Download (ii) Three-phase transformer

LECTURER- 16
Basic construction, working and types of transformer
Electrical transformer is a static electrical machine which transforms electrical power from
one circuit to another circuit, without changing the frequency. Transformer can increase or
decrease the voltage with corresponding decrease or increase in current.
Working principle of transformer
The basic principle behind working of a transformer is the phenomenon of mutual induction
between two windings linked by common magnetic flux. Basically a transformer consists of
two inductive coils; primary winding and secondary winding. The coils are electrically
separated but magnetically linked to each other. When, primary winding is connected to a
source of alternating voltage, alternating magnetic flux is produced around the winding. The
core provides magnetic path for the flux, to get linked with the secondary winding. Most of
the flux gets linked with the secondary winding which is called as 'useful flux' or main 'flux',
and the flux which does not get linked with secondary winding is called as 'leakage flux'. As
the flux produced is alternating (the direction of it is continuously changing), EMF gets
induced in the secondary winding according to Faraday's law of electromagnetic induction.
This emf is called 'mutually induced emf', and the frequency of mutually induced emf is same
as that of supplied emf. If the secondary winding is closed circuit, then mutually induced
current flows through it, and hence the electrical energy is transferred from one circuit
(primary) to another circuit (secondary).
Basic construction of transformer
Basically a transformer consists of two inductive windings and a laminated steel core. The
coils are insulated from each other as well as from the steel core. A transformer may also
consist of a container for winding and core assembly (called as tank), suitable bushings to
take our the terminals, oil conservator to provide oil in the transformer tank for cooling
purposes etc.
In all types of transformers, core is constructed by assembling (stacking) laminated sheets of
steel, with minimum air-gap between them (to achieve continuous magnetic path). The steel
used is having high silicon content and sometimes heat treated, to provide high permeability
and low hysteresis loss. Laminated sheets of steel are used to reduce eddy current loss. The
sheets are cut in the shape as E,I and L. To avoid high reluctance at joints, laminations are
stacked by alternating the sides of joint. That is, if joints of first sheet assembly are at front
face, the joints of following assemble are kept at back face.
Types of transformers
Transformers can be classified on different basis, like types of construction, types of cooling
etc.
(A) On the basis of construction, transformers can be classified into two types as; (i) Core
type transformer and (ii) Shell type transformer.
(i) Core type transformer
In core type transformer, windings are cylindrical former wound, mounted on the core limbs
as shown in the figure above. The cylindrical coils have different layers and each layer is
insulated from each other. Materials like paper, cloth or mica can be used for insulation. Low
voltage windings are placed nearer to the core, as they are easier to insulate.
(ii) Shell type transformer
The coils are former wound and mounted in layers stacked with insulation between them. A
shell type transformer may have simple rectangular form as shown in above fig, or it may
have a distributed form.
(B) On the basis of their purpose
1. Step up transformer: Voltage increases (with subsequent decrease in current) at
secondary.
2. Step down transformer: Voltage decreases (with subsequent increase in current) at
secondary.
(C) On the basis of type of supply
1. Single phase transformer
2. Three phase transformer
(D) On the basis of their use
1. Power transformer: Used in transmission network, high rating
2. Distribution transformer: Used in distribution network, comparatively lower rating
than that of power transformers.
3. Instrument transformer: Used in relay and protection purpose in different instruments
in industries
o
Current transformer (CT)
o
Potential transformer (PT)
(E) On the basis of cooling employed
1. Oil-filled self cooled type
2. Oil-filled water cooled type
3. Air blast type (air cooled)
LECTURER-17
(Turn’s ratio and transformer emf equation)
Voltage Transformation Ratio
In a transformer, source of alternating current is applied to the primary winding. Due to this,
the current in the primary winding (called as magnetizing current) produces alternating flux
in the core of transformer. This alternating flux gets linked with the secondary winding, and
because of the phenomenon of mutual induction an emf gets induced in the secondary
winding. Magnitude of this induced emf can be found by using the following EMF equation
of the transformer.
EMF equation of the Transformer
Let,
N1 = Number of turns in primary winding
N2 = Number of turns in secondary winding
Φm = Maximum flux in the core (in Wb) = (Bm x A)
f = frequency of the AC supply (in Hz)
As, shown in the fig., the flux rises sinusoidally to its maximum value Φm from 0. It reaches
to the maximum value in one quarter of the cycle i.e in T/4 sec (where, T is time period of the
sin wave of the supply = 1/f).
Therefore,
average rate of change of flux = Φm /(T/4) = Φm /(1/4f)
Therefore,
average rate of change of flux = 4f Φm
....... (Wb/s).
Now,
Induced emf per turn = rate of change of flux per turn
Therefore, average emf per turn = 4f Φm ..........(Volts).
Now, we know, Form factor = RMS value / average value
Therefore, RMS value of emf per turn = Form factor X average emf per turn.
As, the flux Φ varies sinusoidally, form factor of a sine wave is 1.11
Therefore, RMS value of emf per turn = 1.11 x 4f Φm = 4.44f Φm.
RMS value of induced emf in whole primary winding (E1) = RMS value of emf per turn X
Number of turns in
primary winding
E1 = 4.44f N1 Φm
............................. eq 1
Similarly, RMS induced emf in secondary winding (E2) can be given as
