© 2014 IJIRT | Volume 1 Issue 6 | ISSN : 2349-6002
STUDY OF IDEAL TRANSFORMER AND
PRACTICAL TRANSFORMER
RUBY DHANKAR, SAPNA KAMRA,VISHAL JANGRA
Abstract- This paper proposes the study of real and
ideal transformer. It also explains load, no-load
conditions and phasor diagrams of ideal and practical
transformer. For a better understanding and an easier
explanation of a practical transformer, certain
idealizing assumptions are made which are close
approximations for a practical transformer.
Keywords- transformer , flux , reactance ,
magnetizing current, primary and secondary
windings.
I.
INTRODUCTION
Michel faraday introduced the principle of
electro –magnetic induction in 1831.it estates that a
voltage appear across the terminals of an electric coil
when the flux linked with the coil changes. The
magnitude of the induced voltage is corresponding to
the rate of change of flux linkages.This finding
forms the basic for many magneto electric
devices.The primitive use of this phenomenon was in
the development of induction coils. These coils were
used to develope high voltage pulses to ignite the
explosive charges in the mines. As the d.c. power
arrangement was in use at that time, very little of
transformer principle was made use of. In the d.c.
supply system the generating station and the load
centre have to be necessarily close to each other due
to the demand of economic transmission of power.
Also the d.c. generators cannot be scaled up due to
the drawback of the commutator.
2. Discussion
As discussed earlier the transformer is a
static device working on the principle of faraday’s
law of induction. It estates that a voltage appears
across the terminals of an electric coil when the flux
correlated with the same changes.this emf is
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proportional to the rate of variation of flux linkages.
Establishing mathematically.
Here the common constuctional aspects alone are
explained.
1. Ideal transformer
2. practical transformer
2.1. ideal transformer
Earlier it is seen that a voltage is induced in
a coil when the flux correlated with the same
changes. It posses certain essential features of a real
transformer but some details of minor significance
are ignored which will be introduced step-by-step
while analyzing a transformer. The idealizing
assumptions made are as follows
1) No winding resistance
2) No magnetic leakage
3) No iron loss and zero magnetizing current
2.1.1. Ideal transformer equations: using faraday’s law of
induction
. . (1)
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2.2.2. Winding resistances in transformer
. . . (2)
Combining ratio of (1) & (2) turns ratio will be
equal to
. . . (3)
Where
for step-down transformers, a > 1
for step-up transformers, a < 1
By law of Conservation of Energy, apparent ,
real and reactive power are each conserved in the
input and output
Primary and secondary windings made up of
copper wire. Every conductor has its own resistance.
So primary and secondary side both have resistance.
Primary resistance R1 and secondary resistance R2 are
in series with the respect to the windings. Figure 1
shows that the resistance of the windings on both
sides. Due to winding resistance when current passes
through the windings there will voltage drop IR and
creates power loss. thus
the E1 < V1 and V2 <
E 2.
. . . (4)
Combining (3) & (4) with this endnote yields the
ideal transformer identity
. . . (5)
By Ohm's Law and ideal transformer identity
. . . (6)
Apparent load impedance Z'L (ZL referred to the
primary)
. . . (7)
2.2.3.Leakage reactance in transformer
Mutual
Flux Φ Links With Both The
Primary And Secondary Side. Primary Current
I1Creates Individual Flux Φ1 In Primary Side And
Secondary Current I2 Produces Flux Φ2 In Secondary
Side, Those Two Fluxes Is Not Common In Both
Sides. Thus The Flux Φ1 And Flux Φ2 Are Known
As Leakage Flux in Transformer. Figure 2 Shows
Leakage Fluxes
2.2. practical transformer
Ideal transformer has no losses although practical
transformer have



Iron loss
Magnetic leakage
Winding resistances
2.2.1. Iron losses in transformer
Alternating flux Φ passes through iron
kernel. It produces eddy current and hysteresis loss in
it. Two of these losses called either iron loss or core
loss. Iron loss rely upon the core volume, supply
frequency, maximum flux density etc. Magnitude of
iron loss is small in practical transformer.
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The way of leakage flux is through the air
mainly. The effect of primary leakage flux Φ 1
generates an inductive reactance X1 series in primary
winding and secondary leakage flux Φ2 introduces an
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inductive reactance X2 in series with the secondary
winding shown in figure 1.
Primary leakage inductance, L1 = primary
leakage flux linkages / primary current = N1 Φ1 /
I1Primary leakage reactance, is X1 = 2πf L1
In the same manner secondary leakage
inductance, L2 = N2 Φ2 / I2 Secondary leakage
reactance, is X2 = 2πf L2 .There is No power loss
occurs due to leakage reactance. But it changes the
power factor as well there is voltage loss due
to IX drop. Flux leakage is absolutely small about 5%
of mutual flux Φ in a transformer. But it cannot be
avoided or ignored.

