৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

A transformer works on alternating current (a.c.) only, not on direct current (d.c.). This is because a transformer relies on the changing magnetic field produced by the alternating current to induce a voltage in the secondary coil. Direct current does not create a changing magnetic field, so the transformer cannot function with it.

The primary and secondary windings of a transformer are coupled magnetically, not electrically. This coupling is achieved through the core of the transformer, typically made of soft iron or silicon steel, which provides a path for the magnetic flux generated by the alternating current in the primary coil. The varying magnetic flux induces a current in the secondary coil. Since no direct electrical connection exists between the primary and secondary windings, this form of coupling allows for electrical isolation between the two.
Book Reference:
For more on this, see "Principles of Electrical Engineering" by A. Hughes 

In a transformer, the winding with more turns is typically the high-voltage winding. This is based on the principle of turns ratio, which determines the voltage ratio between the primary and secondary coils. According to the transformer equation:
V₂​/V_1​​=N_2​/N_1​​
Where V₁​ and V₂​ are the voltages across the primary and secondary windings, and N​₁andN₂​ are the number of turns in the primary and secondary windings. If the number of turns in the secondary winding is greater than the primary, it will step up the voltage. Thus, the high-voltage side will have more turns compared to the low-voltage side. "Electrical Machines" by V.K. Mehta:

The efficiency of a transformer is primarily due to the principle of inductive coupling, which is the correct answer.

A transformer operates on the principle of electromagnetic induction. When an alternating current (a.c.) flows through the primary winding of the transformer, it creates a varying magnetic field. This varying magnetic field then induces a current in the secondary winding through the process of inductive coupling.

Here’s a breakdown of why this process makes transformers efficient:

Inductive Coupling (Correct Answer):
In inductive coupling, the energy is transferred from one coil to another without direct electrical contact, through the magnetic field produced by the alternating current in the primary coil. This is an efficient energy transfer method because it minimizes energy loss in the form of heat and allows the transformer to step up or step down voltages effectively. The core material (often iron or steel) enhances the magnetic coupling, further improving efficiency.
Why Other Options Are Incorrect:

(i) Static Device:
While it's true that transformers are static (they don’t have moving parts), the reason transformers are efficient is not because of their static nature. The efficiency comes from the electromagnetic process used for energy transfer, not the fact that they don’t move.
(iii) Uses Capacitive Coupling:
Capacitive coupling is not the principle used by transformers. It involves the transfer of energy through electric fields between two conductors. However, transformers rely on magnetic fields, not electric fields, to transfer energy. Therefore, capacitive coupling doesn’t apply to transformers.
(iv) Uses Electric Coupling:
Electric coupling involves the direct transmission of electrical signals between conductors, but in transformers, the energy is transmitted via magnetic fields. Electric coupling does not accurately describe the operation of a transformer. "Electrical Machines" by V.K. Mehta:

When a transformer core has air gaps, the reluctance (resistance to magnetic flux) of the magnetic path increases significantly. This is because the core material, typically made of iron or steel, has a much lower reluctance compared to air. Air, being non-magnetic, offers a higher reluctance to the flow of magnetic flux. As a result, the overall reluctance of the magnetic path increases when air gaps are introduced.

Why Magnetizing Current Increases:
In a transformer, the magnetizing current is the current that flows through the primary winding to establish the magnetic flux in the core. This current is proportional to the reluctance of the magnetic path. When the reluctance increases (due to air gaps), the transformer needs to draw more magnetizing current to establish the same level of magnetic flux. The relationship is given by the formula:\

Magnetizing Current ∝ Voltage /Reluctance

As the reluctance of the core increases (due to air gaps), the magnetizing current must increase to overcome this higher reluctance and establish the required magnetic field. This increased magnetizing current is not ideal, as it results in additional power losses and reduced efficiency of the transformer.

Effects of Increased Magnetizing Current:
Higher Losses:
The increased magnetizing current leads to more energy losses, particularly in the form of core losses such as hysteresis loss and eddy current loss. The increased magnetizing current can also result in the transformer drawing more current than needed, affecting its efficiency.

Reduced Performance:
Transformers with air gaps in the core often exhibit higher magnetizing currents, which can lead to lower overall performance. The increased magnetizing current can cause saturation issues in the core, reducing the transformer’s ability to transfer energy efficiently.
Effect on Voltage Regulation:
Since the magnetizing current is related to the magnetizing flux, higher magnetizing current can affect the transformer’s voltage regulation, making it less stable.

Why Other Options Are Incorrect:
(i) Reluctance of magnetic path is decreased:
This is incorrect because air gaps increase the reluctance of the magnetic path.
(ii) Hysteresis loss is decreased:
This is incorrect. Hysteresis loss actually increases because the air gap disrupts the smooth flow of the magnetic flux, leading to greater energy dissipation in the core material.
(iv) Eddy current is increased:
This is also not correct. While air gaps may affect eddy currents in some cases, they are not the direct cause of increased eddy current losses. Eddy currents are typically more affected by the material properties and thickness of the core, not the presence of air gaps. "Electrical Machines" by V.K. Mehta:

In a single-phase transformer, the magnetic flux in the core is purely alternating.