E2 = 4.44f N2 Φm.
............................ eq 2
from the above equations 1 and 2,
𝐸1
𝑁1
=
𝐸2
𝑁2
=4.44f Φm
This is called the emf equation of transformer, which shows, emf / number of turns is same
for both primary and secondary winding.
For an ideal transformer on no load, E1 = V1 and E2 = V2
.
where, V1 = supply voltage of primary winding
V2 = terminal voltage of secondary winding
Turns ratio (n) is the inverse of Voltage Transformation Ratio (K)
As derived above,
𝐸1
𝑁1
=
𝐸2
𝑁2
=K i.e n=1/K
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-up transformer
Q-1) A transformer has a primary voltage of 480 volts and a secondary voltage of 120
volts. If the primary windings have 700 turns, how many turns are in the secondary
windings?
Cross multiply:
Q-2) A transformer has 20 primary windings and 100 secondary windings. If the secondary
voltage is 25 v, find the primary voltage.
ANS.
100/20=25/X
CROSS MULTIPLY
100X=20*25
100X=500
X=500/100
X=5 VOLTS IN THE PRIMARY.
Q-3) If the primary to secondary voltage ratio is 4:1 for a transformer, what is the
secondary voltage if the primary voltage is 460 volts?
ANS.
primary/secondary=4/1=460/x
> 4x=460
So
--> x=115
So the secondary voltage is 115 Volts
LECTURER-18
Open circuit and Short circuit Test on transformer
These two transformer tests are performed to find the parameters of equivalent circuit of
transformer and losses of the transformer. Open circuit test and short circuit test on
transformer are very economical and convenient because they are performed without
actually loading of the transformer.
Open circuit or No load test on Transformer
Open circuit test or no load test on a transformer is performed to determine 'no load loss (core
loss)' and 'no load current I0'. The circuit diagram for open circuit test is shown in the
figure below.
Usually high voltage (HV) winding is kept open and the low voltage (LV) winding is
connected to its normal supply. A wattmeter (W), ammeter (A) and voltmeter (V) are
connected to the LV winding as shown in the figure. Now, applied voltage is slowly
increased from zero to normal rated value of the LV side with the help of a variac. When the
applied voltage reaches to the rated value of the LV winding, readings from all the three
instruments are taken.
The ammeter reading gives the no load current I0. As I0 itself is very small, the voltage drops
due to this current can be neglected.
The input power is indicated by the wattmeter (W). But, as the other side of transformer is
open circuited, there is no output power. Hence, this input power only consists of core losses
and copper losses. But as described above, short circuit current is so small that these copper
losses can be neglected. Hence, now the input power is almost equal to the core losses. Thus,
the wattmeter reading gives the core losses of the transformer. Sometimes, a high resistance
voltmeter is connected across the HV winding. Though, a voltmeter is connected, HV
winding can be treated as open circuit as the current through the voltmeter is negligibly small.
This helps in to find voltage transformation ration (K).
The two components of no load current can be given as,
Iμ = I0sinΦ0 and
Iw = I0cosΦ0.
cosΦ0 (no load power factor) = W / (V1I0). ... (W = wattmeter reading)
From this, shunt parameters of equivalent circuit parameters of equivalent circuit of transformer
(X0 and R0) can be calculated as
X0 = V1/Iμ and R0 = V1/Iw.
(These values are referring to LV side of the transformer.)
Hence, it is seen that open circuit test gives core losses of transformer and shunt parameters
of the equivalent circuit.
Q-4) An ideal transformer with a 300 turn primary connected to a 480 V, 60 Hz supply
line needs to output 120 V from the secondary. If a 100 Ω resistor is connected across
the secondary, determine: A) How many turns the secondary must have to output the
desired voltage. B) The current drawn through the primary. C) The maximum flux in
the core of the transformer
Solution:
A):The ratio of primary voltage to secondary voltage is directly proportional to the
ratio of number of turns on the primary to number of turns on the secondary:
Where
Voltage across primary,
of turns in primary,
Voltage across secondary,
Number
Number of turns in secondary
To solve for the number of turns required for the secondary, the equation is rearranged
solving for
:
B):The ratio of primary current to secondary current is inversely proportional to the
ratio of number of turns on the primary to number of turns on the secondary:
Where
Current in primary,
in primary,
Current in secondary,
Number of turns
Number of turns in secondary
Rearranging to solve for
:
C)The induced emf of the secondary can be calculated by:
Solving
Where
for
,
we
can
max flux in core,
Number of turns in secondary
calculate
the
maximum
Voltage across secondary,
flux
in
the
core:
Frequency of line,
LECTURER-19
Short circuit or Impedance test on Transformer
This test is conducted to determine R01 (or R02), X01 (or X02) and full-load copper losses of
the transformer. The connection diagram for short circuit test or impedance test on
transformer is as shown in the figure below. The LV side of transformer is short circuited and
wattmeter (W), voltmere (V) and ammeter (A) are connected on the HV side of the
transformer. Voltage is applied to the HV side and increased from the zero until the ammeter
reading equals the rated current. All the readings are taken at this rated current.
The ammeter reading gives primary equivalent of full load current (Isc).
The voltage applied for full load current is very small as compared to rated voltage. Hence,
core loss due to small applied voltage can be neglected. Thus, the wattmeter reading can be
taken as copper loss in the transformer.
Full load Cu loss, PC = Wattmeter reading = WS
Applied voltage = Voltmeter reading = VSC
F.L. primary current = Ammeter reading = I1
PC = I21R1 + I21R’2 = I21R01
OR; R01 = PC /
I21
where R01 is the total resistance of transformer referred to primary.
Total impedance referred to primary,Z01 =VSC/I1
