Experiment 2
Single Phase Transformer (I)
Objectives
•
•
•
To determine the turns ratio of a single phase transformer.
To determine the parameter values of the excitation branch that model the core and magnetization
losses in a real transformer.
To determine the parameter values of components that model the copper losses and leakage flux in a
real transformer.
Turns Ratio Test
Introduction
From Figure 2.1 the turn’s ratio of a single phase transformer is defined as the ratio of the primary
winding voltage V1 to the secondary winding voltage V2 when no load is applied as follow
a=
V1 VP
=
V2 VS
(2.1)
ip
is
Ns
Np
+
+
v
vs
p
-
ip
is
Ns
Np
+
v
+
vs
p
-
-
Figure 2.1: Schematic symbol of a transformer
In an ideal transformer, there is no leakage in the magnetic field (Φ) flowing through the iron core and
since Faradays law states that "If a flux passes through a turn of a coil of wire, a voltage will be induced
in the turns of wire that is directly proportional to the rate of change in the flux with respect to time" so
that
eind = -N
dφ
dt
(2.2)
then, it is clear that
−
dφ V1 V2
=
=
dt N1 N 2
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(2.3)
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Experiment 2
Single Phase Transformer (I)
thus
a=
V1 N1
=
V2 N 2
(2.4)
Moreover, the apparent power consumed in both sides of the transformer is the same such that
S1 = V1I1 = V2 I 2 = S2
(2.5)
then, it is clear that
a=
V1 I 2
=
V2 I1
(2.6)
The exact equivalent circuit of a real transformer is shown in Figure 2.2 where V1 is the supply voltage
(primary), V2 is the load voltage (secondary), R1 and R2 are the winding internal resistances, X1 and X2
are the winding leakage reactances, Rm is the core loss resistance, Xm is the magnetizing reactance, and
a is transformer turns ratio. From Figure 2.2, the following relationship can be obtained
ZL =
VL VS V2
=
=
IL
IS
I2
(2.7)
then
Z′L = a 2 ⋅ Z L
(2.8)
where Z′L is the apparent impedance of the primary circuit of the transformer.
(a)
(b)
(c)
Figure 2.2: (a) Transformer equivalent circuit, (b) Transformer model referred to its primary voltage,
and (c) Transformer model referred to its secondary voltage
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Experiment 2
Single Phase Transformer (I)
Figure 2.3: Real photo of lab equipments needed for the experiment
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Experiment 2
Single Phase Transformer (I)
Procedures
Using the equipments shown in Figure 2.3, connect the circuit show in Figure 2.6.
Calculate the transformer turns ratio "a" from the measurement of primary and secondary voltages
V1 [V]
a=
V2 [V]
V1
V2
Open-Circuit Test
Introduction
In the open-circuit test, a transformers secondary windings is open circuited, and its primary windings is
connected to a full-rated line voltage. (Refer to the equivalent circuit in Figure 2.4). Under the
conditions described, all the input current must be flowing through the excitation branch of the
transformer. The series elements RP and XP are too small in comparison to RC and XM to cause a
significant voltage drop, so essentially all the input voltage is dropped across the excitation branch.
(a)
(b)
Figure 2.4: (a) Transformer model referred to its primary voltage, and (b) Transformer model referred to
its secondary voltage
The open-circuit test connections are shown in Figure 2.5. Full line voltage is applied to the primary of
the transformer, and the input voltage, input current, and input power to the transformer are measured.
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Experiment 2
Single Phase Transformer (I)
From this information, it's possible to determine the power factor of the input current and therefore both
the magnitude and the angle of the excitation impedance.
