Chapter 2

Introduction:

One of the main advantages of a.c transgression and distribution

is the ease with which an a Hernating voltage can be increase or

reduced. For instance the general practice is to generate at

voltages about 22 kv, then step up by means of transformers to

higher voltages for the transmission lines. At suitable point,

other transformers are introduced to step the voltage down to

values suitable for motors, lamps, heaters etc. a medium size

transformer has a full efficiency of about 97-98 percent cent, so

that the transformer loss at each point of transformation is

small (although 20% of 100 mw is not insignificant) since there

are no moving parts in a transformer, the amount of supprvision

is practically negligible.

Although transformers are generally associated with power system

applications, they also occur in many low – power applications

including electronic circuit. However it is best to consider the

common power system transformer.

2.1 WORLING PRINCIPLES OF A TRANSFORMER

A transformer is a static (or stationary) piece apparatus by

means of which electric power in one circuit is transformed to

electric power of the same frequency in another circuit. It can

raise or lower the voltage in a circuit but with a decreasing or

increase in current.

The physical basis of a transformer is mutual induction between

two circuits linked by a common magnetic flux. In a simplest

form, it consist of two inductive coils which are electrically

separate but magnetically linked through a path of low reluctance

as shown in the fig 2.1.

The two coils possess high mutual inductance. If one coil is

connected to the source of alternating voltage, an alternating

flux is set up in the laminated core, most of which is linked

with the other coil in which it produces mutually induce emf.

(according to faraday’s Law of electric magnetic induction ie e =

mdi/dt). If the second coil circuit energy is transferred

entirely magnetically from the first coil to the second coil.

The first coil, in which electric energy is fed from the a.c

supply mains, is called primary winding and the other, from which

energy is fed from which energy is drawn out, is called secondary

winding. In brief, a transformer is a device that

1. Transfers electric power from one circuit to another

2. Does so without change of frequency

3. Accomplishes this by electromagnetic induction and

4. Where the two electric circuit are linked by mutual induction.

2.2 TRANSFORMER CONSTRUCTION

The simple element of a transformer consist of two coils having

mutual inductance and a laminated core. The two coild are

insulated from each other and from the steel core. Other

necessary parts are: some suitable container for the assembled

core and winding: a suitable medium for insulating the core and

its windings and bringing out the terminals of the winding from.

Its container; suitable busting (either of porcelain, oil filled

or capacitor – type) for insulating and bringing out the terminal

of the winding from the tank. In all types of transformers, the

core is constructed of transformer sheet steel laminations

assembled to provide a continues magnetic path with minimum of

air-gap included. The steel used is of high silicon content heat

treated to produce high permeability and low hysteresis loss at

the usually operating flux density. The lamination being

insulated from each other by light coat of core plate varnish or

by an oxide layer on the surface. The thinness of laminations

varies from 0.35mm for a frequency of 50HZ to 0.5mm for a

frequency of 25HZ. The core laminations (in the form of strips)

are joined as shown in fig 2.2. It is seen that the joined in

the alternate layers are staggered in order to avoid the presence

of narrow gaps right through the section staggered joints are

said to be imbricate.

Constructional, the transformer are of

two general types,

distinguished from each other merely by

the manner in which the primary and

secondary coils are placed around the

laminated steel core. The two types are

known as (i) core-type and (ii) shell

type.

Another recent development is spiral core or wound – core type,

the trade name being spira kore transformer.

In the so called core type transformer, the windings surround a

considerable part of the core where as in shell type

transformers, the core surrounds a considerable portion of the

winding as show in fig 2.3 (a) and (b) respectively

In a simplified diagram for the core-type transformer, in fig 2.3

(a) the primary and secondary windings are shown located on the

opposite legs (or limbs) of the core, but in actual construction,

these are always inter leaved in order to reduce leakage flux.

As show in fig 2.3 (a) half the primary and half the secondary

windings have been placed side by side or concentrically on each

limb, not primary on one limb (or leg) and the secondary on the

other.

EMF EQUATION OF A TRANSFORMER

Let N1 = No. of turns in primary

N2 = No. of turns in secondary

Qm = Maximum flux in the core in webers = Bm

x A.

F = Frequency of a.c input in hertz (Hz)

As show in fig 2.4, the flux increase from its zero value to

maximum value Qm in one quarter of the cycle ie ¼ f seconds.

