The document describes the key characteristics and equations of an ideal transformer and how they differ from a practical transformer. 1) An ideal transformer has no winding resistance, no leakage flux, and no core losses. It transfers power efficiently between its primary and secondary windings according to turns ratio. 2) A practical transformer has winding resistance, leakage flux, and core losses that reduce its efficiency compared to an ideal model. It draws a small magnetizing current to produce flux and a power loss component to account for iron losses. 3) Faraday's law of induction and Lenz's law explain how an alternating voltage is induced in transformer windings via a changing magnetic flux. In practice, some flux leaks and does not

1. 1
EEE-233
Electrical Machines-1
Lecture 5-14

2. 2
Ideal Transformer
• In order to have better understanding of the behavior of the transformer, initially
certain idealizations are made and the resulting ‘ideal’ transformer is studied. These
idealizations are as follows:
• 1. Magnetic circuit is linear and has infinite permeability. The consequence is that a
vanishingly small current is enough to establish the given flux. Hysteresis loss is
negligible. As all the flux generated confines itself to the iron, there is no leakage flux.
• 2. Windings do not have resistance. This means that there are no copper losses,
nor there is any ohmic drop in the electric circuit.
• In fact the practical transformers are very close to this model and hence no major
departure is made in making these assumptions.
 Assumptions of an Ideal Transformer

3. 3
Ideal Transformer
• No winding resistance, no leakage flux and no iron losses in the core.
• 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°.
• 𝑰𝒎 produces an alternating flux Φ which is proportional to and in phase with it.
• The alternating flux Φ induces e.m.f. 𝑬𝟏 in the primary and e.m.f. 𝑬𝟐 in the
secondary.
• 𝑬𝟏 is equal to and in opposition to V1 (Lenz’s law) and both 𝑬𝟏and 𝑬𝟐 lag behind
flux Φ by 90°.

4. 4
Ideal Transformer
 E.M.F EQUATION OF AN IDEAL TRANSFORMER
• Let, an alternating voltage V1 of frequency f is applied to the primary
• The sinusoidal flux, 𝝓 produced by the primary can be represented as:
𝝓 = 𝝓𝒎𝒔𝒊𝒏𝝎𝒕 𝒘𝒆𝒃𝒆𝒓
• The induced emf in primary winding is:
𝒆𝟏 = −𝑵𝟏
𝒅𝝓
𝒅𝒕
= −𝑵𝟏𝝓𝒎𝝎𝒄𝒐𝒔𝝎𝒕 = 𝑵𝟏𝝓𝒎𝝎𝒔𝒊𝒏 𝝎𝒕 − 𝟗𝟎𝟎
Where, N1 is the number of turns in primary winding, Φm, the maximum (peak) flux,
and f the frequency of the supply voltage.
• It is clear from above equation that e.m.f. e1 induced in the primary lags behind
the flux Φ by 90°
• Here, The maximum value of the induced e.m.f in primary winding is:
𝑬𝒎𝟏 = 𝑵𝟏𝝓𝒎𝝎 = 𝟐𝝅𝒇𝑵𝟏𝝓𝒎

5. 5
Ideal Transformer
• The r.m.s value of the induced e.m.f in primary winding is:
𝑬𝟏 =
𝑬𝒎𝟏
𝟐
=
𝟐𝝅𝒇𝑵𝟏𝝓𝒎
𝟐
𝟒. 𝟒𝟒𝑵𝟏𝝓𝒎𝒇
• Similarly, r.m.s value of the induced e.m.f in secondary winding,
𝑬𝟐 = 𝟒. 𝟒𝟒𝑵𝟐𝝓𝒎𝒇
Where, 𝑵𝟐 is the number of turns in secondary winding.
From the induced e.m.f equations
∴
𝑬𝟏
𝑬𝟐
=
𝑵𝟏
𝑵𝟐
= 𝒂 = 𝑻𝒖𝒓𝒏𝒔 𝑹𝒂𝒕𝒊𝒐
If, 𝒂 > 𝟏, 𝑺𝒕𝒆𝒑 𝒅𝒐𝒘𝒏 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒆𝒓
𝒂 < 𝟏, 𝑺𝒕𝒆𝒑 𝒖𝒑 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒆𝒓
 E.M.F EQUATION OF AN IDEAL TRANSFORMER (Contd.)

6. 6
Ideal Transformer
• For an ideal transformer,
𝑬𝟏 = 𝑽𝟏 and 𝑬𝟐 = 𝑽𝟐
as there is no voltage drop in the windings
∴
𝑬𝟏
𝑬𝟐
=
𝑽𝟏
𝑽𝟐
=
𝑵𝟏
𝑵𝟐
= 𝒂
• As there are no losses the volt-amperes input
to the primary are equal to the output volt-
amperes
i.e. 𝑽𝟏𝑰𝟏 = 𝑽𝟐𝑰𝟐
∴
𝑰𝟐
𝑰𝟏
=
𝑽𝟏
𝑽𝟐
= 𝒂
 E.M.F EQUATION OF AN IDEAL TRANSFORMER (Contd.)

7. 7
Winding Polarity
 THE DOT CONVENTION
• If the primary voltage is positive at the dotted end of the winding with respect to
the undotted end, then the secondary voltage will be positive at the dotted end also.
Voltage polarities are the same with respect to the dots on each side of the core.
• If the primary current of the transformer flows into the dotted end of the primary
winding, the secondary current will flow out of the dotted end of the secondary
winding.