Total leakage reactance referred to primary, X01 = (Z201 - R201 )
Short-circuit p.f, cosf2 = PC/ VSCI1
These, values are referred to the HV side of the transformer. Hence, it is seen that the short
circuit test gives copper losses of transformer and approximate equivalent resistance and
reactance of the transformer.
Why Transformers are rated in kVA?
From the above transformer tests, it can be seen that Cu loss of a transformer depends on
current, and iron loss depends on voltage. Thus, total transformer loss depends on voltampere (VA). It does not depend on the phase angle between voltage and current, i.e.
transformer loss is independent of load power factor. This is the reason that transformers are
rated in kVA.
LECTURER-20
Equivalent circuit of Transformer
In a practical transformer
(a) Some leakage flux is present at both primary and secondary sides. This leakage gives rise
to leakage reactances at both sides, which are denoted as X1 and X2 respectively.
(b) Both the primary and secondary winding possesses resistance, denoted as R1 and R2
respectively. These resistances causes voltage drop as, I1R1 and I2R2 and also copper loss
I12R1 and I22R2.
(c) Permeability of the core can not be infinite, hence some magnetizing current is needed.
Mutual flux also causes core loss in iron parts of the transformer.
We need to consider all the above things to derive equivalent circuit of a transformer.
Equivalent circuit of transformer
Resistances and reactances of transformer, which are described above, can be imagined
separately from the windings (as shown in the figure below). Hence, the function of
windings, thereafter, will only be the transforming the voltage.
The no load current I0 is divided into, pure inductance X0 (taking magnetizing components Iμ)
and non induction resistance R0 (taking working component Iw) which are connected into
parallel across the primary. The value of E1 can be obtained by subtracting I1Z1 from V1. The
value of R0 and X0 can be calculated as, R0 = E1 / Iw and X0 = E1 / Iμ.
But, using this equivalent circuit does not simplifies the calculations. To make calculations
simpler, it is preferable to transfer current, voltage and impedance either to primary side or to
the secondary side. In that case, we would have to work with only one winding which is more
convenient.
From the voltage transformation ratio, it is clear that,
E1 / E2 = N1 / N2 = K
Now, lets refer the parameters of secondary side to primary.
Z2 can be referred to primary as Z2'
where, Z2' = (N1/N2)2Z2 = K2Z2. ............where K= N1/N2.
that is, R2'+jX2' = K2(R2+jX2)
equating real and imaginary parts,
R2' = K2R2 and X2' = K2X2
.And V2' = KV2
The following figure shows the equivalent circuit of transformer with secondary
parameters referred to the primary.
Now, as the values of winding resistance and leakage reactance are so small that, V 1 and E1
can be assumed to be equal. Therefore, the exciting current drawn by the parallel combination
of R0 and X0 would not affect significantly, if we move it to the input terminals as shown in
the figure below.
Now, let R1 + R2' = R'eq and X1 + X2' = X'eq
Then the equivalent circuit of transformer becomes as shown in the figure below
Approximate equivalent circuit of transformer
If only voltage regulation is to be calculated, then even the whole excitation branch (parallel
combination of R0 and X0) can be neglected. Then the equivalent circuit becomes as shown
in the figure below
Transformer with resistance and leakage reactance
Magnetic leakage
In a transformer it is observed that, all the flux linked with primary winding does not get
linked with secondary winding. A small part of the flux completes its path through air rather
than through the core (as shown in the fig at right), and this small part of flux is called as
leakage flux or magnetic leakage in transformers. This leakage flux does not link with both
the windings, and hence it does not contribute to transfer of energy from primary winding to
secondary winding. But, it produces self induced emf in each winding. Hence, leakage flux
produces an effect equivalent to an inductive coil in series with each winding. And due to this
there will be leakage reactance.
(To minimize this leakage reactance, primary and secondary windings are not placed on
separate legs, refer the diagram of core type and shell type transformer from construction of
transformer.)
Practical Transformer with resistance and leakage reactance
In the following figure, leakage reactance and resitance of the primary winding as well as
secondary winding are taken out, representing a practical transformer.
Where, R1 and R2 = resistance of primary and secondary winding respectively
X1 and X2 = leakage reactance of primary and secondary winding resp.
Z1 and Z2 = Primary impedance and secondary impedance resp.
Z1 = R1 + jX1 ...and Z2 = R2 + jX 2
.The impedance in each winding lead to some voltage drop in each winding. Considering this
voltage drop the
voltage equation of transformer can be given as
V1 = E1 + I1(R1 + jX1 ) --------primary side
V2 = E2 - I2(R2 + jX2 ) --------secondary side
where, V1 = supply voltage of primary winding
V2 = terminal voltage of secondary winding
E1 and E2 = induced emf in primary and secondary winding respectively. (EMF equation of a
transformer
LECTURER-21
Transformer - Losses and Efficiency
Losses in transformer
In any electrical machine, 'loss' can be defined as the difference between input power and
output power. An electrical transformer is an static device, hence mechanical losses (like
windage or friction losses) are absent in it. A transformer only consists of electrical losses
(iron losses and copper losses). Transformer losses are similar to losses in a DC machine,
except that transformers do not have mechanical losses.
Losses in transformer are explained below (i) Core losses or Iron losses
Eddy current loss and hysteresis loss depend upon the magnetic properties of the material
used for the construction of core. Hence these losses are also known as core losses or iron
losses.

Hysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in
the transformer core. This loss depends upon the volume and grade of the iron,
frequency of magnetic reversals and value of flux density. It can be given by,
Steinmetz formula:
Wh= ηBmax1.6fV (watts)
where, η = Steinmetz hysteresis constant
V = volume of the core in m3

Eddy current loss in transformer: In transformer, AC current is supplied to the
primary winding which sets up alternating magnetizing flux. When this flux links
with secondary winding, it produces induced emf in it. But some part of this flux also
gets linked with other conducting parts like steel core or iron body or the transformer,
which will result in induced emf in those parts, causing small circulating current in
them. This current is called as eddy current. Due to these eddy currents, some energy
will be dissipated in the form of heat.

Eddy current loss= ke f 2B2m t2 watts /m3
(ii) Copper loss in transformer
Copper loss is due to ohmic resistance of the transformer windings. Copper loss for the
primary winding is I12R1 and for secondary winding is I22R2. Where, I1 and I2 are current in
primary and secondary winding respectively, R1 and R2 are the resistances of primary and
secondary winding respectively. It is clear that Cu loss is proportional to square of the
current, and current depends on the load. Hence copper loss in transformer varies with the
load.
Efficiency of Transformer
Just like any other electrical machine, efficiency of a transformer can be defined as the
output power divided by the input power. That is efficiency = output / input .
Transformers are the most highly efficient electrical devices. Most of the transformers have
full load efficiency between 95% to 98.5% . As a transformer being highly efficient, output
and input are having nearly same value, and hence it is impractical to measure the efficiency
of transformer by using output / input. A better method to find efficiency of a transformer is
using, efficiency = (input - losses) / input = 1 - (losses / input).
Efficiency = Output/Input
On specified Power factor and load, the Transformer efficiency can be found by dividing its
output on Input (Similar to other Electrical Machines i.e. motors, generators etc). But the
values of both input and Output should be same in unites (i.e. in Watts, kilowatts, megawatts
etc)
But note that a transformer has very high efficiency because the losses occur in transformer is
very low. Since the Input and Output almost equal, therefore measurement of input and
output is not possible practically. The best way to find the transformer efficiency is that, first
determine the losses in transformer and then calculate the transformer efficiency with the
help of these losses.
Efficiency = η= Output / Input
Efficiency = η= Output / (Output + Losses) …….… (As Input = Output +Losses)
Efficiency = η= Output / (Output +Cupper Losses + Iron Losses)
You may also find the Efficiency by the following formula
Efficiency = η= Output / Input
Efficiency = η = (Input – Losses) / Input …….… (As Output = Input – Losses)
Efficiency = η = 1 – (Losses /Input)
As we know that the rating of Transform is expressed in kVA not in kW. But the efficiency
doesn’t depend on VA i.e. it would be expressed in Power Watts (kW) not in kVA. Although,
the Losses are directly proportional to VA (Volt-Amperes), thus, efficiency depends on
Power factor on every kind of VA load. And the efficiency would be maximum on unity (1)
Power factor.
Condition for Maximum Efficiency of Transformer
We know that,
Copper Loss = WC = I12. R1or I22R2
Iron Loss=Wi = Hysteresis Loss + Eddy Current Loss
W I = WH + WE
Suppose to Primary Side…
Primary Input = P1 = V1I1 Cosθ1
Efficiency = η = Output / Input
Efficiency = η = (Input – Losses) / input ….. (As Output = Input – Losses)
Efficiency = η = (Input – Copper losses – Iron Losses)/Input
Efficiency = η = (P1 - WC - WI) / P1
Efficiency = η = (V1 I1 Cosθ1 - I12. R1 - WI)/ V1 I1 Cosθ1
Taking LCM
Efficiency = η = 1- (I12. R1 /V1I1 Cosθ1) –(WI/ V1 I1Cosθ1)
Or
Efficiency = η = 1- (I1. R1 /V1Cosθ1) – (WI/ V1 I1 Cosθ1)
Differentiate both sides with respect to I1
dη/ dI1 = 0 – ( R1 /V1 Cosθ1) + (WI/V1 I12 Cosθ1)
dη/ dI1= – ( R1 /V1 Cosθ1) + (WI/V1 I12 Cosθ1)
For Maximum Efficiency, the value of (Dη/ dI1) should be Minimum i.e.
dη/ dI1 = 0
The above Equation can be written as
R1 / (V1 Cosθ1) = (WI/V1 I12 Cosθ1)
Or
WI = I12. R1
or
I22R2
Iron Loss = Copper Loss
The value of Output current (I2) on which Maximum efficiency can be gained
I2 = √ (WI/ R2)
The value of Output current (I2) is the actor who equals the value of Copper Loss and Iron
Loss (i.e. Copper Loss = Iron Loss)
Doing so, the maximum efficiency can be gained. Therefore, with proper designing,
maximum efficiency can be attained at any desired load i.e. Copper loss and Iron Loss can be
equaled.
All Day Efficiency of Transformer
As we know that the commercial or typical efficiency of a transformer is the ratio of Output
and Input in watts
Efficiency = Output (in Watts)/Input (in Watts)
But there are number of transformers whose performance can’t be monitored according the
above general efficiency.
Those distribution transformers which supply electrical energy to lighting and other general
circuits, their primary energize for 24 hours, but the secondary windings does not energize all
the time. In other words, Secondary windings only energize at the night time when they
supply electrical energy to lighting circuits. I.e. secondary windings supply eclectic power for
very small load or no load for maximum time in 24 hours. It means that core loss occurs for
24 hours regularly but copper loss occurs only when transformer is on loaded.
Therefore it realizes the necessity to design a transformer in which the core loss should be
low. As copper loss depends on load, therefore, they should be neglected. In this type of
transformers, we can track their performance only by all day efficiency. All day efficiency
may be also called “Operational efficiency”. On the base of usable energy, we estimate the all
day efficiency for a specific time (During the 24 hours =one day). And we can find it by the
following formula
All Day Efficiency = Output (in kWh)/Input (in kWh)
To understand about the all day efficiency, we must know about the load cycle i.e. how much
load is connected, and for how much time (in 24 hours).
LECTURER-22
Three Phase Transformer
Usually power is generated and distributed in three phase system, and therefore it is obvious
that we would need three phase transformers to step up and step down voltages. Although,
it is practically possible to use three suitably interconnected 'single phase transformers'
instead of one 'three phase transformer. A three-phase transformer can be built in two ways
viz., (i) by suitably connecting a bank of three single-phase transformers or (ii) by
constructing a three-phase . transformer on a common magnetic structure. In either case, the
windings may be connected in Y - Y, D - D,
Y - D or D - Y.
(i) Bank of three single-phase transformers
Three similar single-phase transformers can be connected to form a three-phase transformer.
The primary and secondary windings may be connected in star (Y) or delta (D) arrangement.
The above Fig. shows a Y - D connection of a threephase transformer. The primary windings
are connected in star and the secondary windings are connected in delta. The primary and
secondary windings shown parallel to each other belong to the same single-phase
transformer. The ratio of secondary phase voltage to primary phase voltage is the phase
transformation ratio K.
𝑁2
Secondary phase voltage
Phase transformation ratio, K =𝑁1 =
Primary phase voltage
(ii) Three-phase transformer
A three-phase transformer can be constructed by having three primary and three secondary
windings on a common magnetic circuit. The basic principle of a 3- phase transformer is
shown bellow.
The three single-phase coretype transformers, each with windings (primary and secondary)
on only one leg have their unwound legs combined to provide a path for the returning flux.
The primaries as well as secondaries may be connected in star or delta. If the primary is
energized from a 3-phase supply, the central limb (i.e., unwound limb) carries the fluxes
produced by the 3-phase primary windings. Since the phasor sum of three primary currents at
any instant is zero, the sum of three fluxes passing through the central limb must be zero.
Hence no flux exists in the central limb
and it may, therefore, be eliminated. This
modification gives a three leg coretype 3-phase transformer.
In this case, any two legs will act as a return path for the flux in the third leg. For example, if
flux is f in one leg at some instant, then flux is f/2 in the opposite direction through the other
two legs at the same instant. All the connections of a 3-phase transformer are made inside the
case and for delta-connected winding three leads are brought out while for star connected
winding four leads are brought out.
The advantages of three phase transformers are 
One 'three phase transformer' occupies less space than a gang of three 'single phase
transformers'.