Here’s why:

Alternating Current (AC): The transformer operates on alternating current, which means the current in the primary coil alternates direction continuously. This alternating current creates a magnetic flux that also alternates (changes direction) with the same frequency as the applied AC.
Flux Behavior: The magnetic flux generated by the AC current in the primary winding is alternating because it follows the AC's cyclical nature. The direction and magnitude of the flux change as the AC supply fluctuates, creating an alternating magnetic field in the core of the transformer.
No Rotating Flux: In a transformer, the flux doesn't rotate like in a rotating machine (such as a motor or generator). The flux in a transformer is sinusoidal and oscillates back and forth with the changing direction of the current. This is what makes it a purely alternating flux.

Why Other Options Are Incorrect:
(ii) Purely rotating one:
This is incorrect because, in a transformer, the flux is not rotating. The magnetic field oscillates back and forth, but it doesn’t rotate as in rotating machines like motors.

(iii) Partly alternating and partly rotating:
This is incorrect as well. The flux in a transformer is entirely alternating; it doesn’t involve any rotation. The alternating nature of the current in the primary coil directly leads to the alternating flux in the core.

(iv) None of the above:
This is incorrect because option (i) accurately describes the flux in a transformer.


Source: "Electric Machinery Fundamentals" by Stephen J. Chapman.

A transformer is designed to work with alternating current (AC), and it relies on the principle of electromagnetic induction. When a D.C. supply is applied to the primary winding of a transformer, several issues arise, leading to the possibility of damage or even burnout. Let’s go through the options to understand why this is the case:

D.C. Current and Induction:
In a transformer, the changing magnetic field generated by the alternating current (AC) in the primary coil induces a voltage in the secondary coil. However, direct current (D.C.) does not produce a changing magnetic field. The current in the primary winding of a transformer with a DC supply is constant, meaning the magnetic field in the core does not fluctuate. This lack of change in the magnetic field prevents induction from occurring in the secondary coil, making the transformer ineffective in transferring power.

Constant D.C. Current:
When a D.C. supply is connected, the current starts to flow through the primary winding, and since there's no changing magnetic field, the only thing that limits the current is the resistance of the winding and the magnetizing impedance of the core. D.C. current is not limited by reactance as AC current is, and thus, the primary winding may draw a large, uncontrolled current. This can lead to excessive heating and eventually cause the primary winding to burn out. "Electrical Machines" by V.K. Mehta:

Why Other Options are Incorrect:
Primary draws small current:
This is incorrect. In the case of a D.C. supply, the current drawn by the primary coil can be very large because there is no inductive opposition (no changing magnetic field). The primary will continue to draw current until it is limited by the resistance of the winding or until it burns out.

Primary leakage reactance is increased:
This is incorrect. Leakage reactance is a result of inductance in the coils, which depends on the changing magnetic flux. With D.C., the flux does not change, and the inductive reactance is effectively zero. The primary winding does not generate a fluctuating magnetic field, so the leakage reactance is not relevant in the case of D.C. supply.

Core losses are increased:
This is also incorrect. Core losses, which include hysteresis loss and eddy current loss, are typically related to the frequency of the alternating current. In the case of D.C., there is no alternating magnetic flux, so core losses (due to hysteresis and eddy currents) are essentially zero. The primary problem with D.C. is that the transformer is not designed to operate with a constant magnetic field.

To solve this problem, we need to use Ohm's Law and the properties of the transformer.

Step 1: Use Ohm’s Law to find the current in the secondary coil.
Ohm's Law states: I = V​/R

Where:

Iis the current,
V is the voltage,
R is the resistance.
We know the secondary voltage is 24 V (since the transformer steps down 120 V to 24 V), and the resistance of the lamps is given as 9.6 ohms.

Now, applying Ohm’s Law for the secondary circuit:

I_second = 24/9.6 = 2.5 A.

 We shall assume the power factor is unit.
V₁ I ₁cosΦ = 4*10^3
⇒ I₁ = 40A

No-load voltage ratio of the transformer: 6000 V / 250 V
Frequency (f): 50 Hz
Maximum flux Φmax: 0.06 Wb (Webers)

V = 4.44 × f × N × Φmax​

Where:

V is the voltage,
f is the frequency,
N is the number of turns,
Φmax is the maximum flux in the core.

Np = 6000/(4.44*50*0.06) = 450 turns.

Given:
Transformer rating: 1600 kVA
Primary voltage: 6600 V
Secondary voltage: 440 V
Frequency: 50 Hz
Maximum flux density (Bmax​): 12 Wb/m²
Induced voltage per turn: 30 V

V = 4.44 × f × N × Φmax​

Where:

V is the voltage,
f is the frequency,
N is the number of turns,
Φmax is the maximum flux in the core.

