Lecture 9: Transformer
Department of Electrical Engineering
National Institute of Technology Rourkela
EE1000: Basic Electrical Engineering
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Outline
1. Goals of the lesson
2. Introduction
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Goals of the lesson
After going through the lesson, you shall get a broad idea of the following:
• two winding ideal transformer, its properties and working principle under no load
condition as well as under load condition.
• How to draw phasor diagram under no load and load conditions.
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Introduction
• The transformer is a device, or machine, that transfers electrical energy from one
electrical circuit to another electrical circuit through the medium of magnetic field and
without a change in the frequency.
• The electric circuit which receives energy from the supply mains called primary winding
and the other circuit which delivers electric energy to the load is called secondary winding.
• If the secondary winding has more turns than the primary winding, then the secondary
voltage is higher than the primary voltage and the transformer is called a step-up
transformer.
• In case the secondary winding has less turns than the primary winding, then the secondary
voltage is lower than the primary voltage and the transformer is called a step-down
transformer.
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A typical Transformer
Figure 1: A typical Transformer
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Principle of Transformer Action
• A transformer works on the principle of electromagnetic induction. According to this
principle, an emf is induced in a coil if it links a changing flux.
• The primary winding P is connected to an alternating voltage source, therefore, an
alternating current Iϕ starts flowing through N1 turns.
• The alternating mmf N1 Iϕ sets up alternating flux ϕ which is confined to the high
permeability iron path as indicated in the Fig.
• The alternating flux induces voltage E1 in the primary P and E2 in the secondary S. If
the load is connected across the secondary, a load current starts flowing.
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Ideal Transformer
In the beginning, a transformer is assumed to be a ideal one. For a transformer to be ideal
one, the various assumptions are as follows
• Primary and secondary windings has no resistance.
• All the flux produced by the primary links the secondary winding i,e., there is no leakage
flux.
• Permeability µr of the core is infinitely large. In other words, to establish flux in the core
vanishingly small (or zero) current is required.
• Core loss comprising of eddy current and hysteresis losses are neglected.
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Analysis of Ideal Transformer
• Let the voltage V1 applied to the primary of a
transformer, with secondary open circuited, be
sinusoidal.
• Then the current Iϕ , due to applied voltage V1 ,
will also be a sine wave. Then mmf N1 Iϕ and
therefore, the core flux ϕ will follow the variations
of Iϕ very closely.
• Therefore, if the applied voltage V1 has sine
waveform, the flux ϕ must also have a sine
waveform. Let the sinusoidal variation of flux ϕ
be expressed as ϕ = ϕmax sin ωt, where ϕmax is
the maximum value of the magnetic flux in weber
and ω = 2πf , is the angular frequency in rad/sec
and f is the supply frequency in Hz.
Figure 2: Ideal Transformer
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Analysis of Ideal Transformer
• The emf e1 in volts, induced in the primary N1 turns by the alternating flux is given by
dϕ
dt
= −N1 ωϕmax cos ωt
e1 = −N1
= N1 ωϕmax sin(ωt −
π
π
) = E1max sin(ωt − )
2
2
where E1max = N1 ωϕmax .
• RMS value of emf E1 induced in primary winding is given by
E1max
E1 = √
2
2πfN1 ϕmax
√
=
2
√
= 2πfN1 ϕmax
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Analysis of Ideal Transformer
• The current Iϕ in the primary is assumed to flow along the path abcda. The emf e1
induced in N1 turns must be in such a direction as to oppose the cause, i.e., Iϕ ; as per
Lenz’s Law.
• Therefore, the direction of e1 is as shown by the arrows in the primary N1 turns and it is
seen to oppose v1 . Since primary winding resistance is negligible, e1 at every instant,
must be equal and oppositive to v1 .
v1 = −e1 = N1
dϕ
dt
orV1 = −E1
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Analysis of Ideal Transformer
• The emf induced in the secondary is
dϕ
dt
= −N2 ωϕmax cos ωt
e2 = −N2
= N2 ωϕmax sin(ωt −
π
π
) = E2max sin(ωt − )
2
2
where E2max = N2 ωϕmax .