8. 8
• The real power 𝑷𝒊𝒏 supplied to the transformer by the primary circuit is given by
the equation
𝑷𝒊𝒏 = 𝑽𝟏𝑰𝟏𝒄𝒐𝒔𝜽𝟏
where 𝜽𝟏 is the angle between the primary voltage and the primary current.
• The real power 𝑷𝒐𝒖𝒕 supplied by the transformer secondary circuit to its loads is
given by the equation
𝑷𝒐𝒖𝒕 = 𝑽𝟐𝑰𝟐𝒄𝒐𝒔𝜽𝟐
where 𝜽𝟐 is the angle between the secondary voltage and the secondary current.
• Since voltage and current angles are unaffected by an ideal transformer,
𝜽𝟏 = 𝜽𝟐 = 𝜽
• The primary and secondary windings of an ideal transformer have the same
power factor.
Power in an Ideal Transformer
 POWER FACTOR

9. 9
• The output power of a transformer is:
𝑷𝒐𝒖𝒕 = 𝑽𝟐𝑰𝟐𝒄𝒐𝒔𝜽
• Applying the turns-ratio equations gives 𝑽𝟐 = 𝑽𝟏 𝒂 and 𝑰𝟐 = 𝑰𝟏𝒂,
𝑷𝒐𝒖𝒕 = 𝑽𝟏 𝒂 𝑰𝟏𝒂 𝒄𝒐𝒔𝜽
𝑷𝒐𝒖𝒕 = 𝑽𝟏𝑰𝟏𝒄𝒐𝒔𝜽 = 𝑷𝒊𝒏
• Thus, the output power of an ideal transformer is equal to its input power.
• The same relationship applies to reactive power Q and apparent power S:
𝑸𝒊𝒏 = 𝑽𝟏𝑰𝟏𝒔𝒊𝒏𝜽 = 𝑽𝟐𝑰𝟐𝒔𝒊𝒏𝜽 = 𝑸𝒐𝒖𝒕
and
𝑺𝒊𝒏 = 𝑽𝟏𝑰𝟏 = 𝑽𝟐𝑰𝟐 = 𝑺𝒐𝒖𝒕
Power in an Ideal Transformer
 COMPARISON OF INPUT AND OUTPUT POWER

10. 10
• The losses that occur in real transformers have to be accounted for in any
accurate model of transformer behavior. The major items to be considered in the
construction of such a model are
• Iron losses: Iron losses depend upon the supply frequency, core volume and
maximum flux density in the core. Magnitude of iron losses is quite small in a
practical transformer. It has following two components:
i. Eddy current losses. Eddy current losses are resistive heating losses in the
core of the transformer. They are proportional to the square of the voltage applied
to the transformer.
ii. Hysteresis losses. Hysteresis losses are associated with the rearrangement of
the magnetic domains in the core during each half-cycle. They are a complex,
nonlinear function of the voltage applied to the transformer.
Practical Transformer

11. 11
• Copper (PR) losses. When current flows through the windings, there will be
resistive heating losses, also known as copper losses, in the primary and
secondary windings of the transformer. They are proportional to the square of the
current in the windings.
• There will be a loss in voltage due to IR drop in the windings. They are
represented by resistances R1 and R2 in series with the primary and secondary
windings. Hence, E1 will be less than V1 whereas V2 will be less than E2.
Practical Transformer
• Leakage flux. Primary current produces some flux Φ1 not linked with the
secondary winding. Secondary current produces some flux Φ2 not linked with the
primary winding. The fluxes Φ1, and Φ2 which escape the core and pass through
only one of the transformer windings are leakage fluxes. The primary and
secondary leakage flux Φ1, Φ2 introduce inductive reactances X1 and X2 in series
with the primary and secondary windings. The effects of this leakage reactances
must be accounted for.

12. 12
Practical Transformer

13. 13
Practical Transformer
• At no load, the primary draws a small no-load
current, 𝑰𝒐. It has two components:
• Magnetization current 𝑰𝒎(or 𝑰𝝁): The magnetization
current 𝑰𝒎 is a current proportional (in the unsaturated
region) to the voltage applied to the core and lagging
the applied voltage by 90°. It is required to set-up the
magnetic field (or the flux in the iron core, Φ). This can
be modeled by a reactance 𝑿𝑴 (or 𝑿𝒐) connected
across the primary voltage source.
• Active (working) or core loss component, 𝑰𝑾 (or 𝑰𝑪):
This component of current is in phase with the applied
voltage 𝑽𝟏 and it mainly accounts for iron loss
(hysteresis and eddy current losses) and small
quantity of primary Cu-loss. This can be modeled by a
resistance 𝑹𝑪 (or 𝑹𝒐) connected across the primary
voltage source.
 Transformer on No Load

14. 14
Practical Transformer
• Total No-load current, 𝑰𝒐 is phasor sum of 𝑰𝒎 and 𝑰𝑾.
• Hence 𝑰𝒐 = 𝑰𝒎
𝟐 + 𝑰𝑾
𝟐
• Where, 𝑰𝒎 = 𝑰𝒐𝒔𝒊𝒏∅𝒐 and 𝑰𝑾 = 𝑰𝒐𝒄𝒐𝒔∅𝒐
• No load PF, 𝒄𝒐𝒔∅𝒐 =
𝑰𝒘
𝑰𝒎
• Hence, primary no load current 𝑰𝒐 is not 90° behind the applied voltage 𝑽𝟏 but
lags it by an angle ∅𝒐 < 𝟗𝟎𝒐 .
• No load primary copper loss (i.e. 𝑰𝒐
𝟐
𝑹𝟏), being very small, may be neglected,
hence:
No load input power, 𝑾𝒐 = 𝑽𝟏𝑰𝒐𝒄𝒐𝒔∅𝒐 = 𝑰𝒓𝒐𝒏 𝒍𝒐𝒔𝒔
• At no load, there is no secondary current so 𝑽𝟐 = 𝑬𝟐 and on the primary side,
the drops in 𝑹𝟏 and 𝑿𝟏are very small, hence 𝑽𝟏 = 𝑬𝟏.
 Transformer on No Load (contd,)

15. 15
• According to Faraday’s law of
electromagnetic induction,
𝒆𝒊𝒏𝒅 = −𝑵
𝒅𝝓
𝒅𝒕
(1)
Where, 𝒆𝒊𝒏𝒅 = voltage induced in the coil,
𝑵 = number of turns of wire in coil, 𝝓 =
flux passing through coil.
• The minus sign in the equations is an
expression of Lenz’s law. Lenz’s law states
that the direction of the voltage buildup in
the coil is such that if the coil ends were
short circuited, it would produce current
that would cause a flux opposing the
original flux change. Since the induced
voltage opposes the change that causes it,
a minus sign is included in the equation.
Practical Transformer
 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS
Figure: The meaning of Lenz’s
law: (a) A coil enclosing an
increasing magnetic flux; (b)
determining the resulting voltage
polarity.