30 = 4.44 × 50 × Φmax​ = 0.135


The flux in the core is related to the core area by the formula:

Φmax​ = B_max​ × A_c​Where:

A_c​ is the core area in m²,
B_max​ is the maximum flux density in Wb/m².
2
A_c​ = 0.135/1.2 =1125 cm²

Primary turns (N₁) = 400
Secondary turns (N₂) = 1000
Core area (Aₐ) = 60 cm² = 60×10−4 m260 \times 10^{-4} \, \text{m}^260×10−4m2 (converted to m²)
Primary voltage (V₁) = 500 V
Frequency (f) = 50 Hz

The induced voltage per turn in the transformer is related to the maximum flux and frequency by:

V1=4.44×f×N1×Φmax
 Where:

V1 is the primary voltage,
f is the frequency,
N1​ is the number of primary turns,
Φmax is the maximum flux in the core.
We can rearrange the formula to find the maximum flux (Φmax​):

Φmax= 500/(4.44 * 50 * 400)
Φmax​=5.63×10⁻³Wb


Now, we can calculate the peak flux density using the relationship between flux, flux density, and core area:
Φmax​=Bmax​×Acore​

Bmax = 5.63×10⁻³ /6 ×10^-4  Wb = 0.94 Wb/m²

Given:
Frequency (f) = 50 Hz
Core dimensions: The core is square with a side of 20 cm, so the area of the core is Acore=(20 cm)² = 400 cm²  = 0.04 m²
Permissible flux density (B) = 1 Wb/m²
Stacking factor (k) = 0.9

Φmax​=B×Acore​ = 0.04 Wb

The voltage induced per turn in a transformer is given by:

V_turn = 4.44 × f × Φmax × k

V_turn​ = 4.44 × 50 × 0.04 × 0.9 = 8

In a transformer, the no-load primary current I0I_0I0​ is the current that flows in the primary winding when the transformer is connected to a voltage source, but the secondary circuit is open (no load). This current is primarily responsible for magnetizing the core and overcoming core losses (hysteresis and eddy currents).

The full-load primary current I_FL on the other hand, is the current that flows in the primary winding when the transformer is supplying its rated load on the secondary side.

The relationship between no-load and full-load currents:
The no-load current is typically much smaller than the full-load current. This is because the no-load current only needs to overcome the losses in the transformer core, while the full-load current must supply the total power required by the secondary load.
For most transformers, the no-load primary current is about 3-5% of the full-load primary current. This is because at no load, the transformer only draws the magnetizing current, which is small relative to the full-load current.
The correct answer is (i) 3-5%. This range is typical for most transformers, especially under normal operating conditions.

Why other options are incorrect:
(ক) 15-30% and
(গ) 30-40% are much too high for no-load current in most transformers, as they would represent a transformer with very high core losses or poor efficiency.
(ঘ) Above 40% is extremely high for no-load current and is not typical for well-designed transformers.

Source: "Electrical Machines" by V.K. Mehta:

The no-load current (I0I_0I0​) of a transformer is the current that flows in the primary winding when the secondary is open (no load). It is made up of two components:

Magnetizing current: This is the current that is responsible for producing the magnetic flux in the core of the transformer. It is generally lagging the voltage and is responsible for the inductive nature of the transformer.
Core loss current (iron loss current): This current accounts for the losses in the transformer core, including hysteresis loss and eddy current loss. It is a real current and is in phase with the voltage.
The no-load power factor of the transformer is typically low because:

The magnetizing current is primarily inductive, meaning it lags the voltage by 90 degrees. Therefore, the current is largely reactive, contributing little to the real power.
The iron loss current is much smaller in comparison, and its phase is in line with the voltage, contributing to the real power. However, this current is relatively small compared to the magnetizing current.
Thus, the no-load power factor is low because most of the current is magnetizing current, which is highly reactive and leads to a poor power factor.

Source: "Electrical Machines" by V.K. Mehta.

When a transformer is operating at no-load, the only significant power losses are due to the core losses (iron losses), which include hysteresis loss and eddy current loss.

Iron Loss: The iron loss occurs in the core of the transformer due to the alternating magnetic flux. This loss includes:

Hysteresis Loss: Losses due to the continuous magnetization and demagnetization of the core material.
Eddy Current Loss: Losses caused by circulating currents induced in the core material by the changing magnetic flux.
These two losses are responsible for most of the no-load input power. On the other hand, copper losses occur when the transformer is under load and result from the resistance in the windings.

At no-load, the copper losses are negligible because no current is flowing through the secondary winding, and only a small current flows through the primary winding (the magnetizing current)."Electrical Machines" by V.K. Mehta:

he no-load current I_0​ is made up of two components:

Magnetizing current I_m​: This is the current responsible for magnetizing the transformer core.
Iron loss current ​: This current is responsible for the core losses (hysteresis and eddy current losses) in the transformer.
The iron loss current I_iron is in phase with the voltage, while the magnetizing current Im​ is in quadrature with the voltage (90 degrees out of phase).

Where θ is the phase angle between the voltage and the iron loss current. Since the iron loss current is in phase with the voltage, the cosine of the phase angle cos⁡(θ) = 1.

Therefore, the equation simplifies to:
P_{no-load ​}= V₁​×I_iron​
64 W = 200 V×I_iron​
I_iron​ = 64/200 = 0.32 A

Core loss, W₀ = VIcosφ = 287.5

The primary copper loss is the loss due to the resistance of the primary winding, and it is given by:

P_cu = I₀² × R₁

P_Cu = (6.5)² × 0.06 = 2.535 W

The total no-load power absorbed by the transformer is the sum of the core loss (iron loss) and the copper loss:

P_no-load = P_core + P_cu   

P_core = 187- 2.535 = 184.465W

Ampere-turns and voltage per turn:
Ampere-turns: Ampere-turns are the product of the current and the number of turns in a coil. In a transformer, the ampere-turns for the primary winding are equal to the ampere-turns for the secondary winding. This is because the transformer is designed in such a way that the magnetizing force (ampere-turns) must be balanced to produce the magnetic flux.