• RMS value of emf E2 induced in secondary winding is given by
E2max
E2 = √
2
2πfN2 ϕmax
√
=
2
√
= 2πfN2 ϕmax
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Analysis of Ideal Transformer
•
E1
N1
=
E2
N2
√
E1
E2
∴
=
= 2πf ϕmax
N1
N2
∴
emf per turn in primary= emf per turn in the secondary
• Assuming no loss,
V2
V1
=
N1
N2
V1
N1
=
=k
V2
N2
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Analysis of Ideal Transformer
∴ V2 =
V1
k
k < 1 ⇒ step-up transformer and k > 1 ⇒ step-down transformer. when a load is connected
across the secondary winding, a current I2 flows. In an ideal transformer losses are neglected
and a transformer is considered to be 100% efficient.
Hence primary and secondary volt-amperes are equal, i.e.,
V1 I1 = V2 I2
I2
N1
V1
= =
=k
∴
V2
I1
N2
The rating of transformer is stated in terms of the volt-amperes that it can transform with out
overheating. Transformer rating is either V1 I1 or V2 I2 (where I2 is the full-load secondary 13 / 93
Example Problem
• An ideal transformer has turns ration 8 : 1 and
the primary current is 3 Amp when it is supplied
at 240 volt. Calculate the secondary voltage and
current.
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Example Problem
• A 5 kVA, 1ϕ transformer has turns ration 10 : 1
and is fed from a 2.5kV supply. Neglecting losses,
determine (1) the full-load secondary current (2)
the minimum load resistance which can be
connected across the secondary winding to give
full load kVA. (3) primary current at full load
kVA.
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Phasor Diagram of Ideal Transformer
• Iϕ and ϕ they are in same phase.
• Here N1 and N2 are assumed equal for
convenience and therefor E1 = E2 .
• From the equations, they are lagging behind the
current Iϕ by 90◦ .
• The applied voltage V1 is equal and opposite to
E1 and it is drawn opposite to E1 .
Figure 3: Phasor diagram
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No Load Phasor Diagram for a Practical Transformer
• The no load current I0 , taken by the primary
consists of two components:
• A reactive or magnetizing component, Im ,
producing the flux and therefore in phase with the
ϕ.
• An active or power component, Ic , supplying the
hysteresis and eddycurrent losses in the core and
the negligible I 2 R loss in the primary winding.
Component Ic is in phase with the applied
voltage, i.e.Ic V1 = core loss. This component is
usually very small compared with Im , so that the
no-load power factor is very low.
• Power factor on no load is cos ϕ0 = IIc .
Figure 4: Phasor diagram
0
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Example Problem
• The emf per turn for a single phase, 2310/220V,
50 Hz transformer is approximately 13V.
Calculate (a) the number of primary and
secondary turns (b) the net cross-sectional area of
core, for a maximum flux density of 1.4T.
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Transformer under loaded condition
• In Fig., if the switch S is closed, a load
impedance ZL gets connected across the
secondary terminals. Since the secondary winding
resistance is zero, V2 = E2 .
• When load is connected at secondary side, due to
induced voltage, E2 current I2 will flow and it can
be determined by the magnitude and phase angle
of the impedance ZL .
• As soon as current I2 starts flowing in the
secondary current, by the Lenz’s law the induced
current opposes the cause that produces it. That
is, it will establish a flux which is opposite to the
flux ϕ established by the coil 1.
Figure 5: Magnetic circuit
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Transformer under loaded condition
• As a result of which, the total flux linking coil 1 will become ϕm = ϕ1 − ϕ2 . The total
flux linking coil 1 will reduce. Hence E1 will reduce.
• For an ideal transformer, V1 = −E1 and, therefore, mutual flux ϕm in the core must
remain constant.
• However there is no impedance between the applied voltage and the coil. There will be
current I1 drawn from the source of such magnitude so that the induced voltage E1 of coil
1 always balances the applied voltage V1 .
• ϕm = ϕ1 − ϕ2 ⇒ N1 I´1 = N2 I2 ( To keep the net flux ϕm constant).
• Here I´1 is called the load component of primary current I1 .
• Assuming I2 to lag behind V2 by an angle θ2 , the phaser diagram under load for an ideal
transformer can be drawn as shown in Fig.
