MAKERERE UNIVERSITY

COLLEGE OF ENGINEERING, DESIGN, ART AND TECHNOLOGY

SCHOOL OF ENGINEERING

DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING

MEC 2201: ELECTRICAL ENGINEERING FOR MECHANICAL ENGINEERS II LECTURE NOTES

2016/2017

CHAPTER THREE: TRANSFORMERS

Instructor: Thomas Makumbi

BSc. Eng (MUK, Uganda)
MSc. RET (MUK, Uganda)
MSc. SEE (HIG, Sweden)

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 TRANSFORMERS
A transformer is an electrical machine that has no moving parts but is able to transform
alternating voltages and currents from high to low values and vice versa. Transformers are used
extensively in all branches of electrical engineering and range from the largest power
transformer employed on the national grid to the smallest signal transformer of an electronic
amplifier.
Principle of operation
The transformer operates on the principle of mutual induction. When an alternating voltage is
applied to a concentrated coil wound on a ferromagnetic core, a back emf is induced in the coil
due to the continual alternation of the self flux linkage. The emf is due to the rate of change of
flux linkage within the coil. If another coil is placed around the core, this coil will have an emf
induced in it due to the effect of mutual induction. However, the two coils are insulated from
each other and are therefore electrically separate.

Figure 1.1: Simple Transformer

Assuming no losses occur, the induced emf of the primary winding must equal the applied
voltage. This is termed the primary emf E1. Each turn provides its own proportion of the total
voltage and if there are N1 turns in the primary winding, the induced emf per turn is E1/N1. Each
turn on the secondary winding will also have the same emf induced in it i.e. E1/N1. If there are
N2 turns in the secondary winding, the total induced emf E2 is given by;
E1 N1
 ………………………….. (1)
E2 N 2
E2 is termed the secondary emf. The ratio of the emfs and thus the voltage in the two windings is
therefore seen to be the same as the ratio of the turns. A simple means is thus provided of
transforming from one voltage to another. Generally, the applied voltage is sinusoidal and for the
ideal transformer, the flux produced in the core will also be sinusoidal. Let the flux be
represented by;
  m sin t
The instantaneous emf induced in a coil of N turns linked by this flux φ is given by;
d d
e N
dt dt
d (m sin t )
eN  Nm cost
dt
e  2fNm sin(t  90o )

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This represents an induced emf of maximum value 2fNm , which leads the flux by 90o.

Em  2fNm
2fNm
E 
2
E  4.44 fNm …………………………… (2)

Thus E1  4.44 fN1m

and E2  4.44 fN 2m

E1 N1
  as before
E2 N 2
If a load is connected across the secondary winding, the secondary emf E2 will cause a current I2
to flow in the resulting circuit. Such a current flow gives rise to an mmf in the ferromagnetic
core, the mmf being in phase with the current and is given by;
F2  I 2 N 2
The loaded transformer circuit is shown in Figure 1.2;

Figure1.2: Transformer on load

Such an mmf will alter the flux in the core from its original value. However the flux in the core
must remain unchanged from the previous no load condition because this is the flux , which,
when linking the primary winding and varying at the supply frequency, gives rise to the primary
back emf equal to the supply voltage . This equality must be maintained at all times in order to
comply with Kirchoff’s second law. In order to conserve this equality, an additional current
must flow in the primary winding to give rise to an additional primary mmf equal in magnitude
but opposite in direction of action to the secondary mmf. Let the primary current be I1.
F1  I1 N1
but F1  F2
I1 N1  I 2 N 2
N1 I 2
  …………………… (3)
N 2 I1
Equations (1) and (3) may be combined to give the relation below;

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 E1 N1 I 2
  ………….. (4)
E 2 N 2 I1
Ideal Transformer
The ideal transformer operates on the principle of mutual induction and has negligible losses
either in the electric circuits or in the magnetic circuit. In addition, it has the important property
that it requires no current to magnetize the core. It has been noted that when a transformer is
loaded, an mmf balance is set up between the primary and the secondary circuits in order that the
core flux remains unchanged. If the inductance of the primary winding is infinite, then zero
current is required to create the required core flux. It follows that when a transformer is loaded,
all of the primary current is used to balance the secondary current in accordance with equation
(3). Relation (4) therefore applies to an ideal transformer. The phase diagram for such a
transformer is as shown below;