Average rate of change of flux = Qm = ¼ f

= 4fQm wb/s.

Now, rate of change of flux per turn means induced e.m.f. in

volts.

Average emf induced/turn = 4Qm volts.

If flux Q varies sinusoidally, then r.m.s value induced e.m.f is

obtained by multiphying the average value with form factors.

Form factor = r.m.s value = 1.11Average value

r.m.s value of e.m.f Hurn = 1.11 x 4fQm

4.44fQm volts.

Now r.m.s value of e.m.f in the whole of primary winding

= (induced e.m.f/turn) x No. of primary turns

E1 = 4.44fN1Qm

Similarly, r.m.s value of induced e.m.f in secondary is E2 =

4.44fQN2

2.3 VOLTAGE RATIO (K)

From the above equation, weget

E2 = V2 = N2 = KE1 V2 N1

This constant k is known as voltage transformation ration.

(i) If N2 > N1 ie, K > 1 then transformer is called step up

transformer

(ii) If N2 < N1 ie K < 1, then transformer is known as step-down

transformer.

Example: The no load ratio of a 50HZ, single phase transformer is

6,000/250v. estimate the number of turns in each winding if the

maximum flux is 0.06wb in the core.

Solution

Using the transformer e.m.f equation,

E1 = 4.44fN1 Qm

6,000 = 4.44 x 50 x 0.60 x N

N1 = 6,000 = 450 terns 4.44 x 50 x 0.06

Similarly E2 = 4.44fN2Qm

= 250 = 4.44 x 50 x N2 x 0.06

N2 = 250 = N2 = 194 turns 4.44 x 50 x 0.06

Example 2: A 200KVA, 3300/240 volts, 50HZ single – phase

transformer has 80 turns on the secondary winding. Assuming Qm

ideal transformer, calculate (i) primary and secondary, currents

on full load, (ii) the maximum value of flux (iii) the number of

primary turns.

Solution

A 3300/340v transformer is one whose primary and secondary

voltages are 3300v and 240v respectively.

K = 240/3300 = 12/165 =

(i) N2 = 80 = 12 N1 = 1100 turns N1 N1 165

(ii) KVA = V2 I2

I2 = KVA = 2000,00 = I2 = 833A V2 240

I1 = KI2 = (12/165) x 833 = 60.6A

(iii) E1 = 4.44fN2Qm 3300 = 4.44 x 50 x 110 x Qm;

Qm = 13.5mwb.

2.4 THREE PHASE TRANSFORMERS

Large Scale generation of electric power is usually 3-phase at

generated voltages of 132kv or some what higher. Transmission is

generally accomplished at higher voltage of 66, 110, 132, 220 and

a 275kv for which purpose 3-phase transformers are necessary to

step up the generated voltage to that of the transmission line.

Next at no load centers, transmission voltages are reduced to

distribution voltages or 6600, 4600 and 2300 volts. Further at

most of the consumers, the distribution voltages of 440, 220 or

110 volts.

Years ago, it was a common practice to use suitable

interconnected three single phase transformers instead of a

single 3-phase transformer. But these days, the single – phase

are gaining popularity partly because of improvement in design

and manufacture but principally because of butter aquitance of

operating men with three-phase type.

As compare to bank of 3 single phase transformers, the main

advantages of a single 3 – phase transformer are that it occupies

less floor space for equal ratings, weighs less, cost about 15%

less and further, that only one unit is to be handled and

connected. Like single phase transformer, the three phase

transformers are also of the core type or shell type.

PRINCIPLE OF OPERATIO

2.5 EQUIVALENT CIRCUIT

The behavior of a transformer may be conveniently considered by

assuming it to be equivalent to an ideal transformer, in a

transformer having no losses and no magnetic leakage and a

ferromagnetic core of infinite permeability requiring no

magnetizing current, and then allowing for the imperfections of

the actual transformer by means of additional circuits or

impedances inserted between the secondary and load. Thus in fig

2.5 P and S represent the primary and secondary windings of the

ideal transformer, R1 and R2 are resistance equal to the

resistance of the primary and secondary windings of the actual

transformer. Similarly, inductive reactance X1 and X2 represents

the reactance’s of the winding due to leakage flux in the actual

transformer.

Fig 2.5 EQUIVALENT CIRCUIT OF A TRANSFORMER

The inductive reactor X is such that it takes a reactive current

equal to the magnetizing current IOM of the actual transformer.