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Transformer under loaded condition
• Since F1 and F2 tend to magnetize the core in
opposite directions, they are shown in phase
opposition.
• The total primary current I1 is the phasor sum of
I´1 and Iϕ , i.e., I1 = I´1 + Iϕ .
• The power factor on the primary side of ideal
transformer is cos θ1 .
• If the magnetizing current Iϕ is neglected, the
following can be written I1 N1 = I2 N2 .
• I1 = ( N2 )I2 = I´2 =Reflected Load Current
N1
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Transformer under loaded condition
Figure 6: Phasor Diagram
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Phasor diagram for a loaded transformer
• With this assumption, it follows that the
secondary terminal voltage V2 is the same as the
e.m.f. E2 induced in the secondary, and the
primary applied voltage V1 is equal to the e.m.f.
E1 induced in the primary winding. Also, if we
again assume equal number of turns on the
primary and secondary windings, then E1 = E2 .
• Let us consider the general case of a load having
a lagging power factor cos ϕ2 ; hence the phasor
representing the secondary current I2 lags V2 by
an angle ϕ2 .
Figure 7: Phasor diagram for a loaded
transformer having negligible voltage drop in
windings
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Phasor diagram for a loaded transformer
• Phasor I´2 represents the component of the primary current to neutralize the
demagnetizing effect of the secondary current and is drawn equal and opposite to I2 .
Here I´2 is described as ‘I2 referred’; I0 is the no-load current of the transformer.
• The phasor sum of I´2 and I0 gives the total current I1 taken from the supply, and the
power factor on the primary side iscos ϕ1 , where ϕ1 is the phase difference between V1
and I1
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Modelling of a Practical Transformer
The behaviour of a transformer may be conveniently considered by assuming it to be
equivalent to an ideal transformer, i.e. a transformer having no losses and no magnetic leakage
and a ferromagnetic core of infinite permeability requiring no magnetizing current.
Now allowing for the imperfections of the actual transformer by means of additional circuits or
impedances inserted between the supply and the primary winding and between the secondary
and the load.
Winding resistances: Since the windings consist of copper conductors, it immediately follows
that both primary and secondary will have winding resistance. The primary resistance R1 and
secondary resistance R2 act in series with the respective winding.
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Equivalent Circuit of a Transformer considering only winding
resistance
Figure 8: Equivalent Circuit considering only winding resistance
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Equivalent Circuit of a Transformer considering leakage
reactance
• In practical transformer, the flux of each limb has to return through air. Since the flux of
each limb is linked only with the winding by which it is produced, it is referred to as
leakage flux and is responsible for inducing an e.m.f. of self-inductance in the winding
with which it is linked.
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Equivalent Circuit of a Transformer considering leakage
reactance
Figure 9: Equivalent Circuit considering only winding resistance
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Equivalent Circuit of a Transformer considering leakage
reactance
• Let ϕℓ1 and ϕℓ2 are the leakage flux in the primary and secondary coil respectively.
• Total flux linking in coil 1: ϕ1 = ϕm1 + ϕℓ1 − ϕm2 .
• Similarly, total flux linking coil 2: ϕ2 = ϕm1 − ϕm2 − ϕℓ2 .
• Voltage induced in coil 1:
N1
d
dϕℓ1
dϕ1
= N1 (ϕm1 − ϕm2 ) + N1
dt
dt
dt
dϕℓ1
= e1 + N1
dt
• Voltage induced in coil 2:
N2
dϕ2
dϕℓ2
= e2 − N2
dt
dt
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Equivalent Circuit of a Transformer considering leakage
reactance
• Considering
N1 I1
Rℓ1
N2 I2
ϕℓ2 =
Rℓ2
ϕℓ1 =
• We can write
dϕℓ1
di1
N 2 di1
= 1
= ℓ1
dt
Rℓ1 dt
dt
2
dϕℓ2
N di2
di2
N2
= 2
= ℓ1
dt
Rℓ2 dt
dt
N1
Here,
N12
= ℓ and
N22
= ℓ are the leakage inductance at primary and secondary
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Equivalent Circuit of a Transformer considering leakage
reactance
• Using these
di1
dϕ1
= e1 + ℓ1
dt
dt
dϕ2
di2
N2
= e2 − ℓ2
dt
dt
N1
• Finally, considering the winding resistance and leakage inductances
di1
+ e1
dt
di2
e2 = v2 + R2 i2 + ℓ2
dt
v1 = R1 i1 + ℓ1
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Equivalent Circuit of a Transformer: winding resistance and
leakage inductances
Figure 10: Equivalent Circuit
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Equivalent Circuit of a Transformer considering leakage
reactance
• The no-load current of the transformer I0 consists of two components: i) Magnetization
current Im produces flux in the transformer core ii) Core loss current Ic to make up the
hysteresis and eddy current losses.