Single 'three phase' unit is more economical

The overall bus-bar structure, switchgear and installation of 'three phase transformer'
is simpler.
A disadvantage of the three-phase transformer lies in the fact that when one phase becomes
defective, the entire three-phase unit must be removed from service. When one transformer in
a bank of three single-phase transformers becomes defective, it may be removed from service
and the other two transformers may be reconnected to supply service on an emergency basis
until repairs can be made
LECTURER-23
Three-Phase Transformer Connections
A three-phase transformer can be built by suitably connecting a bank of three single-phase
transformers or by one three-phase transformer. (i) star-star, (ii) delta-delta, (iii) star-delta,
(iv) delta-star, (v) open delta and (vi) Scott connection. These configurations are explained
below. In any of these configurations, there will be a phase difference of 120° between any
two phases
Star-star (Y-Y)

Star-star connection is generally used for small, high-voltage transformers. Because of
star connection, number of required turns/phase is reduced (as phase voltage in star
connection is 1/√3 times of line voltage only). Thus, the amount of insulation required
is also reduced.

The ratio of line voltages on the primary side and the secondary side is equal to the
transformation ratio of the transformers.

Line voltages on both sides are in phase with each other.

This connection can be used only if the connected load is balanced.
Delta-delta (Δ-Δ)

This connection is generally used for large, low-voltage transformers. Number of
required phase/turns is relatively greater than that for star-star connection.

The ratio of line voltages on the primary and the secondary side is equal to the
transformation ratio of the transformers.

This connection can be used even for unbalanced loading.
Another advantage of this type of connection is that even if one transformer is disabled,
system can continue to operate in open delta connection but with reduced available capacity.
Star-delta OR wye-delta (Y-Δ)

The primary winding is star star (Y) connected with grounded neutral and the
secondary winding is delta connected.

This connection is mainly used in step down transformer at the substation end of the
transmission line.

The ratio of secondary to primary line voltage is 1/√3 times the transformation ratio.

There is 30° shift between the primary and secondary line voltages.
Delta-star OR delta-wye (Δ-Y)

The primary winding is connected in delta and the secondary winding is connected in
star with neutral grounded. Thus it can be used to provide 3-phase 4-wire service.

This type of connection is mainly used in step-up transformer at the beginning of
transmission line.