Figure 1.3: Phasor diagram for an ideal transformer

In the ideal case, the applied voltage V1 is equal to the primary back emf E1 and similarly E2=V2,
where V2 is the voltage applied to the load by the secondary. From equation (4);
V1 E1 I 2
 
V2 E2 I1

V1I1  V2 I 2
 S1  S2
But the primary and secondary voltages are in phase with one another and the primary and
secondary currents are also in phase with each other hence;
cos1  cos2  cos
S1 cos1  S2 cos2
P1  P2 ……………… (5)
Thus, the input power is equal to the output power which is in agreement with the result to be
expected from an ideal transformer.
Practical Transformer
The practical transformer differs from the ideal transformer in two ways;
i. It requires a current to magnetize it
ii. It has energy losses

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From the first point, there must always be a current in the primary winding to magnetize the
core. When a load is connected to the secondary winding of the transformer, an additional
current flows in the primary to maintain the mmf balance in the core. Thus, the total primary
current now consists of two components i.e. the current to magnetize the core and the load
current. Since the primary current is designated I1, and I 2' the current required to maintain the
mmf balance. From equation (3);
N1 I 2
 ………………………… (5)
N 2 I 2'
The current that magnetizes the core exists at all times and remains virtually constant regardless
of the secondary load. Because it can be observed on its own when there is no load connected to
the secondary winding, it is termed the no-load current Io.
I1  I o  I 2' …………….. (6)
The no-load current has in its turn, two components. The first of these is the magnetizing current
Iom, which gives rise to the core flux. This is a purely reactive current and is in phase with the
flux that creates it. The second component is an active one due to the hysteresis and eddy current
losses in the core. It is designated Iol. The total no-load current is given by the phasor sum of the
components.
I o  I om  I ol ………..(7)
It should be noted that while the existence of the no-load current should be acknowledged, it
does not always play a part in the solution of the accompanying problems. In many transformers,
particularly power transformers, the omission of the no-load current will not materially affect the
validity of any on-load solution due to the relative smallness of the no-load current.
Provided that there are no losses in the windings, E1=V1 and E2=V2. If there was a voltage drop
within the winding due to either the resistance of the winding conductor or leakage reactance
created by imperfect coupling magnetically of the windings, these relations would not hold. The
phasor diagram is as shown in Figure 1.4 with the assumption that there are no voltage drops in
the windings. It should be noted that E1 and E2 are shown to be in phase since this is the normal
method of winding connection;

Figure 1.4: Phasor diagram for a practical transformer without winding voltage drop
Magnetization of core
If the applied voltage across a winding wound on a ferromagnetic core is sinusoidal, then the rate
of change of the core flux must also be sinusoidal. It follows that the flux must vary sinusoidally.
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However, it does not follow that the current which creates this flux must vary sinusoidally. If the
B/H curve is redrawn as φ/i curve, the relation between flux and current can be derived as shown
in Figure 1.5;

Figure 1.5: Waveform of magnetizing current

The change of scale for the axes is permissible since   BA and similarly i  Hl N , where the
core length l and the number of turns on the coil N are constant. Using this φ/i curve, by plotting
instantaneous values of flux against current, the current waveform can be derived. The
significance of this is that since the current waveform is not sinusoidal, it cannot be measured
using moving iron instruments such as an ammeter.
Another factor is the effect of the saturation of the core on the back emf. Consider a
ferromagnetic cored coil that also has the same winding resistance. If the core has an ideal B/H
characteristic as shown in Figure 1.6, then provided the flux density of the knee point is not
exceeded, a back emf will oppose the applied voltage at all times. However, if this flux density
value is exceeded, the flux in the core can only increase if the coil current increases by a very
large amount. This increase is limited by the coil resistance and so the core flux is unable to
change at the required rate. This result in the collapse of the back emf and the appropriate wave
forms are shown in Figure 1.6. In order to avoid this effect, transformers and similar
ferromagnetic cored machines are designed so that the core flux density does not exceed the
value appropriate to the knee point.

