtained using below three assumptions:

(i) The electrical resistance of the primary coil and secondary coil are negligible.
(ii) There are very few fluxes escapes from the core or the flux linkage to both the primary and the secondary coil is the same.
(iii) The secondary current is very small.

If the transformer is ideal or \(100\%\) efficient (no energy losses), then the power input will be equal to the power output.
\({i_p}{V_p} = {i_s}{V_s}\) ……..(6)
Combining Eqs. (5) and (6), we have
\(\frac{{{i_p}}}{{{i_s}}} = \frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}} = K\) ……..(7)
In the above equation, \(K\) is known as the turn ratio.
That is, if the secondary coil has a larger number of turns than the primary coil \((N_s > N_p)\), the voltage is stepped up \((V_s > V_p)\). This type of arrangement is called a step-up transformer.
If the secondary coil has less number of turns than the primary coil \((N_s < N_p)\), then we will have a step-down transformer.

Efficiency of Transformer

The efficiency of a Transformer can be given as:-
\({\text{Efficiency}} = \frac{{{\text{Output}}\,{\text{Power}}}}{{{\text{Input}}\,{\text{Power}}}} = \frac{{{\text{Output}}\,{\text{Power}}}}{{{\text{Output}}\,{\text{Power + Losses}}}}\)

Losses in Transformer

In the above equations, we considered ideal transformers (without any energy losses). But in actual transformers, some energy losses do occur due to the following reasons:
(i) Flux Leakage: Some flux leak from the core, so not all of the flux due to primary passes through the secondary coil. This happens due to the poor design of the core or the air gaps in the core. It can be reduced by winding the primary coil and secondary coil one over the other. It can also be reduced by the good design of the core.
(ii) Resistance of the Windings: The wire used for the windings has some electrical resistance, so energy is lost due to heat produced in the windings. In high current, low voltage windings, these are minimized by using thick wire of high conductive material.
(iii) Eddy Currents: The alternating magnetic flux induces eddy currents in the iron core and causes energy losses in the form of heating. The effect is reduced by putting a laminated core.
(iv) Hysteresis Loss: The magnetization of the core is reversed by the alternating magnetic field in each \(AC\) cycle. The loss of energy in the core appears due to hysteresis loss appears as heat and is kept to a minimum by using a magnetic material that has a low hysteresis loss. A soft iron core is used to reduce the losses.

Application of Transformers

The major application of transformers are:-

1. Increasing or decreasing the voltage level in the circuit of an \(AC\) circuit for the proper function of the different electrical components in the circuit.
2. It prevents the flow of \(DC\) from one circuit to another circuit.
3. It isolates two electric circuits.
4. Stepping up the voltage level at the site of the electric power plant before the transmission and distribution can take place.

Common places where an electrical transformer is used are pumping stations, railways, industries, commercial establishments, windmills, and power generation units.

Sample Problems

Q 1. The number of turns in the secondary coil of a \(22\,\rm{KVA}\), \(2200\,\rm{V}/220\,\rm{V}\) single-phase transformer is \(50\), then find the:-
a. Number of primary turns
b. Primary full load current
c. Secondary full load current
Neglect all kinds of losses in the transformer.

Ans: The value of the turns ratio is
\(\frac{{{V_p}}}{{{V_s}}} = \frac{{2200}}{{220}} = 10 = K\)
a. Number of primary turns
The value of the primary turns can be determined as:-
\(\frac{{{N_p}}}{{{N_s}}} = K\)
\(\frac{{{N_p}}}{{50}} = 10\)
\(N_p = 500\)
b. Primary full load current
The value of the primary full load current is,
\({I_p} = \frac{{{\text{Power}}}}{{{\text{Voltage}}}} = \frac{{22 \times {{10}^3}}}{{2200}} = 10\,{\text{A}}\)
c. Secondary full load current
The value of the Secondary full load current is
\({I_s} = \frac{{{\text{Power}}}}{{{\text{Voltage}}}} = \frac{{22 \times {{10}^3}}}{{220}} = 100\,{\text{A}}\)

Summary

A Transformer increase or decrease the voltage level (or current level) on its input winding to another value on its output winding using a magnetic field. It consists of two electrically isolated coils called primary coil and secondary coil. It operates on Faraday’s principle of “mutual induction”, in which an EMF is induced in the transformer secondary coil by the magnetic flux generated by the applied voltages in the primary coil winding.
Both the primary coil and secondary coil windings are wrapped around a common soft iron core made of the individual laminated sheet to reduce eddy current and power losses. The primary winding of the transformer is connected to the \(AC\) power source, while the secondary winding supplies electrical power to the load.
For an ideal transformer (no power loss), \(\frac{{{i_p}}}{{{i_s}}} = \frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}\). Here \(p\) represents primary coil, and \(s\) represents secondary coil.

FAQs

Q.1. Which transformer is used in homes?
Ans: In distribution networks, the step-down transformer is commonly used to convert the high grid voltage to low voltage that can be used for home appliances.

Q.2. Do transformers use AC or DC?
Ans: The transformer uses AC because it works on the principle of mutual induction. Transformers do not pass direct current DC.

Q.3. Are Transformers 100% efficient?
Ans: An ideal transformer would have no losses and would therefore be 100% efficient. But in the actual transformer, some losses are there. Transformers are, in general, highly efficient, and large power transformers (around 100 MVA and larger) may attain an efficiency as high as 99.75%.

Q.4. Will the transformer work when the output and input of the transformer will be interchanged?
Ans: Theoretically, it will work and will have inversed coil ratio. But it may cause some practical design problems like failure of insulation in very high coil ratio transformer.

We hope this detailed article on Transformer helps you in your preparation. If you get stuck do let us know in the comments section below and we will get back to you at the earliest.