• The magnetizing current Im ∝ V1 and lagging V1 by 90◦ . It can be modelled by a
reactance X0 connected across the V1 .
• Core loss current Ic ∝ V1 in phase with V1 . It can be modelled as resistance R0
connected across the V1 .
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Equivalent Circuit of a Transformer
Figure 11: Equivalent Circuit
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Phasor Equations
V1 = R1 I1 + jX1 I1 + E1 = I1 Z1 + E1
E2 = R2 I2 + jX2 I2 + V2 = I2 Z2 + V2
V2 = ZL I2
E1
Im =
jX0
E1
Ic =
R0
N2
E2 =
E1
N1
N2
I´2 =
I2
N1
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Phasor Diagram
Figure 12: Phasor diagram for lagging power factor of the load
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Equivalent Circuit of a Transformer
• Circuit model of transformer shown in Fig. 11 is not suitable for applying any known
circuit analysis technique since the ideal transformer is present between primary and
secondary.
• Hence we cannot write the KVL/KCL considering the entire loop of the transformer, since
we don’t know the voltage drop across the ideal transformer.
• Therefore, it will be more convenient if we eliminate this ideal transformer and this can be
done by referring the secondary of the ideal transformer to the primary or referring
primary of the ideal transformer to the secondary.
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Equivalent Circuit of a Transformer: referring the secondary of
the ideal transformer to the primary
• By referring , we mean that we want to connect an equivalent circuit across the terminals
of the ideal transformer at primary side such that the current drawn by the equivalent
′
circuit is same as I2 as it is for ideal transformer.
• The current
I2 =
E2
ZL + R2 + jX2
′
2
• Now considering, I2 = N
N I2 , we can write
1
′
I2 =
N2
N1 E2
2 2 N1
(N
N1 ) N2 E2
=
ZL + R2 + jX2
ZL + R2 + jX2
E1
= N
( N1 )2 (ZL + R2 + jX2 )
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Equivalent Circuit of a Transformer (Contd..)
• The last equation indicates that the voltage across the two terminals of the primary side
′
is E1 and the current drawn from this terminals is I2 .
′
1 2
• The same current I2 will flow provide the impedance of ( N
N2 ) (ZL + R2 + jX2 ) is connected
across the primary side.
• Referred parameters at primary side:
′
′
′
R2 =
X2 =
ZL =
N1
N2
2
N1
N2
2
N1
N2
2
R2
X2
ZL
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Equivalent Circuit:referring the secondary of the ideal
transformer to the primary
Figure 13: Equivalent Circuit
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Is the equivalent circuit preserves the power relation?
• Actual power loss in the secondary circuit is I22 R2 . The power loss with referred variables
′
′
I22 R2 . We can write
N2
′
I2 =
I2
N1
2
N1
′
R2 =
R2
N2
′
′
• ∴ I22 R2 = I22 R2 . Hence power relationship is preserved.
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Is the equivalent circuit preserves the kVA relation of load
circuit?
′
• Referred secondary voltage V2
′
′
′
V2 = I2 ZL =
=
′
N2
N1
I2
N1
N2
2
ZL
N1
V2
N2
′
N1
2
• Now I2 = ( N
N )I2 and V2 = ( N )V2 . Therefore,
1
′ ′
2
• Therefore, V2 I2 = V2 I2 . So the kVA consumed by the referred load circuit is same as the
KVA original load circuit.
