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gpmacademics.weebly.com Transformers

G.Purushotham, Asst.Professor, Dept of EEE @SVCE, TPT 1

TRANSFORMERS

Syllabus: Principles of operation, Constructional Details, Losses and efficiency, Regulation of

Transformer, Testing: OC & SC test.

TRANSFORMER: It is a static device which transfers electric energy from one electric circuit to another

electric circuit without any change in frequency. It is used either for raising or lowering the voltage of an a.c.

supply with a corresponding decrease or increase in current. It is a constant power device.

CONSTRUCTION OF A TRANSFORMER:

The transformer consists of mainly two parts:

Magnetic core

Windings or coils.

It essentially consists of two windings, the primary

and secondary, wound on a common laminated

magnetic core as shown in Fig. The winding connected to the a.c. source is called primary winding (or

primary) and the one connected to load is called secondary winding.

(i) The core is made of silicon steel which has low hysteresis loss and high permeability.

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 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.

Based on construction the transformers are classified into two types.

Core type transformer.

Shell type transformer.



Core type:

The coils used are form-wound and are of cylindrical type. The general

form of these coils may be circular or oval or rectangular. In core type

transformer, half of the primary winding and half of the secondary

winding are placed round each limb as shown.

Shell type:

The coils are form-wound but are multi-layer disc type .The

different layers of such multi-layer discs are insulated from each

other by paper. The complete winding consists of stacked discs with

insulation space between the coils. This method of construction

involves the use of a double magnetic circuit. Both the windings are

placed round the central lime, the other two limbs acting simply as a

low-reluctance flux path.

Comparison between core-type and shell-type transformers:

PRINCIPLE of OPERATION:- A transformer works on the principle of Mutual Induction.

When an alternating voltage V1 is applied to the primary, an alternating flux Ø is set up in the core. This

alternating flux links both the windings and induces e.m.f.s E1 and E2 in them according to Faraday’s laws of

electromagnetic induction. The e.m.f. E1 is termed as primary e.m.f. and e.m.f. E2 is termed as secondary

e.m.f. This induced e.m.f. E2 in the secondary causes a secondary current I2. Consequently, terminal voltage

V2 will appear across the load.



Magnitudes of E2 and E1 depend upon the number of turns on the secondary and primary respectively. If load

is connected across the secondary winding, the secondary e.m.f. E2 will cause a current I2 to flow through the

load. Thus, a transformer enables us to transfer a.c. power from one circuit to another with a change in

voltage level.

If N2 > N1, then E2 > E1 (or V2 > V1) and we get a step-up transformer.

If N2 < N1, then E2 < E1 (or V2 < V1) and we get a step-down transformer.

If N2 = N1, then E2 =E1 (or V2 =V1) and we get an isolation or one-to-one transformer.

The following points may be noted carefully:

(i) The transformer action is based on the laws of electromagnetic induction.

(ii) There is no electrical connection between the primary and secondary. The a.c. power is transferred from

primary to secondary through magnetic flux.

(iii) There is no change in frequency i.e., output power has the same frequency as the input power.

(iv) The losses that occur in a transformer are:

(a) Core losses—eddy current and hysteresis losses

(b) Copper losses—in the resistance of the windings

A transformer has very high efficiency. Because the losses are very small so that output power is

nearly equal to the input primary power.



IDEAL TRANSFORMER:

An ideal transformer is one that has

No winding resistance

No leakage flux i.e., the same flux links both the windings

No iron losses (i.e., eddy current and hysteresis losses) in the core

Consider an ideal transformer on no load i.e., secondary is open-circuited as shown in Fig. (i). under such

conditions, the primary is simply a coil of pure inductance. When an alternating voltage V1 is applied to the

primary, it draws a small magnetizing current Im which lags behind the applied voltage by 90°.

This alternating current Im produces an alternating flux Φ which is proportional to and in phase with it. The

alternating flux f links both the windings and induces e.m.f. E1 in the primary and e.m.f. E2 in the secondary.

