1.

TRANSFORMER EQUIVALENT CIRCUIT

COMPONENTS
• The performance of a transformer can be analyzed using
its equivalent circuit (fig. 32.41).

• The key parameters in the equivalent circuit include:

o equivalent resistance (R₀₁ or R₀₂): represents the
combined resistance of primary and secondary
windings.
o Equivalent leakage reactance (X₀₁ or X₀₂):
represents the leakage flux in the transformer.
o Core-loss conductance (G₀) or core resistance
(R₀): models the power lost in the core due to
hysteresis and eddy currents.
o Magnetizing susceptance (B₀) or magnetizing
reactance (X₀): represents the magnetizing current
required to establish the core flux.
2. EQUIVALENT CIRCUIT EXPLANATION (FIG. 32.41)

• Fig. 32.41 (a): detailed equivalent circuit

o Primary winding is represented with components:
▪ R₁ and X₁ (resistance and leakage reactance of
primary).
o Core components:
▪ R₀ (core loss resistance) and X₀ (magnetizing
reactance).
o Secondary winding is reflected to the primary side
with:
▪ R₂ and X₂ (resistance and reactance of
secondary).
o Total equivalent impedance (z e1 = r e1 + jx e1)
models the entire winding impedance.
2. EQUIVALENT CIRCUIT EXPLANATION (FIG. 32.41)

• Fig. 32.41 (b): simplified equivalent circuit

o The circuit is reduced by combining primary and
secondary winding resistances and reactances
into:
▪ Re1 (equivalent resistance).
▪ Xe1 (equivalent reactance).
o The transformer is simplified to a single
impedance (z e1) in series with the load 𝒁𝑳 .
o The current i₁ = i₂, meaning the transformer
behaves like an impedance in series with the load.
3. TRANSFORMER TESTS FOR PARAMETER DETERMINATION

• Open-circuit test
o Determines core loss parameters (R₀, X₀).
o Conducted by leaving the secondary open and applying rated voltage to the primary.
o Measures no-load current (i₀) and power loss.

• Short-circuit test
o Determines equivalent impedance (R e1, X e1).
o Conducted by shorting the secondary and applying a small voltage to the primary.
o Measures short-circuit current (𝑰𝒔𝒄 ) and impedance losses.
4. PRACTICAL IMPORTANCE OF THESE TESTS

The tests are economical and convenient since

they do not require full-load operation.

These methods are also applied to large ac

machines.

The equivalent circuit helps predict efficiency,

voltage regulation, and losses of a transformer.
1. PURPOSE OF THE TEST

• To determine no-load loss (core loss) and no-load current (I₀) of the transformer.

• Helps in finding the core resistance (r₀) and magnetizing reactance (x₀) of the transformer equivalent circuit.
2. TEST SETUP (REFER TO FIG. 32.43)

• Primary (low voltage side) is connected to the

power supply:

o A voltmeter (V) measures the applied voltage V₁.

o An ammeter (a) measures the no-load current

(i₀) drawn by the transformer.

o A wattmeter (w) measures the power input,

which is mainly the core loss.

• Secondary (high voltage side) is left open:

o Since there is no load, the secondary current is

zero.

o The secondary voltage E₂ = V₂ (induced emf) can

be measured to determine the transformation
ratio (a).
3. OBSERVATIONS AND CALCULATIONS

• Wattmeter reading (W): represents the core loss

(iron loss), which consists of:

o Hysteresis loss (due to repeated magnetization

cycles).

o Eddy current loss (due to circulating currents in

the core).