Figure 2.5: Open-circuit equivalent circuit
The easiest way to calculate the values of RC and XM is to look first at the admittance of the excitation
branch. The conductance of the core-loss resistor is given by:
GC =
1
RC
(2.9)
And the susceptance of the magnetizating inductor is given by:
BM =
1
XM
(2.10)
Since these two elements are in parallel, their admittances add, and the total excitation admittance is
YE = G C - jB M =
1
1
-j
R C XM
(2.11)
The magnitude of the excitation admittance (referred to the primary circuit) can be found from the opencircuit test voltage and current as follows
YE =
I OC
VOC
(2.12)
The angle of admittance can be found from the knowledge of the circuit power factor. The open circuit
power factor (PF) is given by
⎛ POC ⎞
⎟⎟
(2.13)
PF = cos θ = ⎜⎜
⎝ I OC ⋅ VOC ⎠
and the power factor angle θ is given by
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Experiment 2
Single Phase Transformer (I)
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θ = COS (
POC
)
IOC ⋅ VOC
(2.14)
The power factor is always lagging for a real transformer, so the angle of the current always lags the
angle of the voltage by θ . Therefore the admittance YE is
YE =
IOC
∠COS-1PF
VOC
(2.15)
Using the above equations, it is possible to determine the values of RC and XM directly from the opencircuit test data.
Procedures
Using the lab equipments shown in Figure 2.3, do the following:
1. Connect the circuit shown in Figure 2.6.
2. Vary the applied voltage on the low voltage side from 220 to 150V.
3. Record the readings required in Table 2.1 and then calculate the power factor, I R C and I XM at no
load.
V1 [V]
IM [A]
V2 [V]
Wo [W]
PF =
COS(θo)
Calculated
I R C [A]
Calculated
I X M [A]
220
200
175
150
Table 2.1: Open circuit test measurements and calculations.
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Experiment 2
Single Phase Transformer (I)
Figure 2.6.a: Open circuit test wiring diagram
Figure 2.6.b: Open circuit test schematic circuit
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Experiment 2
Single Phase Transformer (I)
Short-Circuit Test
Introduction
In the short circuit test, the secondary terminals of the transformer are short-circuited, and the primary
terminals are connected to fairly low-voltages source, as shown in Figure 2.7.
Figure 2.7: Connection for transformer short-circuit test
The input voltage is adjusted until the current in the short-circuited windings is equal to its rated value
(be sure to keep the primary voltage at a safe level). Since the input voltage is too low during the short
circuit test, negligible current flows through the excitation branch. If the excitation current is ignored,
then all the voltage drop in the transformer can be attributed to the series elements in the circuit. The
magnitude of the series impedances referred to the primary side of the transformer is
ZSE =
VSC
ISC
(2.16)
The power factor of the current is given by
⎛ PSC ⎞
PF = COSθ = ⎜
⎟
⎝ ISC ⋅ VSC ⎠
(2.17)
and is lagging. The current angle is thus negative, and the overall impedance angle θ is positive as
PSC
)
ISC ⋅ VSC
(2.18)
VSC ∠0o VSC o
ZSE =
=
∠θ
ISC ∠-θ o ISC
(2.19)
ZSE = R eq + jXeq = (R p + a 2 R s ) + j(X p + a 2 Xs )
(2.20)
-1
θ = COS (
therefore,
The series impedance ZSE is equal to
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Experiment 2
Single Phase Transformer (I)
It is possible to determine the total series impedance referred to the primary side by using this technique,
but there is no easy way to split the series impedance into primary and secondary components.
Fortunately, such separation is not necessary to solve normal problems.
These same tests may also be performed on the secondary side of the transformer if it is more
convenient to do so because of voltage levels or other reasons. If the tests are performed of the
secondary side, the results will naturally yield the equivalent circuit impedances referred to the
secondary side of the transformer instead if the primary side.
Procedures
Using the lab equipments shown in Figure 2.3, do the following:
1. Connect the circuit shown in Figure 2.8.
2. Increase the supply voltage gradually from zero until rated current is measured in the primary
winding.
3. Do all measurements and calculations needed to complete Table 2.2.
I1(SC) [A]
V1(SC) [V]
W1(SC) [W]
PF=
COS(θ o)
Zeq [Ώ]
Req [Ώ]
Xeq [Ώ]
1.5
1.2
0.9
0.6
0.3
Table 2.2: Short circuit test measurements and calculations
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Experiment 2
Single Phase Transformer (I)
Figure 2.8.a: Short circuit test wiring diagram
Figure 2.8.b: Short circuit test schematic circuit
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