Transformers Pune University MCQs
Transformers Pune University MCQs
This set of Transformers Questions and Answers for Campus interviews focuses on “Basic Materials Used in Transformer Parts”.
1. The majority of power transformers in use throughout the world are oil filled using a mineral oil.
a) True
b) False
Answer: a
Explanation: In majority power transformers dielectric material used is the oil, which serves the dual purpose of providing insulation and as a cooling medium to conduct away the losses which are produced in the transformer in the form of heat.
2. Dielectric mineral oil is used in ____________
a) Small transformers
b) Medium transformers
c) Large transformers
d) In all transformers
Answer: c
Explanation: Because of the fire hazard associated with mineral oil, it has been the practice to use designs for smaller transformers which do not contain oil. It is usual, therefore, to locate transformers with mineral oil, out of doors where a fire is more easily dealt with and consequentially the risks are fewer.
3. The purpose of the transformer core is to provide ____________
a) High reluctance path
b) Low reluctance path
c) High inductive path
d) High capacitive path
Answer: b
Explanation: The purpose of a transformer core is to provide a low-reluctance path for the magnetic flux linking primary and secondary windings. In doing so, the core experiences iron losses due to hysteresis and eddy currents flowing within it which, in turn, show themselves as heating of the core material.
4. Transformer core is designed to reduce ______________
a) Hysteresis loss
b) Eddy current loss
c) Hysteresis loss and Eddy current loss
d) Cannot be determined
Answer: c
Explanation: Hysteresis loss and eddy current loss are the losses which take place in core of the transformer thus they are also termed as core losses. While other losses take place in winding or in air gap which can’t be dealt with core design.
5. Transformers windings are generally made of __________
a) Steel
b) Iron
c) Copper
d) Steel iron alloy
Answer: c
Explanation: In order to avoid losses due to loading current, winding materials must be chosen wisely. Winding conductors are thus made of copper or more precisely saying they are made of high conductivity copper by some industrial processes.
6. Before using oil in transformers, insulation material was _________
a) Asbestos
b) Cotton
c) Low grade pressboard in air
d) Kraft paper
Answer: d
Explanation: At the time of discovery of transformer, people were using asbestos, cotton, low grade pressboard in air for insulation purpose. Further, Kraft paper was invented which became much popular insulation material.
7. Which transformer insulation material is best compare to Kraft paper?
a) Oil
b) Asbestos
c) Low grade pressboard
d) Cotton
Answer: a
Explanation: Newly developed oil-filled transformers have capabilities much greater than those transformers which used Kraft paper as dielectric material. Also, electrical properties of Kraft paper depend on physical and chemical properties of paper.
8. Which of the following is not the property of oil that should be fulfilled before using in transformer?
a) Low viscosity
b) High flash point
c) Low electrical strength
d) High chemical stability
Answer: c
Explanation: There are various important parameters that oil must follow for its use in oil cooled transformer. These parameters include low viscosity, high stability, high flash point, high electrical strength, low pour point.
9. Transformer ratings are given in _____________
a) kW
b) kVAR
c) HP
d) kVA
Answer: d
Explanation: There are two types of losses in a transformer, Copper Losses and Iron Losses or Core Losses or Insulation Losses. Copper losses (I 2 R) depends on current passing through transformer winding while Iron losses or Core Losses or Insulation Losses depends on Voltage. That’s why the rating of Transformer is in kVA.
10. Function of transformer is to _________________
a) Convert AC to DC
b) Convert DC to AC
c) Step down or up the DC voltages and currents
d) Step down or up the AC voltages and currents
Answer: d
Explanation: A Transformer does not work on DC and operates only on AC, therefore it Step up of Step down the level of AC Voltage or Current, by keeping frequency of the supply unaltered on the secondary side.
11. What is the dielectric strength of a transformer oil?
a) 1 kV
b) 35 kV
c) 100 kV
d) 330 kV
Answer: b
Explanation: For mineral oil, an accepted minimum dielectric strength is 30 kV for transformers with a high-voltage rating of 230 kV and above and 27 kV for transformers with a high-voltage rating below 230 kV. New oil should pass the condition of a minimum dielectric strength of 35 kV by ASTM methods of testing.
12. Which of the following is not a part of transformer installation?
a) Conservator
b) Breather
c) Buchholz relay
d) Exciter
Answer: d
Explanation: Conservator, breather, Buchholz relay are the parts which are much important in transformer construction in order to maintain temperature of the transformer and to work transformer with good efficiency.
13. The insulating material that can withstand the highest temperature safely is _______________
a) Cellulose
b) Asbestos
c) Mica
d) Glass fibre
Answer: c
Explanation: Mica is extremely stable when it is exposed to moisture and extreme temperatures to maintain superior electrical properties as an insulator. The mechanical properties of mica allow it to be cut, punched, stamped and machined to close tolerances along with maintenance of a high thermal conductivity.
14. The part of a transformer which is visible from outside _______________
a) Bushings
b) Core
c) Primary winding
d) Secondary winding
Answer: a
Explanation: Core, primary winding, secondary winding of a transformer are generally kept in closed container filled with an oil so that, oil acts as a coolant and provides electrical neutrality also. Thus, only bushings are visible from outside.
This set of Transformers Multiple Choice Questions & Answers focuses on “Transformer Construction ”.
1. Transformer core is generally made of ___________
a) Single block of core material
b) By stacking large number of sheets together
c) Can be made with any of the above method
d) Cannot be determined
Answer: b
Explanation: Transformer core experiences eddy current losses when transformer is in the operations. In order to reduce eddy current losses, it is advisable to use large number of sheets laminated from each other are stick together than using one single block.
2. Transformer core is constructed for ______________
a) Providing least effective magnetic linkage between two windings
b) Providing isolation between magnetic linkages of one coil from another
c) Providing most effective magnetic linkage between two windings
d) Cannot be determined
Answer: c
Explanation: Transformer core is so chosen that it will provide low reluctance path and will transfer maximum amount of flux from one winding to other, providing most effective magnetic linkage between two windings.
3. Which of the following statements is/are correct?
a) High frequency power supplies are light weight
b) Transformer size gets reduced at high frequency
c) Transformer size is more at higher frequency
d) High frequency power supplies are light weight and transformer size gets reduced at high frequency
Answer: d
Explanation: From the induced emf equation of transformer emf is given by E ∝ φf. For same emf, φf = constant φ 1 f 1 = φ 2 f 2 i.e. B 1 A 1 f 1 = B 2 A 2 f 2 . For constant flux density B 1 = B 2 . A 1 f 1 = A 2 f 2 . For high frequency f 2 > f 1 , A 2 < A 1 . Thus, at high frequencies transformer size get reduced and also light weight.
4. Transformer operating at 25-400 Hz frequency contain core made of _____________
a) Highly permeable iron
b) Steel alloy
c) Air core
d) Highly permeable iron and Steel alloy
Answer: d
Explanation: When core is made of highly permeable iron or steel alloy . This transformer is generally called an iron-core transformer. Transformers operated from 25–400 Hz are invariably of iron-core construction.
5. In various radio devices and testing instruments we use ______________
a) Iron core transformer
b) Air core transformer
c) W/O core transformer
d) Any transformer can be used
Answer: a
Explanation: In special cases, the magnetic circuit linking the windings may be made of nonmagnetic material, in which case the transformer is referred to as an air-core transformer. The air-core transformer is of interest mainly in radio devices and in certain types of measuring and testing instruments.
6. Which type of flux does transformer action need?
a) Constant magnetic flux
b) Increasing magnetic flux
c) Alternating magnetic lux
d) Alternating electric flux
Answer: c
Explanation: The energy transfer in a transformer, is from one winding to another, entirely through magnetic medium it is known as transformer action. Therefore, transformer action requires an alternating or time varying magnetic flux in order to transfer power from primary side to secondary side. Since induced emf in the winding is due to flux linkage.