• Copper loss negligible:

o Since I₀ is small (2-10% of rated current), copper

loss in the primary is very small.

o Copper loss in the secondary is zero (since no

current flows).
3. OBSERVATIONS AND CALCULATIONS

• Power factor calculation:

o The no-load power factor cos φ₀ is determined

𝑾
as: 𝐜𝐨 𝐬 𝝋𝟎 =
𝑽𝟏 𝑰𝟎

• Separation of components of I₀:

o Magnetizing current (𝑰µ ): 𝑰µ = 𝑰𝟎 𝐬𝐢 𝐧 𝝋𝟎

o Core loss component (𝑰𝒘 ): 𝑰𝒘 = 𝑰𝟎 𝐜𝐨 𝐬 𝝋𝟎

• Calculation of equivalent circuit parameters:

𝑽𝟏
o Magnetizing reactance (X₀): 𝑿𝟎 =
𝑰µ

𝑽𝟏
o Core resistance (R₀): 𝑹𝟎 =
𝑰𝒘
4. PRACTICAL CONSIDERATIONS

• A high-resistance voltmeter may be used across the

secondary to measure E₂, helping determine the
transformation ratio a.

𝑰𝟎
• The exciting admittance (y₀) is given by: 𝒀𝟎 =
𝑽𝟏

• Exciting conductance (g₀) and susceptance (b₀):

𝑾
o Conductance: 𝐆₀ =
𝑽𝟐𝟏

o Susceptance: 𝐁₀ = 𝒀𝟐𝟎 − 𝐆₀𝟐

5. IMPORTANCE OF THE OPEN-CIRCUIT TEST

• Helps determine core losses at rated voltage, which

remain constant at all load levels.

• Provides values for x₀ and r₀, which are needed for

transformer equivalent circuit modeling.

• Essential for efficiency and regulation calculations

of transformers.
 During the no-load test of a single-phase transformer, the following
test data were recorded:
Example 32.27 • Primary voltage: 220 V

• Secondary voltage: 110 V

• Primary current: 0.5 A

• Power input: 30 W

• Primary winding resistance: 0.6 Ω

Determine the following:

1. The turns ratio of the transformer.

2. The magnetizing component of the no-load current.
3. The working (loss) component of the no-load current.
4. The iron loss of the transformer.
Additionally, draw the no-load phasor diagram to scale.
Example 32.28
A 5 kva, 200/1000 V, 50 hz single-phase transformer was tested, yielding the following results:

Open-circuit test (L.V. Side): Short-circuit test (H.V. Side):

o Voltage: 200 V o Voltage: 50 V
o Current: 1.2 a o Current: 5 A
o Power: 90 W o Power: 110 W
Determine the following:

1. The equivalent circuit parameters referred to the low-voltage (L.V.) Side.

2. The secondary output voltage when supplying 3 kw at 0.8 power factor lagging, given that the primary
voltage is 200 V. Also, compute the percentage voltage regulation.
KEY NOTES ON SEPARATION OF CORE LOSSES

• The total core loss in a

transformer consists of:
• Hysteresis loss (wh): due to
Core loss magnetization and
components demagnetization of the core.
• Eddy current loss (we): due to
circulating currents induced
within the core laminations.
KEY NOTES ON SEPARATION OF CORE LOSSES

• Hysteresis loss:
• 𝑾𝒉 = 𝑷𝑩𝟏.𝟔
Max 𝒇
Mathematical • Eddy current loss:
expressions • 𝑾𝒆 = 𝑸𝑩𝟐Max 𝒇𝟐
• Total core loss:
• 𝑾𝒊 = 𝑾𝒉 + 𝑾𝒆 𝑷𝑩𝟏.𝟔
Max 𝒇 + 𝑸𝑩 𝟐
Max 𝒇 𝟐
KEY NOTES ON SEPARATION OF CORE LOSSES

o By conducting two experiments at different

frequencies but with the same maximum flux
Determining density (Bmax), we can determine the values
of P and Q.
Constants P o Once P and Q are known, the hysteresis and

eddy current losses can be computed

and Q separately.
• This method allows us to analyze and optimize
transformer core design by minimizing core
losses, which improves efficiency.
Example 32.29