The primary e.m.f. E1 is, at every instant, equal to and in opposition to V1 (Lenz’s law). Both e.m.f.s E1 and

E2 lag behind flux Φ by 90°. However, their magnitudes depend upon the number of primary and secondary

turns. Fig. (ii) Shows the phasor diagram of an ideal transformer on no load. Since flux Φ is common to both

the windings, it has been taken as the reference phasor. The primary e.m.f. E1 and secondary e.m.f. E2 lag

behind the flux Φ by 90°. Note that E1 and E2 are in phase. But E1 is equal to V1 and 180° out of phase with

it.

EMF EQUATION:

Let

Ø = Flux

Øm=Max. value of fluxØ= rate of change of flux

N1= No.of turns on Primary Winding

N2= No.of turns on Secondary Winding

f = Supply frequency Fig. 1



E1= RMS value of induced e.m.f on primary winding

E2= RMS value of induced e.m.f on secondary winding

According to faraday’s law of electromagnetic induction the average induced e.m.f per turn is proportional to

the rate of change of flux.

Average e.m.f per turn =Ø

Consider the ¼ Th cycle of the flux as shown in fig.1.time required for one complete cycle is 1/f sec.

In ¼th time period, the change in flux is from 0 to ØmØ = ∅as dt for ¼ Th time period is

= 4 ∅ /∴Average e.m.f per turn = 4 ∅For sinusoidal quantityForm factor = . . . = 1.11∴ R.M.S. value = 1.11 X Average value∴R.M.S. value of e.m.f per turn = 1.11 X 4 ∅= 4.44 ∅There are N no.of turns on winding hence the R.M.S. value of induced emf is ,= 4.44 ∅If N1 turns on primary the R.M.S. value of induced emf is = 4.44 ∅If N2 turns on primary the R.M.S. value of induced emf is = 4.44 ∅

Voltage Transformation Ratio (K): From the above equations of induced e.m.f., we have= = = =Assuming to be an ideal transformer where:

(i) E1 = V1 and E2 = V2 as there is no voltage drop in the windings.

(ii) There are no losses. Therefore, volt-amperes input to the primary are equal to the output volt-amperes

i.e. V1I1 = V2I2

Hence, currents are in the inverse ratio of voltage transformation ratio. This simply means that if we raise the

voltage, there is a corresponding decrease of current.

LOSSES AND EFFICIENCY:

There two types of power losses that occur in a transformer. They are:

1) Core loss 2) Copper loss



1) CORE LOSS: This is the power loss that occurs in the core part. they are also called as iron loss because

the core is made of iron material which is magnetic in nature. This loss is due to the alternating frequency of

the emf. Iron loss in further classified into two other losses.

a) Eddy current loss b) Hysteresis loss

a) Hysteresis Loss: This is the loss in the iron core, due to the magnetic reversal of the flux in the core,

which results in the form of heat in the core. This loss is directly proportional to the supply frequency.

Hysteresis loss can be minimized by using the core material having high permeability.

b) Eddy Current Loss: This power loss is due to the alternating flux linking the core, which will induced an

emf in the core called the eddy emf, due to which a current called the eddy current is being circulated in the

core. As there is some resistance in the core with this eddy current circulation converts into heat called the

eddy current power loss. Eddy current loss is proportional to the square of the supply frequency.

Eddy current loss can be minimized by using the core made of thin sheets of silicon steel material, and each

lamination is coated with varnish insulation to suppress the path of the eddy currents.

Both hysteresis and eddy current losses depend upon:

(i) Maximum flux density Bmax in the core and

(ii) Supply frequency f. Since transformers are connected to constant-frequency, constant voltage

supply, both f and Bm are constant. Hence, core or iron losses are practically the same at all loads.

Iron or Core losses, WC = Hysteresis loss (Wh) + Eddy current loss (We) = Constant losses (Wc)

2) COPPER LOSS: This is the power loss that occurs in the primary and secondary coils when the

transformer is on load. This power is wasted in the form of heat due to the resistance of the coils. This loss is

proportional to the sequence of the load hence it is called the Variable loss.

