1
Single and Three Phase Transformers and Power Flow
Caution: In these experiments high voltages are involved, which could be harmful or fatal
if one is exposed to them. Adhere to electrical safety rule at all times. Make sure that all of
your connections are correct before turning on the power.
A transformer is an electrical device that is used to raise or lower the voltage (or current) level. A
single-phase transformer has two electrically isolated windings. The winding that is connected to
the electrical power source is called the “primary” winding and the winding that is used to draw
electrical power is called the “secondary” winding. Fig. 1 shows a schematic diagram of a singlephase transformer.
Φm
Figure 1: Schematic diagram of a single phase transformer.
Let us examine the ideal transformer shown in Fig. 1 where,
EP = induced voltage on the primary side (RMS)
ES = induced voltage on the secondary side (RMS)
NP = Number of turns on the primary windings
NS = Number of turns on the secondary windings
Syed A. Rizvi
Summer’12-Revised Spring’16
ENS 441/ELT 437
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Φm
= magnetic flux created by the source voltage E
When the primary winding is connected to an AC power source it produces an instantaneous
magnetic flux “Φm (t)” linking the primary as well as secondary windings of the transformer. The
flux linkage induces an instantaneous voltage “Ep (t)” in the primary winding in accordance with
the Faraday’s law of electromagnetic induction. That is,
𝐸𝑝 (𝑡) = 𝑁𝑝
𝑑∅𝑚 (𝑡)
𝑑𝑡
(1)
Since the voltage applied at the primary is sinusoidal, therefore, the magnetic flux Φm (t) is also
sinusoidal and mathematically it can be expressed as
∅𝑚 (𝑡) = ∅𝑚𝑎𝑥 sin 𝜔𝑡
(2)
𝑑∅𝑚 (𝑡)
𝑑(sin 𝜔𝑡)
= 𝑁𝑝 × ∅𝑚𝑎𝑥
𝑑𝑡
𝑑𝑡
(3)
therefore,
𝐸𝑝 (𝑡) = 𝑁𝑝
or
𝐸𝑝 (𝑡) = 𝑁𝑝 × ∅𝑚𝑎𝑥 × 𝜔 cos 𝜔𝑡.
(4)
Substituting cos 𝜔𝑡 = sin(𝜔𝑡 + 90𝑜 ) and 𝜔 = 2𝜋𝑓 in Eq. 3 we get
𝐸𝑝 (𝑡) = 2𝜋𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥 sin(𝜔𝑡 + 90𝑜 ).
(5)
Eq. 4 shows that the induced voltage Ep (t) leads the magnetic flux Φm (t) by 90o. The RMS value
of Ep (t), denoted by Ep is given by
𝐸𝑝 = 𝑅𝑀𝑆[𝐸𝑝 (𝑡)] =
|𝐸𝑝 (𝑡)|𝑚𝑎𝑥
√2
=
2𝜋𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥
(6)
√2
or
𝐸𝑝 = 4.44 𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥 .
(7)
Since the magnetic flux Φm (t) also links the secondary winding that has Ns turns, the RMS value
of the induced voltage at the secondary side of the transformer, denoted by Es is given by
𝐸𝑠 = 4.44 𝑓 × 𝑁𝑠 × ∅𝑚𝑎𝑥 .
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Comparing Eqs. (7) and (8) yields
𝐸𝑝
𝐸𝑠
=
4.44 𝑓 × 𝑁𝑝
4.44 𝑓 × 𝑁𝑠
(9)
𝐸𝑝
𝐸𝑠
=
𝑁𝑝
𝑁𝑠
(10)
𝐸𝑝
𝑁𝑝
=
=𝑎
𝐸𝑠
𝑁𝑠
(11)
𝐸𝑝 = 𝑎 × 𝐸𝑠
(12)
or
or
where, “a” is called the turn ratio of the transformer.
Figure 2: An ideal transformer with a load on the secondary side.