Voltage per turn: The voltage induced per turn in the primary winding is equal to the voltage induced per turn in the secondary winding, assuming an ideal transformer. The voltage per turn depends on the number of turns and the applied voltage.
Therefore, ampere-turns and voltage per turn are the same for both primary and secondary windings.

Resistances and leakage reactances:
The resistances of the primary and secondary windings typically differ because the windings are made of different wire gauges and the current through each winding is different.
Leakage reactances are also generally different for the primary and secondary windings. The leakage reactance is proportional to the number of turns and the geometry of the windings.

Currents and induced voltages:
The currents in the primary and secondary windings are different due to the voltage transformation ratio.
The induced voltages in the primary and secondary are different because they are scaled according to the turns ratio.

Transformers operate using the principle of electromagnetic induction, and the performance of a transformer is affected by frequency, particularly when operating at very low frequencies. Here's an explanation of each option:

ক Permeability of core is increased:Permeability of the core material is a property that determines how easily the material can be magnetized. While the permeability of the core does affect transformer performance, it decreases with frequency rather than increases. At very low frequencies, the core material may not respond efficiently to the changing magnetic field, leading to poor performance. Therefore, this option is not correct.

খ Magnetizing current is abnormally high:
At low frequencies, the magnetizing current increases because the transformer needs to establish a magnetic field, but the changing magnetic flux takes longer (due to the lower frequency). Since the voltage is being applied over a longer time period, the magnetizing current becomes abnormally high, which results in inefficiency and poor operation of the transformer. This is the correct reason.

 গ  Primary reactance is too much increased: Primary reactance is proportional to the frequency, and at very low frequencies, the reactance of the primary winding becomes very large. This leads to high impedance, reducing the current that can flow through the primary winding. This makes the transformer inefficient at low frequencies, which is part of the reason for poor operation. However, while this is a contributing factor, the more significant reason for poor performance is the high magnetizing current (option ii)."Electrical Machines" by V.K. Mehta:

When a transformer operates at very high frequencies, several things happen that impact its performance:

ক Primary reactance is too much increased:
Reactance is directly proportional to frequency (X = 2πfL), where f is the frequency and L is the inductance. At very high frequencies, the primary reactance increases significantly, limiting the current that can flow through the primary winding. This high reactance impedes the normal operation of the transformer and reduces the transformer’s ability to transfer power efficiently. This statement is correct, but does not fully capture the most significant issue.

খ Primary will draw large power:
At very high frequencies, the impedance of the primary winding increases due to high reactance. This increase in reactance results in a reduction in current flow and power drawn by the primary winding. The primary will not draw excessive power; in fact, it may draw less power due to the higher impedance. Therefore, this statement is incorrect.
 
 গ Core losses will be excessive:
Core losses in a transformer consist of hysteresis loss and eddy current loss. At very high frequencies, eddy current losses become excessive because the changing magnetic flux in the core induces circulating currents. These losses increase as frequency increases, and the core material is unable to handle the high-frequency flux efficiently. Hysteresis losses also increase because the core material has to constantly switch polarity at a faster rate. Therefore, core losses are indeed excessive at very high frequencies. "Electric Machinery Fundamentals" by Stephen J. Chapman

In a transformer, leakage flux refers to the portion of the magnetic flux that does not link both the primary and secondary windings. Instead, it only links the primary winding or the secondary winding, but not both. This flux is responsible for certain inefficiencies and effects on the transformer’s performance.

Leakage flux increases the impedance of the transformer, which can lead to a voltage drop across the windings. This is because the current, being affected by the leakage flux, creates a voltage drop proportional to the leakage reactance.

Source: "Electric Machinery Fundamentals" by Stephen J. Chapman.

In a transformer, mutual flux is the magnetic flux that is shared between the primary and secondary windings, and it plays a key role in the transformer’s operation. The mutual flux remains constant at all load conditions for the following reasons:

(ক) Applied voltage and frequency are constant:
Constant voltage and frequency are essential because they control the rate at which the magnetic field changes. If the voltage and frequency applied to the transformer are constant, the magnetic flux generated in the core remains constant under ideal conditions. The primary current changes with load, but the magnetizing current, which generates the mutual flux, remains largely unaffected by the load as long as the supply voltage and frequency are stable. This is correct.

(খ) Leakage flux is small:
Leakage flux refers to the portion of the magnetic flux that does not link both the primary and secondary windings. It does not contribute to the mutual flux. While leakage flux can affect the transformer’s efficiency, it does not affect the constancy of mutual flux. So, while leakage flux is small, it is not the reason why the mutual flux remains constant at all load levels. This is incorrect.

(গ) Iron core is used:
Iron core provides a highly efficient path for the magnetic flux, reducing the core losses and ensuring that the mutual flux remains strong and stable. While the use of an iron core improves efficiency, the constant mutual flux is more directly related to the applied voltage and frequency rather than the material of the core. This is a contributing factor but not the main reason.

(ঘ) Losses are small:
Losses (such as iron losses and copper losses) can affect the efficiency of a transformer, but they do not directly influence the constancy of mutual flux. The mutual flux remains constant as long as the applied voltage and frequency remain stable, regardless of the losses. This is incorrect.