7. Different core construction is required for core type and shell type transformer.
a) True
b) False
Answer: a
Explanation: In the “closed-core” type transformer, the primary and secondary windings are wound outside and surround the core ring. In the “shell type” transformer, the primary and secondary windings pass inside the steel magnetic circuit which forms a shell around the windings.
8. There is only one magnetic flux path in the circuit. The transformer is definitely ________________
a) Core type
b) Shell type
c) Can be any of the above
d) Depends on other parameters
Answer: a
Explanation: In core type transformer, winding is placed on two core limbs, while in case of shell type transformer, winding is placed on mid arm of the core. Other limbs will be used as mechanical support. Core type transformers have only one magnetic flux path.
9. Which of the following is correct statement?
a) Core type transformer has more output than shell type
b) Core type transformer has higher efficiency compare to shell type
c) Core type transformer has lower efficiency than shell type
d) Can’t predict
Answer: c
Explanation: In core type winding is surrounded with considerable part of core whereas in shell type core is surrounded with considerable part of winding of transformer. In core type output is less, because of losses. In shell type transformer output is high because of less loss, thus efficiency will be more in case of shell transformer.
10. Core type transformer is with ____________________
a) Large size
b) Small size
c) High voltage
d) Everywhere
Answer: a
Explanation: Core type is very useful when we need large size of the transformer with operation at low voltage. While shell type transformer is very useful when we need small size high voltage. Cooling is more in core type.
11. Which of the following is the correct statement?
a) Shell type has more mechanical protection
b) Cooling is more in shell type
c) In core type sandwiched wiring is used
d) In core type concentric winding is used
Answer: d
Explanation: Shell type has less mechanical protection to coil while Core type has better mechanical protection to coil. Core type is easy to repair and maintain. In core type transformer concentric cylindrical winding are used. In shell type transformer sandwiched winding are used.
12. What is the purpose of providing an iron core in a transformer?
a) Provide support to windings
b) Reduce hysteresis loss
c) Decrease the reluctance of the magnetic path
d) Reduce eddy current losses
Answer: c
Explanation: Iron core is used in a transformer to carry flux from one winding to another winding, so there should be minimum opposition to flux passing through iron core. Hence, transformer function is to decrease the reluctance of magnetic path.
13. What is the thickness of laminations used in a transformer?
a) 0.1 mm to 0.5 mm
b) 4 mm to 5 mm
c) 14 mm to 15 mm
d) 25 mm to 40 mm
Answer: a
Explanation: Laminations are made to reduce the eddy currents and is made of thin strips. Generally, the steel transformer lamination range for 50 Hz varies from 0.25mm to 0.5mm, if it is a 60 Hz transformer then it ranges from 0.17–0.27mm.
This set of Transformers Multiple Choice Questions & Answers focuses on “Transformer Construction ”.
1. In the transformer which of the following winding has got more cross-sectional area?
a) Copper winding
b) Steel winding
c) Aluminium winding
d) Iron winding
Answer: a
Explanation: The wire used for carrying current in a transformer winding is either copper or aluminium. While aluminium wire is lighter and less expensive than copper wire, a larger cross-sectional area of conductor must be used to carry the same amount of current as with copper.
2. Primary winding of a transformer ______________
a) Is always a high voltage winding
b) Is always a low voltage winding
c) Could either be a low voltage or high voltage winding
d) Cannot be determined
Answer: c
Explanation: Primary winding used in a transformer, can be at higher or lower voltage potential, depending on the number of turns with secondary winding. For step up and step-down transformers primary winding will be at lower and higher potential respectively.
3. Which winding has more number of turns?
a) Low voltage winding
b) High voltage winding
c) Primary winding
d) Secondary winding
Answer: b
Explanation: High voltage winding always has a large number of turns, as voltage is directly proportional to the number of turns. If large numbered winding is present on primary side then the transformer is step down transformer.
4. Part of the transformer which undergoes most damage from overheating is ___________
a) Iron core
b) Copper winding
c) Winding insulation
d) Frame or case
Answer: c
Explanation: Copper windings carry current through them. The heat loss producing in any winding carrying current is proportional to the square of the current passing through it multiplied with resistance. For large transformers, current is very high, so heating causes most of the damage to insulation material.
5. If a transformer is continuously operated the maximum temperature rise will occur in ___________
a) Core
b) Windings
c) Tank
d) Cannot be determined
Answer: b
Explanation: Copper windings carry currents in a transformer. The loss in form of heat in copper winding carrying current is proportional to the square of the current passing through them multiplied by the resistance of the winding. This loss is dissipated in heat and corresponding temperature rise.
6. If secondary number of turns are higher then, transformer is called _________
a) Step-down
b) Step-up
c) One-one
d) Autotransformer
Answer: b
Explanation: When secondary number of turns are higher compare to primary, voltage induced in secondary windings will obviously high. Thus, this transformer is used for stepping up the output voltage by keeping frequency constant.
7. If primary number of turns are higher then, transformer is called _________
a) Step-down
b) Step-up
c) One-one
d) Autotransformer
Answer: a
Explanation: When primary number of turns are higher compare to secondary, voltage induced in secondary windings will obviously low compare to primary. Thus, this transformer is used for stepping down the output voltage by keeping frequency constant.
8. If a transformer is having equal number of turns at primary and secondary then transformer is called as _______________
a) Step-down
b) Step-up
c) One-one
d) Autotransformer
Answer: c
Explanation: A transformer is having equal number of turns at primary and secondary then transformer is called as one-one transformer. This transformer have turns ratio equal to 1, so is the voltage ratio for the one-one transformer.
9. One to one transformers are used because ______________
a) To isolate any part of circuit electrically
b) To get more voltage at secondary
c) To get less voltage at secondary
d) To reduce losses, present in circuit
Answer: a
Explanation: In one to one transformers, we have same number of turns in primary and in secondary. So, increasing/ reducing voltage is not possible. They are generally used to isolate one part of circuit from another part of circuit, electrically.
10. Same type and kind of insulations are not used in all types of transformers.
a) True
b) False
Answer: a
Explanation: The windings of huge power transformers use conductors with heavier insulation and are assembled with greater mechanical support and the winding layers are insulated from each other, this is known as minor insulation for which pressed board or varnished cloth is used. While for major insulation and insulating cylinders, they are made of specially selected pressed board or synthetic resin bounded cylinders, is used between LV and core and LV and HV.
11. Sandwiched type of winding is used in ____________
a) In all transformers
b) In core type transformers
c) In shell type transformers
d) In all transformers except shell and core type transformers
Answer: c
Explanation: According to the construction of transformers core type transformers don’t require sandwiched wiring. While, in the shell type transformer, the primary and secondary windings pass inside the steel magnetic circuit which forms a shell around the winding.
This set of Transformers Multiple Choice Questions & Answers focuses on “Cooling Techniques for Transformer”.
1. Which is the most common, famous and adopted method of cooling of a power transformer?
a) Air blast cooling
b) Natural air cooling
c) Oil cooling
d) Any of the above method can be used
Answer: c
Explanation: Oil acts as a best coolant material, for transformer cooling. Due to its high efficiency as a coolant it is most widely used in transformers. Not only same but Oil with suitable properties can be used for various power transformers according to their ratings.
2. Function of conservator in an electrical transformer is __________
a) Supply cooling oil to transformer in time of need
b) Provide fresh air for cooling the transformer
c) Protect the transformer from damage when oil expends due to heating
d) Cannot be determkned
Answer: c
Explanation: When transformer is loaded and when ambient temperature rises, the volume of oil inside transformer increases as oil expands. A conservator tank properly installed on transformer provides required space to this expanded transformer oil. It performs another function as a reservoir for transformer insulating oil.