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Equivalent Circuit:referring the primary of the ideal transformer
to the secondary
Figure 14: Equivalent Circuit
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Equivalent Circuit:referring the primary of the ideal transformer
to the secondary
• The secondary voltage
E2 = I2 (ZL + R2 + jX2 )
′
′
′
′
E2 = V1 − I1 (R1 + jX1 )
′
′
′
′
1
• Therefore, I2 (ZL + R2 + jX2 ) = V1 − I1 (R1 + jX1 ). Multiplying both side by ( N
N ),
2
N1
N1 ′ ′
N1
′
−
I (R + jX1 ) =
I2 (ZL + R2 + jX2 )
N2
N2 1 1
N2
2
N1 ′ ′
N2
N1
N1
′
′
(ZL + R2 + jX2 )
V1 −
I1 (R1 + jX1 ) =
I2
N2
N2
N1
N2
h ′
i
′
′
′
= I2 ZL + R1 + jX1 ) = E1 = V1 − I1 (R1 + jX1 )
′
V1
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Equivalent Circuit:referring the primary of the ideal transformer
to the secondary
• Now, comparing
N1
N2
′
V1 −
N1
N2
′
′
′
I1 (R1 + jX1 ) = V1 − I1 (R1 + jX1 )
′
′
N1
2
• from the above, the following relation can be written V1 = N
V1 or V1 = N
N V1
′
′
′
2
1
• Power consumed by R1 and R1 must be same I12 R1 = I12 R1 . This will be ensured if
N1
′
I1 =
I1
N2
2
N1
′
R1 =
R1
N2
2
′
N1
• Similarly, X1 = N
X1 .
2
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Equivalent Circuit:referring the primary of the ideal transformer
to the secondary
2
2
R0
0
• Power balance to be maintained at the shunt branches E2′ = ER1 or R0 =
E2
N2
E1 = N1 .
′
• R0 =
N2
N1
′
E2
E1
2
R0 . But
2
R0 .
2
′
N2
• Similarly, X0 = N
X0 .
1
′
′
′
N1
N1
1
• Also, I0 = N
I
,
I
=
I
and
I
=
0
c
c
m
N2
N2
N2 Im
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Summary
Secondary referred to primary
′
• I2 =
N2
N1
I2
′
1
• V2 = N
N2 V2
2
′
1
• R2 = N
R2
N2
2
′
1
• X2 = N
X2
N2
2
′
1
• ZL = N
ZL
N2
Primary referred to Secondary
′
2
I1 and V1 = N
N1 V1
′
′
N1
1
• I0 = N
=
I
and
I
0
c
N2
N2 Ic
′
1
• Im = N
N2 Im
2
′
2
• R1 = N
R1
N1
2
′
N2
• X1 = N
X1
1
2
′
2
• X0 = N
X0
N1
2
′
2
• R0 = N
R0
N1
′
• I1 =
N1
N2
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Problem
Figure 15: Single Phase Transformer
48 / 93
Problem
Figure 16: Equivalent Circuit
Find V2 , I2 , I1 , I0 .
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Solution
Figure 17:
50 / 93
Solution
Figure 18: Equivalent Circuit referred to primary
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Solution
Figure 19: Thevenin’s Equivalent Circuit
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Few Salient Features of Equivalent Circuit
• Applied Voltage is 230∠0◦ and VTH = 229.1∠0◦ . Is this very different from the applied
voltage?
• The actual impedance at primary is 0.75Ω + j0.8Ω. The Thevenin’s equivalent resistance
is ZTH = 0.7461 + j0.7978.
• Therefore, it is possibly make sense in order to reduce the computational complexity to
obtain this Thevenin’s equivalent circuit simply omitting the parallel path and connecting
across the supply voltage. In fact that is done in most practical cases. Now, the exact
equivalent circuit is simplified by approximate equivalent circuit.
