Transformer MCQ Quiz - Objective Question with Answer for Transformer - Download Free PDF
Last updated on Jun 19, 2025
Latest Transformer MCQ Objective Questions
Transformer Question 1:
Copper losses in a transformer are proportional to:
Answer (Detailed Solution Below)
Transformer Question 1 Detailed Solution
Copper Losses in a Transformer
Definition: Copper losses in a transformer refer to the power losses that occur due to the resistance of the winding conductors when current flows through them. These losses are one of the primary contributors to the overall efficiency reduction in a transformer. Copper losses are also known as I2R losses, as they are directly related to the current flowing through the winding and the resistance of the winding material.
Correct Option Analysis:
The correct answer is:
Option 1: Copper losses in a transformer are proportional to the square of the load current.
Explanation:
Copper losses in a transformer are mathematically expressed as:
Power Loss = I2 × R
Where:
- I is the current flowing through the winding (load current).
- R is the resistance of the winding conductor.
From the formula above, it is evident that copper losses are proportional to the square of the current (I2). This relationship arises because the power dissipated as heat in the winding resistance is dependent on the square of the current. Therefore, as the load current increases, the copper losses increase quadratically.
In transformers, the load current varies with the load connected to the transformer. As the load current increases, the copper losses increase significantly due to the squared relationship. This is why transformers are typically designed to operate efficiently at their rated load, as excessive copper losses can lead to overheating and reduced efficiency.
Important Considerations:
- Copper losses are directly related to the winding material's resistance. High-quality materials with low resistivity (such as copper or aluminum) are used to minimize these losses.
- While copper losses depend on the load current, they do not depend on the voltage, frequency, or turns ratio of the transformer.
- Reducing the winding resistance by using thicker conductors or reducing the length of the winding can help minimize copper losses.
Additional Information:
To further understand the analysis, let’s evaluate the other options:
Option 2: Copper losses are proportional to the square of the frequency.
This option is incorrect. Copper losses are not dependent on the frequency of the electrical supply. While frequency does affect core losses (hysteresis and eddy current losses), copper losses are solely dependent on the load current and the winding resistance.
Option 3: Copper losses are proportional to the square of the turns ratio.
This option is also incorrect. The turns ratio of a transformer determines the voltage and current transformation between the primary and secondary windings but does not directly influence copper losses. Copper losses depend on the load current flowing through the windings, not on the turns ratio itself.
Option 4: Copper losses are proportional to the square of the voltage.
This option is incorrect. Copper losses are not directly related to the voltage. While the voltage determines the current flowing through the load (based on the load's impedance), copper losses are specifically proportional to the square of the load current, not the voltage.
Conclusion:
Copper losses in a transformer are a critical factor affecting its efficiency and performance. Understanding that these losses are proportional to the square of the load current is essential for designing and operating transformers effectively. By minimizing winding resistance and optimizing the transformer for its rated load, these losses can be reduced, improving the overall efficiency of the transformer.
Transformer Question 2:
Which of the following is a significant feature of Scott connection transformers?
Answer (Detailed Solution Below)
Transformer Question 2 Detailed Solution
Explanation:
Scott Connection Transformers
Definition: A Scott connection transformer, also known as a Scott-T transformer, is a type of transformer connection used to convert a three-phase electrical power supply into a two-phase power supply and vice versa. This connection is primarily employed in industrial applications where two-phase power is required, such as in the operation of certain types of electrical motors or systems.
Working Principle:
The Scott connection transformer operates by using two transformers: the main transformer and the teaser transformer. The primary winding of the main transformer is connected across two lines of a three-phase supply, while the primary winding of the teaser transformer is connected between the third line and the center-tap of the main transformer’s primary winding. By carefully selecting the turns ratio of the windings, a balanced two-phase output can be obtained. The two-phase output has a 90° phase difference, which is the defining characteristic of two-phase systems.
Advantages of the Scott Connection:
- Efficient conversion between three-phase and two-phase power, enabling the use of two-phase equipment in systems powered by three-phase supply.
- Provides a balanced load on the three-phase supply, reducing the risk of unbalanced current and associated issues.
- Simplifies the integration of legacy two-phase systems into modern three-phase infrastructure.
Applications: Scott connection transformers are widely used in applications such as:
- Supplying power to two-phase motors and other two-phase equipment.