The ratio of secodary to primary line voltage is √3 times the transformation ratio.
There is 30° shift between the primary and secondary line voltages
Open delta (V-V) connection
Two transformers are used and primary and secondary connections are made as shown in the
figure below. Open delta connection can be used when one of the transformers in Δ-Δ bank is
disabled and the service is to be continued until the faulty transformer is repaired or replaced.
It can also be used for small three phase loads where installation of full three transformer
bank is un-necessary. The total load carrying capacity of open delta connection is 57.7% than
that would be for delta-delta connection.
LECTURE-25
Construction of a 3 phase induction motor
Just like any other motor, a 3 phase induction motor also consists of a stator and a rotor.
Basically there are two types of 3 phase IM - 1. Squirrel cage induction motor and 2. Phase
Wound induction motor (slip-ring induction motor). Both types have similar constructed
rotor, but they differ in construction of rotor. This is explained further.
Stator
 It consists of a steel frame which encloses a hollow, cylindrical core made up of thin
laminations of silicon steel to reduce hysteresis and eddy current losses.
 A number of evenly spaced slots are provided on the inner periphery of the
laminations.
 The insulated connected to form a balanced 3-phase star or delta connected circuit.
 The 3-phase stator winding is wound for a definite number of poles as per
requirement of speed. Greater the number of poles, lesser is the speed of the motor
and vice-versa.
 When 3-phase supply is given to the stator winding, a rotating magnetic field of
constant magnitude is produced. This rotating field induces currents in the rotor by
electromagnetic induction.
Rotor
The rotor, mounted on a shaft, is a hollow laminated core having slots on its
outer periphery. The winding placed in these slots (called rotor winding) may be
one of the following two types:
(i) Squirrel cage type (ii) Wound type
Squirrel Cage Rotor
 It consists of a laminated cylindrical core having parallel slots on its outer periphery.
One copper or aluminum bar is placed in each slot.
 All these bars are joined at each end by metal rings called end rings . This forms a
permanently short-circuited winding which is indestructible.
 The entire construction (bars and end rings) resembles a squirrel cage and hence the
name.
 The rotor is not connected electrically to the supply but has current induced in it by
transformer action from the stator. Those induction motors which employ squirrel
cage rotor are called squirrel cage induction motors.
 Most of 3-phase induction motors use squirrel cage rotor as it has a remarkably simple
and robust construction enabling it to operate in the most adverse circumstances.
 However, it suffers from the disadvantage of a low starting torque. It is because the
rotor bars are permanently short-circuited and it is not possible to add any external
resistance to the rotor circuit to have a large starting torque.
Rotor slots are slightly skewed to achieve following advantages -
1. it reduces locking tendency of the rotor, i.e. the tendency of rotor teeth to remain under
stator teeth due to magnetic attraction.
2. increases the effective transformation ratio between stator and rotor
3. increases rotor resistance due to increased length of the rotor conductor
Phase Wound Rotor
 It consists of a laminated cylindrical core and carries a 3- phase winding, similar to
the one on the stator.
 The rotor winding is uniformly distributed in the slots and is usually star connected.
 The open ends of the rotor winding are brought out and joined to three insulated slip
rings mounted on the rotor shaft with one brush resting on each slip ring. The three
brushes are connected to a 3-phase star-connected rheostat.

At starting, the external resistances are included in the rotor circuit to give a large
starting torque. These resistances are gradually reduced to zero as the motor runs up
to speed.
 The external resistances are used during starting period only. When the motor attains
normal speed, the three brushes are short-circuited so that the wound rotor runs like a
squirrel cage rotor.
LECTURE-26
Principle of Operation of Induction motor
Consider a portion of 3-phase induction motor .The operation of the motor can be explained
as under:
(i) When 3-phase stator winding is energized from a 3-phase supply, a rotating magnetic
field is set up which rotates round the stator at synchronous speed Ns (= 120 f/P).
(ii) The rotating field passes through the air gap and cuts the rotor conductors, which as yet,
are stationary. Due to the relative speed between the rotating flux and the stationary rotor,
e.m.f.s are induced in the rotor conductors. Since the rotor circuit is short-circuited, currents
start flowing in the rotor conductors.
(iii) The current-carrying rotor conductors are placed in the magnetic field produced by the
stator. Consequently, mechanical force acts on the rotor conductors. The sum of the
mechanical forces on all the rotor conductors produces a torque which tends to move the
rotor in the same direction as the rotating field which can be explained by Lenz’s law.
According to this law, the direction of rotor currents will be such that they tend to oppose the
cause producing them. Now, the cause producing the rotor currents is the relative speed
between the rotating field and the stationary rotor conductors. Hence to reduce this relative
speed, the rotor starts running in the same direction as that of stator field and tries to catch it.
LECTURE-27
Concept of Slip and theory behind torque-slip characteristics
The difference between the synchronous speed Ns of the rotating stator field and the actual
rotor speed N is called slip. It is usually expressed as a percentage of synchronous speed i.e.,
% age slip, s =
N s−𝑁
Ns
x 100
(i) The quantity Ns - N is sometimes called slip speed.
(ii) When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %.
(iii) In an induction motor, the change in slip from no-load to full-load is hardly 0.1% to 3%
so that it is essentially a constant-speed motor
Rotor Torque
The torque T developed by the rotor is directly proportional to:
(i) rotor current
(ii) rotor e.m.f.
(iii) power factor of the rotor circuit
T ∝ ɸ I2 cosɸ2
OR
T = k ɸ I2 cosɸ2
.
where, ɸ = flux per stator pole,
I2 = rotor current at standstill,
ɸ2 = angle between rotor emf and rotor current,
k = a constant.
Now, let E2 = rotor emf at standstill
we know, rotor emf is directly proportional to flux per stator pole, i.e. E2 ∝ ɸ.
therefore, T ∝ E2 I2 cosɸ2
OR
T =k1 E2 I2 cosɸ2.
Starting torque
The torque developed at the instant of starting of a motor is called as starting torque. Starting
torque may be greater than running torque in some cases, or it may be lesser.
We know, T =k1 E2 I2 cosɸ2.
let, R2 = rotor resistance per phase
X2 = stand still rotor reactance
Rotor impedance/phase
Rotor current/phase
Rotor p.f.,
Therefore, starting torque can be given as,
The constant k1 = 3 / 2πNs
Condition For Maximum Starting Torque
If supply voltage V is kept constant, then flux ɸ and E2 both remains constant. Hence,
Hence, it can be proved that maximum starting torque is obtained when rotor resistance is
equal to standstill rotor reactance. i.e. R22 + X22 =2R22 .
Torque Under Running Condition
T ∝ ɸ Ir cosɸ2 .
where, Er = rotor emf per phase under running condition = sE2. (s=slip)
Ir = rotor current per phase under running condition
reactance per phase under running condition will be = sX2
therefore,
as, ɸ ∝ E2.
LECTURE-31
Split-Phase Induction Motor
 The stator of a split-phase induction motor is provided with an auxiliary or
starting winding S in addition to the main or running winding M.
 The starting winding is located 90° electrical from the main winding and
operates only during the brief period when the motor starts up.
 The two windings are so designed that the starting winding S has a high
resistance and relatively small reactance while the main winding M has
relatively low resistance and large reactance as .Consequently, the currents
flowing in the two windings have reasonable phase difference (25° to 30°)
Operation
(i) When the two stator windings are energized from a single-phase supply, the
main winding carries current Im while the starting winding carries current Is.
(ii) Since main winding is made highly inductive while the starting winding
highly resistive, the currents Im and Is have a reasonable phase angle a (25° to 30°)
between them . Consequently, a weak revolving field approximating to that of a 2phase machine is produced which starts the motor. The starting torque is given by;
Ts= K Im Is sinα
where k is a constant whose magnitude depends upon the design of the motor.
(i)
When the motor reaches about 75% of synchronous speed, the centrifugal
switch opens the circuit of the starting winding. The motor then operates
as a single-phase induction motor and continues to accelerate till it reaches
the normal speed. The normal speed of the motor is below the synchronous
speed and depends upon the load on the motor.
Characteristics
(i) The starting torque is 1.5 to 2 times the full-loud torque mid (lie starting
current is 6 to 8 times the full-load current.
(ii) Due to their low cost, split-phase induction motors are most popular singlephase motors in the market.
(iii) Since the starting winding is made of fine wire, the current density is high
and the winding heats up quickly. If the starting period exceeds 5 seconds, the
winding may burn out unless the motor is protected by built-in-thermal relay. This
motor is, therefore, suitable where starting periods are not frequent.
(iv) An important characteristic of these motors is that they are essentially
constant-speed motors. The speed variation is 2-5% from no-load to full- load.
(v) These motors are suitable where a moderate starting torque is required and
where starting periods are infrequent e.g., to drive: (a) fans (b) washing machines
(c) oil burners (d) small machine tools etc.
The power rating of such motors generally lies between 60 W and 250 W.
Capacitor-Start Motor
 The capacitor-start motor is identical to a split-phase motor except that the
starting winding has as many turns as the main winding.
 Moreover, a capacitor C is connected in series with the starting winding.
The value of capacitor is so chosen that Is leads Im by about 80° (i.e., α ~
80°) which is considerably greater than 25° found in split-phase motor.