When secondary side is open (no load condition) under ideal conditions no current flows through
the primary winding and the primary current “IP” is zero. When a load is connected to the
secondary side (see Fig. 2), a current, Is, flows out of the secondary windings. The secondary
current, Is, creates a magnetic flux “Φs” in the transformer’s core. Note that the source voltage, E,
is unchanged, which mean the flux, Φm, must remain the same as it was before the load was
connected. Therefore, the new flux (Φs) due to the secondary the current, Is, must be countered
by an equal and opposite flux. The counter flux “Φp” is produced by causing a current, Ip, to flow
into the primary windings. Under ideal conditions, the fluxes Φs and Φp cancel each other and the
total flux through the core remains Φm. Therefore, the primary and secondary induced voltages
(Ep and Es) remain unchanged.
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Summer’12-Revised Spring’16
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The flow of the primary current, Ip, represents the power transfer from primary side to the
secondary side of the transformer to support the load Zs. Under ideal conditions there is no loss
of power during the power transfer from the primary side to the secondary side of the
transformer. Therefore, the total power, Sp, on the primary side of the transformer must be equal
to the total power, Ss, on the secondary side of the transformer, where
𝑆𝑝 = 𝐸𝑝 × 𝐼𝑝
(13)
𝑆𝑠 = 𝐸𝑠 × 𝐼𝑠
(14)
𝐸𝑝 × 𝐼𝑝 = 𝐸𝑠 × 𝐼𝑠 .
(15)
𝑎 × 𝐸𝑠 × 𝐼𝑝 = 𝐸𝑠 × 𝐼𝑠
(16)
and
but Sp = Ss, therefore,
Combining Eqs. 12 and 14 yeilds
or
𝐼𝑝 =
1
× 𝐼𝑠 .
𝑎
(17)
The load impedance as seen by the primary side, denoted by Zp, can be given by
𝑍𝑝 =
𝐸𝑝 𝑎 × 𝐸𝑠
𝐸𝑠
=
= 𝑎2 ×
1
𝐼𝑝
𝐼𝑠
𝑎 × 𝐼𝑠
(18)
𝐸𝑠
𝐼𝑠
(19)
but
𝑍𝑠 =
therefore,
𝑍𝑝 = 𝑎2 × 𝑍𝑠 .
(20)
Equation 20 shows an important property of the transformer; that is, it can be used to change the
effective impedance of any load and thus can also act as an impedance transformer. Equation 20
is used to express impedances on the secondary side as the equivalent impedances on the primary
side to greatly simplify the circuit analysis involving transformers (see Fig. 3).
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Figure 3: An ideal transformer with the load shifted to the primary side.
So far we have considered an ideal transformer with no primary current when there was no load
attached to the secondary side. That is, the ideal transformer is lossless. In a real transformer,
however, the primary current is not zero under no-load condition due to the “iron losses” and
“magnetizing reactance” of the transformer. The iron losses are due to eddy currents and
hysteresis losses (described below).
Eddy Current Losses:
When time-varying magnetic flux passes through a solid metal core, the solid metallic core acts
like a bundle of concentric conductors and a voltage is induced in each of these sections of the
metallic core. Note that in a solid metal plate each of these sections would behave like a shortcircuited conductor and, therefore, the induce voltages would cause a strong currents which swirl
back and forth in the core. These currents are called eddy currents. Fig. 4 shows a cross-section
of a solid metal core with circulating eddy currents. Eddy currents are a major source of
generating heat in transformers. In a solid metal core these currents can be so strong that they
render the core red hot. To minimize eddy current, cores are designed with thin laminated stripes
instead of using a solid piece of metal. The lamination insulates the strips from each other, thus
greatly reducing the length of the conducting path, which in turn, significantly reduces the
magnitudes of the induced voltages in the core thereby diminishing eddy current losses. In
general, eddy current losses are reduced proportional to the square of the number of strips used
in the core.