16. 16
• Equation (1) assumes that exactly the same flux is present in each turn of the
coil. Unfortunately, the flux leakage out of the core into the surrounding air
prevents this from being true.
• Considering if leakage is quite high or if extreme accuracy is required, the
magnitude of the voltage in the i-th turn of the coil is given by,
𝒆𝒊 =
𝒅 𝝓𝒊
𝒅𝒕
--------(2)
Practical Transformer
 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS (Countd.)

17. 17
• If there are 𝑵 turns in the coil of wire, the total voltage on the coil is,
𝒆𝒊𝒏𝒅 = 𝒊=𝟏
𝑵
𝒆𝒊 = 𝒊=𝟏
𝑵 𝒅 𝝓𝒊
𝒅𝒕
=
𝒅
𝒅𝒕 𝒊=𝟏
𝑵
𝝓𝒊 --------(3)
• The term in parentheses in eqn (3) is called the flux linkage 𝝀 of the coil, and
Faraday’s law can be rewritten in terms of flux linkage as,
𝒆𝒊𝒏𝒅 =
𝒅𝝀
𝒅𝒕
-------(4)
Where, 𝝀 = 𝒊=𝟏
𝑵
𝝓𝒊 -------(5); the unit of flux linkage is weber-turns.
• Now, the average flux per turn is given by
𝝓 =
𝝀
𝑵
-------(6); and 𝒆𝒊𝒏𝒅 = 𝑵
𝒅𝝓
𝒅𝒕
-------(7)
Practical Transformer
 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS (Countd.)

18. 18
• The flux in the primary coil of the transformer can thus be divided into two
components:
i. a mutual flux, which remains in the core and links both windings, and
ii. a small leakage flux, which passes through the primary winding but returns through
the air, bypassing the secondary winding:
𝝓𝑷 = 𝝓𝑴 + 𝝓𝑳𝑷
• Where, 𝝓𝑷 = total average primary flux, 𝝓𝑴 = flux component linking both primary
and secondary coils, 𝝓𝑳𝑷 = primary leakage flux
Similarly, The flux in the secondary coil of the transformer can be divided as
𝝓𝑺 = 𝝓𝑴 + 𝝓𝑳𝑺
• Where, 𝝓𝑺 = total average secondary flux, 𝝓𝑴 = flux component linking both
primary and secondary coils, 𝝓𝑳𝑺 = secondary leakage flux
 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS (Countd.)
Practical Transformer

19. 19
• Faraday’s law for the primary circuit can be re-expressed as,
𝒗𝒑 𝒕 = 𝑵𝑷
𝒅𝝓𝑷
𝒅𝒕
= 𝑵𝑷
𝒅𝝓𝑴
𝒅𝒕
+ 𝑵𝑷
𝒅𝝓𝑳𝑷
𝒅𝒕
= 𝒆𝑷 𝒕 + 𝒆𝑳𝑷 𝒕
• The voltage on the secondary coil of the transformer can also be expressed in
terms of Faraday’s law as,
𝒗𝑺 𝒕 = 𝑵𝑺
𝒅𝝓𝑺
𝒅𝒕
= 𝑵𝑺
𝒅𝝓𝑴
𝒅𝒕
+ 𝑵𝑺
𝒅𝝓𝑳𝑺
𝒅𝒕
= 𝒆𝑺 𝒕 + 𝒆𝑳𝑺 𝒕
• From these two relationship,
𝒆𝑷 𝒕
𝑵𝑷
=
𝒆𝑺 𝒕
𝑵𝑺
=
𝒅𝝓𝑴
𝒅𝒕
∴
𝒆𝑷 𝒕
𝒆𝑺 𝒕
=
𝑵𝑷
𝑵𝑺
= 𝒂
Practical Transformer
 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS (Countd.)

20. 20
• Since much of the leakage flux is through air, and since air has a constant
reluctance than the core reluctance, the flux 𝝓𝑳𝑷 and 𝝓𝑳𝑺 are directly proportional to
the primary current 𝒊𝑷 𝒕 and secondary current 𝒊𝑺 𝒕 respectively.
𝝓𝑳𝑷 =
𝑵𝑷 𝒊𝑷
𝓡
= 𝓟𝑵𝑷 𝒊𝑷
𝝓𝑳𝑺 =
𝑵𝑺 𝒊𝑺
𝓡
= 𝓟𝑵𝑺 𝒊𝑺
Where, 𝓟 = permeance of flux path
𝒆𝑳𝑷 𝒕 = 𝑵𝑷
𝒅𝝓𝑳𝑷
𝒅𝒕
= 𝑵𝑷
𝒅
𝒅𝒕
𝓟𝑵𝑷 𝒊𝑷 = 𝑵𝑷
𝟐
𝓟
𝒅 𝒊𝑷
𝒅𝒕
= 𝑳𝑷
𝒅 𝒊𝑷
𝒅𝒕
𝒆𝑳𝑺 𝒕 = 𝑵𝑺
𝒅𝝓𝑳𝑺
𝒅𝒕
= 𝑵𝑺
𝒅
𝒅𝒕
𝓟𝑵𝑺 𝒊𝑺 = 𝑵𝑺
𝟐
𝓟
𝒅 𝒊𝑺
𝒅𝒕
= 𝑳𝑺
𝒅 𝒊𝑺
𝒅𝒕
Where, 𝑳𝑷 = 𝑵𝑷
𝟐
𝓟 is the leakage inductance of the primary coil and 𝑳𝑺 = 𝑵𝑺
𝟐
𝓟 is
the leakage inductance of the secondary coil. Therefore, the leakage flux will be
modeled by primary and secondary inductors.
Practical Transformer
 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS (Countd.)