In a transformer, the core loss is measured as 52 W at 40 hz and 90 W at 60 hz, both at the same
peak flux density. Determine the hysteresis loss and eddy current loss at 50 hz.
Example 32.30
In a power loss test conducted on a 10 kg specimen of sheet steel laminations, the maximum flux
density and waveform factor remain constant. The following results were recorded:
Frequency (Hz) Total Loss (Watt)
25 18
40 33.5
50 50
60 66
80 104
Determine the eddy current loss per kg at 50 hz.
Example 32.32

A transformer is connected to a 1000-V, 50-hz supply, resulting in a core loss of 1000 W, with 650 W
due to hysteresis loss and 350 W due to eddy current loss.

If the applied voltage is increased to 2000 v and the frequency to 100 hz, determine the new core
losses.
Example 32.33
A transformer operating at rated voltage has a flux density of 1.4 wb/m² and experiences core losses
of 1000 W due to eddy currents and 3000 W due to hysteresis.

Determine the new core losses under the following conditions:

(A) increasing the applied voltage by 10% while maintaining the rated frequency.
(B) reducing the frequency by 10% while keeping the applied voltage unchanged.
(C) increasing both the applied voltage and frequency by 10%.
Example 32.34

A transformer is supplied with 2200 V at 40 hz, resulting in a core loss of 800 W, where 600 W is due
to hysteresis and the remaining loss is attributed to eddy currents.

Determine the new core loss when the supply voltage and frequency are increased to 3300 v and 60
hz, respectively.
The short-circuit test is a cost-effective method used to determine key transformer parameters,
including:

1. Equivalent Impedance (𝑍𝑒1 𝑜𝑟 𝑍𝑒2 ), Leakage Reactance (𝑋𝑒1 𝑜𝑟 𝑋𝑒2 ​), and Total Resistance (𝑅𝑒1 𝑜𝑟 𝑅𝑒2 ​)—all
referenced to the winding where the measuring instruments are connected.
2. Copper Loss at Full Load—which is essential for efficiency calculations at different load conditions.
3. Voltage Regulation—by using the equivalent impedance to compute the total voltage drop in the transformer.
As illustrated in Fig. 32.45, the low-voltage winding is
short-circuited using a thick conductor or an ammeter
(which also measures the rated load current). A reduced
voltage is applied to the high-voltage winding, just
enough to circulate the rated current. The voltmeter (𝑉)
measures the applied voltage (𝑉1 ​), the wattmeter (𝑊)
records the power loss, and the ammeter (𝐴) measures
the current (𝐼0 ​). The results from this test help in
evaluating transformer performance under load
conditions.
Equivalent circuit under short-circuit conditions

Fig. 32.46 illustrates the equivalent circuit of a

transformer during a short-circuit test:
The left diagram represents the detailed
equivalent circuit with primary resistance (𝑅1 ​),
leakage reactance (𝑋1 ​), secondary resistance (𝑅2′ ​),
and secondary leakage reactance (𝑋2′ ).
The right diagram simplifies this into the
equivalent short-circuit impedance (𝑍01 ​),
consisting of:
o Equivalent resistance (𝑅01 ​)
o Equivalent reactance (𝑋01 ​)
Using the measured values from the test:

• Impedance Calculation:
𝑉𝑠𝑐
𝑍01 =
𝐼1
• Copper Loss Calculation:
𝑊
𝑊 = 𝐼12 𝑅01 𝑅01 =
𝐼12
• Leakage Reactance Calculation:

2 2
𝑋01 = 𝑍01 − 𝑅01

This test is particularly useful for evaluating voltage regulation and determining the internal impedance
of the transformer, which affects its performance under different load conditions.
VECTOR DIAGRAM FOR SHORT-CIRCUIT TEST
the short-circuit test measures the impedance of the transformer, and fig. 32.47 provides a vector
(phasor) representation of the voltage drops across the transformer impedances.
Analysis of Fig. 32.47(a)
• This vector diagram represents the short-circuit condition
where all parameters are referred to the primary side.
• The applied voltage 𝑽𝑺𝑪 is consumed entirely in the
impedance drop of both primary and secondary windings.
• The components:
o 𝑰𝟏 𝑹𝟏 : Voltage drop across the primary resistance.
o 𝑰𝟏 𝑿𝟏 : Voltage drop across the primary reactance.
𝑰𝟏 𝑹′𝟐 : Voltage drop across the referred secondary
resistance.
o 𝑰𝟏 𝑿′𝟐 : Voltage drop across the referred secondary
reactance.
o The total short-circuit voltage is given by:
𝑽𝑺𝑪 = 𝑰𝟏 𝒁𝟎𝟏
Key takeaways
1. The short-circuit voltage 𝑽𝑺𝑪 is entirely used to overcome the impedance drop.
2. Total impedance 𝒁𝒆𝟏 can be broken down into primary and secondary components.
3. Using this test, we can determine:
o Equivalent impedance (𝒁𝑒1 ​)
o Equivalent resistance (𝑹01​)
o Equivalent reactance (𝑿0 1​)
o Voltage regulation of the transformer.
This analysis is critical for understanding transformer performance under load conditions and
predicting voltage drop and efficiency.
WHY IS THE TRANSFORMER RATING IN KVA INSTEAD OF KW?
The power rating of a transformer is given in kva (kilovolt-amperes) rather than kw (kilowatts)
because of the way transformer losses occur. Let's break this down:

1. Transformer losses depend on voltage and current, not power factor

• Copper losses (𝑰𝟐 𝑹): these depend on the current flowing through the windings.
• Iron (core) losses: these depend on the applied voltage.
• The total loss is the sum of copper and iron losses, both of which depend only on
voltage (v) and current (a), not on the phase angle or power factor.
WHY IS THE TRANSFORMER RATING IN KVA INSTEAD OF KW?
The power rating of a transformer is given in kva (kilovolt-amperes) rather than kw (kilowatts)
because of the way transformer losses occur. Let's break this down:

2. Power Factor Varies with the Type of Load

• The actual power delivered by a transformer depends on the power factor (PF) of the
connected load.
• If the power factor is high (close to 1), most of the apparent power (𝑽𝑨) is converted into
real power (𝑾).
• If the power factor is low (e.g., inductive loads), more current is needed to deliver the
same real power, but transformer losses remain based on current and voltage.
• Since transformers supply power to various types of loads (resistive, inductive,
capacitive), the power factor is not fixed.
WHY IS THE TRANSFORMER RATING IN KVA INSTEAD OF KW?
The power rating of a transformer is given in kva (kilovolt-amperes) rather than kw (kilowatts)
because of the way transformer losses occur. Let's break this down:

3. kVA Rating Ensures Universal Applicability

• If a transformer were rated in kW, it would assume a fixed power factor, which is not
practical since different loads have different power factors.
• By specifying the rating in kVA, manufacturers ensure that the transformer can be used
for any type of load without assuming a specific power factor.
KEY FORMULA RELATIONSHIPS

1. Apparent Power (S):𝑺 = 𝑽 × 𝑰 (in VA or kVA)

2. Real Power (P):𝑷 = 𝑽 × 𝑰 × 𝐜𝐨 𝐬 𝛉 (in W or kW)
o This depends on power factor (𝒄𝒐 𝒔 𝜽).
3. Reactive Power (Q):𝑸 = 𝑽 × 𝑰 × 𝐬𝐢 𝐧 𝛉 (in VAR or kVAR)
o This represents reactive power due to inductive or capacitive loads.
CONCLUSION

Since transformer losses depend only on voltage and current, and not on the power
factor of the load, the transformer is rated in kVA instead of kW. This allows the transformer
to be used with any type of load without making assumptions about power factor.
PRACTICAL EXAMPLE