"Electric Machinery Fundamentals" by Stephen J. Chapman.

The core design of a transformer plays a crucial role in its efficiency and performance. Large transformers often use a circular cross-section for the core to address various loss mechanisms. Let’s go over each option:

Copper losses occur due to the resistance of the windings and are proportional to the square of the current. The shape of the core does not directly affect copper losses, as copper losses are more related to the winding design (size and resistance) and the current passing through the windings. 

In a transformer, the primary and secondary windings are typically wound in such a way that they are closely coupled to reduce several losses and improve efficiency. The option that best explains why the windings are wound on top of each other is leakage reactance.

(ক) Iron losses:
Iron losses are caused by the magnetic flux in the core material and are not directly influenced by how the windings are arranged. The arrangement of the windings does not reduce iron losses. Therefore, this option is incorrect.

(খ) Copper losses:
Copper losses occur due to the resistance in the windings, and they are proportional to the square of the current (Pcu=I₂R) The arrangement of the windings (primary and secondary on top of each other) does not directly affect copper losses, so this option is incorrect.

(গ) Leakage reactance:
Leakage reactance arises from the portion of the magnetic flux that does not link both the primary and secondary windings. By winding the primary and secondary windings on top of each other, the magnetic coupling between the windings improves. This reduces the amount of leakage flux and, as a result, decreases the leakage reactance. This is the primary reason for the arrangement of windings on top of each other, so this option is correct.

(ঘ) Winding resistance:
The winding resistance depends on the material, length, and cross-sectional area of the winding wire. The arrangement of the windings does not directly affect the resistance of the windings themselves, so this option is incorrect.

In a transformer, the amount of copper used in the primary and secondary windings depends on the current and voltage ratings of both sides. The relationship between the copper in the primary and secondary is influenced by the current that flows through the windings.

For a given transformer, the current in the primary and the current in the secondary are related through the turns ratio (based on the voltage ratio):

I2​/I1 ​​= N1​/N2 ​= V2​/V1​​ 

Where:

I1 and I2​ are the currents in the primary and secondary windings, respectively.
V1​ and V2 are the voltages in the primary and secondary windings, respectively.
N1 and N2​ are the number of turns in the primary and secondary windings, respectively.

Because the current in the primary is inversely proportional to the voltage (for a given power), and the current in the secondary is proportional to the voltage, the amount of copper used in both windings can be similar, considering that thicker wires are used in the secondary for higher current, while thinner wires are used in the primary due to lower current. 

"Electrical Machines" by V.K. Mehta:

The open-circuit test (also known as the no-load test) is typically performed on a transformer to determine its core losses (iron losses), such as hysteresis and eddy current losses, and the no-load current. During this test, the secondary winding is left open (no load connected), and the primary winding is connected to a voltage source.

The open-circuit test is always conducted on the low-voltage winding for the following reasons:

Safety: It's safer to perform the test on the low-voltage side since the current is smaller, reducing the risks associated with handling higher voltage.
Convenience: It's more practical to apply a low voltage to the primary winding, making it easier to measure and handle the test setup. 

"Electrical Machines" by V.K. Mehta.

The open-circuit test (also known as the no-load test) is typically used to determine certain parameters of a transformer that are related to core losses and the characteristics of the core material. During this test:

The secondary winding is open (no load is connected). The primary winding is supplied with a voltage, usually at rated voltage, and the current and power are measured. The primary current in the open-circuit test is mainly composed of:

Magnetizing current: Responsible for creating the magnetic field in the transformer core.

Core loss current: Responsible for the iron losses (hysteresis and eddy current losses).
The open-circuit test allows you to measure the core losses of the transformer, which are iron losses in the core due to hysteresis and eddy currents. These losses are not related to the load and are typically quite small compared to copper losses, but they can be measured directly from the open-circuit test.

However, the open-circuit test does not directly provide information about copper losses, the total equivalent resistance, or leakage reactance. 

Source: "Electrical Machines" by V.K. Mehta.

In a short-circuit test (also called the sc-test) on a transformer, the secondary (low-voltage) winding is typically short-circuited, while the primary (high-voltage) winding is connected to a low-voltage supply. This test is performed to determine the impedance of the transformer, which includes both the resistance and the leakage reactance of the windings.

Here’s why we short-circuit the low-voltage winding:

Safety and Convenience: Short-circuiting the low-voltage winding is safer and more practical because the voltage across the low-voltage winding is typically much lower than the high-voltage winding, making it safer to handle.

Impedance Measurement: In the short-circuit test, the current is increased by applying a reduced voltage to the primary winding, causing a small voltage drop across the windings. This allows us to measure the impedance (resistance and reactance) of the transformer under load conditions.

Why Not the High-Voltage Winding?: Short-circuiting the high-voltage winding would require a large amount of current to flow, which could be dangerous and impractical. The low-voltage winding is chosen for the test to limit the current and make the test more manageable. 

Source: "Electrical Machines" by V.K. Mehta.

The open-circuit test (also known as the no-load test) is conducted on a transformer to measure the core losses, which are also known as iron losses. During this test:

The secondary winding is kept open (no load is connected), and the primary winding is connected to the supply.
The primary current is measured, which is very small compared to the full-load current.
The power absorbed by the transformer is also measured during this test, and this power is primarily used to overcome core losses (hysteresis and eddy current losses in the core material).
Since copper losses depend on the current passing through the windings and occur mainly under load conditions, copper losses are not measured during the open-circuit test. Similarly, friction losses are not directly related to the open-circuit test.