The core losses due to hysteris is and edely current are allowed

for by a resistor R of such value that it takes acurent IO1 equal

to the core – loss component of the primary, ie I2O1R is equal to

the core loss of the actual transfore.

The resultant of Iom and Io1 is Io, namely the current which the

transformer takes on no load.

2.6 PHASOR DIAGRAM

For convenience let us assume an equal number of turns on the

primary and secondary windings, so that E1 = E2. As shown in the

Fig 2.7 E1 leads the flux by 900, and represents the voltage

across the primary of the ideal transformer.

Let us also assume the general case of a load having a lagging

power factor; consequently, in the fig 2.7 in has been drawn

lagging E2 by about 450.

Then

I2R2 = voltage drop due to secondary resistance

I2 X2 = voltage drop due to secondary

Leakage reactance

and I2Z2 = voltage drop due to secondary impedance

Fig 2.7

PHASOR DIAGRAM FOR TRANSFORMER ON NO LOAD

The secondary terminal voltage V2 is the phasor difference of E2

and I2Z2 in other words, V2 must be such that the phasor sum of V2

and I2Z2 is E2, and the derivation of the phasor representing V2 is

evident from fig 2.7. the power factor of the load is cosQz,

where Q2 is the phase difference between V2 and I2, I2 represents

the component of the primary current to neutralize the

demagnetizing effect of the secondary current and is drawn equal

and opposite to I2. Here Io is the no – load current of the

transformer. The phasor sum of I2 and Io gives the total current

I1 taken from the supply.

I1R1 = voltage drop due to primary resistance

I1X1 = voltage drop due to primary leakage reactance

I1Z1 = voltage drop due to primary impedance and V1 = E1

And I1Z1, = supply voltage.

If Q1 is the phase difference between V1 and I1, then cos Q1 is

the power factor on the primary side of the transformer. Fig 2.7

is the phasor representing the no load current and the primary

and secondary voltage drop are, for clearness, shown for larger

relative to the other phasors than they are in an actual

transformer.

2.7 VOLTAGE REGULATION

When a transformer is loaded, its secondary terminal voltage

falls (for lagging p.f) provided the applied primary voltage V1

is held constant. This variation in secondary terminal voltage

from no-load to full load (f.l) is called the (voltage)

regulation of the transformer and is usually expressed as a

percentage of the secondary no-load voltage.

Percentage regulation = Sec. volt. (Nol) – Sec. colt (f.l) x 100Sec. Volt (N.L)

V2 = Secondary terminal voltage on no load

V2 = Secondary terminal voltage on load

Then % regulation = oV2 – V2 x 100oV2

Since opV2 – V2 = I2 (Ro2 cosQ + Xoz sinQ)

% voltage regulation = I2 (Roz cosQ = Xo2 sinQ)oV2

example: A 100KVA transformer has 400 terns on the primary and 80

turns on the secondary. The primary and secondary resistance are

0.3π and 1.01π respectively, and the corresponding leakage

reactance are 1.1π and 0.035π respectively. The supply coltage

is 2200v. calculate:

(a) The equivalent impedance reffered to the primary

circuit;

(b) The voltage regulation and the secondary terminal

voltage for full load having a power factor of

(i) 0.8 lagging and (ii) 0.8 leading.

Solution

(a) equivalent resistance referred to the primary

Re = R2 +

Re = 0.3 + 0.01(40080 )2 = 0.55π

equivalent leakage reactance referred to primary is

Xe = X

Xe = 1.1 + 0.035 (40080 )2 = 1.975π

equivalent impedance referred to the primary

is Ze = √ℜ2+Xe2

= √¿¿ = 2.05π

(b) (i) since cos Q2 = 0.8

sinQ = 0.6

full load primary current = 100x10002200 = 45.45 A

voltage regulation for the power factor of 0.8 lagging

= I1 (Re cosQ2 + Xe cos Q2)

V1

= 45.45(0.5x0.8+1.975x0.6)2200 = 100

ANS = 3.36%

Secondary terminal voltage on no load is

V1 x N2N1

= 2200 x 80400 = 440v

Therefore decrease of secondary terminal voltage between no –

load and full load is

Secondary terminal voltage x per unit regulation

= 440 x 0.0336 = 14.8v

Therefore secondary terminal voltage on full load

440 – 14.8 = 425.2v

(ii) voltage regulation for power factor 0.8 leading

45.45 (0.55 x 0.8 – 1.975 x 0.6) 2200

= -0.0154 per unit

= -1.54%

Increase of secondary terminal voltage between no load and full

load is

440 x 0.0154 = 6.78v

Therefore secondary terminal voltage on full load is

440 + 6.78 = 446.8v

2.7 OPEN-CIRCUIT OR NO – LOAD TEST

The main purpose of this test is to determine no-10-ad losses.