3. Natural oil cooling method have some limitations due to which it is adopted for transformers up to a rating of ____________
a) 3000 kVA
b) 1000 kVA
c) 500 kVA
d) 250 kVA
Answer: a
Explanation: For the transformers n higher kVA ratings can be used with this cooling method. While transformers having capacity beyond 5 MVA, due to some improper limitations forced cooling is used. Natural cooling is based on the important phenomenon seen in fluids that when oil is heated up, moves in upward direction.
4. What is the function of spacers?
a) To insulate the coils from each other
b) To provide free passage to the cooling oil
c) To insulate coils and provide free passage
d) Cannot be determined
Answer: b
Explanation: The winding layers of transformer are separated by spacers. One or more spacers are provided here, along with at least one integrated electrical discharge barrier extending off the central body of the spacer in the vicinity of the area where the spacer is in contact with a winding.
5. Which of the following is the most important quality required for chemical in breather, so that it can be used perfectly in an electrical transformer?
a) Ionizing air
b) Absorbing moisture
c) Cleansing the transformer oil
d) Cooling the transformer oil
Answer: b
Explanation: Most of the power generation plants use silica gel breathers fitted to the conservator of oil filled transformers. The purpose of silica gel breathers is to absorb the moisture in the air sucked in by the transformer during the breathing process.
6. Which chemical is used in breather?
a) Asbestos fibre
b) Silica sand
c) Sodium chloride
d) Silica gel
Answer: d
Explanation: In order to absorb moisture from air while breathing process, breather chemical is used. So, breather chemical should possess the required ability of absorbing moisture. In all chemicals available as on today, silica gel is most perfect and best material that can be used for such process.
7. A transformer oil used in an electrical transformer must be free from ________
a) Gases
b) Odour
c) Sludge
d) Moisture
Answer: d
Explanation: Transformer oil serves the purpose of cooling and it also acts as an insulator between primary and secondary winding, thus it must be free from moisture else it will conduct electric current through it, leading to failure of a transformer.
8. On which of the following transformer, Buchholz’s relay can be fixed on?
a) Auto-transformers
b) Air-cooled transformers
c) Welding transformers
d) Oil cooled transformers
Answer: d
Explanation: Buchholz relay is used in transformers for protection against all kinds of faults. Buchholz relay is a famous and mostly used gas-actuated relay, which is installed to serve its best in oil-immersed transformers. It gives an alarm, via its electrical circuitry, if any fault occurs in the transformer.
9. Gas is liberated due to temperature limit and due to dissociation of transformer oil after ___________
a) 50°C
b) 80°C
c) 100°C
d) 150°C
Answer: d
Explanation: Gas is usually not liberated due to dissociation of transformer oil. But when the oil temperature exceeds 1500, it dissociates and liberates. It is found that hydrogen H 2 and methane CH 4 are produced in large quantity if internal temperature of transformer rises up to 150 o C to 300 o C due to abnormal thermal stresses.
10. Buchholz’s relay will give warning and protection against ___________
a) Electrical fault inside the transformer itself
b) Electrical fault outside the transformer in outgoing feeder
c) For both outside and inside faults
d) Cannot be determined
Answer: a
Explanation: Buchholz relay is used in transformers for protection against all kinds of faults which are tend to happen inside a transformer. It is most famous gas-actuated relay which is installed in an oil-immersed transformer.
11. Which of the following listed component will see and perform according to changes in volume of transformer cooling oil due to variation of atmospheric temperature during day and night?
a) Conservator
b) Breather
c) Bushings
d) Buchholz relay
Answer: a
Explanation: Conservator is an additional tank provided with transformer which stores oil when it gets expanded due to temperature rise. It also serves another important purpose that is, as a reservoir of transformer oil. Thus, at all temperature variations of day and night transformer can work without any problem.
12. What should be ideal volatility and ideal viscosity of the transformer oil?
a) Low, low
b) High, high
c) Low, high
d) High, low
Answer: a
Explanation: Transformer oil has a low viscosity, high flash point, high dielectric strength, high resistivity. It has a low pour point and low volatility with good gas absorbing properties, while It resists oxidation, sludging and emulsification with water.
13. What is the function of breather in a transformer?
a) To provide oxygen inside the tank
b) To cool the coils during reduced load
c) To cool the transformer oil
d) To arrest flow of moisture when outside air enters the transformer
Answer: d
Explanation: Most of the power generation stations use silica gel breathers fitted to conservator of oil filled transformers. The most used purpose of these silica gel breathers is to arrest the moisture when the outside air is sucked in by the transformer during the breathing process.
14. Natural air cooling method can’t be adopted because of some unavoidable effects, beyond _______
a) 1.5 MVA
b) 5 MVA
c) 15 MVA
d) 50 MVA
Answer: a
Explanation: Smaller size transformers are immersed in a tank containing transformer oil. The transformer oil because temperature properties, which is surrounding the core and windings gets heated, expands and moves upwards. It then flows downwards by the inside of tank walls which cause it to drop temperature and oil goes down to the bottom of the tank from where it rises once again completing the circulation cycle.
This set of Transformers Questions and Answers for Entrance exams focuses on “No Load Operation of Transformer”.
1. What is the no-load current drawn by transformer?
a) 0.2 to 0.5 per cent
b) 2 to 5 per cent
c) 12 to 15 per cent
d) 20 to 30 per cent
Answer: b
Explanation: The no load current is about 2-5% of the full load current and it accounts for the losses in a transformer. These no-load losses include core losses, which contains eddy current losses & hysteresis losses and the copper losses due to the no Load current.
2. Purpose of no-load test on a transformer is ___________
a) Copper loss
b) Magnetising current
c) Magnetising current and loss
d) Efficiency of the transformer
Answer: c
Explanation: No-load current is little bit greater than actual magnetizing current. Total no-load current supplied from the source has two components, one is magnetizing current which is utilized for magnetizing the core and other component is consumed for compensating the core losses in transformer.
3. No-load current in a transformer ________________
a) Lags behind the voltage by about 75°
b) Leads the voltage by about 75°
c) Lags behind the voltage by about 15°
d) Leads the voltage by about 15°
Answer: a
Explanation: No-load current lags behind the voltage by an angle which is near to 900. Thus, angle between no-load current and magnetizing current is very small. No-load current has another component which is in phase with voltage.
4. Which of the following statement is true for no-load current of the transformer?
a) has high magnitude and low power factor
b) has high magnitude and high power factor
c) has small magnitude and high power factor
d) has small magnitude and low power factor
Answer: d
Explanation: Since no-load current lags voltage by the angle of nearly 900, power factor being equal to cosine of the angle between current and voltage, it will be equal to value which is near to 0. Thus, power factor will be low.
5. In no-load test we keep secondary terminals __________
a) Shorted
b) Shorted via fixed resistor
c) Open
d) Shorted via variable resistors
Answer: c
Explanation: In no-load test, as we don’t require any load, we are not allowed to connect any resistor to the transformer secondary. We don’t short the secondary terminals either.
6. Maximum value of flux established in a transformer on load is equal to _________
a) E1/ (4.44*f*N 1 )
b) E1/ (4.44*f*N 2 )
c) E2/ (4.44*f*N 1 )
d) Cannot define
Answer: a
Explanation: E1/ (4.44*f*N 1 ). The emf induced in the primary due to applied voltage to primary winding is equal to change in flux with respect to time multiplied by number of turns in the primary. So, by solving this equation we get, E1= (4.44*f*φ*N 1 ).
7. Induced emf in the primary of transformer is equal to terminal voltage applied at primary.
a) True
b) False
Answer: a
Explanation: Induced emf in the primary is approximately equal to the applied voltage. Ideally there lies a very small difference in the values, but it is neglected because winding resistance in the transformer is of very small order.