The open-circuit test measures iron losses, which are caused by the alternating magnetic flux in the core of the transformer. 

Source: "Electrical Machines" by V.K. Mehta.

When the secondary of a transformer is short-circuited, it affects the current and the overall impedance of the transformer.

Primary Inductance:
The primary inductance is related to the transformer's ability to oppose the change in current through the magnetic field created by the primary winding. However, the inductance of the primary is not purely constant; it also depends on the load conditions, particularly the secondary current.

Effect of Short-Circuiting the Secondary:
When the secondary winding is short-circuited, the current in the secondary winding increases to the point where the transformer operates under high current conditions for the primary winding as well.
The leakage reactance increases under these conditions because the flux linkage between the primary and secondary is affected.
The primary inductance typically decreases because the flux in the core is affected by the high secondary current. The presence of a short-circuit creates a situation where the core's magnetization is dominated by the large current in the secondary, causing a reduction in the effective inductance of the primary winding.

Source: "Electrical Machines" by I.J. Chapman.

In a transformer, there are different types of losses:

Eddy current losses: These are caused by the changing magnetic flux in the core. Eddy currents circulate in the core material and produce heat, but they are largely independent of the load and depend on the frequency and the core material.

Hysteresis losses: These losses are also in the core material and are related to the magnetization and demagnetization of the core material due to the alternating magnetic field. Hysteresis losses are primarily independent of load but depend on the properties of the core material and frequency.

Copper losses: These occur in the windings due to the resistance of the copper wires. Copper losses are directly proportional to the square of the current (P_cu = I²R), so when the load on the transformer decreases, the current decreases, which in turn decreases copper losses.

Friction losses: These are generally very small in transformers and are related to mechanical losses like the movement of cooling fans, etc. However, these are usually negligible compared to the other types of losses.

Source: "Electrical Machines" by I.J. Chapman.

Eddy current losses occur in the core material of a transformer due to the alternating magnetic flux, which induces circulating currents (called eddy currents) in the core. These eddy currents flow in loops within the core material and cause energy dissipation in the form of heat.

The amount of eddy current loss is influenced by several factors, including:

Thickness of the core lamination: Eddy current loss is inversely proportional to the square of the thickness of the core lamination. The core is typically made of thin laminated sheets to reduce eddy current losses. Thicker laminations create larger paths for eddy currents, increasing the losses. Therefore, the thinner the lamination, the lower the eddy current loss.

Supply frequency: Eddy current loss is directly proportional to the square of the supply frequency. As the supply frequency increases, the rate at which the magnetic field changes increases, inducing more eddy currents and thereby increasing the losses.

Flux density in the core: Eddy current loss is directly proportional to the square of the flux density. The higher the flux density, the greater the induced voltage in the core, which leads to more eddy currents and higher losses.

Source: "Electrical Machines" by I.J. Chapman.

The efficiency of a transformer is given by the formula:
η = P_out/ P_out + P_loss

The transformer efficiency will be maximum when the losses are minimized. The two main types of losses in a transformer are:

Copper losses: These losses are caused by the resistance of the windings and are proportional to the square of the current (I²R)

Constant losses (also called core losses or iron losses): These losses are primarily caused by the alternating magnetic field in the core and are constant at all load levels.
For maximum efficiency, the copper losses and core losses should be equal, because when these two losses are equal, the total losses are minimized, and thus, the transformer operates at optimal efficiency.

Source: "Principles of Electrical Machines and Power Electronics" by V.K. Mehta and Rohit Mehta.

The efficiency of a transformer is defined as the ratio of the output power to the input power and is typically very high for large transformers. This is because large transformers are designed to operate efficiently, with minimal losses in both the core and the winding.

For large transformers, the efficiency usually falls in the range of 95% or higher, due to the relatively low losses (especially copper losses and core losses) compared to the power being transferred. The higher the transformer rating, the better the efficiency, as the relative losses decrease with the size of the transformer.Electrical Machines" by A.E. Fitzgerald, Charles Kingsley, and Stephen D. Umans

In a practical transformer, the total losses can be divided into copper losses and core losses:

Copper losses occur due to the resistance of the windings and are proportional to the square of the current (I²R). Copper losses are higher when the transformer is under full load or near full load.
Core losses (also known as iron losses) occur due to the alternating magnetic field in the core material, primarily due to hysteresis and eddy currents. These losses are relatively constant and are primarily a function of the core material and the supply frequency.
For a practical transformer under full-load conditions, copper losses generally account for a significant portion of the total losses, typically around 85% of the total losses. This is because the copper losses increase with the load, and they are typically higher than the core losses.

The core losses remain relatively constant and are a smaller portion of the total losses. The percentage of copper losses depends on the design of the transformer, but it can be as high as 85% under full-load conditions."Electrical Machines, Drives, and Power Systems" by Theodore Wildi

In a core-type transformer, the core is designed to enclose the windings and form the primary and secondary magnetic circuits. The core typically has a closed loop shape, and the magnetic flux travels through the core, linking both the primary and secondary windings. The design of the core directly impacts the magnetic path.