Figure 4: Eddy currents in a solid metallic core.
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Summer’12-Revised Spring’16
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Hysteresis Losses
The magnetic flux is caused by magnetic field strength, H. The resulting flux density, B, in any
material is related to the magnetic field strength (H) by the following equation:
𝐵 = 𝜇𝑜 × 𝜇𝑟 × 𝐻
(21)
where,
B = magnetic flux density (weber/m2 or tesla)
H = magnetic field strength (A-turns/m)
µo = permeability of free space = 4𝜋 × 10−7
µr = relative permeability of the material
Note that the value of µr changes with flux and, therefore, the relationship between B and H is
expressed in terms of a B-H curve. Under an AC current the iron core is continuously
magnetized in one direction, demagnetized, and then magnetized again in the opposite direction.
However, the relationship between B and H is not linear. Fig. 5 shows the change in flux density
with the magnetic field intensity for one complete cycle of AC current. Initially, B reaches its
peak (let’s assume the positive peak) as H reaches the maximum field strength (at maximum
current in the positive direction). After that H decrease with the decreasing current and so does
B; however, B decreases at a slower rate than H following the path a-b. The result is that as the
current reduces to zero bringing H to zero, there still remains some residual flux density in the
core (“+Br” in Fig. 5). As the current starts increasing in the negative direction, producing a
negative magnetic field strength, B continue to decrease following the path b-c and eventually
reduces to zero at a field strength of “-Hc.” As the current continue to decrease in negative
direction B starts to decrease in negative direction following the path c-d and eventually reach its
maximum value in the negative direction (-Bmax) as H reaches its negative peak. As the current
starts decreasing again reducing H, B decreases as well, but with a slower rate following the path
d-e. Consequently, when the current and H reduce to zero, B still maintains a negative flux of –
Br in the core. B eventually reduces to zero at when H = +Hc following the path e-f. B then
increases to its peak value following the path f-a, as H continues to increase and eventually peaks
(point “a” in Fig. 5).
The closed B-H curve shown in Fig. 5 is called a hysteresis loop and the area under this curve
represents the energy that changes into heat during each cycle (joules per cubic meters). To
minimize the hysteresis losses materials with relatively narrow hysteresis loop are used to build
transformer cores. Since the iron losses (eddy current and hysteresis losses) only generate heat
and do not affect the induced voltage they can be represented as resistance (Rm) in parallel to the
primary terminals of the transformer.
Furthermore, a current must flow through the primary windings to generate the magnetic flux
Φm. This current is called magnetizing current and depends on the permeability of the material
used to build the core. Higher permeability means lower magnetic current and vice versa.
Permeability can be represented by a parallel inductance (Xm) to the primary terminal because
magnetizing current does not affect the amount of induced voltage. This inductance is called
magnetizing inductance.
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Summer’12-Revised Spring’16
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Figure 5: Hysteresis loop.
In a real transformer, the winding of the transformer has a finite resistance that must be taken
into account while analyzing or modeling a practical transformer. Since these winding
resistances result in a potential drop when current passes through the windings, they are
represented as the series resistors (Rp and RS), both at the primary as well as at the secondary
side of the transformer.
In the analysis of the ideal transformer we assumed the flux due to the load current (Φs) and the
counter flux (Φp) generated by the primary remain completely inside the core. However, in a real
transformer some of the flux Φs and Φp does leak in the air. This flux leakage induces voltages
both in the primary winding and secondary windings, which is counter to the voltages induced by
the flux Φm. Consequently, it results in a voltage drop both at the primary as well as the
secondary side. Accordingly, the effect of the leakage flux is represented as the series
impedances (Xlp and Xls) at the primary as well as the secondary side of the transformer of the
model of a practical transformer. Fig. 6 shows the complete model of practical transformer that
incorporate the iron losses, magnetizing current, primary and secondary winding resistances, and
the effect of the leakage flux under load conditions.