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Approximate equivalent circuit: Secondary Referred to primary
Figure 20: Approximate equivalent circuit
Figure 21: Approximate equivalent circuit
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Approximate equivalent circuit: Primary Referred to Secondary
2
′
′
2
• Req = N
Req = R2 + R1 =
N1
2
N2
R2 + N
R1
1
2
′
′
N2
• Xeq = N
Xeq = X2 + X1 =
1
2
2
X2 + N
X1
N1
2
2
′
′
N2
N2
• R0 = N
R
;
X
=
X0
0
0
N1
1
′
′
N2
1
• V1 = N
V1 ; V2 = N
N V2
1
′
2
′
N2
1
• I1 = N
N I1 ; I2 = N I2
2
1
Figure 22: Approximate equivalent circuit
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Testing of Single Phase Transformer
• The transformer equivalent circuit whether represented by secondary referred to primary
or primary referred to secondary, useful only if one knows the values of the parameters.
• Now, therefore it is important to be able to determine transformer equivalent circuit
parameter either from design data or carry out certain test/ experiment.
• It is possible to estimate the parameters of the transformer from the dimensions and the
design data. However, most cases the users will not have access such design data and
also the design data gives a first approximation of the parameters. In order to get the
accurate values, user must conduct certain test on transformer.
• There are two tests normally used to find out the equivalent circuit parameters of the
transformer.
• No load Test: which finds out the shunt branch paraments R0 , X0 .
• Short Circuit Test: which finds the series branch parameters Req , Xeq .
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No Load Test/ Open Circuit Test
• In order to do the no load test, a voltmeter, wattmeter and an ammeter are connected on
the low voltage side of the transformer.
• The high voltage side is left open circuited.
• The variac is started from zero voltage position and the voltage supplied by the variac is
slowly raised till the voltmeter connected across the primary reads the rated volatge of
L.V. winding. For example 110/220 Volt transformer, LV side voltage 110 volt and HV
side voltage 220 volt.
• Under this condition, take the readings: V1 , I0 and W0 . Where voltmeter readings gives
V1 gives LV side applied voltage, ammeter reading I0 gives no load current and wattmeter
reading W0 gives core loss or iron loss.
• Now, how to use no load test data to find out shunt branch parameters?
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Circuit diagram for open circuit test
Figure 23: Circuit diagram for open circuit test
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At no load the transformer equivalent circuit and phasor diagram
Figure 24: equivalent circuit
Figure 25: Phasor diagram
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No Load Test/ Open Circuit Test
From equivalent circuit,
V1
• R0 = VI 1 = I cos
ϕ (using phasor diagram)
c
0
0
V1
• X0 = VI 1 = I sin
ϕ0 (using phasor diagram)
m
0
• Now, W0 = V1 I0 cos ϕ0 . ∴ cos ϕ0 = VWI0 and sin ϕ0 =
1 0
p
1 − cos2 ϕ0 .
• Therefore
R0 =
X0 =
V1
V2
= 1
I0 cos ϕ0
W0
V1
; sin ϕ0 =
I0 sin ϕ0
s
1−
W02
.
V12 I02
• As the test is performed at L.V. side, the values of R0 and X0 will be referred to L.V. side.
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Short Circuit Test
• Using this test we can find out the series branch parameters Req , Xeq .
• The low voltage side of the transformer is short circuited and the instruments are placed
on the high voltage side.
• The applied voltage is adjusted by variac, to circulate rated current in the high voltage
side.
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Circuit diagram for short circuit test
Figure 26: Circuit diagram for short circuit test
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Short Circuit Test
• How do we find out the series branch parameters from the short circuit test. Let us draw
the approximate equivalent circuit for short circuit test.
• Even Vs is at rated voltage , I0 is 5 to 7% of the rated current, Is .
• Therefore, when we reduce Vs in order to have Is and short the L.V. side and f Is is
restricted to its rated value , the voltage Vs is required to send I0 will be very small/
negligible current.
• Hence, magnetization branch is neglected during short circuit test.
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Approximate equivalent circuit of short circuit test
Figure 27: Equivalent Circuit and phasor diagram
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Short Circuit Test
• Then we have very simple relation:
Zs =
q
|Vs |
2 + X2 =
Req
eq
|Is |
• The power loss during short circuit test is entirely due to the series resistance. Therefore,
Ws = Is2 Req
Ws
Req = 2
I
qs
2
Xeq = Zs2 − Req
• These values are referred to HV side.