Consequently, starting torque (Ts = k Im Is sin α) is much more than that
of a split-phase motor Again, the starting winding is opened by the
centrifugal switch when the motor attains about 75% of synchronous
speed. The motor then operates as a single-phase induction motor and
continues to accelerate till it reaches the normal speed.
Characteristics
(i) Although starting characteristics of a capacitor-start motor are better than
those of a split-phase motor, both machines possess the same running
characteristics because the main windings are identical.
(ii) The phase angle between the two currents is about 80° compared to about 25°
in a split-phase motor. Consequently, for the same starting torque, the current in
the starting winding is only about half that in a split-phase motor. Therefore, the
starting winding of a capacitor start motor heats up less quickly and is well suited
to applications involving either frequent or prolonged starting periods.
Capacitor-Start Capacitor-Run Motor
This motor is identical to a capacitor-start motor except that starting winding is
not opened after starting so that both the windings remain connected to the supply
when running as well as at starting. Two designs are generally used.
(i)
In one design, a single capacitor C is used for both starting and running.
This design eliminates the need of a centrifugal switch and at the same
time improves the power factor and efficiency of the motor.
(ii)
In the other design, two capacitors C1 and C2 are used in the starting
winding . The smaller capacitor C1 required for optimum running
conditions is permanently connected in series with the starting winding.
The much larger capacitor C2 is connected in parallel with C1 for
optimum starting and remains in the circuit during starting. The starting
capacitor C1 is disconnected when the motor approaches about 75% of
synchronous speed. The motor then runs as a single-phase induction
motor.
Characteristics
(i) The starting winding and the capacitor can be designed for perfect 2-phase
operation at any load. The motor then produces a constant torque and not a
pulsating torque as in other single-phase motors.
(ii) Because of constant torque, the motor is vibration free and can be used in:
(a) hospitals (6) studios and (c) other places where silence is important.
Shaded-Pole Motor
 The shaded-pole motor is very popular for ratings below 0.05 H.P.
because of its extremely simple construction.
 It has salient poles on the stator excited by single-phase supply and a
squirrel- cage rotor .
 A portion of each pole is surrounded by a short-circuited turn of copper
strip called shading coil.
Operation
(i) The operation of the motor can be understood by referring to Fig. (i) which shows one
pole of the motor with a shading coil. During the portion OA of the alternatingcurrent cycle , the flux begins to increase and an e.m.f. is induced in the shading coil.
The resulting current in the shading coil will be in such a direction (Lenz’s law) so as
to oppose the change in flux. Thus the flux in the shaded portion of the pole is
weakened while that in the unshaded portion is strengthened as shown in fig(ii)
(ii) During the portion AB of the alternating-current cycle, the flux has reached almost
maximum value
and is not changing. Consequently, the flux distribution across the pole is
uniform [Fig. (iii)] since no current is flowing in the shading coil. As the flux decreases
(portion BC of the alternating current cycle), current is induced in the shading coil so as to
oppose the decrease in current. Thus the flux in the shaded portion of the pole is strengthened
while that in the unshaded portion is weakened as shown in Fig. (iv).
(iii) The effect of the shading coil is to cause the field flux to shift across the pole face from
the unshaded to the shaded portion. This shifting flux is like a rotating weak field moving in
the direction from unshaded portion to the shaded portion of the pole. The rotor is of the
squirrel-cage type and is under the influence of this moving field. Consequently, a small
starting torque is developed. As soon as this torque starts to revolve the rotor, additional
torque is produced by single-phase induction-motor action. The motor accelerates to a speed
slightly below the synchronous speed and runs as a single-phase induction motor.
Characteristics
(i) The salient features of this motor are extremely simple construction
and absence of centrifugal switch.
(ii) Since starting torque, efficiency and power factor are very low, these
motors are only suitable for low power applications e.g., to drive:
(a) small fans (6) toys (c) hair driers (d) desk fans etc.
The power rating of such motors is upto about 30 W
LECTURE-32
Single-Phase Motors
As the name suggests, these motors are used on single-phase supply. Single- phase motors are
the most familiar of all electric motors because they are extensively used in home appliances,
shops, offices etc.
Types of Single-Phase Motors
Single-phase motors are generally built in the fractional-horsepower range and may be
classified into the following four basic types:
1. Single-phase induction motors
(i) split-phase type,
(ii) capacitor type,
(iii) shaded-pole type
2. A.C. series motor or universal motor
3. Repulsion motors
(i) Repulsion-start induction-run motor
(ii) Repulsion-induction motor
4. Synchronous motors
(i) Reluctance motor
(ii) Hysteresis motor
Single-Phase Induction Motors
A single phase induction motor is very similar to a 3-phase squirrel cage induction motor. It
has
(i) a squirrel-cage rotor identical to a 3-phase motor and
(ii) a single-phase winding on the stator