Figure 6: Model of a practical transformer.
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Summer’12-Revised Spring’16
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The model of the practical transformer shown in Fig. 6 becomes a lot easier to analyze if the
elements on the secondary side are shifted to the primary side using equation (20). Figure 7
shows the model of the practical transformer when everything is viewed from the primary side.
We will use the model in Fig. 7 in our analysis of practical transformers.
Figure 7: Model of a practical transformer when the circuit is viewed from the primary side.
Let’s now examine the model for the practical transformer shown in Fig. 7 for two special load
conditions: (1) No-load condition and (2) full-load condition.
No-load Condition:
Under no-load condition the secondary is open and, therefore, no current flows through the
secondary terminals. Consequently, I/p = 0, which means that Rs and Xls can be eliminated from
the circuit. Also, Ip = Io, that is the branch containing Rp and Xlp now appear in series with the
parallel combination of Rm and Xm. Note that both Rp and Xlp are significantly smaller than Rm
and Xm that they can safely be ignored when compared to Rm and Xm. This simplifies the model
of the practical transformer under no-load condition to the one given in Fig. 8.
Full-load Condition:
Note under the full-load condition 𝐼𝑝 ≫ 𝐼𝑜 and, therefore, we can ignore Rm and Xm and the
model for the practical transformer reduces to the one shown in Fig. 9. The model of the
practical transformer can further be simplified by adding the resistances and inductances as
follows:
𝑅𝑒𝑝 = 𝑅𝑝 + 𝑎2 × 𝑅𝑠
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Figure 8: Model for the practical transformer under no-load condition.
Figure 9: Model of a practical transformer under full-load condition.
and
𝑋𝑒𝑝 = 𝑋𝑙𝑝 + 𝑎2 × 𝑋𝑙𝑠 .
(23)
Figure 10 shows the simplified circuit for the practical transformer with full load. Furthermore,
Rlp and Xlp can be represented by the complex impedance Zp as follows:
𝑍𝑝 = 𝑅𝑒𝑝 + 𝑗𝑋𝑒𝑝
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Figure 10: Simplified circuit of the practical transformer under full-load condition.
Zp is referred to as transformer impedance and is an important parameter that is used to compute
the internal voltage drop of a transformer for varying load conditions (as long as Ip ≥ 10 Io).
Example: Specifications of a power transformer are as follows:
Sn
Enp
Ens
Inp
Ins
Rp
Rs
(KVA)
(V)
(V)
(A)
(A)
(Ω)
(Ω)
1000
69000 6900
14.5
145
27.2
0.25
Xlp
Xls
Xm (Ω)
151
1.5
505000 432000 0.210
(Ω)
(Ω)
Rm (Ω)
Io (A)
Compute the full-load Vs, Is, as well as heat dissipation in the transformer using the simplified
model of Fig. 10. Also, compute Iron losses using the model under no-load condition of Fig. 8.
Solution: Let’s assume it’s a pure resistive load and full-load secondary voltage is the same as
Vns and full-load secondary current is the same as Ins. Now we can calculate the load resistance,
RL, as follows:
𝑅𝐿 =
𝐸𝑛𝑠
𝐼𝑛𝑠
=
6900
14.5
= 47.6 Ω
The turn ratio is given by
𝑎=
𝐸𝑛𝑝 69000
=
= 10
𝐸𝑛𝑠
6900
Now we can compute Rep and Xep
𝑅𝑒𝑝 = 𝑅𝑝 + 𝑎2 × 𝑅𝑠 = 27.2 + 102 × 0.25 = 52.2 Ω
𝑋𝑒𝑝 = 𝑋𝑝 + 𝑎2 × 𝑋𝑠 = 151 + 102 × 1.5 = 301 Ω
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The resistance shifted to primary, RLp, is given by
𝑅𝐿𝑝 = 𝑎2 × 𝑅𝐿 = 102 × 47.6 = 4760 Ω.