21. 21
Loading of Transformer
• At no load, when an alternating voltage V1 is applied to the primary of a
transformer, it draws a small magnetizing current Im which lags behind the applied
voltage by 90°.
• Im produces an alternating flux Φ which is proportional to and in phase with it.
• The alternating flux Φ induces e.m.f. E1 in the primary and e.m.f. E2 in the
secondary.
• E1 is equal to and in opposition to V1 (Lenz’s law) and both E1 and E2 lag behind
flux Φ by 90°.
 IDEAL TRANSFORMER ON LOAD

22. 22
Loading of Transformer
• When load is connected to the transformer, the
secondary e.m.f. 𝑬𝟐 will cause a current 𝑰𝟐 to
flow through the load.
𝑰𝟐 =
𝑬𝟐
𝒁𝑳
=
𝑽𝟐
𝒁𝑳
• 𝑰𝟐 sets up an m.m.f. 𝑵𝟐𝑰𝟐 which produces a
flux 𝝓𝟐, in the opposite direction to the main flux
𝝓 (set up in the primary by the magnetizing
current), changing the main flux in the core and
back e.m.f 𝑬𝟏 tends to be reduced.
 IDEAL TRANSFORMER ON LOAD (CONTD.)

23. 23
Loading of Transformer
• Momentarily the 𝑽𝟏 gains upper hand over 𝑬𝟏
and causes more current to flow in the primary
and the primary develops an m.m.f which exactly
counterbalances the secondary m.m.f. 𝑵𝟐𝑰𝟐.
• The primary current 𝑰𝟏flows such that:
𝑵𝟏𝑰𝟏 = 𝑵𝟐𝑰𝟐
𝑰𝟐
𝑰𝟏
=
𝑵𝟏
𝑵𝟐
= 𝒂
• Thus as the secondary current increases, the
primary current 𝑰𝟏 also increases to keep the
mutual flux 𝝓 constant. The power input
automatically increases with increase in output.
 IDEAL TRANSFORMER ON LOAD (CONTD.)

24. 24
Loading of Transformer
 Ideal Transformer on Load (Contd.)
Phasor Diagram
• The value of ‘𝒂’ has been assumed unity so
that primary phasors are equal to secondary
phasors.
• Secondary current 𝑰𝟐 lags behind 𝑽𝟐 (or 𝑬𝟐)
by 𝝋𝟏 causing a primary current 𝑰𝟏 = 𝑰𝟐 𝒂
which is in anti-phase with it.
i. 𝛗𝟏 = 𝛗𝟐 or, 𝐜𝐨𝐬𝛗𝟏 = 𝐜𝐨𝐬𝛗𝟐
ii. Since there are no losses in an ideal
transformer, input primary power is equal to
the secondary output power i.e.,
𝑽𝟏𝑰𝟏𝒄𝒐𝒔𝝋𝟏 = 𝑽𝟐𝑰𝟐𝒄𝒐𝒔𝝋𝟐

25. 25
Loading of Transformer
1. Transformer With No Winding Resistance and Leakage Reactance
• With this assumption, 𝑽𝟐 = 𝑬𝟐 and 𝑽𝟏 = 𝑬𝟏,
• Take the case of inductive load causing the secondary current 𝑰𝟐 to lag the
secondary voltage 𝑽𝟐 by 𝝓𝟐.
• The no-load current 𝑰𝟎 to meet the iron losses in the transformer and provides flux
in the core.
 Practical Transformer on Load

26. 26
Loading of Transformer
1. Transformer with No Winding Resistance and Leakage Reactance (contd.)
• The total primary current 𝑰𝟏 must supply:
a) The no-load current 𝑰𝟎 to meet the iron losses in the transformer and provides
flux in the core.
b) A current 𝑰𝟐
′
to counteract the demagnetizing effect of secondary current 𝑰𝟐 and
magnitude of 𝑰𝟐
′
will be such that:
𝑵𝟏𝑰𝟐
′
= 𝑵𝟐𝑰𝟐
𝑰𝟐
𝑰𝟐
′ =
𝑵𝟏
𝑵𝟐
= 𝒂
• The total primary current I1 is the phasor sum of I'2 and I0 i.e.,
𝑰𝟏 = 𝑰𝟐
′
+ 𝑰𝟎; where, 𝑰𝟐
′
= − 𝑰𝟐 𝒂
• Current 𝑰𝟐
′
is 180° out of phase with 𝑰𝟐.
 Practical Transformer on Load (Contd.)

27. 27
Loading of Transformer
1. Transformer With No Winding Resistance and Leakage Reactance (contd.)
Phasor diagram
 Practical Transformer on Load (Contd.)
𝒂=1
• Both E1 and E2 lag behind the mutual flux Φ by 90°.
• The primary current 𝑰𝟐
′
neutralizes the
demagnetizing effect of secondary current 𝑰𝟐.
• 𝑰𝟐
′
= − 𝑰𝟐 𝒂 and is anti-phase with 𝑰𝟐.
• The phasor sum of 𝑰𝟐
′
and no load current 𝑰𝟎 gives
the total primary current 𝑰𝟏.
• The value of a is assumed to be unity
• Primary p.f. = 𝒄𝒐𝒔𝝋𝟏
• Secondary p.f. = 𝒄𝒐𝒔𝝋𝟐
• Primary input power = 𝑽𝟏𝑰𝟏𝒄𝒐𝒔𝝋𝟏
• Secondary output power = 𝑽𝟐𝑰𝟐𝒄𝒐𝒔𝝋𝟐

28. 28
Loading of Transformer
2. Transformer with Resistance and Leakage Reactance
• Due to voltage drop in 𝑹𝟏 and 𝑿𝟏 the primary e.m.f. 𝑬𝟏 is less than 𝑽𝟏.
• Due to voltage drop in 𝑹𝟐 and 𝑿𝟐 the secondary terminal voltage 𝑽𝟐 is less than
the secondary e.m.f. 𝑬𝟐.
• The inductive load causes the secondary current 𝑰𝟐 to lag behind the secondary
voltage 𝑽𝟐 by 𝝓𝟐.
 Practical Transformer on Load (Contd.)