Figure 1.6: Effect of excessive flux in a ferromagnetic core used for an ac machine

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 Impedance Transformation
Consider the effect of the secondary load as seen from the primary winding. This may be
analyzed by considering the arrangement shown in Figure 1.7;

Figure 1.7: Transformer supplying a secondary load

V2
Z2 
I2
V1
Z1 
I1
2
N 1 N N 
Z1  V2 . 1 . . 1   1  .Z 2 ……………………. (8)
N2 I2 N2  N2 
Since the transformer is ideal;
2
N 
Z1 cos  Z cos   1  .Z 2 cos
'
2
 N2 
2
N 
R1  R   1  .R2
'
2
 N2 
2
N 
Also, X 1  X   1  .X 2
'
2
 N2 
Transformer Losses
There are three (3) sources of power loss in a transformer. These are;
i. Hysteresis loss
ii. Eddy current loss
iii. Winding loss (i2R)
The hysteresis losses are minimized by using quality steel with high silicon content; the eddy
current loss is minimized using laminated steel cores while the winding loss accounts for 90% of
the total losses and little can be done to counteract it.

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 Flux leakage in a Transformer
Consider again the magnetic circuit of a simple transformer as shown in Figure 1.8;

Figure 1.8: Magnetic circuit of a simple transformer

The major portion of the magnetizing flux set up by the no-load current passes through both
primary and secondary windings linking them. However, not all of the flux remains within the
ferromagnetic path; some is diverted into the surrounding medium, generally either air or oil.
This is because the surrounding medium also has a definite permeability although it is much less
than that of the core. Thus, a little of the flux traverses this external path. This flux termed
leakage flux serves no useful purpose since it fails to link the secondary winding to the primary
winding.
When a transformer is operated on load, the flux pattern is modified to that shown in Figure 1.8.
Again there is leakage flux emanating from the primary winding. This is termed the primary
leakage flux. However, the secondary current sets up an mmf which opposes the main flux and
causes a portion of the flux to be diverted in secondary leakage paths. Thus, there is formed a
flux which links only the secondary winding. This is termed the secondary leakage flux.

The effect of these leakage fluxes is purely self inductive, hence both windings appear to have
self inductive reactance. This serves no useful purpose but acts as an effective voltage drop
within the windings. The main useful flux decreases very slightly as the load increases, but the
leakage fluxes are practically proportional to the currents in the respective windings. The effect
of flux leakage upon the ratio of transformation is thus to reduce the secondary terminal voltage
for a given applied primary voltage. To minimize the leakage, transformers are constructed in
one of the following designs;
Core type construction: This is a modification of the simple transformer and has a single
magnetic circuit as shown in Figure 1.9. It is usual to wind one half of each winding on one limb,
the low voltage winding being innermost for mechanical strength. The low voltage winding
carries the higher currents and therefore experiences the greatest repulsion between its current
carrying conductors. The placing of the windings in this manner reduces the flux leakage.

Figure1.9: Core type transformer

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Shell type construction: This method of construction involves the use of a double magnetic
circuit as shown in Figure1.10. The windings are wound concentrically but are complete on the
central limb. Finally it should be noted that the leakage flux forms only a very small part of the
total flux. Modern transformers generally have coefficients of coupling in excess of 0.99.

Figure1.10: Shell type transformer

Problems
1. Describe the operation of an ideal single phase transformer on no load and on load. State
for each condition where the effective losses occur in a practical transformer and explain
how these losses occur.
2. A 20 kVA, single phase transformer has a turns ratio of 44:1. If the primary winding has
4000 turns and is connected to a 11 kV, 50 Hz sinusoidal supply, calculate for full load;
i) The primary and secondary currents (Ans. 1.8A; 80 A)
ii) The maximum value of the core flux (Ans. 12.4 mWb)
3. A single phase transformer has a ratio of 1:10 and a secondary winding of 1000 turns.
The primary winding is connected to a 25V sinusoidal supply. If the maximum core flux
is 2.25mWb, determine;
i) The secondary voltage on open circuit (Ans. 250V)
ii) The number of primary turns (Ans. 100 turns)
iii) The supply frequency (Ans. 25 Hz)
4. A single phase transformer is connected to a 240V sinusoidal supply. When the
secondary supplies a current of 50A at a power factor of 0.8 leading, the primary current
is 4.62 A at a power factor 0.974 leading. When the secondary supplies a current of 50A
at a power factor 0.8 lagging, the primary current is 6.72 A at a power factor of 0.67
lagging. Calculate the secondary voltage if the transformer may be considered ideal.
(Ans. 24.2 V)