- Traction systems in electric railways, where two-phase power is sometimes required.
- Converting power in industrial facilities with a mix of three-phase and two-phase systems.
Correct Option Analysis:
The correct option is:
Option 3: They provide a 2-phase output from a 3-phase input.
This option accurately describes the primary feature of Scott connection transformers. These transformers are designed to convert three-phase electrical power into two-phase power, which is essential for operating two-phase equipment in systems powered by three-phase electricity. The Scott connection ensures a balanced two-phase output with a 90° phase difference, maintaining the integrity of the power supply and enabling efficient operation of two-phase loads.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1: They provide high voltage output.
This statement is incorrect in the context of Scott connection transformers. While transformers, in general, can step up or step down voltage, the Scott connection specifically refers to the method of converting three-phase power into two-phase power. The voltage level of the output depends on the turns ratio of the transformer windings, but this is not the primary feature of a Scott connection transformer.
Option 2: They convert 1-phase power into 3-phase.
This option is incorrect as it describes a different type of power conversion. Converting single-phase power into three-phase power typically requires a phase converter or a rotary transformer, not a Scott connection transformer. The Scott connection is specifically designed for converting three-phase power into two-phase power or vice versa.
Option 4: They are used for stepping down voltage only.
This option is misleading as it limits the functionality of Scott connection transformers. While these transformers can step down voltage if designed to do so, their primary purpose is to convert three-phase power into two-phase power. Voltage transformation is a secondary consideration and depends on the specific application and design of the transformer.
Conclusion:
The Scott connection transformer is a specialized device used for converting three-phase power into two-phase power, making it invaluable in applications where two-phase equipment needs to operate within a three-phase system. By understanding its working principle and advantages, it is clear that the correct option is Option 3, as it accurately describes the main feature of Scott connection transformers. Evaluating the other options highlights common misconceptions and underscores the importance of recognizing the specific function of Scott connection transformers in electrical systems.
Transformer Question 3:
What does the use of a single winding in an auto transformer result in?
Answer (Detailed Solution Below)
Transformer Question 3 Detailed Solution
Explanation:
Use of Single Winding in an Auto Transformer
Definition: An auto transformer is a type of electrical transformer where the primary and secondary windings share a common single winding. Unlike conventional transformers that have separate primary and secondary windings, an auto transformer utilizes a single winding for both input and output, with part of the winding serving as both the primary and secondary. This shared winding configuration significantly reduces the copper consumption compared to conventional transformers.
Working Principle: In an auto transformer, electrical energy is transferred from the input (primary) side to the output (secondary) side through electromagnetic induction. The shared winding allows for direct electrical connection between portions of the primary and secondary circuits. By tapping the winding at different points, varying voltage levels can be achieved. The reduced number of windings results in less material usage, which is one of the primary advantages of an auto transformer.
Advantages:
- Reduced copper consumption due to the use of a single winding for both primary and secondary circuits.
- Compact design with smaller size and lighter weight compared to conventional transformers.
- Higher efficiency due to lower losses in the winding.
- Cost-effective solution for voltage regulation applications.
Disadvantages:
- Limited isolation between primary and secondary circuits, which can be a drawback for applications requiring complete electrical isolation.
- Not suitable for applications where the primary and secondary windings need to handle significantly different power levels.
- Susceptible to fault propagation between the primary and secondary circuits due to the shared winding design.
Applications: Auto transformers are commonly used in applications such as voltage regulation in power systems, motor starting, and laboratory equipment where compactness, efficiency, and cost savings are crucial.
Correct Option Analysis:
The correct option is:
Option 2: Reduced copper consumption
The use of a single winding in an auto transformer significantly reduces the amount of copper required for its construction. In conventional transformers, separate primary and secondary windings are made, which require more copper. Auto transformers share a single winding for both primary and secondary, minimizing the need for additional windings and, consequently, reducing copper consumption. This reduction in material usage results in lower manufacturing costs and contributes to the compact size and lighter weight of auto transformers.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1: Higher losses
This option is incorrect because the single winding design of an auto transformer typically results in lower losses compared to conventional transformers. With fewer windings, the resistive losses in the winding are reduced, leading to higher efficiency. Additionally, the compact design minimizes leakage flux, further improving performance.