8. For a linear B-H relationship, which option is correct?
a) The exciting current is equal to core loss current
b) The exciting current is equal to magnetizing current
c) The exciting current is equal to de-magnetizing current
d) The exciting current is equal to cross-magnetizing current
Answer: b
Explanation: For a linear B-H relationship it is assumed that, there are no losses present in the core like eddy current losses and hysteresis losses are neglected. Thus, core loss current is equal to 0, which ultimately confirms exciting current is purely magnetizing one.
9.Third harmonic current in transformer at no-load is ______________
a) 3% of exciting current
b) 10% of exciting current
c) 25% of exciting current
d) 35% of exciting current
Answer: d
Explanation: The effect of saturation nonlinearity is to create a family of odd-harmonic components in the exciting current, the predominant being the third harmonic; this may constitute as large as 35–40% of the exciting current.
10. Ii in no-load test is responsible for ______________
a) Production of flux
b) Reactive power drawn from the supply
c) Active power drawn from the supply
d) No significance
Answer: c
Explanation: It will be assumed here that the current Io and its magnetizing component Im and its core-loss component Ii are sinusoidal on equivalent rms basis. In other words, Im is the magnetizing current and is responsible for the production of flux, while Ii is the core-loss current responsible for the active power being drawn from the source to provide the hysteresis and eddy-current loss.
11. The parallel circuit model is drawn because _________________
a) Conductance Gi accounts for core-loss current
b) Inductive susceptance Bm accounts for magnetizing current
c) Gi for core – loss current and Bm for magnetizing current
d) Cannot say
Answer: c
Explanation: The parallel circuit model of exciting current can be easily imagined wherein conductance Gi accounts for core-loss current Ii and inductive susceptance Bm for magnetizing current Im. Both these currents are drawn at induced emf E1 = V1 for resistance-less, no-leakage primary coil; even otherwise E1 =V1.
This set of Transformers Multiple Choice Questions & Answers focuses on “Ideal Transformer”.
1. A transformer cannot work on the DC supply because __________________
a) There is no need to change the DC voltage
b) A DC circuit has more losses
c) Faraday’s laws of electromagnetic induction are not valid since the rate of change of flux is zero
d) Cannot be determined
Answer: c
Explanation: For DC supply the direction and the magnitude of the supply remains constant, produced flux will be constant. Thus, rate of change of flux through the windings will be equal to zero. As a result, voltage at secondary will always be equal to 0.
2. An ideal transformer has infinite primary and secondary inductances.
a) True
b) False
Answer: b
Explanation: The primary and secondary windings have zero resistance. It means that there is no ohmic power loss and no resistive voltage drop in the ideal transformer. An actual transformer has finite but small winding resistances.
3. In a transformer the resistance between its primary and secondary is ______________
a) Zero
b) Very small
c) Cannot be predicted
d) Infinite
Answer: d
Explanation: Since the primary and secondary windings are not connected to each other, one can say there exists the resistance of infinite ohms. These windings are connected to each other magnetically not electrically.
4. Identify the correct statement relating to the ideal transformer.
a) no losses and magnetic leakage
b) interleaved primary and secondary windings
c) a common core for its primary and secondary windings
d) core of stainless steel and winding of pure copper metal
Answer: a
Explanation: There is no leakage flux so that all the flux is confined to the core and links both the windings. An actual transformer does have a small amount of leakage flux which can be accounted in detailed analysis by appropriate circuit modelling.
5. An ideal transformer will have maximum efficiency at a load such that _____________
a) copper loss = iron loss
b) copper loss < iron loss
c) copper loss > iron loss
d) cannot be determined
Answer: a
Explanation: Maximum efficiency of a transformer is defined at the that values when, copper losses become completely equal to the iron losses. In all other cases the efficiency will be lower than the maximum value.
6. Which of the following statement regarding an ideal single-phase transformer is incorrect? Transformer is having a turn ratio of 1: 2 and drawing a current of 10 A from 200 V AC supply is incorrect?
a) It’s a step-up transformer
b) Its secondary voltage is 400 V
c) Its rating is 2 kVA
d) Its secondary current is 20 A
Answer: d
Explanation: Since turns ratio is equal to 1:2 the transformer will give higher voltage at secondary with respect to the primary voltage, and current in secondary thus will be halved. In last option current is doubled which is opposite to the ratings given.
7. Ideal transformer core has permeability equal to _____
a) Zero
b) Non-zero finite
c) Negative
d) Infinite
Answer: d
Explanation: The core has infinite permeability so that zero magnetizing current is needed to establish the requisite amount of flux in the core. The core-loss is considered zero.
8.Turns ratio of the transformer is directly proportional to ____________
a) Resistance ratio
b) Currents ratio
c) Voltage ratio
d) Not proportional to any terms
Answer: c
Explanation: According to the voltage expression, emf induced in the primary is directly proportional to the change in the flux with respect to the time and number of turns of the primary winding. Similarly, for secondary winding.
9. Which of the following statement is correct regarding turns ratio?
a) Current ratio and turns ratio are inverse of each other
b) Current ratio is exactly same to the voltage ratio
c) Currents ratio is exactly same to the turns ratio
d) Voltage ratio and turns ratio are inverse of each other
Answer: a
Explanation: Voltage ratio of transformer winding is exactly similar to the turns ratio of transformer, while voltage ratio and turns ratio is exactly inverse of the currents ratio. Hence, by knowing any of these quantities on can identify the type of transformer.
10. Which of the following is the expression for emf induced in primary with voltage applied to primary of an ideal transformer?
a) e=V
b) V= √2*e*cos ωt
c) e= √2*V*cos ωt
d) Cannot say
Answer: c
Explanation: For an ideal transformer having a primary of N1 turns and a secondary of N2 turns on a common magnetic core. The voltage of the source to which the primary is connected is v = √2 V cos wt. while the secondary is initially assumed to be an open circuited. As a consequence, flux f is established in the core such that e = v = N1 dφ/dt.
11. Which of the following is the wrong expression?
a) i 1 N 1 =i 2 N 2
b) i 1 v 1 =i 2 v 2
c) i 1 N2=i 2 N 1
d) v 2 N 1 =v 1 N 2
Answer: c
Explanation: According to the transformation ratio, current flowing through the transformer is inversely proportional to the number turns of winding and voltage applied across it. While, voltage applied is directly proportional to the number of turns.
12. For transformer given, turns ratio is equal to a, what will be the impedance of primary with respect to secondary?
a) a 2 times the secondary impedance
b) a times secondary impedance
c) secondary impedance/a
d) secondary impedance/a 2
Answer: d
Explanation: The ratio of impedances on primary to the secondary is directly proportional to the inverse of square of turns ratio of transformer. Hence primary impedance to the secondary impedance ratio will be 1/ a 2 .
13. Power transformed in the ideal transformer with turns ratio a is _______
a) a 2 times primary
b) a times primary
c) primary power/ a 2
d) primary power
Answer: d
Explanation: In an ideal transformer, voltages are transformed in the direct ratio of turns, currents in the inverse ratio and impedances in the direct ratio squared; while power and VA remain unaltered. Thus, primary power= secondary power.
14. For a transformer with primary turns 100, secondary turns 400, if 200 V is applied at primary we will get ___________
a) 80 V at secondary
b) 800 V at secondary
c) 1600 V at secondary
d) 3200 V at secondary
Answer: b
Explanation: Voltage in the primary of the transformer will get modified in the transformer secondary, according to the number of turns. Thus, turns are modified with 4 times the primary, we’ll get 4 times higher voltage at secondary.