29. 29
Loading of Transformer
2. Transformer with Resistance and Leakage Reactance (contd.)
• The total primary current I1 must meet two requirements:
a) Supply the no-load current 𝑰𝟎 to meet the iron losses in the transformer and to
provide flux in the core.
b) Supply a current 𝑰𝟐
′
to counteract the demagnetizing effect of secondary current
𝑰𝟐.
• The magnitude of I'2 will be such that:
𝑵𝟏𝑰𝟐
′
= 𝑵𝟐𝑰𝟐
𝑰𝟐
𝑰𝟐
′ =
𝑵𝟏
𝑵𝟐
= 𝒂
• The total primary current 𝑰𝟏 will be the phasor sum of 𝑰𝟐
′
and 𝑰𝟎
𝑰𝟏 = 𝑰𝟐
′
+ 𝑰𝟎; where, 𝑰𝟐
′
= − 𝑰𝟐 𝒂
𝑽𝟏 = −𝑬𝟏 + 𝑰𝟏 𝑹𝟏 + 𝒋𝑿𝟏 = −𝑬𝟏 + 𝑰𝟏𝒁𝟏; where, 𝑰𝟏 = 𝑰𝟎 + − 𝑰𝟐 𝒂
𝑽𝟐 = 𝑬𝟐 − 𝑰𝟐 𝑹𝟐 + 𝒋𝑿𝟐 = 𝑬𝟐 − 𝑰𝟐𝒁𝟐
 Practical Transformer on Load (Contd.)

30. 30
Loading of Transformer
2. Transformer with Resistance and Leakage Reactance (contd.)
Phasor diagram
 Practical Transformer on Load (Contd.)
• Both E1 and E2 lag the mutual flux ϕ by 90°.
• The primary current I'2 neutralizes the demagnetizing effect
of secondary current I2.
• I'2 = I2/a and is opposite to I2.
• The phasor sum of I'2 and no load current I0 is the total
primary current I1.
• Back e.m.f. E1 opposes the applied voltage V1.
• If I1R1 (in phase with I1) and I1X1 (90° ahead of I1) is added to
- E1, we get the applied primary voltage V1.
• The phasor E2 represents the induced e.m.f. in the
secondary by the mutual flux Φ.
• The secondary terminal voltage V2 is shown after
subtracting I2R2 and I2X2 from E2.

31. 31
Loading of Transformer
• Load power factor = cos ϕ2
• Primary power factor = cos ϕ1
• Input power to transformer, P1 = V1I1 cos ϕ1
• Output power of transformer, P2 = V2I2 cos ϕ2
 Practical Transformer on Load (Transformer with resistance and leakage reactance)
2. Transformer with Resistance and Leakage Reactance (contd.)
Phasor diagram (contd.)

32. 32
Impedance Transformation Through A Transformer
• The impedance of a device or an element is defined as
the ratio of the phasor voltage across it to the phasor
current flowing through it. If the secondary current of a
transformer is 𝑰𝟐 and the secondary voltage 𝑽𝟐, then the
impedance of the load is given by 𝒁𝑳 =
𝑽𝟐
𝑰𝟐
• The primary voltage of a transformer can be expressed as 𝑽𝟏 = 𝒂𝑽𝟐
• The primary current of a transformer can be expressed as 𝑰𝟏 =
𝑰𝟐
𝒂
• The apparent impedance of the primary circuit of the transformer is 𝒁𝑳
′
𝒁𝑳
′
=
𝑽𝟏
𝑰𝟏
=
𝒂𝑽𝟐
𝑰𝟐
𝒂
= 𝒂𝟐
𝑽𝟐
𝑰𝟐
= 𝒂𝟐𝒁𝑳
• With a transformer, it is possible to match the magnitude of a load impedance to a
source impedance simply by picking the proper turns ratio.

33. 33
Percent Impedance of a Transformer
 PERCENT IMPEDANCE OF A TRANSFORMER
%𝒁 =
𝒁𝟎𝟏
𝑽𝟏𝒓𝒂𝒕𝒆𝒅
𝑰𝟏𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎 =
𝒂𝟐
𝒁𝟎𝟐
𝑽𝟏𝒓𝒂𝒕𝒆𝒅
𝑰𝟏𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎 =
𝒂𝑰𝟏𝒓𝒂𝒕𝒆𝒅𝒁𝟎𝟐
𝑽𝟏𝒓𝒂𝒕𝒆𝒅
𝒂
× 𝟏𝟎𝟎
∴ %𝒁 =
𝑰𝟐𝒓𝒂𝒕𝒆𝒅𝒁𝟎𝟐
𝑽𝟐𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎 =
𝒁𝟎𝟐
𝑽𝟐𝒓𝒂𝒕𝒆𝒅
𝑰𝟐𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎
∴ %𝒁 is the same whether referred to primary or secondary.
• Similarly,
%𝑹 =
𝑹𝟎𝟏
𝑽𝟏𝒓𝒂𝒕𝒆𝒅
𝑰𝟏𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎 =
𝑹𝟎𝟐
𝑽𝟐𝒓𝒂𝒕𝒆𝒅
𝑰𝟐𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎
%𝑿 =
𝑿𝟎𝟏
𝑽𝟏𝒓𝒂𝒕𝒆𝒅
𝑰𝟏𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎 =
𝑿𝟎𝟐
𝑽𝟐𝒓𝒂𝒕𝒆𝒅
𝑰𝟐𝒓𝒂𝒕𝒆𝒅
× 𝟏𝟎𝟎
%𝒁 = %𝑹 + 𝒋%𝑿

34. 34
SHIFTING IMPEDANCES IN A TRANSFORMER
• The resistance and reactance of one winding can be transferred to the other to
make the analysis of the transformer simple.

35. 35
SHIFTING IMPEDANCES IN A TRANSFORMER
𝐑𝟏
𝐗𝟏
𝐑𝟐
′
= 𝐚𝟐
𝐑𝟐
𝐗𝟐
′
= 𝐚𝟐
𝐗𝟐
 REFERRED TO PRIMARY
• When secondary resistance / reactance is transferred to primary, called equivalent
secondary resistance / reactance referred to primary and is denoted by 𝑹𝟐
′
or 𝑿𝟐
′
, it is
multiplied by 𝒂𝟐
.