The component Iw is admitted as iron loss or
active or working component. It is in phase with
the enforced voltage V1. It provides a very small
primary copper loss and iron loss.
Iw = I0 cos Φ0
Here It is clear that I0 is the phasor sum of Im and Iw.
I0 = √(Im2 + Iw2)
No load power factor, cos Φ0 = Iw / I0
At no load practical transformer primary
copper loss I02R is quite small and this loss may be
ignored . Hence, primary no load input power of
practical transformer is equal to the iron loss in
transformer.
No load input power is , W0 = Iron loss
2.2.4.No load practical transformer
As primary loss in practical transformer is quite small
so it may be written at no load, V1 = E1.
Here is no load in secondary so E2 = V2.
Practical transformer on no load phasor diagram
Figure 4 is the phasor of practical transformer on no
load condition. Primary small current I0 is phasor
sum of Im and Iw.
A practical transformer diagram is shown in
figure 3, there is no load in secondary terminal it is
open circuited. When ac source is coupled in
primary a small current I0 flows through the primary.
It occurs a very small extent of copper loss and iron
loss in the primary. In order that the primary no load
current I0 is not 90˚ behind the applied voltage V1 but
lags it by angle Φ0 < 90˚.
Primary no load current I0 lags by V1 voltage by an
angle Φ0 < 90˚.
We can deduce the magnetizing and iron
loss current using above equations now we will solve
a math.
Here No load input power, W0 = V1 I0 cos Φ0
In primary side for I0 we get two components Iw and
Im.

The component Im is admitted as magnetizing
component. This component creates mutual flux
Φ in the core. Im lagging behind V1 by 90˚.
Im = I0 sin Φ0
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For practical transformer on load we will take two
cases
Case 1: when the transformer has no winding
resistance and leakage flux is also zero.
Case 2: in this when the transformer has winding
resistance and leakage flux
Case 1: No winding resistance and no leakage flux
Phasor diagram of practical transformer on load
Figure 1 :represent a practical transformer on load.
We have to suppose it has no winding resistance
and no leakage flux. Another word winding
resistance and leakage flux is neglected . For this
supposition V1 = E1 and V2 = E2 .
Take the load at secondary is inductive load which
causes the secondary current I2 to lag the secondary
voltage V2 by Φ2. Primary current I1 must meet two
conditions
i) Primary current must supply the no load current
I0 to accommodate the iron losses in the transformer
and to provide flux in the core.
ii) Primary current must supply a current I2′ to
counteract the demagnetizing influence of secondary
current I2. The magnitude of I2′ will be
Figure 2 : represent the phasor diagram of practical
transformer on load for inductive load. Here E1and
E2 are lagging behind by common flux Φ by 90˚.
Phasor sum of I0 and I2′ is the primary current I1. I2′
is anti phase with I2. The value of K is supposed to be
unity so primary phasor is equal to secondary phasor.
Primary power factor = cos Φ1
Secondary power factor = cos Φ2
Primary input power = V1 I1 cos Φ1
Secondary input power = V2 I2 cos Φ2
Case 2: Transformer with resistance and leakage reactance
N 1 I 2′ = N 2 I 2
Or I2′ = I2 N2/N1 = K I2
The phasor sum of I2′ and I0 is the total primary
current I1.
I 1 = I 2′ + I 0
Whereas I2′ = – K I2
I2′ is 180˚ out of phase with I2 current.
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Figure 3 :shows a practical transformer with winding
resistance and leakage resistance. This is real
condition which exists in a practical transformer.
Voltage drop R1 and X1 occurs in primary side so
V1 > E1 and voltage drop R2 and X2 occurs in
secondary side so V2 < E2.
Assume
an inductive load which causes the
secondary current I2 to lag behind the secondary
voltage V2 by Φ2. Primary current I1 must follow the
two conditions
i) Primary current must supply no load current I0 to
meet the iron losses in the transformer and to provide
flux in the core.
ii) It must supply a current I2′ to counteract the
demagnetizing influence of secondary current I2. The
magnitude of I2′ will be
Primary input power = V1 I1 cos Φ1
Secondary input power = V2 I2 cos Φ2
I 1 = I 2′ + I 0
Where I2′ = – K I2
3. Conclusion
This paper concludes that An ideal
transformer is the one whose windings do not
have any ohmic resistance and whose core does not
have any leakage flux and eddy current losses.
A real transformer is the one whose windings do have
some amount of ohmic resistance and the core also
have some leakage flux and eddy current losses.
References
1.
2.
N 1 I 2′ = N 2 I 2
J.B. Gupta Electromechanical
conversion, S.K. kataria & Sons.
ece.colorado.edu/bart/book
energy
Or I2′ = I2 N2/N1 = K I2
The phasor sum of I2′ and I0 is the total primary
current I1.
Phasor diagram of practical transformer
resistance and reactance:
with
Figure 4: represents the phasor diagram of a
practical transformer for the usual case of inductive
load. Here E1 and E2 are lagging behind by mutual
flux Φ by 90˚.
Primary power factor = cos Φ1
Load power factor = cos Φ2
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