15. For a transformer with primary turns 400, secondary turns 100, if 20A current is flowing through primary, we will get ___________
a) 80A at secondary
b) 5A at secondary
c) 800A at secondary
d) 40A at secondary
Answer: a
Explanation: Current in the primary of the transformer will get modified in the transformer secondary, according to the number of turns, in inverse proportion. Thus, turns are modified with 1/4 times the primary, we’ll get 4 times higher current at secondary.
This set of Transformers Questions and Answers for Aptitude test focuses on “Real Transformer and Equivalent Circuit”.
1. When does transformer breath in?
a) load on it increases
b) load on it decreases
c) load remains constant
d) cannot be determined
Answer: b
Explanation: Transformer in the low loading condition, also called extreme condition , oil inside contracts and then air is taken inside thus breath in to main via the balloon like structure through silica gel breather.
2. A transformer transforms ________________
a) voltage
b) current
c) power
d) frequency
Answer: c
Explanation: Since, in a transformer voltage and current is changed according to the number of turns simultaneously, we call that power is transformed, though the magnitude remains same. Frequency is kept constant.
3. Greater the secondary leakage flux ___________
a) less will be the secondary induced emf
b) less will be the primary induced emf
c) less will be the primary terminal voltage
d) cannot be determined
Answer: a
Explanation: Since emf induced in the transformer coils is directly proportional to the change in the flux with respect to time, we can say that if flux reduces the change in flux after some time will be less which will induce less voltage in secondary.
4. Which of the following is not the purpose of iron core in a step-up transformer?
a) to provide coupling between primary and secondary
b) to increase the magnitude of mutual flux
c) to decrease the magnitude of magnetizing current
d) to provide all above features
Answer: c
Explanation: In real transformers, the two coils are generally wound onto the same iron core. The purpose of the iron core is to provide the path for the magnetic flux generated by the current flowing around the primary coil, so that as much of it as possible also links the secondary coil, with minimum losses.
5. In a transformer the tappings are generally provided on
a) Primary side
b) Secondary side
c) Low voltage side
d) Can be connected to any side
Answer: d
Explanation: The turns ratio is different with different tappings and hence different voltages are obtained with different tappings adjustment. The tappings are placed either on high voltage or low voltages or sometimes on both high and low voltage windings to get required output.
6. Helical coils can be used at _____________
a) low voltage side of high kVA transformers
b) high frequency transformers
c) high voltage side of small capacity transformer
d) high voltage side of high kVA rating transformers
Answer: a
Explanation: Helical winding is used for low voltage and high current winding of large generator transformers. Due to its distinct design that is spiral form, small number of turns and high current, some additional eddy-current losses may happen in winding.
7. In real transformer, primary winding has _________
a) Infinite resistance
b) Zero resistance
c) Some finite resistance
d) Cannot say
Answer: c
Explanation: For a real transformer on load, both the primary and secondary have finite resistances which are uniformly spread throughout the winding. These resistances give rise to associated copper (I 2 R) losses.
8. Both resistances and leakage reactances of the transformer windings are __________
a) Series effects
b) Parallel effects
c) Series-parallel effects
d) Cannot say
Answer: a
Explanation: Both resistances and leakage reactances of the transformer windings are series effects and for low operating frequencies at which the transformers are commonly employed , these can be regarded as lumped parameters.
9. To convert an ideal transformer into a practical transformer we add ____________
a) Primary winding resistance and secondary winding resistance
b) Primary winding leakage reactance and secondary winding leakage reactance
c) Primary winding resistance, leakage and secondary winding leakage reactance
d) Cannot be determined
Answer: c
Explanation: We consider all resistances i.e. of primary and secondary as series parameters of equivalent circuit of transformer, while all leakage reactances are also connected into the circuit as series parameters.
10. Parallel parameters in a transformer equivalent circuit contains ___________
a) G i and B m
b) R 1 and X 1
c) R 2 and X 2
d) Cannot be determined
Answer: a
Explanation: Primary and secondary resistances and leakage reactances are the series parameters in the transformer equivalent circuit. So, these are not included in parallel parameters. Parallel parameters contain Gi and Bm in which current Ii and Im flows respectively.
11. When does capacitor is included in equivalent circuit of transformer?
a) Transformer of very high VA rating
b) Transformer with very high frequency operation
c) Transformer with less VA
d) Never
Answer: b
Explanation: The passive lumped T-circuit representation of a transformer discussed above is adequate for most power and radio frequency transformers. In transformers operating at higher frequencies, the interwinding capacitances are often significant and must be included in the equivalent circuit.
12. The size of a transformer core will depend on _____________
a) frequency
b) area of the core
c) flux density of the core material
d) frequency and area of the core
Answer: d
Explanation: According to the frequency of transformer size of the core of transformer changes. While area of core also depends upon many parameters like operating voltage, capacity of transformer, hence all these contribute to the size of the core.
13. A single phase transformer has specifications as 250 KVA, 11000 V/415 V, 50 Hz. What are the values of primary and secondary currents?
a) Primary current = 602.4A, Secondary current = 22.7A
b) Secondary current = 202.7A, Primary current = 602.4A
c) Primary current = 22.7A, Secondary current = 602.4A
d) Primary current = 11.35A, Secondary current = 301.2A
Answer: c
Explanation: Primary current is defined as the ratio of rated capacity of transformer to the rated primary voltage of the transformer. Rated primary current= Rated power/voltage= 250000/11000= 22.7 A. Similarly calculating for secondary current gives secondary current = 602.4A.
14. A 25 KVA transformer is constructed to a turns ratio of N1/N2 = 10. The impedance of primary winding is 3+j5 ohms and of secondary winding is 0.5+j0.8 ohms. What will be the impedance of transformer when referred to primary?
a) 53j + 85 ohms
b) 53 + 85j ohms
c) 3.5 + 5.8j ohms
d) Can’t be calculated
Answer: b
Explanation: Given turns ratio is 10. Thus, secondary resistance when referred to the primary is equal to k 2 *Z 2 , so net impedance on primary is equal to primary impedance + k^2*Z2 . Substituting all the terms we get net impedance = 53 + 85j ohms.
This set of Transformers Question Bank focuses on “Approximate Circuit of Transformer”.
1. If R is the resistance of secondary winding of an electrical transformer and K is the transformation ratio then the equivalent secondary resistance referred to primary will be _________
a) R/VK
b) R/K 2
c) R 2 /K 2
d) R 2 /K
Answer: b
Explanation: Resistances or more precisely impedances are transformed in the ratio of inverse square of the transformation ratio or turns ratio. Thus, primary to secondary resistance is equal to reciprocal of square of turns ratio.
2. The use of higher flux density in the transformer design ________________
a) reduces weight per kVA
b) reduces iron losses
c) reduces copper losses
d) increases part load efficiency
Answer: a
Explanation: If a material is having higher flux density it will store and transfer maximum amount of flux from primary to secondary, which will be very helpful as less core material will be required and weight per KVA will get reduced.
3. The value of flux involved in the emf equation of a transformer is _______________
a) average value
b) rms value
c) maximum value
d) instantaneous value
Answer: c
Explanation: In the emf equation flux involved is maximum flux. Thus, here we can conclude that as flux increase/decrease emf at the secondary also increases/decreases. Emf varies according to the AC wave input.
4. Which winding of the transformer has less cross-sectional area?
a) Primary winding
b) Secondary winding
c) Low voltage winding
d) High voltage winding
Answer: d
Explanation: Winding having less cross-sectional area may be primary or secondary winding. For high voltage winding cross sectional area is less while for low voltage winding cross sectional are is more, due to inverse proportionality.