The total copper losses: 2221

21 RIRIW cu

012

1 RIW cu 0222 RI

It is clear that copper losses vary as the square of load current.

Total losses in a transformer = Constant losses + Variable losses.

Wt = Wc + Wcu

EFFICENCY: It is the ratio of the output power to the input power of a transformer.

Ƞ =

Input = Output + Total losses

= Output + Iron loss + Copper loss.

22max fBKW ee

fBKW hh6.1

max



Efficiency =

Where, V2 is the secondary (output) voltage,

I2 is the secondary (output) current and

Cosϕ is the power factor of the load.

The transformers are normally specified with their ratings as KVA,

Therefore, Ƞ = x 103x 103 + +Since the copper loss varies as the square of the load the efficiency of the transformer at any desired load

(or fraction of load) n is given byȠ = VA rating x 103 cosΦVA rating x 103x cosΦ + Wcore+ 2 WcopperWhere Wcopper is the copper loss at full load

Wcopper = I2R watts

CONDITION FOR MAXIMUM EFFICIENCY:

In general for the efficiency to be maximum for any device the losses must be minimum. Between the

iron and copper losses the iron loss is the fixed loss and the copper loss is the variable loss. When these two

losses are equal and also minimum the efficiency will be maximum.

Therefore the condition for maximum efficiency in a transformer is

Iron loss = Copper loss (whichever is minimum)

TRANSFORMER on NO-LOAD:

For an ideal transformer, we have assumed that there are no

core losses and copper losses. For practical transformers,

these two losses cannot be neglected. At no-load condition,

the primary current is not fully reactive and it supplies:

(i) Iron loss in the core, that is, hysteresis loss and eddy

current loss.

(ii) Very small amount of copper loss in the primary. There is copper loss in the secondary because it is an

open circuit.

The no-load current lags behind V1 by an angle Φ0, which is less than 90° (around 80°–85°).

coppercore WWIV

IV

copperlosscorelossroutputpowe

routputpowe

cos

cos

22

22



The no-load input power is given by: W0=V1I0Cos Φ0

Where CosΦ0 is the no-load power factor.

Figure shows the no-load phasor diagram of a practical transformer.

From Figure the no-load primary current (I0) has the following two

components:

One component of I0, that is Ic = I0 CosΦ0 is in phase with V1.

Since Ic supplies the iron loss and primary copper loss at no

load, it is known as active or working or core loss component.

The other component of I0 that is, Im= I0 sin Φ0 is in quadrature

with V1. It is known as magnetizing component. Its function is to sustain the alternating flux in the

core and it is wattless.

From phasor diagram we have:

Io= +Since I0 is very small, the no-load copper loss is negligible. Hence, no-load input is practically equal to the

iron loss in the transformer.

TRANSFORMER ON LOAD:

Transformer on load, which means load, is connected

to the secondary terminals. Consider, transformer having core

loss but no copper loss and leakage reactance. Whenever load

is connected to the secondary winding, load current will start

to flow through the load as well as secondary winding. This

load current depends upon the characteristics of the load and

also secondary voltage of the transformer.

This current is called secondary current or load current;

here it is denoted as I2. As I2 is flowing through the secondary

windings, a self mmf in secondary winding will be produced.

Here it is N2I2, where, N2 is the number of turns of the

secondary winding of transformer. This mmf or magneto

motive force in the secondary winding produces flux φ2. This

φ2 will oppose the main magnetizing flux and momentarily

weakens the main flux and tries to reduce primary self

induced emf E1. If E1 falls down below the primary source



voltage V1, there will be extra current flows from source to

primary winding. This extra primary current I2′ produces extra

flux φ′ in the core which will neutralized the secondary counter

flux φ2. Hence the main magnetizing flux of core, Φ remains

unchanged irrespective of load.

i.e., N1I2'=N2I2´ =´ = KI2 : this is opposite to that of I2.

So total current, this transformer draws from source can be divided into two components, first one is

utilized for magnetizing the core and compensate the core loss i.e.

Io. It is no-load component of the primary current. Second one is

utilized for compensating the counter flux of the secondary

winding. It is known as load component of the primary current.