Core-Type Transformer Design:
Magnetic Path: The magnetic flux in a core-type transformer follows a path through the core material, which provides a low reluctance path for the flux. This path is determined by the geometry of the core and the arrangement of the windings.

Length of Magnetic Path: In a core-type transformer, the magnetic path is relatively shorter compared to a shell-type transformer, where the magnetic path is longer due to the winding arrangement and core design.

Length per Turn: The magnetic path length is shorter in a core-type transformer, which reduces the average length per turn of the windings. This also reduces the copper losses and improves efficiency.

Source: "Principles of Electrical Machines and Power Electronics" by V.K. Mehta and Rohit Mehta

Transformers with ratings less than 5 kVA are generally small transformers used in various applications like residential or small industrial settings. The cooling method for such small transformers is typically natural air cooling because:

Oil cooling is typically used for transformers with higher ratings (typically 5 kVA and above) where there is more heat generation due to larger current and voltage levels. Oil helps in efficiently transferring heat away from the transformer and provides insulation.

Natural air cooling is more suitable for smaller transformers. In this case, the transformer relies on the surrounding air to dissipate heat. The design of small transformers allows sufficient heat dissipation via air, making oil or water cooling unnecessary.

Water cooling is typically used for very large transformers (in the range of hundreds of kVA or MVA) and is not commonly used for small transformers like those under 5 kVA.

The temperature rise of a transformer is primarily related to the losses occurring in the transformer, which are mainly due to copper losses (resistive losses in the windings) and core losses (iron losses in the core).

Copper losses (Pcu = I²R) depend on the current flowing through the windings, and the current increases with the load on the transformer. The more the transformer is loaded, the higher the current, and thus the higher the copper losses and temperature rise.

Apparent power (S) is the combination of both real power (P) and reactive power (Q), and it is related to the current in the transformer. When the transformer carries more apparent power, the current increases, which results in higher copper losses and hence greater temperature rise.

Reactive power refers to the power that does not perform any useful work (it is exchanged between the source and the transformer but doesn't contribute to the output power). While it affects the total current in the transformer, it is not directly responsible for heating as it does not contribute to real power dissipation in the form of losses. Therefore, reactive power is not directly proportional to temperature rise.

Leakage reactance affects the impedance of the transformer and influences how much voltage is dropped under load conditions, but it doesn't directly affect the power dissipation (and hence temperature rise).

Source: "Electrical Machines" by I.J. Chapman.

In an impedance-matching transformer, the turns ratio is related to the square root of the ratio of the impedances. The formula for the turns ratio (Ns/Np​​) is:

(Ns/Np​​)² = Zp/Zs

Zp = 6250
Zs = 10 ohm

Np/Ns = √(6250/10) = 25

Source: "Electrical Machines" by I.J. Chapman.

In an impedance-matching transformer, the turns ratio is related to the square root of the ratio of the impedances. The formula for the turns ratio (Ns/Np​​) is:

(Ns/Np​​)²=Zp/Zs

Where:

Zp is the primary impedance,
Zs​ is the secondary impedance,
Np​ and Ns​ are the number of turns in the primary and secondary windings, respectively.

Np/Ns =√(5/5) =1

Source: "Electrical Machines" by I.J. Chapman.

Given:
Transformer voltage ratio: 6600/600 V
Load impedance ZL = 4 + j3 Ω (complex impedance)
Load power to be calculated

Is = Vs/Z_L
= 6000/(4+j3)

|Z_L| =√(4²+3²) = 5 ohm

So, the magnitude of the current is:
|Is| = 600/2 = 120

The phase angle θ  can be found from the impedance:
θ = tan^-1(3/4) = 36.87°

cos(36.87∘) ≈ 0.8

P = Vs​⋅Is​⋅cos(θ)
= 600×120×0.8
= 57.6×10³W

    Given:
    - Transformer rating: 10 kVA, 2000/400 V (primary/secondary voltage ratio)
    - Primary resistance (Rp) = 5 ohms
    - Primary inductive reactance (Xp) = 12 ohms
    - Secondary resistance (Rs) = 0.2 ohms
    - Secondary inductive reactance (Xs) = 0.48 ohms

    We are asked to find the equivalent impedance of the transformer referred to the primary side.

    Step 1: Find the turns ratio (n)
    The turns ratio of the transformer (n) is the ratio of the primary voltage to the secondary voltage:
    n = Vp / Vs = 2000 / 400 = 5

    Step 2: Convert the secondary impedance to the primary side
    The impedance on the secondary side needs to be referred to the primary side using the turns ratio (n). 
    The impedance referred to the primary side (Zp) is calculated by multiplying the secondary impedance (Zs) by n^2:
    
    For resistance:
    Rp = n² × Rs = 5² × 0.2 = 25 × 0.2 = 5 ohms

    For reactance:
    Xp = n² × Xs = 5² × 0.48 = 25 × 0.48 = 12 ohms

    Step 3: Add the primary and secondary components
    The total resistance referred to the primary side is the sum of the primary resistance and the referred secondary resistance:
    Rtotal = Rp + 5 = 5 + 5 = 10 ohms

    The total reactance referred to the primary side is the sum of the primary reactance and the referred secondary reactance:
    Xtotal = Xp + 12 = 12 + 12 = 24 ohms

    Step 4: Find the total equivalent impedance
    The total equivalent impedance (Ztotal) on the primary side is the combination of resistance and reactance:
    Ztotal = sqrt(Rtotal² + Xtotal²) = sqrt(10² + 24²) = sqrt(100 + 576) = sqrt(676) ≈ 26 ohms

 Conclusion:
    The equivalent impedance of the transformer referred to the primary side is approximately 26 ohms, so the correct answer is:
    (iv) 26 ohms.