Now we can compute the total impedance of the model, ZTp, as follows
𝑍𝑇𝑝 = 𝑅𝑒𝑝 + 𝑅𝐿𝑝 + 𝑗𝑋𝑒𝑝 = 52.2 + 4760 + 𝑗301 = 4812.2 + 𝑗301 = 4821.6 ⟨0.0186𝑜 Ω
Now we can compute primary current, Ip, primary voltage, Ep, secondary current,
Is, and secondary voltage, Es as follows
𝐼𝑝 =
𝐸𝑛𝑝 69000
=
= 14.31 𝐴
𝑍𝑇𝑝 4821.6
𝐸𝑝 = 𝐼𝑝 × 𝑍𝑇𝑝 = 14.31 × 4821.6 = 68118.4 𝑉
𝐸𝑠 =
𝐸𝑝 68118.4
=
= 6811.84 𝑉
𝑎
10
𝐼𝑠 =
𝐸𝑠 6811.84
=
= 143.1 𝐴
𝑅𝐿
47.6
Now we can compute the heat loss due to the load as follows
𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 = 𝐼𝑝2 × 𝑅𝑒𝑝 = 14.312 × 52.2 = 10689.3 𝑊 = 10.69 𝐾𝑊
Finally, we’ll compute iron losses at no load and at full load as follows
2
𝐸𝑛𝑝
690002
𝐼𝑟𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑡 𝑛𝑜 𝑙𝑜𝑎𝑑 =
=
= 11020.83 𝑊 = 11.02 𝐾𝑊
𝑅𝑚 432000
𝐼𝑟𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑡 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 =
𝐸𝑝2 68118.42
=
= 10741 𝑊 = 10.74 𝐾𝑊
𝑅𝑚
432000
Total power loss at full load = 10.69 + 10.74 = 21.43 KW.
This power loss produces a significant amount of heat that must be removed from the
transformer to avoid any damage to the transformer. There are a number of techniques that are
used in the transformers to remove heat, such as through the natural circulation of air (type AA),
through forced air circulation (type AFA), oil-immersed self-cooled (OA), and oil immersed selfcooled/forced air cooled (type OA/FA) to name a few. Figure 11 shows the front view of an
OA/FA type power transformer and Fig. 12 shows the rear view of the same transformer where
cooling fans can be seen, which remove heat from the transformer through forced air circulation.
A transformer is called a “step-down” transformer if the secondary voltage, ES, is lower than the
primary voltage, EP. On the hand, if the secondary voltage is higher than the primary voltage, it
is called a “step-up” transformer.
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Summer’12-Revised Spring’16
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Figure 11: Front view of an OA/FA type power transformer.
Figure 12: Rear view of an OA/FA type power transformer. The cooling fans can be in, which
are used for forced air circulation to remove heat from the power transformer.
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Terminal Marking on Power Transformers:
In power transformers, the high voltage (HV) terminals are marked by symbols H1 and H2; the
low voltage (LV) terminals are marked by symbols X1 and X2. These terminals are mounted on
the transformer tank in a way that they either have additive polarity or they have subtractive
polarity. The transformer is said to have additive polarity when the terminal H1 is mounted
diagonally opposite to the terminal X1. On the other hand, if the terminal H1 is mounted directly
opposite to the terminal X1, it is said to have subtractive polarity (see Fig. 13). When you face
the voltage side of a power transformer, the terminal at the extreme right is H1, and H2, and so
on. The location of the low voltage side terminal would depend on whether the transformer has
an additive or subtractive polarity. Subtractive polarity is standard on transformer above 200
KVA with HV rating of above 8660 V. Otherwise the transformer would have additive polarity.
Note that it is imperative to know the correct polarity of the transformers if you are connecting
them in parallel or configuring a 3-phase transformer bank using single phase transformers.