Hence total no load primary current I1 of a transformer having no

winding resistance and leakage reactance can be represented as

follows:

I1 = Io + I2′

Where θ2 is the angle between Secondary Voltage and

Secondary Current of transformer.

Here the load is assumed to be a lagging load (inductive

load) that draws lagging power. i.e. the secondary current I2 lags

the secondary terminal voltage V2 by an angle θ2.

EQUIVALENT RESISTANCE

Equivalent Resistance Referred to Primary:

a transformer having primary resistance R1 and secondary

resistance R2, where resistances have been shown external to the

windings. In Figure, it is assumed that there is no leakage of flux.

It is possible to transfer resistance from one winding to another

to simplify the calculation. Let N1 and N2 be the number of turns

of primary and secondary winding respectively. Let the turns

ratio be ‘k’. Let I1 and I2 be the currents in primary and

secondary winding, respectively. Neglecting I0,



Total copper loss = += += + ∵ =

`Let the referred value of R2 be R2' when it is transferred to primary. The copper loss of secondary

is I22R2 when R2 is in secondary. The copper loss across R2

' is I12R2

' when R2 has been transferred to primary.

These two losses must be equal.

I22R2=I1

2R2'

i.e. , = =The total resistance referred to as primary becomes

R01= R1 + R2'

= R1 + .

This is also known as equivalent or effective resistance of the transformer referred to as primary and is

denoted by R01.

Similarly reactance & impedance X01= X1+ X2’,, =

Z01= Z1+ Z2’,, =

Equivalent Resistance Referred to Primary:

If R1 is transferred to secondary, having referred value R1',

we have I12R1=I2

2R1'

, = =The total resistance referred to as secondary becomes

R02 = R2 + R1'

= R2 +K2R1.

Similarly the magnetic flux leakage in the transformer is represented as reactance as shown in figure.



Similarly reactance & impedance X02= X2+ X1’,, = 2 1

Z02= Z2+ Z1’,, = 2 1

VOLTAGE REGULATION:

The voltage regulation of a transformer is defined as the change in the secondary terminal voltage between

no load and full load, expressed as a percentage of the no-load terminal voltage. Assuming that primary

voltage and frequency to be constant. That is% = − 100also the regulation can be expressed as percentage of full-load terminal voltage.% = − 100Voltage regulation is a measure of the change in the terminal voltage of a transformer between No load and

Full load. A good transformer has least value of the regulation of the order of ±5%.

Expression for Voltage Regulation of Transformer:

An electrical power transformer is open circuited means load is not connected with secondary terminals. In

this situation the secondary terminal voltage of the transformer will be its secondary induced emf E2.

Whenever full load is connected to the secondary terminals of the transformer, rated current I2 flows through

the secondary circuit and voltage drops occurs at load. So, primary winding will also draw equivalent full

load current from source. The voltage drop in the secondary is I2Z2 where Z2 is the secondary impedance of

transformer. If now, at this loading condition the voltage between secondary terminals V2 across load

terminals will be obviously less than no load secondary voltage E2 and this is because of I2Z2 voltage drop in

the transformer.

% Voltage regulation = x 100Where E2 is the no-load secondary voltage of the transformer.

Voltage Regulation of Transformer for lagging

Power Factor:

the expression of voltage regulation,

lagging Power Factor of the load = Cosθ2,

that means angle between secondary current and

voltage is θ2 and the current I2 lags V2.

Here, from the diagram,

OA = V2



AE = I2R02

ED = I2X02

Angle BAE = θ2 = angle EDF

OD = E2

From the diagram

OC = OA + AB + BC

Here, OA = V2

Here, AB = AE Cosθ2 = I2R2 Cosθ2

and, BC =EF= DE Sinθ2 = I2X2 Sinθ2

Angle between OC & OD may be very small so it can be neglected and OD is considered nearly equal to OC

i.e.