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Problem
• A 20kVA, 2500/250V, 50Hz, single phase transformer gave the following test result
• Open circuit test (on LV side): 250V, 1.4 A, 105 watts
• Short circuit test (on HV side): 104V, 8A, 320 watts
Compute the parameters of the approximate equivalent circuit referred to high-voltage
and low voltage sides. Also draw the exact equivalent circuit referred to low-voltage side.
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Solution
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Efficiency of transformer
• Certain performance parameters are important from users point of view. For transformer,
Power
the efficiency(η) is defined as η = Output
Input Power .
• Now, why there will be difference in output power and input power? It is due to losses.
• In a practical transformer we have seen mainly two types of major losses namely core and
copper losses occur. These losses are wasted as heat and temperature of the transformer
rises. Therefore output power of the transformer will be always less than the input power
drawn by the primary from the source.
• Core Loss: These are hysteresis and eddy current losses. Core loss is constant for
transformer operated at constant voltage and frequency.
• Copper Loss (I 2 R-Loss): This loss occurs in winding resistances when the transformer
carries the load current. Copper loss varies as the square of the loading expressed as ratio
of the full load.
• Major Losses i) Core loss (or Iron Loss)⇒ Pi = constant loss. ii) Copper loss ⇒ Pcu =
variable loss.
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Efficiency of transformer
• For practical transformer, efficiency(η) is defined as:
Output power in KW
Output power in KW + Losses
Output power in KW
=
Output power in KW + Core Loss+ Copper Loss
η=
• It can be shown that from no load to the full load condition the core loss, Pi remains
practically constant since the level of flux remains practically same.
• On the other hand we know that the winding currents depend upon the degree of loading
and copper loss directly depends upon the square of the current and not a constant from
no load to full load condition.
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Efficiency of transformer
′
• Copper loss in the two windings are: Cu-Loss = (I2 )2 R1 + (I2 )2 R2 .

!2 
′
I
′ 2
R1  I22
(I2 ) R1 + (I2 )2 R2 = R2 + 2
I2
"
2 #
N2
R1 I22
= R2 +
N1
′
= Req I22
′
Here, Req is the equivalent resistance referred to secondary.
• Let, x = I I2 , where I2,fl is the full load current at secondary side.
2,fl
• I2 = xI2,fl . (full load is meant the load (obviously, on secondary) which would make
transformer transfer it’s rated power from primary to secondary.)
′
′
2 R ′ = x2 P . Where P
2
• Now, copper loss=Req (xI2,fl )2 = x2 I2,fl
cu
cu = I2,fl Req = full-load
eq
cu-loss. Pcu is the wattmeter reading during short-circuit test.
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Efficiency of transformer
• Output power of the transformer
= V2 I2 cos ϕ
= xV2 I2,fl cos ϕ
= xPfl
where Pfl = V2 I2,fl cos ϕ= full load output. Efficiency at any fractional load x,
xPfl
xPfl + Pi + x2 Pcu
x(kVA)RATED cos ϕ
=
x(kVA)RATED cos ϕ + Pi + x2 Pcu
ηx =
• Here, Pi is the wattmeter reading during no load test.
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When Efficiency will be maximum?
• For efficiency to be maximum, d(Denominator)
is set to zero and we get
dx
d
Pi
Pfl +
+ xPcu = 0
dx
x
Pi
− 2 + Pcu = 0
x
Pi
x2 =
Pcu
q
• The fractional load at which the efficiency will be maximum x = PPi . Thus we see that
cu
for a given
q power factor, transformer will operate at maximum efficiency when it is loaded
to core
Pi
Pcu (kVA)RATED .
• For transformers intended to be used continuously at its rated kVA, should be designed
such that maximum efficiency occurs at x = 1. That is, core-loss=cu-loss. Power
transformers fall under this category.
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Discussion
• However for transformers whose load widely varies over time, it is not desirable to have
maximum efficiency at x = 1. Distribution transformers fall under this category and the
typical value of x for maximum efficiency for such transformers may between 0.75 to 0.8.
• Figure shows a family of efficiency Vs. degree of loading curves with power factor as
parameter. It canq
be seen that for any given power factor, maximum efficiency occurs at
a loading of x =
Pi
Pcu .