The single-phase stator winding produces a magnetic field that pulsates in strength in
a sinusoidal manner.

The field polarity reverses after each half cycle but the field does not rotate.
Consequently, the alternating flux cannot produce rotation in a stationary squirrelcage rotor.

However, if the rotor of a single-phase motor is rotated in one direction by some
mechanical means, it will continue to run in the direction of rotation. As a matter of
fact, the rotor quickly accelerates until it reaches a speed slightly below the
synchronous speed.

Once the motor is running at this speed, it will continue to rotate even though singlephase current is flowing through the stator winding.
Double-Field Revolving Theory
 The double-field revolving theory is proposed to explain this dilemma of no
torque at start and yet torque once rotated.
 This theory is based on the fact that an alternating sinusoidal flux (ɸ = ɸm cos
ωt) can be represented by two revolving fluxes, each equal to one-half of the
maximum value of alternating flux (i.e.,
ɸm/2 ) and each rotating a
synchronous speed (Ns = 120 f/P, ω = 2∏f) in opposite directions.
 The instantaneous value of flux due to the stator current of a single-phase
induction motor is given by; ɸ = ɸm cos ωt
 Consider two rotating magnetic fluxes ɸ 1 and ɸ2 each of magnitude ɸ m/2 and
rotating in opposite directions with angular velocity ω .
Let the two fluxes start rotating from OX axis at t = 0. After time t seconds, the angle
through which the flux vectors have rotated is at. Resolving the flux vectors along-Xaxis and Y-axis, we have,
Total X-component=
Total Y-component=
ɸm
2
ɸm
2
ɸm
cos ωt +
sin ωt -
2
ɸm
2
cos ωt= ɸm cos ωt
sin ωt= 0
So resultant flux= ɸm cos ωt
Thus the resultant flux vector is ɸm cos ωt along X-axis. Therefore, an alternating
field can be replaced by two relating fields of half its amplitude rotating in opposite
directions at synchronous speed.
The operation of single-phase induction motor by double-field revolving theory
(ii)
Rotor at standstill
Consider the case that the rotor is stationary and the stator winding is connected to
a single-phase supply. The alternating flux produced by the stator winding can be
presented as the sum of two rotating fluxes f1 and f2, each equal to one half of the
maximum value of alternating flux and each rotating at synchronous speed (Ns =
120 f/P) in opposite directions
Let the flux
ɸ1 rotate in anti clockwise direction and flux ɸ2 in clockwise direction. The flux ɸ1
will result in the production of torque T1 in the anti clockwise direction and flux
ɸ2 will result in the production of torque T2 In the clockwise direction. At
standstill, these two torques are equal and opposite and the net torque developed is
zero. Therefore, single-phase induction motor is not self-starting.
(iii)
Rotor running
Now assume that the rotor is started by spinning the rotor or by using auxiliary
circuit, in say clockwise direction. The flux rotating in the clockwise direction is
the forward rotating flux (ɸf) and that in the other direction is the backward
rotating flux (ɸb). The slip w.r.t. the forward flux will be
sf =
N s−𝑁
Ns
=s
where Ns = synchronous speed
N = speed of rotor in the direction of forward flux
The rotor rotates opposite to the rotation of the backward flux. Therefore, the
slip w.r.t. the backward flux will be
sb =
N s−(−𝑁)
Ns
=
N s+𝑁)
Ns
=
2N s−N s+𝑁)
Ns
2N s
=Ns −
(N s−𝑁)
Ns
= 2-s
so; sb= 2-s
Thus for forward rotating flux, slip is s (less than unity) and for backward rotating
flux, the slip is 2 - s (greater than unity). Since for usual rotor resistance/reactance
ratios, the torques at slips of less than unity arc greater than those at slips of more
than unity, the resultant torque will be in the direction of the rotation of the
forward flux. Thus if the motor is once started, it will develop net torque in the
direction in which it has been started and will function as a motor.
Applications
Capacitor-start motors are used where high starting torque is required and where
the starting period may be long e.g., to drive: (a) compressors (b) large fans (c)
pumps (d) high inertia loads
The power rating of such motors lies between 120 W and 7-5 kW.