Option 3: Lower voltage regulation
This option is incorrect because auto transformers generally offer better voltage regulation due to their efficient design and low losses. The shared winding allows for precise voltage adjustment, making auto transformers suitable for applications requiring stable voltage output.
Option 4: Increased size and weight
This option is incorrect because the use of a single winding in an auto transformer results in a more compact design with reduced size and weight compared to conventional transformers. The reduced copper consumption and simplified construction contribute to the smaller dimensions and lighter weight of auto transformers.
Option 5: (Not applicable in this case)
There is no relevant information provided for Option 5 in this context, and it does not align with the operational characteristics of an auto transformer.
Conclusion:
The use of a single winding in an auto transformer is a significant design feature that leads to reduced copper consumption. This reduction not only makes the auto transformer more cost-effective but also contributes to its compact size and higher efficiency. While the shared winding design offers numerous advantages, it also comes with certain limitations, such as reduced electrical isolation. Understanding these trade-offs is essential for selecting the appropriate transformer type for specific applications.
Transformer Question 4:
Core losses in a transformer are proportional to:
Answer (Detailed Solution Below)
Transformer Question 4 Detailed Solution
Core Losses in a Transformer
Definition: Core losses (also known as iron losses) in a transformer are the losses that occur in the magnetic core of the transformer. These losses are primarily caused by two phenomena: hysteresis loss and eddy current loss. Core losses are independent of the load and depend solely on the supply voltage and frequency.
Working Principle:
The magnetic core of a transformer is subjected to alternating magnetic flux due to the alternating current in the primary winding. This alternating flux causes two types of losses in the core:
- Hysteresis Loss: This loss is caused by the repeated magnetization and demagnetization of the core material as the magnetic field alternates. The energy lost in this process depends on the hysteresis loop area of the core material, the volume of the core, and the frequency of the magnetic flux.
- Eddy Current Loss: This loss is caused by circulating currents induced in the core material due to the changing magnetic flux. These currents flow in closed loops within the core and generate heat, leading to energy loss. Eddy current loss depends on the square of the frequency and the square of the magnetic flux density.
Mathematical Representation:
The total core loss (Pcore) in a transformer can be expressed as the sum of hysteresis loss (Ph) and eddy current loss (Pe):
Pcore = Ph + Pe
The hysteresis loss is proportional to the frequency (f), while the eddy current loss is proportional to the square of the frequency (f2). Thus, the total core loss is strongly dependent on the frequency.
Correct Option Analysis:
The correct option is:
Option 1: The square of the frequency.
Core losses in a transformer are proportional to the square of the frequency because of the dominance of eddy current loss, which varies directly with f2. Although hysteresis loss is proportional to frequency (f), the overall dependence of core loss on frequency is heavily influenced by the f2 relationship of eddy current loss. This makes option 1 the most accurate answer.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 2: The square of the stator current.
This option is incorrect. Core losses are independent of the load, and hence they are not influenced by the stator current. Stator current affects copper losses (I2R losses) in the winding rather than core losses.
Option 3: The square of the voltage.
While core losses are dependent on the supply voltage, they are not directly proportional to the square of the voltage. Instead, core losses are primarily influenced by the frequency of the alternating magnetic flux, as explained earlier.
Option 4: The square of the load current.
This option is also incorrect. Core losses are independent of the load current. Load current affects copper losses in the transformer windings but does not impact the core losses, which are constant and dependent on the supply voltage and frequency.
Conclusion:
Core losses in a transformer are a crucial factor in determining its efficiency and performance. These losses are caused by hysteresis and eddy currents in the core, and their dependence on frequency makes it essential to carefully select the operating frequency of the transformer. Among the given options, option 1 is correct because core losses are proportional to the square of the frequency, primarily due to the contribution of eddy current loss.
Transformer Question 5:
What does the flux created by the load current in the primary winding, which links only with the primary winding, represent?
Answer (Detailed Solution Below)
Transformer Question 5 Detailed Solution
The correct answer is option 3.
Leakage flux
- The flux created by the load current in the primary winding, which links only with the primary winding, represents leakage flux.
- This flux doesn't pass through the core to link with the secondary winding and instead, flows through the winding insulation and insulating oil. This leakage flux contributes to the primary leakage reactance.
- This flux does not contribute to energy transfer between windings and is considered a form of energy loss.