36. 36
 REFERRED TO SECONDARY
• When primary resistance / reactance is transferred to secondary, called equivalent primary
resistance / reactance referred to secondary and is denoted by 𝑹𝟏
′
or 𝑿𝟏
′
, it is divided by 𝒂𝟐
.
SHIFTING IMPEDANCES IN A TRANSFORMER
(Contd.)
𝐑𝟐
𝐗𝟐
𝑹𝟏
′
=
𝑹𝟏
𝒂𝟐 𝑿𝟏
′
=
𝑿𝟏
𝒂𝟐
𝐍𝟏
𝐍𝟐
= 𝐚

37. 37
• The parallel circuit 𝑹𝟎 − 𝑿𝟎 is no-load equivalent circuit.
• The resistance 𝑹𝟎 represents the core losses supplied by the current 𝑰𝒘.
• The inductive reactance X0 represents a loss-free coil which passes the magnetizing current
𝑰𝒎.
• The phasor sum of 𝑰𝒘 and 𝑰𝒎 is the no-load current 𝑰𝟎 of the transformer.
EQUIVALENT CIRCUIT OF TRANSFORMER
𝐈𝟐 𝐚

38. 38
• Equivalent circuit has created two circuits separated by an ideal transformer whose function
is to change values according to the equation:
𝑬𝟏
𝑬𝟐
=
𝑵𝟏
𝑵𝟐
=
𝑰𝟐
′
𝑰𝟐
= 𝒂
• When the transformer is on no-load, there is no current in the secondary winding, however,
the primary draws a small no-load current 𝑰𝟎.
• When load 𝒁𝑳 is connected to the secondary circuit the voltage 𝑬𝟐 induced in the
secondary by mutual flux will produce a secondary current 𝑰𝟐.
𝑽𝟐 = 𝑬𝟐 − 𝑰𝟐 𝑹𝟐 + 𝒋𝑿𝟐 = 𝑬𝟐 − 𝑰𝟐𝒁𝟐
EQUIVALENT CIRCUIT OF TRANSFORMER
(Contd.)

39. 39
• When the transformer is loaded to carry the secondary current I2 , the primary current
consists of no-load current I0 to provide
– magnetizing current,
– the current required to supply the core losses
– primary current I'2 (= K I2 ) required to supply the load on the secondary.
• Since the transformer is ideal, the primary induced voltage E1 can be calculated from the
relation:
𝑬𝟏
𝑬𝟐
=
𝑵𝟏
𝑵𝟐
• If we add I1R1 and I1X1 drops to E1 , we get the primary input voltage V1
𝑽𝟏 = −𝑬𝟏 + 𝑰𝟏 𝑹𝟏 + 𝒋𝑿𝟏 = −𝑬𝟏 + 𝑰𝟏𝒁𝟏; where, 𝑰𝟏 = 𝑰𝟎 + − 𝑰𝟐 𝒂
EQUIVALENT CIRCUIT OF TRANSFORMER
(Contd.)

40. 40
• When the transformer is loaded to carry the secondary current I2 , the primary current
consists of no-load current I0 to provide
– magnetizing current,
– the current required to supply the core losses
– primary current I'2 (= K I2 ) required to supply the load on the secondary.
• Since the transformer is ideal, the primary induced voltage E1 can be calculated from the
relation:
𝑬𝟏
𝑬𝟐
=
𝑵𝟏
𝑵𝟐
• If we add I1R1 and I1X1 drops to E1 , we get the primary input voltage V1
𝑽𝟏 = −𝑬𝟏 + 𝑰𝟏 𝑹𝟏 + 𝒋𝑿𝟏 = −𝑬𝟏 + 𝑰𝟏𝒁𝟏; where, 𝑰𝟏 = 𝑰𝟎 + − 𝑰𝟐 𝒂
EQUIVALENT CIRCUIT OF TRANSFORMER
(Contd.)

41. 41
Simplified Equivalent Circuit of A Loaded Transformer
• The no-load current 𝑰𝟎 is small as compared to the rated primary current, hence, voltage
drops in 𝑹𝟏 and 𝑿𝟏 due to 𝑰𝟎 can be neglected.
• The equivalent circuit can be simplified by transferring the shunt circuit 𝑹𝟎 − 𝑿𝟎 to the
input terminals.
= 𝐈𝟐 𝐚

42. 42
Simplified Equivalent Circuit of A Loaded Transformer
(Contd.)
• If all the secondary quantities are referred to the primary, we get the simplified / approximate
equivalent circuit of the transformer referred to the primary.
• When secondary quantities are referred to primary, impedances are multiplied by 𝒂𝟐,
voltages are multiplied by 𝒂 and currents are divided by 𝒂.
𝑹𝟐
′
= 𝒂𝟐
𝑹𝟐; 𝑿𝟐
′
= 𝒂𝟐
𝑿𝟐; 𝒁𝑳
′
= 𝒂𝟐
𝒁𝑳; 𝑽𝟐
′
= 𝒂𝑽𝟐; 𝑰𝟐
′
= 𝑰𝟐 𝒂
= 𝐚𝟐
𝐙𝐋
𝐈𝟐 𝐚
𝐕𝟐
′
= 𝐚𝐕𝟐
 Equivalent circuit referred to primary

43. 43
Simplified Equivalent Circuit of A Loaded Transformer
(Contd.)
• The equivalent circuit of the transformer can be further reduced
Where, 𝑹𝟎𝟏 = 𝑹𝟏 + 𝑹𝟐
′
; 𝑿𝟎𝟏 = 𝑿𝟏 + 𝑿𝟐
′
; 𝒁𝟎𝟏 = 𝑹𝟎𝟏 + 𝒋𝑿𝟎𝟏
𝐈𝟐 𝐚
= 𝐚𝟐
𝐙𝐋
𝐕𝟐
′
= 𝐚𝐕𝟐
 Equivalent circuit referred to primary (Contd.)