    Given:
    - Transformer rating: 100 kVA
    - Primary voltage (Vp) = 2200 V
    - Secondary voltage (Vs) = 440 V
    - Primary resistance (Rp) = 0.3 Ω
    - Primary reactance (Xp) = 1.1 Ω
    - Secondary resistance (Rs) = 0.01 Ω
    - Secondary reactance (Xs) = 0.035 Ω
    - Efficiency = 100% (ideal transformer)

    We need to calculate the full-load secondary current.

    Step 1: Calculate the apparent power at full load
    The apparent power (S) for a 100 kVA transformer is:
    S = 100 kVA = 100,000 VA

    At full load, the apparent power is equal on both the primary and secondary sides (since the efficiency is 100%).

    Step 2: Use the formula for power to find the full-load current
    The apparent power (S) in the secondary side is related to the secondary voltage (Vs) and the secondary current (Is) by the formula:
    S = Vs * Is
    Rearranging to solve for Is:
    Is = S / Vs = 100,000 / 440 ≈ 227.27 A

Given:
- Secondary current (I2) = 10 A
- Power factor (cos Φ2) = 0.8 (lagging)
- Sine of power factor angle (sin Φ2) = 0.6
- Secondary resistance (R2) = 0.75 Ω
- Primary resistance (R1) = 0.2 Ω
- Secondary reactance (X2) = 1.8 Ω
- Primary reactance (X1) = 0.5 Ω
- Turns ratio (K) = 2

Step 1: Calculate the equivalent secondary resistance (R02):
    R02 = R2 + K^2 * R1
    R02 = 0.75 + (2^2) * 0.2 = 0.75 + 4 * 0.2 = 1.55 Ω

Step 2: Calculate the equivalent secondary reactance (X02):
    X02 = X2 + K^2 * X1
    X02 = 1.8 + (2^2) * 0.5 = 1.8 + 4 * 0.5 = 3.8 Ω

Step 3: Calculate the voltage drop in the secondary:
    Voltage drop = I2 * (R02 * cos Φ2 + X02 * sin Φ2)
    Voltage drop = 10 * (1.55 * 0.8 + 3.8 * 0.6) = 10 * (1.24 + 2.28) = 10 * 3.52 = 35.2 V

 Conclusion:
  The total voltage drop in the secondary is 35.2 V. Thus, the correct answer is:
  35.2 V.

Given:
    - Secondary voltage (Vs) = 90 V
    - Load impedance (Z) = 15∠25° Ω (magnitude = 15 Ω, phase angle = 25°)
    - Core losses = 12 W
    - Copper losses = 16 W

    We need to find the percentage efficiency of the transformer.

    Step 1: Calculate the total output power
    The total output power delivered to the load is given by:
    Pout = Vs × Is × cos(θ)

    Step 2: Calculate the secondary current
    The magnitude of the secondary current (Is) can be found using Ohm’s law:
    Is = Vs / |Z|
    Is = 90 / 15 = 6 A

    Step 3: Calculate the output power
    Now, calculate the real power delivered to the load:
    Pout = 90 × 6 × cos(25°)
    Pout = 90 × 6 × 0.9063 = 488.4 W

    Step 4: Calculate the total losses
    The total losses in the transformer are the sum of the core losses and the copper losses:
    P_losses = P_core + P_copper = 12 + 16 = 28 W

    Step 5: Calculate the input power
    The input power is the sum of the output power and the losses:
    Pin = Pout + P_losses = 488.4 + 28 = 516.4 W

    Step 6: Calculate the efficiency
    The efficiency of the transformer is given by:
    η = (Pout / Pin) × 100
    η = (488.4 / 516.4) × 100 ≈ 94.6%

Given:
  - Primary voltage (Vp) = 120 V
  - Secondary voltage (Vs) = 20 V
  - Rated current = 4 A (for both primary and secondary)
  - Secondary resistance (Rs) = 0.75 Ω
  - Primary resistance (Rp) = 5 Ω
  - Turns ratio = Vp / Vs = 120 / 20 = 6

Step 1: Find the primary current (I1)
  Since the power rating is 4 A on the secondary side and the transformer is ideal (100% efficiency), 
  the primary current (I1) can be calculated as:
  I1 = I2 / Turns ratio = 4 / 6 ≈ 0.67 A

Step 2: Calculate the copper loss in the secondary winding
  Copper loss in the secondary is given by:
  P_secondary = I2^2 * Rs = 4^2 * 0.3 = 16 * 0.3 = 4.8 W

Step 3: Calculate the copper loss in the primary winding
  Copper loss in the primary is given by:
  P_primary = I1^2 * Rp = 0.67^2 * 5 = 0.4489 * 5 ≈ 2.2445 W

Step 4: Calculate the total copper loss
  The total copper loss is the sum of the primary and secondary copper losses:
  P_total_copper_loss = P_primary + P_secondary = 2.2445 W + 4.8 W ≈ 7 W.