• A 100 kVA transformer can supply:

o 100 kW if the power factor is 1 (𝐜𝐨 𝐬 𝛉 = 𝟏)
o 80 kW if the power factor is 0.8 (𝐜𝐨 𝐬 𝛉 = 𝟎. 𝟖)
o 50 kW if the power factor is 0.5 (𝐜𝐨 𝐬 𝛉 = 𝟎. 𝟓)

This proves that specifying a transformer's rating in kW would not be practical because the real power
output depends on the connected load's power factor.
Thus, transformers are universally rated in kVA to ensure their capability is independent of power
factor.
Example 32.35
A 30 kva, 76,000/230 V, single-phase transformer has primary and secondary winding resistances of
10 Ω and 0.016 Ω, respectively. The transformer’s reactance, referred to the primary, is 34 Ω.

Determine:

1. The primary voltage required to circulate full-load current when the secondary is short-circuited.
2. The power factor under short-circuit conditions.
Example 32.36 Determine the equivalent circuit of a 200/400 V, 50 hz, single-phase transformer using the following test data:

Open-circuit test (conducted on the low-voltage side):

o Voltage: 200 V

o Current: 0.7 a

o Power: 70 W

Short-circuit test (conducted on the high-voltage side):

o Voltage: 15 V

o Current: 10 A

o Power: 85 W

Additionally, calculate the secondary voltage when supplying a 5 kw load at 0.8 power factor (lagging), given
that the primary voltage is 200 V.
 Derive the approximate equivalent circuit of a commercial transformer starting from the ideal transformer, where all
parameters are lumped and represented on one side.
Example 32.37
A single-phase transformer has a turn ratio of 6. The primary winding has:

• Resistance: 0.9 Ω

• Reactance: 5 Ω

The secondary winding has:

• Resistance: 0.03 Ω

• Reactance: 0.13 Ω

If a 330 V, 50 hz supply is applied to the high-voltage winding, with the low-voltage winding short-circuited, determine:

1. The current in the low-voltage winding

2. The power factor

Assume the magnetizing current is negligible.

 A single-phase transformer with a rating of 10 kva, 500/250 V, 50 hz has the following parameters:
Example 32.38
• Reactance:

o Primary: 0.2 Ω

o Secondary: 0.5 Ω

• Resistance:

o Primary: 0.4 Ω

o Secondary: 0.1 Ω

• Equivalent exciting circuit parameters (referred to primary):

o Resistance: R₀ = 1500 Ω

o Reactance: X₀ = 750 Ω

• Determine the expected meter reading under the given conditions.

Example 32.39
A 1000-kva, 110/220 V, 50-hz, single-phase transformer has an efficiency of 98.5% at half full load
with a 0.8 power factor (leading) and 98.8% at full load with a unity power factor.

Determine the following:

1. Iron loss
2. Full-load copper loss
3. Maximum efficiency at unity power factor
Example 32.40 A 200/400-V step-up transformer has the following equivalent circuit parameters referred to the low-
voltage side:

• Equivalent resistance: 0.15 ω

• Equivalent reactance: 0.37 ω

• Core-loss component resistance: 600 Ω

• Magnetizing reactance: 300 Ω

If the transformer supplies a 10 A load at a power factor of 0.8 lagging, determine:

1. Primary current
2. Secondary terminal voltage
Example 32.41
A 300-kva, 11,000/2500-V, 50-hz transformer has the following specifications:

• Low-voltage winding: 190 turns, resistance = 0.06 ω

• High-voltage winding: 910 turns, resistance = 1.6 ω

When the low-voltage winding is short-circuited, full-load current is achieved with 550 V applied to
the high-voltage winding. Determine:

1. The equivalent resistance and leakage reactance referred to the high-voltage side.
2. The leakage reactance of each winding.
Example 32.42

A 230/115 V, single-phase transformer is supplying a 5 A load at a power factor of 0.866 lagging.

The no-load current is 0.2 A at a power factor of 0.208 lagging. Determine:

1. The primary current

2. The primary power factor