Figure 13: Terminal marking and polarity of the power transformers.
Voltage Regulation:
Voltage regulation is a measure of variation in the secondary voltage from no-load to full-load
condition when the primary voltage is kept constant. Voltage regulation is expressed in percent
(%) and is defined by the following equation
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =
𝐸𝑠 (𝑛𝑜−𝑙𝑜𝑎𝑑) − 𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)
× 100
𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)
(25)
Power in AC Systems:
In AC systems the power is divided into three different categories:
Apparent Power:
It is the product of the voltage and the complex conjugate of the current and is expressed in
“volt-amperes” or VA. If the current I = IR + j IX then the conjugate of I, denoted by I*, is given
by I*=IR – j IX. The apparent power, PA, is given by
𝑃𝐴 = 𝑉 × 𝐼 ∗
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Real Power:
The real power (PR) is the power utilized by a user and is expressed in watts (W). When the
apparent power is expressed in rectangular form, the real part of the apparent power is the real
power. The real power always flows from the source to the user.
Reactive Power:
The reactive power (PX) is needed to establish a magnetic field in inductors. An electric field in a
capacitor acts as a source of reactive power. Reactive power can be supplied as well as absorbed
by an electric power source. It means that the reactive power can flow from the source to the
circuit or from the circuit to the source. When the reactive power flows from the source to the
circuit it is considered “positive” reactive power. However, it is considered “negative” reactive
power when it flows from the circuit to the source.
Mathematically, the apparent power (PA), the real power (PR), and the reactive power (PX) are
related as follows:
𝑃𝐴 = 𝑃𝑅 ± 𝑗𝑃𝑋
Assignment 1.1: Compute Iron losses and magnetizing reactance of a single-phase transformer.
Also, find the polarity of a single phase transformer and measure its voltage regulation.
Components needed:
Single phase transformer (N-6U): 1
Single phase transformer (VPS230-190): 1
Resistor 10 Ω: 1
Resistor 5 Ω: 1
Resistor 2 Ω: 1
1. Connect the transformer (N-6U) and meters as shown in Fig. 14. Make sure that all of
your connections are correct before turning on the power.
2. Turn the power on and measure P through the power meter. Also measure the current Ip
and Vp. Since there is no load connected to the secondary, P is solely due to the iron
losses.
3. Compute apparent power as follows:
𝑆 = 𝑉𝑝 𝐼𝑝
4. Compute the reactive power, Q, as follows:
𝑄 = √𝑆 2 − 𝑃 2
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5. Compute magnetizing reactance “Xm” as follows:
𝑉𝑝2
𝑋𝑚 =
𝑄
Figure 14: Circuit diagram for measuring iron losses and magnetizing reactance
Voltage Regulation:
1. Build the circuit shown in Fig. 15 using the transformer VPS230-190. Use 17 Ω for the
load. Note at full load Is must be less than 1.6 A. If this is not the case, add more
resistance to the load until Is becomes 1.6 A or less.
2. Measure the voltage across the load. This is Es (full-load).
3. Now disconnect the load and measure the voltage across the secondary terminals. This is
Es (no-load).
Figure 15: Circuit diagram for measuring voltage regulation of a transformer.
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4. Compute the voltage regulation as follows
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =
𝐸𝑠 (𝑛𝑜−𝑙𝑜𝑎𝑑) − 𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)
× 100.
𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)
Polarity of a transformer:
1. Connect the HV side of the transformer VPS230-190 to 50 V AC supply.
2. Connect any two adjacent HV and LV terminals using a jumper.
3. Measure the voltages Ep and Ex (see Fig. 16).
4. If Ex > Ep, the polarity of the transformer is additive.
5. However, if Ex < Ep, the polarity of the transformer is subtractive.
Figure 16: Circuit for polarity test of a transformer.
Syed A. Rizvi
Summer’12-Revised Spring’16
ENS 441/ELT 437