One winding of the transformer which is convenient but usually

high voltage winding is kept open and other is connected to a

supply of normal voltage and frequency. A wattmeter meter (w),

voltmeter and ammeter (A) are connected in the low-voltage

winding ie, primary winding in the present case. Fig 2.8 (a)

show the simplied diagram and fig 2.8 (b) shows the actual,

connection. With normal voltage applied to the primary, normal

flux will be set up in the core, hence normal iron losses will

occur which are recorded by the wattmeter. As the primary no-

load current Io (as measured by ammeter) is small (usually 2 to

10% of rated load current), Cu loss is negligible small in

primary and nil in secondary (being open). Hence the wattmeter

reading represents practically the core-loss under no-load

conditions (which is the same for all loads)

Fig 2.8 (a)

Sometimes a high resistance voltmeter is connected across the

secondary. The reading of the voltmeter gives the induce emf in

the secondary winding. This helps in finding the transformation

ration k.

2.8 SHORT CIRCUIT OR IMPEDANCE TEST

It is an economically method of determining the following:

1. Equivalent impedance (Zo1 or Zo2) leakage reactance (Xo1 or

Xo2) and resistance (Ro1 or Ro2) of the transformer as referred

to the winding in which the measuring instruments are placed.

2. Cu loss at full load (and at any desired load). This loss is

used in calculating the efficiency of the transformer.

3. Knowing Zo1 or Zo2 the total voltage drop in the transformer

as referred to the primary or secondary can be calculated and

hence regulation of the transformer determined.

In this test, one winding – usually the low voltage winding is

solidly short circuited by a thick conductor (or through an

ammeter which may serve the additional purpose of indicating

rated load current) in fig 2.9

The low voltage (usually 5 to 10% of normal primary voltage) at

correct frequency (through for cu losses it is not essential) is

applied to the primary and is cautiously in creased till full-

load current are flowing both in primary and secondary (as

indicated by the respective ammeters).

Since in this test, the applied voltage is a small percentage of

the normal voltage, the mutual flux produced in the core is also

a small percentage of normal value (because flux is proportional

to the voltage. Hence, the core losses are very small with the

result that wattmeter reading represents the full-load cu loss or

I2R loss for the whole transformer ie, both primary cu loss and

secondary cu loss. If Vsc is the voltage required to calculate

rated load current on short circuit

Zo1 = VscI1

W = I12 Ro1 Ro1 = W/R1

2

Xo1 = √Zo1−Ro12

Example:

The ratio turns of a single-phase transformer is 8, the

resistance of the primary and secondary windings are 0.85π and

0.012π respectively and the leakage reactance of these winding

are 4.8π and 0.07π respectively. Determine the voltage to be

applied to the primary to obtain a current of 150A in the

secondary when the secondary terminals are short circuited.

Ignore the magnetizing current.

Solution

Turn ration = N1/N2 = : K = ⅛

Now Ro1 = R1 + R2K2 = 0.85 + 0.012 x 64

= 1.618π

X01 = X1 + X2K2 = 4.8 + 64 x 0.07

= 9.28π

Z01 = √(1.618)2+(9.28)2 = 9.419π

I1 = KI2 = 1508 =

Vsc = I1Z01 = 1508 x 9.419

= 176.64

Example:

The low-voltage winding of a 300KUA, 11, 00/12,200 volts, 50Hz

transformer has 190 turns and a resistance of 0.06π. The high –

voltage winding has 910 turns and a resistance of 1.6π. When

current is obtained with 550 volts applied to hv winding.

Calculate the equivalent resistance and leakage reactance

referred to high – voltage side. Assuming full load efficiercy

of 985%

Solution

Full load primary current

= 300,0000.985x11,000 = 27.7A

Z01 = 55027.7 = 19.8π;

R12 = R2/K2 = 0.06

(910190 )2 = 1.38π

R01 = R1 + R12 = 1.6 + 1.38 = 2.98π

X01 = √Zo12−R012 = √19.82−2.982 = 19.5π

2.9 EFFICIENCY

As is the case with other types of electrical machines, the

efficiency of a transformer at a particular load and power factor

is defined as the output divided by the input, the two being

measured in the same units (eithe watts or kilo watts).

