MODULE: ELECTRICAL ENGINEERING TECHNOLOGY IV/ELECTRICAL

MACHINES & CONTROL II UNIT 3: TRANSFORMER

INTRODUCTION
A transformer may be defined as a static electric device that transfers electrical energy from one circuit
to another circuit at the same frequency but with changed voltage (or current or both) through a
magnetic circuit.
CLASSIFICATION OF TRANSFORMER
Transformer is generally divided into two types:
1. Single phase Transformer
2. Three phase Transformer
PRINCIPLES OPERATION
The principle of operation of a transformer is explained with the aid of Fig.1. A transformer works on the
principle of electromagnetic induction between two or more coupled circuits.
When an alternating voltage V1 is applied to the primary winding of a transformer, a current (termed
exciting current, If) flows through it. The exciting current produces an alternating flux (f) in the core, which
links with both the winding (primary and secondary). According to Faraday’s laws of electromagnetic
induction, the flux will cause self induced emf E1 in the primary and mutually induced emf E2 in the
secondary winding. But according to Lenz’s law, primary induced emf will oppose the applied voltage.
Therefore, emf induced in the primary winding is equal and opposite to the applied voltage.
When a load is connected to the secondary side, current will start flowing in the secondary winding.
Voltage induced in the secondary winding is responsible to deliver power to the load connected to it. In
this way, power is transferred from one circuit (primary) to another (secondary) winding through a
magnetic circuit by electromagnetic induction. This is the working principle of the transformer. The
induced emf in the secondary E2 is also in phase opposition to the applied voltage V1 at primary. If the
secondary is open circuited, terminal voltage V2 at the secondary is equal in magnitude and in phase with
the induced emf at secondary.

Figure 1: Schematic diagram of Single-phase transformer.

CONSTRUCTION OF SINGLE-PHASE TRANSFORMER

A single-phase transformer consists of primary and secondary windings placed on a magnetic core. The
magnetic core is a stack of thin silicon steel laminations. The laminations reduce eddy-current loss and
silicon steel reduces hysteresis loss. There are two general types of transformers, core-type and shell-
type.
Power transformer is design so that it approaches the characteristics of an ideal transformer. To
achieve this, following design features are incorporated:
i) The core is made of silicon steel which has low hysteresis loss and high permeability. Further, core is
laminated in order to reduce eddy current loss. These features considerably reduce the iron losses
and the no load current.
 ii) Instead of placing primary on one limb and secondary on the other, it is a usual practice to wind
one-half of each winding on one limb. This ensures tight coupling between the two windings.
Consequently, leakage flux is considerably reduced.
iii) The winding resistances R1 and R2 are minimized to reduce I2R loss and resulting rise in
temperature and to ensure high efficiency.

i) Core-type transformer.
In a core-type transformer, half of the primary winding and half of the secondary winding are placed round
each limb as shown in Fig. (2). This reduces the leakage flux. It is a usual practice to place the low-voltage
winding below the high-voltage winding for mechanical considerations.

Figure 2

ii) Shell-type transformer.

This method of construction involves the use of a double magnetic circuit. Both the windings are placed
round the central limb (Fig. 3), the other two limbs acting simply as a low-reluctance flux path.

Figure 3
The choice of type (whether core or shell) will not greatly affect the efficiency of the transformer. The
core type is generally more suitable for high voltage and small output while the shell-type is generally
more suitable for low voltage and high output.

TYPES OF TRANSFORMER
1. Autotransformer
An autotransformer has a single winding on an iron core and a part of winding is common to both the
primary and secondary circuits. Fig. (4(i)) shows the connections of a step-down autotransformer
whereas Fig. (4(ii)) shows the connections of a step-up autotransformer. In either case, the winding ab
having N1 turns is the primary winding and winding be having N2 turns is the secondary winding. Note
that the primary and secondary windings are connected
electrically as well as magnetically. Therefore, power from the primary is transferred to the secondary
conductively as well as inductively (transformer action). The voltage transformation ratio K of an ideal
autotransformer is
Note that in an autotransformer, secondary and primary voltages are related in the same way as in a
2-winding transformer.

Figure 4
Applications of Autotransformers
i) Autotransformers are used to compensate for voltage drops in transmission and distribution lines.
When used for this purpose, they are known as booster transformers.
ii) Autotransformers are used for reducing the voltage supplied to a.c. motors during the starting
period. iii) Autotransformers are used for continuously variable supply.
2. Current transformer
A current transformer is a device that is used to measure high alternating current in a conductor. Fig. (7.55)
illustrates the principle of a current transformer. The conductor carrying large current passes through a
circular laminated iron core.
The conductor constitutes a one-turn primary winding. The secondary winding consists of a large number
of turns of much fine wire wrapped around the core as shown. Due to transformer action, the secondary
current is transformed to a low value which can be measured by ordinary meters.