5. In constant frequency power transformers, approximate form is used with ___________
a) π equivalent model
b) T equivalent model
c) π and T equivalent both
d) Another model
Answer: b
Explanation: In constant frequency power transformers, approximate forms of the exact T-circuit equivalent of the transformer are commonly used. Thus, we lump all series parameters of the circuit on either side of T circuit.
6. Exciting current in an electrical transformer will not be affected much if primary is excited with E rather than V.
a) True
b) False
Answer: a
Explanation: Since winding resistances and leakage reactances are very small, V1 = E1 even under conditions of load. Therefore, the exciting current drawn by the magnetizing branch would not be affected significantly by shifting it to the input terminals, i.e. it is now excited by V1 instead of E1.
7. In approximate equivalent circuit of the transformer _______________
a) All resistances and inductances are lumped before magnetizing branch
b) All resistances and inductances are lumped after magnetizing branch
c) Resistances and inductances aren’t changed
d) Any of the above will work
Answer: b
Explanation: Since Io is very small compare to like about 5-10% of full load current, voltage drop can be approximated to very large extent. These all resistances and inductances are in series, combined with each other to give approximate equivalent circuit.
8. Final approximate equivalent circuit contains _________
a) Only equivalent X series branch
b) Only equivalent Z series branch
c) Only equivalent R series branch
d) Any of the above
Answer: a
Explanation: We combine all series parameters of equivalent circuit together to get approximate equivalent circuit of the transformer, where we assume that equivalent resistance is zero as it is negligible, and parallel branch is removed.
This set of Transformers Interview Questions and Answers focuses on “Phasor Diagrams of Equivalent and Approximate Model”.
1. Which of the following is not considered in the standard voltage scale for power supply in India?
a) 11kV
b) 33kV
c) 66kV
d) 122kV
Answer: d
Explanation: All the transformers that are used in power system analysis have their secondary voltages equal to 11kV, 33kV, 66kV, etc. As this voltage ranges are commonly used in transmission lines, 122kV is the wrong option.
2. The maximum load that a power transformer can carry is limited because of its ___________
a) temperature rise
b) dielectric strength of oil
c) voltage ratio
d) copper loss
Answer: c
Explanation: One can increase the dielectric strength of oil, by changing the oil. Similarly, temperature rise and copper losses can also be controlled by using various techniques. The only thing which is constant is voltage ratio which can’t be altered.
3. The voltage transformation ratio of a transformer is defined as ratio of _____________
a) primary turns to secondary turns
b) secondary current to primary current
c) secondary induced emf to primary induced emf
d) secondary terminal voltage to primary applied voltage
Answer: c
Explanation: Voltage transformation ratio is equal to secondary induced emf to the primary induced emf. While the secondary and primary voltage induced are different from emfs as these are emf minus the losses in the line.
4. If a transformer is made to run on to a voltage which is more than the rated voltage _______________
a) its power factor will deteriorate
b) its power factor will increase
c) its power factor will remain unaffected
d) its power factor will be zero
Answer: a
Explanation: Every electric device works in appropriate condition with maximum output and minimum losses when it is operated at rated conditions. Thus, if transformer is made to run at higher operating voltage its power factor will deteriorate.
5. Which of the following equation correctly represents the exact phasor diagram of transformer?
a) V 1 =E 1 +I 1 R 1 +jI 1 X 1
b) V 1 =E 1 +I 1 R 1 +jI 2 X 2
c) V 2 =E 2 +I 1 R 1 +jI 1 X 1
d) V 1 =E 1 -I 1 R 1 +jI 1 X 1
Answer: a
Explanation: According to the primary and secondary equivalent circuits of a transformer equation stated in option 1 correctly suits with the kirchoff’s voltage law for primary side of a transformer, similarly equation for secondary side can also be written down.
6. Approximate phasor diagram of a transformer is based on _____________
a) V 1 =E 1 +I 1 R 1 +jI 1 X 1
b) V 2 =E 2 +I 2 R 2 +jI 2 X 2
c) V 1 =V 2 +I R +jI X
d) V 1 =E 1 +I 1 R 1 +jI 1 X 2
Answer: c
Explanation: Approximate equivalent circuit of a transformer neglects the middle branch of transformer equivalent circuit’ which makes transformer circuit even more simpler. Then, all resistances are summed up as R and similarly all reactance as X.
7. Hysteresis loss and eddy current loss is directly proportional to __________
a) f and f 2
b) f 2 and f
c) f and f
d) f 2 and f 2
Answer: a
Explanation: Hysteresis loss is directly proportional to frequency according to Steinmetz’s formula. While eddy current losses are directly proportional to square of flux density, thickness, frequency. Both losses are load independent.
8. What will happen to hysteresis loss if voltage is doubled, load is doubled and frequency is halved?
a) Will be twice
b) Will be halved
c) Will remain same
d) Will be four times
Answer: c
Explanation: Hysteresis loss is directly proportional to voltage and frequency as well. It is load independent. Thus by doing voltage twice and frequency half of the original value we will get same hysteresis loss.
This set of Transformers Multiple Choice Questions & Answers focuses on “Transformer Losses”.
1. In a given transformer for given applied voltage, which of the following losses remain constant irrespective of load changes?
a) Friction and windage losses
b) Copper losses
c) Hysteresis and eddy current losses
d) Cannot be determined
Answer: c
Explanation: Hysteresis and eddy current losses together called as core-loss in a transformer. These losses remain constant for constant voltage and frequency applied to a transformer, these components remain same irrespective of load.
2. On which of the following degree of mechanical vibrations produced by the laminations of a transformer depends?
a) Tightness of clamping
b) Gauge of laminations
c) Size of laminations
d) Tightness of clamping, gauge and size of laminations
Answer: d
Explanation: Mechanical vibrations produced in a transformer are directly effective due to the tightness of the clamping, gauge og laminations, size of laminations as well. There are various methods in order to reduce their effects.
3. Variations in a hysteresis loss in a transformer ____________
a) Bmax
b) Bmax 1.6
c) Bmax 3.83
d) Bmax/2
Answer: b
Explanation: According to Steinmetz’s formula, the heat energy dissipated due to hysteresis is given by Wh=ηβmax 1.6 , and hysteresis loss is thus given by Ph≈ Whf ≈ηfβmax 1.6 . That exponetital term varies fraom 1.4 -1.8 and is equal to 1.6 for iron.
4. Leakage flux in the transformer depends on _____________________
a) Load current
b) Load current and voltage
c) Load current, voltage and frequency
d) Load current, voltage, frequency and power factor
Answer: a
Explanation: Leakage flux is directly proportional to the current, as if is current increased net value of flux increases thus, flux leakage also increases which further contribute to the losses as it is then not able to link with secondary windings .
5. The full-load copper loss of a transformer is 1600 W. At half-load, the copper loss will be _______
a) 6400 W
b) 1600 W
c) 800 W
d) 400 W
Answer: d
Explanation: Copper losses are defined as I2*R losses many times, as they are directly proportional to the square of current flowing through them. Thus, copper losses will reduce if load is reduced that too in square proportion.
6. Silicon steel used in laminations, because it reduces ________________
a) Hysteresis loss
b) Eddy current losses
c) Copper losses
d) Cannot be determined
Answer: a
Explanation: Electrical steels are also known as lamination steel or silicon steel. The main special thing related to the silicon steel is, its magnetic properties such as small hysteresis area and hence, small energy dissipation per cycle, thus low core loss.
7. If the supply frequency to the transformer is increased, the iron loss will ___________
a) Not change
b) Decrease
c) Increase
d) Cannot be determined
Answer: c
Explanation: As frequency increases, the flux density in the core decreases but as the iron loss is directly proportional to the frequency hence effect of increased frequency will be reflected in increase of the iron losses.