Magnetic flux
- Magnetic flux is defined as the number of magnetic field lines passing through a given closed surface. It provides the measurement of the total magnetic field that passes through a given surface area.
Mutual flux
- Mutual flux is the magnetic flux that links both the primary and secondary windings of a transformer, contributing to its operation.
Induced flux
- Induced flux refers to the change in magnetic flux through a loop of wire, which results in the generation of an electromotive force (EMF) or voltage. This process is a fundamental principle in electromagnetic induction.
Top Transformer MCQ Objective Questions
Which is to be short circuited on performing short circuit test on a transformer?
Answer (Detailed Solution Below)
Transformer Question 6 Detailed Solution
Download Solution PDF- Short circuit test performs on the high-voltage (HV) side of the transformer where the low-voltage (LV) side or the secondary is short-circuited.
- A wattmeter is connected to the high voltage side. An ammeter is connected in series with the high voltage side
Generally the no-load losses of an electrical machine is represented in its equivalent circuit by a
Answer (Detailed Solution Below)
Transformer Question 7 Detailed Solution
Download Solution PDFNo load losses of the electrical machine:
When the load is not connected to the electric machine, the machine draws a low value of current to keep the machine active or running
- The current which is taken by the machine is called no-load current
- Current is drawn by the core of the machine
- The equivalent circuit is represented by a high value of resistance in parallel.
- The power factor of the machine will be very low.
Which of the following statements about the core type transformer compared to shell type transformer is INCORRECT?
Answer (Detailed Solution Below)
Transformer Question 8 Detailed Solution
Download Solution PDFThe differences between core type and shell-type transformers are given below.
Core type |
Shell type |
It has two yoke and two limbs |
It has two yokes and three limbs |
Both the limbs are provided with windings and the core is being surrounded by windings |
Only the middle limb is provided with winding and windings are being surrounded by the core |
Both the limbs have the same cross-sectional area |
Outer limbs and middle limb have a different cross-sectional area |
Leakage flux is more, and power transfer capability is less |
Leakage flux is less, and power transfer capability is more |
More amount of copper is required for windings |
Less amount of copper is required for windings |
Less amount of insulating material is required |
More amount of insulating material is required |
High voltage and small kVA rating transformers are suitable |
Low voltage and large kVA rating transformers are suitable |
Suitable for high voltage power transmission |
Suitable for low voltage power transmission |
The mean length of coil is shorter. | The mean length of coil is longer. |
Confusion PointsCore type transformers have a shorter mean coil length. This is because the low voltage windings are wrapped around the core, closest to it, in a low-high configuration. The low voltage section carries more current and uses more material, so placing it closer to the core which reduces the average winding length and the amount of material needed.
In order to minimize the inrush current in a single-phase transformer, the supply switch should be closed at the instant when:
Answer (Detailed Solution Below)
Transformer Question 9 Detailed Solution
Download Solution PDFMagnetizing inrush current:
- Magnetizing inrush current in transformer is the current which is drawn by a transformer at the time of energizing it.
- To have minimum inrush current in the transformer the switch-on instant should be at maximum input voltage
- This inrush current is transient in nature and exists for few milliseconds.
- The inrush current may be up to 10 times higher than the normal rated current of the transformer.
- Although the magnitude of inrush current is high it generally does not create any permanent fault in the transformer as it exists for a very small time.
- But still, inrush current in a power transformer is a problem, because it interferes with the operation of circuits as they have been designed to function.
- Some effects of high inrush current include nuisance fuse or breaker interruptions, as well as arcing and failure of primary circuit components, such as switches. Another side effect of high inrush is the injection of noise.
A single-phase 111-V, 50-Hz supply is connected to a coil with 200 turns of a coil-core assembly as shown in the given figure. Find the magnitude of maximum flux in the core.
Answer (Detailed Solution Below)
Transformer Question 10 Detailed Solution
Download Solution PDFConcept:
The magnitude of net emf of an ideal transformer is given by the formula:
E = 4.44 × f × N × ϕm
Where E = RMS value of applied voltage.
f = frequency of the transformer.
N = number of turns.
ϕm = the magnitude of maximum flux in the core.