• Efficiencies ηmax1 , ηmax2 andηmax3 are respectively the maximum efficiencies corresponding
to power factors of unity, 0.8 and 0.7 respectively.
• It can easily be shown that for a given load (i.e., fixed x), if power factor is allowed to
vary then maximum efficiency occurs at unity power factor.
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Efficiency VS degree of loading curves
Figure 28: Efficiency VS degree of loading curves
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Voltage Regulation
• In a single phase transformer, the supply voltage at primary is constant with constant
frequency. In most cases the load is like to see the a constant voltage across the load
which is independent of load current I2 .
• However, in a practical transformer usually it is not the case. There will be some variation
in load voltage depending on load current.
• This can be predicted from the approximate equivalent circuit which we referred to load
side (secondary).
′
• Obviously, V2 will not be same as V1 for all values of I2 because there will be internal
voltage drop in the series leakage impedance of the transformer and the magnitude of
which will depend upon the degree of loading as well as on the power factor of the load.
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Approximate equivalent circuit: Primary referred to secondary
Figure 29: Approximate equivalent circuit
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Voltage Regulation
• The knowledge of regulation gives us idea about change in the magnitude of the
secondary voltage from no load to full load condition at a given power factor.
• This can be determined experimentally by direct loading of the transformer. To do this,
primary is energized with rated voltage and the secondary terminal voltage is recorded in
absence of any load and also in presence of full load.
• Suppose the readings of the voltmeters are respectively V2,0 and V2,f ℓ . Therefore change
in the magnitudes of the secondary voltage is V2,0 –V2,f ℓ . This change is expressed as a
percentage of the full load secondary voltage to express regulation.
• Lower value of regulation will ensure lesser fluctuation of the voltage across the loads. If
the transformer were ideal regulation would have been zero.
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Voltage Regulation
• Definition: The voltage regulation of transformer is defined as the net change in the
secondary terminal voltage from no-load to full load, expressed as a percentage of its
rated voltage.
• % voltage regulation= V2,0V–V2,f ℓ × 100.
2,f ℓ
• where V2,0 = secondary voltage when load is thrown off (i.e., V2,0 = E2 ).
• and V2,f ℓ = Full-load secondary voltage ( it is assumed to be adjusted to the rated
secondary voltage.
• voltage regulation is figure of merit of transformer. For an ideal transformer voltage
regulation is zero.
• smaller is the voltage regulation, better is the operation of the transformer.
• In order to calculate the % of regulation, we can neglect the shunt branch of the
approximate equivalent circuit referred to secondary. The equivalent circuit for the
calculation of % regulation is shown next
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Approximate equivalent circuit: Primary referred to secondary
Figure 30: Approximate equivalent circuit
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Case-I: Unity Power Factor Load
Figure 32: Phasor diagram
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Case-II: Lagging Power Factor Load
Figure 33: Phasor Diagram
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Case-II: Lagging Power Factor Load
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Case-II: Lagging Power Factor Load
Figure 35: Phasor diagram
Figure 34: equivalent circuit
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Case-III: Leading Power Factor Load
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Solved Numerical Problem
A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary.
The no-load current drawn is 3A at a power factor 0.2 lagging. Calculate the total primary
current and power factor at primary when the secondary is loaded to a current of 280 A at a
power factor of 0.8 lagging. Assume the voltage drop in the windings to be negligible.
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Solution
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Problem
A 4kVA, 200/400V, 50Hz, single phase transformer gave the following test result
• Open circuit test (on LV side): 200V, 0.7 A, 35 watts
• Short circuit test (on HV side): 30V, 10A, 90 watts
a) Compute the parameters of the approximate equivalent circuit referred to high-voltage and
low voltage sides.
b) Calculate the efficiency at 100% of the load with 0.8 power factor.
c) Calculate the efficiency at 50% of the load with 0.8 power factor.
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Solution
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Problem
• A 500 kVA transformer has an efficiency of 95 % at full-load, and the same efficiency also
at 60 % of full-load; both loads at upf (unity power factor) .
a) Compute the iron and copper losses of the transformer.
b) Determine the efficiency of the transformer at 3/4th of full-load.
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Solution
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Problem
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Solution
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The End
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