E2 = OC = OA + AB + BC

E2 = OC = V2 + I2R2cosθ2 + I2X2sinθ2 .....................1

E2 - V2 = I2R2cosθ2 + I2X2sinθ2

Voltage Regulation of transformer at lagging power factor:% = − x 100 = I R Cosϴ + I X SinϴV x 100Voltage Regulation of Transformer for Leading Power Factor:lagging Power Factor of the load = Cosθ2,

that means angle between secondary current and voltage is θ2 and the current I2 leads V2.

Here, from the diagram,

OC = OA + AB − BC

Here, OA = V2

Here, AB = AEcosθ2 = I2R2cosθ2

And, BC=FE = DEsinθ2 = I2X2sinθ2

Angle between OC & OD may be very small so itcan be neglected and OD is considered nearly equal to OC i.e.

E2 = OC = OA + AB − BC

E2 = OC = V2 + I2R2cosθ2 − I2X2sinθ2 .....................2

E2 - V2 = I2R2cosθ2 - I2X2sinθ2



Voltage Regulation of transformer at leading power factor,

% = − x 100 = I R Cosϴ − I X SinϴV x 100TESTING OF TRANSFORMERS:

Large scale transformers are tested indirectly by using open circuit and short circuit tests. In this the

transformer is not loaded. So no loading device is required and large power is not wasted in testing, but more

calculations are involved. By conducting these tests we can determine:

Parameters of equivalent circuit.

Regulation at any desired load and power factor.

Losses and efficiency at any desired load and power factor.

Open Circuit Test:

The connection diagram for open circuit test on transformer is shown in the figure. A voltmeter,

wattmeter, and an ammeter are connected in LV side of the transformer as shown.

The voltage at rated frequency is applied to that LV side with the help of a variac.

The HV side of the transformer is kept open.

Now with help of variac applied voltage is slowly increase until the voltmeter gives reading equal to the

rated voltage of the LV side.

After reaching at rated LV side voltage, all three instruments reading (Voltmeter, Ammeter and

Wattmeter readings) are recorded.

The ammeter reading gives the no load current Io. As no load current Io is quite small compared to rated

current of the transformer, the voltage drops due to this electric current then can be taken as negligible.

Voltmeter reading V1 can be considered equal to secondary induced voltage of the transformer.

The input power during test is indicated by watt-meter reading. As the transformer is open circuited,

there is no output hence the input power here consists of core losses in transformer and copper loss in

transformer during no load condition. But as said earlier, the no load current in the transformer is quite

small compared to full load current so copper loss due to the small no load current can be neglected.

Hence the wattmeter reading can be taken as equal to core losses in transformer. Let us consider

wattmeter reading is Po.

.



Iron losses,Wi = Wattmeter reading = W0

No load current = Ammeter reading = I0

Applied voltage = Voltmeter reading = V1

Input power, W0 = V1 I0 cos ϕ0

therefore

the open circuit test on transformer is used to determine core losses in transformer and parameters of

shunt branch of the equivalent circuit of transformer.

Short circuit test:

The connection diagram for short circuit test on transformer is shown in the figure. A voltmeter,

wattmeter, and an ammeter are connected in HV side of the transformer as shown.

The voltage at rated frequency is applied to that HV side with the help of a variac

The LV side of the transformer is short circuited.

Now with help of variac applied voltage is slowly increase until the ammeter gives reading equal to the

rated current of the HV side.

After reaching at rated current of HV side, all three instruments reading (Voltmeter, Ammeter and Watt-

meter readings) are recorded.



The ammeter reading gives the primary equivalent of full load current IL.

As the voltage, applied for full load current in short circuit test on transformer, is quite small compared

to rated primary voltage of the transformer, the core losses in transformer can be taken as negligible.

Voltmeter reading is Vsc. The input power during test is indicated by watt-meter reading. As the

transformer is short circuited, there is no output hence the input power here consists of copper losses in

transformer.

Since, the applied voltage Vsc is short circuit voltage in the transformer and hence it is quite small

compared to rated voltage so core loss due to the small applied voltage can be neglected. Hence the

wattmeter reading can be taken as equal to copper losses in transformer. Let us consider wattmeter reading

is Wsc.