Efficiency = output¿put

But a transformer being highly efficient piece of equipment, has

very small losses, hence, it is impractical to try to measure

transformer efficiency by measuring its input and output. These

quantities are nearly of the size. A better method is to

determine the losses and then to calculate the efficiency from

Efficiency = outputoutput+losses

= outputoutput+culoss+ironloss

or

ŋ = input−lossesinput = 1 - losses

¿put

It may be noted here that efficiency is base on power output in

watts and not on volt-amperes, although losses are proportional

to Va.

Hence at any volt-ampere load, the efficiency depends on power

factor, being maximum at a power factor of unity.

Efficiency can be computed by determining core loss from no-load

or open – circuit test and cu loss from short – circuit test.

Example:

The primary and secondary windings of a 500KVA transformer have

resistance of 0.42π voltages are 11000v and 415v respectively and

the core is 2.9kw, assuming the power factor of the load to be

0.8, calculate the efficiency on

(a) Full load,

(b) Half load

Solution

(a) Full load secondary current is

500x1000415 = 1205A

and full load primary current = 500x100011000

The secondary I2R loss on full load is (1205)2 x 0.0019 = 2760w

and

Primary I2R loss on full load is (45.5)2 x 0.42 = 870w

Total I2R loss on full load = 3630w = 3.63kw

and

Total loss on full load 3.63 + 2.9 = 6.53kw

output power on full load = 500 x 0.8 = 400kw

input power on full load

400 + 6.53 = 406.53kw

Efficiency = (1 – 6.53405.53) = 0.9 per unit

= 98.39%

(b) Since the I2R loss varies as the square of the current,

Total I2R loss on half load = 3.63 x (0.5)2

= 0.91kw

Total loss one half load = 0.91 + 2.9 = 3.81kw

Efficiency = (1−3.81203.81 ) = 98.13%

Example: The following results were obtained on a 50KUA

transformer open circuit test– primary voltage, 3300V, secondary

voltage, 415v, primary power, 430w. Short circuit test– primary

voltage; 124v; primary current, 15.3A; primary power, 525w;

secondary current, full load value. Calculate

(a) The efficiencies at full load and at half load for 0.7

power factor.

(b) The voltage regulation (i) lagging (ii) leading

(c) The secondary terminal voltages to (i) and (ii)

Solution

Core loss = 430w

Loss on full load = 525w

Total loss on full load 955w = 0.955kw

Efficiency on full load = 50x0.750x0.7+0.955

= (1-0.95535.95)

= 0.9734 per unit

= 97.34%

Efficiency on half load

Loss on half load = 525 x (0.5)2 = 131w

Total loss on half load = 430 + 131

=

561w

=

0.561kw

And efficiency on half load =

= 25x0.7

(25x0.7 )+0.561

= (1 – 0.56118.06)

= 0.969 per unit

= 96.6%

Voltage Regulation

(b) CosQe = 525124x15.3 = 0.2765

Qe = 730

For cosQ2 = 0.7

Q2 = 450

Regulation at 0.7 p.f tagging

Voltage Regulation = 124cos(73−45)3300

= 0.033 per unit = 3.3%

Regulation for p.f of 0.7 leading

Voltage Regulation = 124cos (73+45)3300

= -0.0185 per unit

= -1.85%

(c) Secondary voltage on open circuit = 415v.

Therefore secondary voltage on full load, p.f 0.7 lagging is

415 (1 – 0.033) = 401.3v

And secondary voltage on full load, p.f 0.7 leading is

415 (1 + 0.0185) = 422.7v

2.30 ALL DAY EFFICIENCY

The ordinary or commercial efficiency of a transformer is given

by the ration

Ŋ = output∈wattsinput∈watts

But there are certain types of transformers cannot be judged by

this efficiency. Transformer used for supplying lighting and

general network ie distribution distribution transformer have

their primaries energized for all the twenty-four (24h), although

their secondary supplies little or no much of the time during the

day except during the house-lighting period. It means where as

core loss occurs through out the day, the cu loss occurs only

when the transformer is loaded. Hence, it is considered a good

practice to design such transformer so that core losses are very

low. The cu losses are relatively less importance of such a

transformer should be judged by all-day efficiency also known as

operational efficiency which is computed on the basis of energy

consumed during certain time period, usually a day of 24hours.