44. 44
Simplified Equivalent Circuit of A Loaded Transformer
(Contd.)
Phasor diagram
• The referred value of load voltage 𝑽𝟐
′
= 𝒂𝑽𝟐
is chosen as the reference.
• The referred value of load current 𝑰𝟐
′
is shown
lagging 𝑽𝟐
′
by angle 𝝋𝟐.
• For a given value of 𝑽𝟐
′
both 𝑰𝟐
′
and 𝝋𝟐 are
determined by the load.
• The voltage drop 𝑰𝟐
′
𝑹𝟎𝟏 is in phase with 𝑰𝟐
′
and the voltage drop 𝑰𝟐
′
𝑿𝟎𝟏 leads 𝑰𝟐
′
by 90°.
 Equivalent circuit referred to primary (Contd.)
• The vector sum of 𝑽𝟐
′
, 𝑰𝟐
′
𝑹𝟎𝟏, and 𝑰𝟐
′
𝑿𝟎𝟏 gives the primary voltage 𝑽𝟏.
• Current 𝑰𝒘 is in phase with 𝑽𝟏 & the current 𝑰𝒎 lags behind 𝑽𝟏 by 90°
• The sum of 𝑰𝒘 and 𝑰𝒎 gives the referred value of no-load current 𝑰𝟎.
• The vector sum of 𝑰𝟎 and𝑰𝟐
′
gives the referred primary current 𝑰𝟏.

45. 45
Simplified Equivalent Circuit of A Loaded Transformer
(Contd.)
• If all the secondary quantities are referred to the secondary, we get the simplified /
approximate equivalent circuit of the transformer referred to the secondary.
• When primary quantities are referred to secondary, impedances are divided by 𝒂𝟐, voltages
are divided by 𝒂 and currents are multiplied by 𝒂.
𝑹𝟎
′
= 𝑹𝟎 𝒂𝟐
; 𝑿𝟎
′
= 𝑿𝟎 𝒂𝟐
; 𝑹𝟏
′
= 𝑹𝟏 𝒂𝟐
; 𝑿𝟏
′
= 𝑿𝟏 𝒂𝟐
;
𝑽𝟏
′
= 𝑽𝟏 𝒂; 𝑰𝟏
′
= 𝒂𝑰𝟏
 Equivalent Circuit Referred To Secondary
𝐕𝟏
′
= 𝐕𝟏 𝐚
𝐚𝐈𝟏

46. 46
Simplified Equivalent Circuit of A Loaded Transformer
(Contd.)
• The equivalent circuit of the transformer can be further reduced
Where, 𝑹𝟎𝟐 = 𝑹𝟐 + 𝑹𝟏
′
; 𝑿𝟎𝟐 = 𝑿𝟐 + 𝑿𝟏
′
; 𝒁𝟎𝟐 = 𝑹𝟎𝟐 + 𝒋𝑿𝟎𝟐
 Equivalent Circuit Referred To Secondary (Contd.)
𝐕𝟏
′
= 𝐕𝟏 𝐚
𝐚𝐈𝟏

47. 47
Simplified Equivalent Circuit of A Loaded Transformer
(Contd.)
Phasor diagram
• The load voltage 𝑽𝟐 is chosen as reference.
• The load current 𝑰𝟐 is shown lagging the load
voltage 𝑽𝟐 by angle 𝝋𝟐.
• The voltage drop 𝑰𝟐𝑹𝟎𝟐 is in phase with 𝑰𝟐
and the voltage drop 𝑰𝟐𝑿𝟎𝟐 leads 𝑰𝟐 by 90°.
• The vector sum of 𝑽𝟐, 𝑰𝟐𝑹𝟎𝟐, and 𝑰𝟐𝑿𝟎𝟐 gives
the referred primary voltage 𝑽𝟏
′
= 𝑽𝟏 𝒂 .
• Current 𝑰𝒘
′
is in phase with 𝑽𝟏
′
& the current
𝑰𝒎
′
lags behind 𝑽𝟏
′
by 90°
• The sum of 𝑰𝒘
′
and 𝑰𝒎
′
gives the referred value
of no-load current 𝑰𝟎
′
.
• The vector sum of 𝑰𝟎
′
and 𝑰𝟐 gives the referred
primary current 𝑰𝟏
′
= 𝒂𝑰𝟏 .
 Equivalent Circuit Referred To Secondary (Contd.)

48. 48
• The voltage regulation of a transformer is the arithmetic difference between the
no-load secondary voltage and the secondary voltage on full-load expressed as
percentage of full-load voltage.
• It is defined by the equation,
𝑽𝑹 =
𝑽𝑺,𝑵𝑳−𝑽𝑺,𝑭𝑳
𝑽𝑺,𝑭𝑳
× 𝟏𝟎𝟎%
• Usually it is a good practice to have as small a voltage regulation as possible. For
an ideal transformer, 𝑽𝑹 = 𝟎 percent.
• It is not always a good idea to have a low-voltage regulation. Sometimes high-
impedance and high-voltage regulation transformers (Power transformers) are
deliberately used to reduce the fault currents in a circuit.
Transformer Voltage Regulation

49. 49
• To determine the voltage regulation of a
transformer, it is necessary to understand the
voltage drops within it. Consider the simplified
transformer equivalent circuit in Fig. The
effects of the excitation branch on transformer
voltage regulation can be ignored, so only the
series impedances need be considered. The
voltage
• In all the following phasor diagrams, the
phasor voltage Vs is assumed to be at an angle
of 0°, and all other voltages and currents are
compared to that reference. By applying
Kirchhoff’s voltage law to the equivalent circuit
in Fig., the primary voltage can be found as
𝑽𝑷
𝒂
= 𝑽𝑺 + 𝑰𝑺 𝑹𝒆𝒒 + 𝒋𝑿𝒆𝒒
Transformer Voltage Regulation

50. 50
Transformer Voltage Regulation
Phasor diagram of a transformer operating at lagging, unity, and leading power factor.
 Phasor diagram

51. 51
• The rated voltage is applied to the low-voltage (primary) winding while the high
voltage side is left open-circuited.
• The applied primary voltage V1 is measured by the voltmeter, the no load current I0 by
ammeter and no-load input power W0 by wattmeter.
• As the normal rated voltage is applied to the primary, the normal iron losses will occur
in the transformer core.
• Wattmeter will record the iron losses and small copper loss in the primary.
• Since no-load current I0 is very small the Cu losses in the primary under no-load
condition are negligible as compared with iron losses.
• Wattmeter reading practically gives the iron losses in the transformer.
 Open-Circuit or No-Load Test
Transformer Tests