8. Which of the following can measure iron loss of a transformer?
a) Low power factor wattmeter
b) Unity power factor wattmeter
c) Frequency meter
d) Any type of wattmeter
Answer: a
Explanation: As the secondary side is open in OC, the entire coil will be purely inductive in nature. So, the power will be lagging due to inductive property of the circuit. So LPF wattmeter is used in open circuit test of transformer.
9. How reduction in core losses and increase in permeability can be obtained simultaneously in a transformer?
a) Core built-up of laminations of cold rolled grain oriented steel
b) Core built-up of laminations of hot rolled sheet
c) Cannot be determined
d) Frequency Meter
Answer: a
Explanation: CRGO is supplied by the producing mills in coil form and has to be cut into laminations, which are then used in transformer core, which is an integral part of any transformer. Grain-oriented steel is used in large power, distribution transformers and in certain audio output transformers also.
10. Losses which occur in rotating electric machines and do not occur in transformer are ______
a) Friction and windage losses
b) Magnetic losses
c) Hysteresis and eddy current losses
d) Copper losses
Answer: a
Explanation: Windage and friction losses occur in rotating parts of a machine generally in rotor of the machine, thus they will never occur in transformer, as transformer does not contain any rotating part at its secondary unlike induction motor.
11. In a given transformer for a given applied voltage, which losses remain constant irrespective of load changes?
a) Hysteresis and eddy current losses
b) Friction and windage losses
c) Copper losses
d) Cannot be determined
Answer: a
Explanation: Hysteresis and eddy current losses are voltage and frequency dependent losses that too from primary side thus, load change will not make any effect on these losses and they will remain constant as long as voltage and frequency is constant.
12. Which of the following loss in a transformer is zero even at full load?
a) Core loss
b) Friction loss
c) Eddy current loss
d) Hysteresis loss
Answer: b
Explanation: Friction losses are involved with rotating parts of a machine. Since in a transformer all parts are stationary, friction losses will always be equal to zero, irrespective of the loading condition.
13. A shell-type transformer has __________
a) High eddy current losses
b) Reduced magnetic leakage
c) Negligible hysteresis losses
d) Cannot be determined
Answer: b
Explanation: Since windings are brought closer in shell type compare to core type transformer, leakage of flux is very less in shell type transformer. Most of the flux gets linked with both of the coils though there is some leakage which can’t be avoided.
This set of Transformers Problems focuses on “OC Test on Transformer”.
1. During open circuit test of a transformer _____________
a) primary is supplied rated kVA
b) primary is supplied full-load current
c) primary is supplied current at reduced voltage
d) primary is supplied rated voltage
Answer: d
Explanation: Open circuit test is normally conducted on rated voltage because any machine is constructed to give maximum efficiency near rated value. Hence, it is operated at rated voltage, and we have to perform the test on machine is to be used.
2. Open circuit test on transformers is conducted so as to get ______________
a) Hysteresis losses
b) Copper losses
c) Core losses
d) Eddy current losses
Answer: c
Explanation: Open circuit test gives the core losses also called as iron losses and shunt parameters of the equivalent circuit of transformer. Open circuit test and short circuit test both provide all the parameters of equivalent circuit.
3. Why OC test is performed on LV side?
a) Simple construction
b) Less voltage is required and parameters can be transformed to HV side
c) It’ll not give losses ig conducted on HV side
d) HV side does not have connections for voltage
Answer: b
Explanation: Open circuit test can be performed on any side but for our convenience and supply voltage available we generally conduct the test on LV side, to get corresponding parameters on HV side we can use transformation ratio.
4. In OC test all the power supplied is utilised for ______
a) Core losses
b) Iron losses
c) Windage losses
d) Cannot be determined
Answer: b
Explanation: In open circuit test all the power supplied is used to overcome iron losses and hence, by taking the reading of input power one can easily do the calculations to find shunt parameters of equivalent circuit of transformer.
5. How shunt branch component Gi is calculated?
a) Po/v1 2
b) V1/Io
c) Io/ V 2
d) Any of the above
Answer:
Explanation: Shunt branch resistance inverse is denoted by Gi. This Gi can be calculated by the power drop taking place in the resistance divided by square of the voltage applied across the resistor. Current by voltage will give net admittance.
6. Which of the following statements is/are correct statements?
a) EMF per turn in LV winding is more than EMF per turn in LV winding
b) EMF per turn in LV winding is less than EMF per turn in LV winding
c) EMF per turn in HV winding is equal to EMF per turn in LV winding
d) Can’t comment
Answer: c
Explanation: In a transformer, primary volt-ampere is equal to secondary volt-ampere and primary ampere turns are also equal to secondary ampere turns So, EMF per turn in both the winding are equal. Total induced emf on both sides depends on the number of turns, flux and frequency.
7. If the applied voltage of a transformer is increased by 50% and the frequency is reduced by 50%, the maximum flux density will _____________
a) Changes to three times the original value
b) Changes to 1.5 times the original value
c) Changes to 0.5 times the original value
d) Remains the same as the original value
Answer: a
Explanation: Magnetic flux density α β/A. Magnetic flux φ α V/f. φ2/ φ1 = V2/V1 * f1/f2. Since voltage is increased by 50%, V2 thus becomes 1.5 times V1 and frequency becomes 0.5 times the original frequency. Thus, maximum flux density changes to 3 times the original value.
8. The total core loss can be termed as ____________
a) Eddy current loss
b) Hysteresis loss
c) Copper loss
d) Magnetic loss
Answer: d
Explanation: The total core loss is due to iron core or any core material used. As iron loss is proportional to magnetic flux density or flux, these are also called as magnetic loss. The total core loss or magnetic loss in any given transformer totally consists of eddy current loss and hysteresis loss.
9. 2 KVA, 230 V, 50 Hz single phase transformer has an eddy current loss of 40 watts. The eddy current loss when the transformer is excited by a dc source of same voltage will be ___________
a) Equal to 40W
b) Less than 40W
c) More than 40W
d) Zero watts
Answer: d
Explanation: Eddy current loss is directly proportional to the frequency^2. So, for DC current frequency is equal to 0 Hz. Thus, eddy current losses being directly proportional to square of frequency they’ll be equal to 0.
10. Which of the following is the correct formula for Bm?
a) Bm=
b) Bm=
c) Bm=
d) Bm=
Answer: a
Explanation: We get the value of Y0 from the no-load current and voltage reading as, Io/V1. Similalry we get the value of Gi from output power and voltage reading as, Po/V1. It then follows that, Bm= .
11. How shunt branch component Y0 is calculated?
a) I0/V1
b) V1/I0
c) P0/V1 2
d) Cannot be determined
Answer: a
Explanation: Shunt branch admittance is defined as inverse of shunt branch impedance. As we know, impedance can be calculated by the simple ohm’s law; admittance is equal to the inverse of the impedance.
This set of Transformers Multiple Choice Questions & Answers focuses on “SC Test on Transformer”.
1. While conducting short-circuit test on a transformer which side is short circuited?
a) High voltage side
b) Low voltage side
c) Primary side
d) Secondary side
Answer: b
Explanation: It’s a common practice to conduct SC test from HV side, while keeping LV side short circuited. Thus, short circuited current is made to flow from shorted low voltage terminals i.e. LV side.
2. During short circuit test why iron losses are negligible?
a) The current on secondary side is negligible
b) The voltage on secondary side does not vary
c) The voltage applied on primary side is low
d) Full-load current is not supplied to the transformer
Answer: c
Explanation: Very small amount of voltage is given to the transformer primary thus the magnetic losses which are dependent on magnetic flux density will get minimum value, hence iron losses are negligible.
3. Short circuit test on transformers is conducted to determine ______
a) Core losses
b) Copper losses
c) Hysteresis losses
d) Eddy current losses
Answer: b
Explanation: Short circuit test is used to determine the copper losses taking place in the transformer under operation, while open circuit test gives us the value of core losses taking place in the transformer.