For example, suppose that IP = 100 A in Fig. (5) and the ammeter is capable of measuring a maximum of

1A. Then,

Figure 5

3. Voltage transformer
It is a device that is used to measure high alternating voltage. It is essentially a step-down transformer
having small number of secondary turns as shown in Fig. (6). The high alternating voltage to be measured
is connected directly across the primary. The low voltage winding (secondary winding) is connected to the
voltmeter. The power rating of a potential transformer is small (seldom exceeds 300 W) since voltmeter is
the only load on the transformer.

Figure 6

4. Power transformer
Transformers with ratings above 500 kVA and used in generating stations and substations for stepping up
or stepping down the voltage are called power transformers. These transformers will be in operation during
the load periods and can be disconnected during light load periods (at least one or two should remain in
operation during light-load periods also in a substation). Hence, power transformers should have maximum
efficiency at nearly 80 percent full load. These transformers are designed to have considerably greater
leakage reactance. For these transformers, the voltage regulation is less important than the current limiting
effect of higher leakage reactance.

APPLICATIONS OF TRANSFORMERS
There are four principal applications of transformers viz.
1. Power Transformers. They are designed to operate with an almost constant load which is equal
to their rating. The maximum efficiency is designed to be at full-load. This means that full-load
winding copper losses must be equal to the core losses.
2. Distribution Transformers. These transformers have variable load which is usually considerably
less than the full-load rating. Therefore, these are designed to have their maximum efficiency at
between 1/2 and 3/4 of full load.
3. Autotransformers. An autotransformer has only one winding and is used in cases where the ratio of
transformation (K), either step-up or step down, differs little from 1. For the same output and
voltage ratio, an autotransformer requires less copper than an ordinary 2-winding transformer.
Autotransformers are used for starting induction motors (reducing applied voltage during starting)
and in boosters for raising the voltage of feeders.
4. Instrument transformers. Current and voltage transformers are used to extend the range of a.c.
instruments.

VOLTAGE TRANSFORMATION RATIO (K)

As derived above,

Where, K = constant
This constant K is known as voltage transformation ratio.
 ▪ If N2 > N1, i.e. K > 1, then the transformer is called step-up transformer.
▪ If N2 < N1, i.e. K < 1, then the transformer is called step-down transformer.
Assuming no losses, E1 =V1 and E2 =V2. Hence,
V1/V2 is called the voltage ratio and N1/N2 the turns ratio, or the ‘transformation ratio’ of the transformer.
When a load is connected across the secondary winding, a current I2 flow. In an ideal transformer loss are
neglected and a transformer is considered to be 100% efficient.
Hence input power = output power, or V1I1 =V2I2, i.e. in an ideal transformer, the primary and
secondary volt amperes are equal.
The rating of a transformer is stated in terms of the voltamperes that it can transform without
overheating. With reference to Figure1(a), the transformer rating is either V1I1 or V2I2, where I2 is the
full load secondary current.
Problem 1. A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240V,
determine the secondary voltage, assuming an ideal transformer.
Solution
For an ideal transformer, voltage ratio = turns ratio, i.e.
V1/V2=N1/N2 Hence 240/V2=500/3000 Thus secondary voltage V2= (3000)(240)/500=1440V or 1.44kV

Problem 2. An ideal transformer with a turns ratio of 2:7 is fed from a 240V supply. Determine its output
voltage. Solution
A turns ratio of 2:7 means that the transformer has 2 turns on the primary for every 7 turns on the
secondary (i.e. a step-up transformer). Thus,
N1/N2=2/7
For an ideal transformer, N1/N2=V1/V2 hence 2/7=240/V2 Thus the secondary voltage V2=
(240)(7)/2=840V

Problem 3. An ideal transformer has a turns ratio of 8:1 and the primary current is 3 A when it is
supplied at 240 V. Calculate the secondary voltage and current.
Solution
N1/N2=V1/V2 or secondary voltage V2=V1(N2/N1)=240(1/8)=30V
Also N1/N2=I1/I2 hence secondary current I2=I1(N1/N2)=3(8/1)=24A

Problem 4. An ideal transformer, connected to a 240V mains, supplies a 12V, 150W lamp. Calculate the
transformer turns ratio and the current taken from the supply.
Solution

Turns ratio = N1/N2=V1/V2=240/12=20

V1/V2=I2/I1, from which, I1=I2(V2/V1)=12.5(12/240)=0.625A