52. 52
• Iron losses, Pi = Wattmeter reading = W0
• No load current = Ammeter reading = I0
• Applied voltage = Voltmeter reading = V1
• Input power, W0 = V1 I0 cos Φ0
 Open-Circuit or No-Load Test
Transformer Tests

53. 53
• Input power, W0 = V1 I0 cos Φ0
• No load p.f., 𝒄𝒐𝒔𝝓𝒐 =
𝑾𝒐
𝑽𝟏𝑰𝒐
• 𝑰𝒘 = 𝑰𝒐𝒄𝒐𝒔𝝓𝒐
• 𝑰𝒎 = 𝑰𝒐𝒔𝒊𝒏𝝓𝒐
• 𝑹𝒐 =
𝑽𝟏
𝑰𝑾
• 𝑿𝒐 =
𝑽𝟏
𝑰𝒎
• Thus open-circuit test enables us to determine iron losses and parameters R0
and X0 of the transformer
 Open-Circuit or No-Load Test
Transformer Tests

54. 54
• The test is conducted to determine R01 (or R02), X01 (or X02) and full-load copper
losses of the transformer.
• In the test, the secondary (usually low-voltage winding) is short circuited by a
thick conductor and variable low voltage is applied to the primary
 Short-Circuit or Impedance Test
Transformer Tests

55. 55
• The input voltage is gradually raised till full-load current I1 flows in the primary
then I2 in the secondary also has full-load value since I1/I2 = N2/N1
• Under this condition, the copper loss in the windings is the same as that on full
load.
• There is no output from the transformer under short-circuit conditions, hence,
input power is all loss and this loss is almost entirely copper loss as iron loss in the
core is negligibly small since the voltage VSC is very small.
• The wattmeter will practically register the full-load copper losses in the
transformer windings.
• The equivalent circuit of a transformer on short circuit has been referred to
primary; the no-load current negligibly small.
 Short-Circuit or Impedance Test
Transformer Tests

56. 56
• Full load Cu loss PC (Wattmeter reading) = 𝑾𝑺𝑪
• Applied voltage (Voltmeter reading) = 𝑽𝑺𝑪
• F.L. primary current 𝑰𝟏 (Ammeter reading) = 𝑰𝑺𝑪
• 𝑷𝑪 = 𝑰𝟏
𝟐
𝑹𝟏 + 𝑰𝟏
𝟐
𝑹𝟐
′
= 𝑰𝟏
𝟐
𝑹𝟎𝟏
• 𝑹𝟎𝟏 =
𝑷𝑪
𝑰𝟏
𝟐
• 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 = 𝒁𝟎𝟏
𝟐
− 𝑹𝟎𝟏
𝟐
• Short-circuit p.f, 𝒄𝒐𝒔𝝓𝟐 =
𝑷𝑪
𝑽𝑺𝑪𝑰𝟏
• Thus short-circuit test gives full-load Cu loss, R01 and X01.
 Short-Circuit or Impedance Test
Transformer Tests

57. 57
• The core or iron losses consist of hysteresis and eddy current losses and occur in
the transformer core due to the alternating flux.
• These can be determined by open-circuit test.
• Iron or Core losses, Pi = Hysteresis loss + Eddy current loss = Constant losses
• The hysteresis loss can be minimized by using steel of high silicon content
• The eddy current loss can be reduced by using core of thin laminations.
 Core or Iron Losses
Transformer Losses

58. 58
• These losses occur in both the primary and secondary windings due to their
resistance.
• The losses can be determined by short-circuit test.
• Total Cu losses, 𝑷𝑪 = 𝑰𝟏
𝟐
𝑹𝟏 + 𝑰𝟏
𝟐
𝑹𝟐
′
= 𝑰𝟏
𝟐
𝑹𝟎𝟏 or 𝑰𝟐
𝟐
𝑹𝟎𝟐
• Copper losses vary as the square of load current, thus, if copper losses are 400 W
at a load current of 10 A, they will be (1/2)2 x 400 = 100 W at a load current of 5A.
• Total losses in a transformer = Pi + PC = Constant losses + Variable losses
• In a transformer, copper losses are about 90% of the total losses.
 Copper Losses
Transformer Losses

59. 59
• Efficiency of a transformer is defined as:
Efficiency =
Output power
Input power
=
Output power
Output power+Losses
• Efficiency can be determined by directly loading the transformer and measuring
the input power and output power, however, such method has the following
drawbacks:
― Since the efficiency of a transformer is very high, even 1% error in each wattmeter
(output and input) may give ridiculous results and the test may give efficiency even
higher than 100%.
― While performing test of transformer on load, considerable amount of power is
wasted.
― It is generally difficult to have a device that can absorb all the output power.
― The test gives no information about the proportion of various losses.
• In practice, open-circuit and short-circuit tests are carried out to find the
efficiency.
EFFECIENCY OF TRANSFORMER

60. 60
• F.L. Iron loss = Pi ... from open-circuit test
• F.L. Cu loss = PC ... from short-circuit test
• Total F.L. losses = Pi + PC
• We can now find the full-load efficiency of the transformer at any p.f. without
actually loading the transformer.
• F.L. efficiency, 𝜼𝑭.𝑳. =
𝑭𝒖𝒍𝒍−𝒍𝒐𝒂𝒅 𝑽𝑨 𝒙 𝒑.𝒇
𝑭𝒖𝒍𝒍−𝒍𝒐𝒂𝒅 𝑽𝑨 𝒙 𝒑.𝒇 +𝑷𝒊+𝑷𝑪
• Also for any load equal to x x full-load,
• Corresponding total losses = Pi + x2 PC
• Corresponding ηx𝜼𝒙 =
𝒙𝒙𝑭𝒖𝒍𝒍−𝒍𝒐𝒂𝒅 𝑽𝑨 𝒙 𝒑.𝒇
𝒙𝒙𝑭𝒖𝒍𝒍−𝒍𝒐𝒂𝒅 𝑽𝑨 𝒙 𝒑.𝒇 +𝑷𝒊+𝒙𝟐𝑷𝑪
EFFECIENCY OF TRANSFORMER

61. 61
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