Full load Cu loss, PC = Wattmeter reading = WSc

Applied voltage = Voltmeter reading = VSC

F.L. primary current = Ammeter reading = I1

Where R01 is the total resistance of transformer referred to primary.



Note: The short-circuit test will give full-load Cu loss only if the applied voltage VSC is such so as to

circulate full-load currents in the windings. If in a short-circuit test, current value is other than full-load

value, the Cu loss will be corresponding to that current value.

Advantages of Transformer Tests:

The above two simple transformer tests offer the following advantages:

(i) The power required to carry out these tests is very small as compared to the full-load output of the

transformer. In case of open-circuit lest, power required is equal to the iron loss whereas for a short-circuit

test, power required is equal to full-load copper loss.

(ii) These tests enable us to determine the efficiency of the transformer accurately at any load and p.f.

without actually loading the transformer.

(iii) The short-circuit test enables us to determine R01 and X01 (or R02 andX02). We can thus find the

totalvoltage drop in the transformer as referred to primary or secondary. This permits us to calculate voltage

regulation of the transformer.

Phasor Diagrams for Transformer on LoadConsider a transformer supplying the load as shown in the Fig. 1.

Fig. 1

The various transformer parameters are,

R1 = Primary winding resistance

X1 = Primary leakage reactance

R2 = Secondary winding resistance

X2 = Secondary leakage reactance

ZL = Load impedance

I1= Primary current

I2 = Secondary current = IL = Load current

now Ī1 = Īo + Ī2'

where Io = No load current

I2'= Load component of current decided by the load

= K I2 where K is transformer component

The primary voltage V1 has now three components,

1. -E1, the induced e.m.f. which opposes V1

2. I1 R1, the drop across the resistance, in phase with I1

3. I1 X1, the drop across the reactance, leading I1 by 90o∴ = − + + phasor sum= − +



The secondary induced e.m.f. has also three components,

1. V2, the terminal voltage across the load

2. I2 R2, the drop across the resistance, in phase with I2

3. I2 X2, the drop across the reactance, leading I2 by 90o

∴ = − − phasor sum= −

∴ = + += +

The phasor diagram for the transformer on load depends on the nature of the load power factor. Let us

consider the various cases of the load power factor.

1. Unity power factor load, (cosΦ2 = 1)As load power factor is unity, the voltage

V2 and I2 are in phase. Steps to draw the phasor

diagram are,

1. Consider flux Φ as reference

2. E1 lags Φ by 90o. Reverse E1 to get -E1.

3. E1 and E2 are inphase

4. Assume V2 in a particular direction

5. I2 is in phase with V2.

6. Add I2 R2 and I2 X2 to to get E2.

7. Reverse I2 to get I2'.

8. Add Io and I2' to get I1.

9. Add I1 R1 and to -E1 to get V1.

Angle between V1 and I1 is Φ1 and cosΦ1 is

primary power factor. Remember that I1X1 leads

I1 direction by 90o and I2 X2 leads I2 by 90o as current through inductance lags voltage across inductance by

90o. The phasor diagram is shown in the Fig.2

Fig. 2 Phasor diagram for unity power factor load



Lagging Power Factor Load, cos Φ2

As load power factor is lagging cosΦ2,

the current I2 lags V2 by angle Φ2. So only

changes in drawing the phasor diagram is to

draw I2 lagging V2 by Φ2 in step 5 discussed

earlier. Accordingly direction of I2 R2, I2 X2,

I2', I1, I1 R1 and I1X1 will change. Remember

that whatever may be the power factor of

load, I2X2 leads I2 by 90o and I1X1 leads I1 by

90o.

The complete phasor diagram is shown

in the Fig. 3.

Fig. 3 Phasor diagram for lagging power factor

Loading Power Factor Load, cos Φ2

As load power factor is leading, the current I2 leads V2 by angle Φ2. So change is to draw I2 leading I2 by

angle Φ2. All other steps remain same as before. The complete phasor diagram is shown in the Fig. 4

Fig. 4 Phasor diagram for leading power factor

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