Calculation:
E = 111 V, f = 50 Hz, N = 200
111 = 4.44 × 50 × 200 × ϕm
ϕm = 111 / (4.44 × 50 × 200)
= 2.5 mWb
An iron cored choke coil when connected to a 15 V DC supply draws a current of 1.5 A. When connected to 230 V, 50 hz supply, it takes 2 A current and consume 60 W. The iron loss in the core is:
Answer (Detailed Solution Below)
Transformer Question 11 Detailed Solution
Download Solution PDFConcept:
When dc is applied inductor behaves as a short circuit.
Total loss = copper loss + iron loss
Copper loss = I2R
Calculation:
Given dc supply = 15 volt
DC current = 1.5 amp
Resistance
Total loss = 60 watt
Current drawn = 2 amp
Copper loss = 22 × 10 = 40 watt
Iron loss = total loss – copper loss = 60 – 40 = 20 watt
For a single phase transformer, wattmeter readings for OC and SC test result are as given below.
Wattmeter reading in OC test - 2.5 KW
Wattmeter reading in SC test - 5 KW
Find maximum efficiency of 5 KVA transformer for unity power factor.
Answer (Detailed Solution Below)
Transformer Question 12 Detailed Solution
Download Solution PDFThe correct answer is option 4): 41.38%
Concept:
The maximum efficiency occurs at the proportion of load x
Where,
x = Fraction of load
S = Apparent power in kVA
Pi = Iron losses
Pcu= Copper losses
Maximum efficiency =
Calculation
Given
S = 5 kVA
No-load loss (Pi) = 2.5kW
Copper loss(Pcu) = 5 Kw
cos ϕ = 1
x =
=
= 0.707
=
= 0.41 38
Voltage regulation of transformer is given by
Answer (Detailed Solution Below)
Transformer Question 13 Detailed Solution
Download Solution PDFVoltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.
V0 = no-load secondary voltage
V = full load secondary voltage
Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.
Additional InformationVoltage regulation can be expressed as a fraction or unit-change of the no-load terminal voltage and can be defined in one of two ways,
- Voltage regulation-down, (Regdown) and
- Voltage regulation-up, (Regup)
- When the load is connected to the second output terminal, the terminal voltage goes down, or when the load is removed, the secondary terminal voltage goes up.
- Hence, the regulation of the transformer will depend on which voltage value is used as the reference voltage, load or non-load value.
Transformer Voltage Regulation as a Percentage Change:
Transformer when connected to load:
%Reg =
Transformer under no load:
%Reg =
Eddy currents are induced by changing magnetic fields. Where do they flow with respect to the magnetic field?
Answer (Detailed Solution Below)
Transformer Question 14 Detailed Solution
Download Solution PDFEddy Current:
- Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction.
- Eddy currents flow in closed loops within conductors, in-plane perpendicular to the magnetic field.
- By Lenz law, the current swirls in such a way as to create a magnetic field opposing the change; for this to occur in a conductor, electrons swirl in a plane perpendicular to the magnetic field.
- Because of the tendency of eddy currents to oppose, eddy currents cause a loss of energy.
- Eddy currents transform more useful forms of energy, such as kinetic energy, into heat, which isn’t generally useful.
- Thus eddy currents are a cause of energy loss in alternating current (AC) inductors, transformers, electric motors and generators, and other AC machinery, requiring special construction such as laminated magnetic cores or ferrite cores to minimize them.
- Eddy currents are also used to heat objects in induction heating furnaces and equipment, and to detect cracks and flaws in metal parts using eddy-current testing instruments.
- The magnitude of the current in a given loop is proportional to the strength of the magnetic field, the area of the loop, and the rate of change of flux, and inversely proportional to the resistivity of the material.
When a V-V system is converted into a Δ-Δ system, the capacity of the system increases by ______.
Answer (Detailed Solution Below)
Transformer Question 15 Detailed Solution
Download Solution PDFConcept:
Open delta or V-V connection of 3 ϕ transformer:
V-V connection has only two windings instead of three windings in both primary and secondary.
It is used when one winding of transformer goes in maintenance.
Power transfer capacity reduces to 57.7%(or 1/√ 3) times of Δ-Δ connection or closed delta connection.
The capacity of the transformer is measured in VA = √3 VLIL
For closed delta, transformer output VA = 3 VLIph
For open delta, transformer output VA =√ 3 VLIph
Calculation:
% increase in capacity =
=
= 73.2 %