Ŋ all day =output∈kwhinput∈kwh (for 24 hours).

To find this all-day efficiency or (as it also so called) energy

efficiency, we have to know the load cycle on the transformer ie,

how much and how long the transformer is loaded during 24 hours.

Practical calculations are facilitated by making use of a load

factor.

Example:

A 100KVA lighting transformer has a full-load loss of 3kw, the

losses being equally divided between iron and copper. During a

day, the transformer operates on full-load for 3 hours, one half-

load for 4hours, the output being negligible for the remainder of

the day. Calculate the all-day efficiency.

Solution

It is usually assumed that lighting transformers have a load p.f

of unity.

Iron loss for 24hours = 1.5 x 24 = 36kwh

Full load – cu loss = 1.5kw

Cu loss for 3hours on full load = 3 x 1.5 = 4.5kwh

Cu loss at half full-load = 1.5/4kw

Cu loss for 4hours at half F.L = 4 x (1.5/4) =

1.5kwh

Total losses in 24hours = 36 + 4.5 + 15

= 42kwh

Total output in 24hours = (100 x 3) + (50 x 4)

= 500kwh

Ŋ all-day = 500x100500+42 = 92.25%

NOTE: used this above information to calculate ordinary

efficiency.

Example:

A 20kw distribution transformer has an efficiency of 0.95 at full

load. The copper losses at full-load equal the iron losses. The

transformer is kept connected to the mains all the time but is

loaded with full – load for 6hour a day the load during the

remaining time being zero. Find the all-day efficiency.

Solution

Input = outputefficiency =

200.95 = 21.052

Full-load loss = 21.052 – 20 = 1.052kw

Full-load cu loss = 1.052/2

= 0.526kw, iron loss = 0.526kw

The cu loss takes place only when transformer is fully or

partially loaded whereas iron loss takes place all time the

transformer is energized.

F.L. cu loss per day for 6 hours = 0.526 x 6 = 3.156kwh

Iron loss per day = 24 x 0.526

= 12.624kwh

Total loss in 24 hours = 3.156 + 12.624 =

15.780kwh

Output in 24 hours = 20 x 6 = 120kwh

All – day efficiency = ŋ = 120x100(120+15.78)

= 88.4%

2.31. PARALLEL OPERATION OF TRANSFORMERS

For supplying a load in excess of the rating of an existing

transformers, two or more transformers may be connected in

parallel with an existing transformer. The transformer are

connected when load on one of the transformers is more them its

capacity. The reliability is increased with parallel operation

than to have single larger unit. The cost associated with

maintain the spare is less when two transformer are connected in

parallel.

It is usually economically to install another in parallel

instead of replacing the existing transformer by a single

large unit. The cost of a spare in the case of two parallel

transformers (of equal rating) is also lower than that of a

single transformer. In addition, it is preferable to have a

parallel transformer for the reason of reliability. With is

this at least half the load can be supplied with one

transformer out of service.

2.32. CONDITION FOR PARALLEL OPERATION OF TRANSFORMER:

For parallel connection of transformers, primary winding of

a transformer are connected to source bus – bars and

secondary winding are connected to the load bus-bars.

Various condition that must be fulfilled for the successful

parallel operation of transformers.

1. Same voltage ration and turns Ration (both primary and

secondary voltage Ratings is same).

2. Same percentage impedance and transformer ration.

3. Identical position of tap changer.

4. Same KVA Ratings

5. Same phase angle shift (vector group are same)

6. Same frequency rating.

7. Same polarity

8. Same phase sequence

Some of these condition are convenient and some are mandatory.

The convenient are: same voltage ration and turn ration,

same percentage impedance, same KVA Rating, same position of

tap changer.

The ma ndary conditions are: same phase angle shift, same

polarity, same phase sequence, and same frequency.

When the convenient conditions are not met parallel

operation is possible but not optimal.

ADVANTAGES OF PARALLEL OPERATIONS

2.32. METHODS OF COOLING

2.33 TAP CHANGING

The simplest type of tap-changing mechanism is a rotary switch

which allows the distribution company to select one of several

tapping (connection) points on the high–voltages side of a

transformer. This enables the turn-ration.