4. When a short circuit test on a transformer is performed at 25 V, 50 Hz, the drawn current is I1. If the test is performed by 25 V and 25 Hz and power drawn current is I2, then
a) I 1 > I 2
b) I 1 < I 2
c) I 1 = I 2
d) Can’t be defined
Answer: b
Explanation: Current by ohm’s law is equal to voltage divided by impedance. So, I=V/Z. Here Z is inductive load, thus Z= 2πfL. So as the frequency decreases the impedance also decreases and ultimately it reduces the denominator term causing increase in current.
5. Why SC test is not conducted on LV side?
a) Difficult to arrange low voltage supply
b) Difficult to arrange high current supply
c) Difficult to arrange low voltage and high current supply to the LV
d) SC test on LV does not give correct results
Answer: c
Explanation: If rated voltages and power is considered we need only 5% of rated voltage to be applied at on HV side, while by calculations current requirement is also less. For the same test on LV side though voltage required is less compare to HV side, current required is very high.
6. SC test gives ______________
a) Series parameters of equivalent circuit
b) Parallel parameters of equivalent circuit
c) Both parameters of equivalent circuit
d) Neither series nor parallel parameter of equivalent circuit
Answer: a
Explanation: Short circuit test gives the copper losses; these losses are taken into consideration by series parameters of the equivalent circuit. While, Open circuit test gives us iron losses; which are shown by parallel components of equivalent circuit.
7. For 200 kVA, 440/6600-V transformer, short circuit test on the LV side would require ______
a) 22V
b) 330V
c) 44V
d) Can’t be calculated
Answer: a
Explanation: For a given transformer SC test is conducted on LV side, thus we’ll use 5% of rated voltage on the low voltage side. Hence, 5% of 440V calculation gives the value of 440*5/100= 22V on LV side.
8. For a transformer given of 100 kVA, 220/6000-V transformer, short circuit test is performed. What current rating is needed?
a) 30A
b) 445A
c) 60A
d) Can’t be calculated
Answer: b
Explanation: For a given transformer here, test is performed on low voltage side, thus we need the value of current on the low voltage side, by dividing the reactive power by the rated voltage value, i.e. 200*1000/100= 445A.
9. What will be the value of voltage and current for a given transformer of 10 MVA, 220/4400-V which we are about to perform the Short circuit test?
a) 220 V, 30 A
b) 220 V, 2.27A
c) 440 V, 30 A
d) 440 V, 2.27 A
Answer: b
Explanation: Since short circuit test is always done on the HV side unless mentioned specifically, thus values are calculated with HV side parameters. Voltage required on HV side = 4400*5/100 = 220 V and 10*1000/4400= 2.27A.
10. We only get copper losses from the short circuit test.
a) True
b) False
Answer: a
Explanation: Since the transformer is excited at very low voltage, the iron-loss is negligible , the power input corresponds only to the copper-loss, i.e. PSC = PC .
11. With the help of short circuit calculations we get value of ____________
a) Individual resistance and inductance of both sides
b) Resistance and inductance of primary side
c) Resistance and inductance of primary side
d) Combined resistance and inductance of both sides
Answer: d
Explanation: Short calculations include the ratio of short circuited voltage to the short-circuited current which gives Z value, similarly the R value is calculated by dividing the Short-circuited power with short circuited current square. Then, X is calculated for whole circuit.
12. Short circuit test is performed on a transformer with a certain impressed voltage at rated frequency. What will happen if the short circuit test is now performed with the same magnitude of impressed voltage, but at frequency higher than the rated frequency?
a) The magnitude of current will increase, but power factor will decrease
b) The magnitude of current will decrease, but power factor will increase
c) The magnitude of current will increase, power factor will increase
d) The magnitude of current will decrease, power factor will decrease
Answer: d
Explanation: Since frequency has been increased, the leakage reactance will increase. Which will increase the impedance. Thus, current will be less due to inverse proportionality and power factor will be poorer.
This set of Transformers Multiple Choice Questions & Answers focuses on “Sumpner’s Test”.
1. Sumpner’s test is conducted on transformers to study effect of ____________
a) Temperature
b) Stray losses
c) All-day efficiency
d) Cannot be determined
Answer: a
Explanation: Sumpner’s test is the test which is used to determine the steady temperature rise if the transformer was fully loaded continuously; this is so because under each of these tests the power loss to which the transformer is subjected is either the core-loss or copper-loss but not both.
2. Which of the following tests are enough to find all the parameters related to a transformer?
a) OC test
b) OC, SC test
c) OC, SC, Sumpner’s test
d) Sumpner’s test
Answer: c
Explanation: While OC and SC tests on a transformer yield its equivalent circuit parameters, these cannot be used for the ‘heat run’ test wherein the purpose is to determine the steady temperature rise if the transformer was fully loaded continuously.
3. Sumpner’s test is performed on _________
a) Single transformer at a time
b) Only two transformers at a time
c) Minimum 2 transformers at a time
d) Many transformers at a time
Answer: b
Explanation: Sumpner’s test is used to determine the effect of transformer on loaded condition. Thus, two transformers are tested simultaneously, where one simply acts as a load to another transformer.
4. In Sumpner’s test __________
a) Primaries can be connected in either way
b) Primaries are connected in parallel with each other
c) Primaries of both transformers are connected in series with each other
d) No need to connect primaries
Answer: b
Explanation: Sumpner’s test is also called as back-to-back test, where two transformers are used where one transformer acts as a load to another transformer. Primaries of both of the transformers used in a test, are connected in parallel with each other.
5. Which test is sufficient for efficiency of two identical transformers under load conditions?
a) Short-circuit test
b) Back-to-back test
c) Open circuit test
d) Any of the above
Answer: b
Explanation: Open circuit test and short circuit test collectively gives the value of all parameters of an equivalent circuit of a transformer. While Sumpner’s back-to-back test gives the heat run effect of machine by considering rise in temperature.
6. In Sumpner’s test _____________
a) Two secondaries are connected in phase opposition
b) Two secondaries are connected in phase addition
c) Can be connected in either way
d) Never connected with each other
Answer: a
Explanation: In conducting Sumpner’s test two primaries are connected in parallel to the rated voltage supply and secondaries are connected in phase opposition. For the secondaries to be in phase opposition rated secondary voltage across the terminals to be zero.
7. When secondaries are connected in phase opposition, power drawn by the circuit is equal to ___________
a) 2*P i
b) Pi 2
c) Pi
d) 2*Pc
Answer: a
Explanation: If V 2 source is assumed shorted, the two transformers appear in open circuit to source V 1 as their secondaries are in phase opposition and therefore no current can flow in them. The current drawn from source V 1 is thus 2I0 and power is 2P0 = 2Pi, twice the core-loss of each transformer.
8.When the AC supply at the primary side of a transformer are shorted, power drawn by the circuit is equal to ___________
a) 2*PC
b) 2*Pi
c) 2*PC + 2*Pi
d) Can’t be determined
Answer: a
Explanation: When the ac supply terminals are shorted, the transformers are series-connected across V2 supply and are short-circuited on the side of primaries. Therefore, the impedance seen at V 2 is 2Z and when V 2 is adjusted to circulate full-load current (If l ), the power fed in is 2PC .
9.Total power required for Sumpner’s test is given by ________
a) PC + Pi
b) PC + 2Pi
c) 2PC + Pi
d) 2
Answer: d
Explanation: In the Sumpner’s test while the transformers are not supplying any load, full iron-loss occurs in their cores and full copper-loss occurs in their windings; net power input to the transformers being . The heat run test could, therefore, be conducted on the two transformers, while only losses are supplied.
This set of Transformers Assessment Questions and Answers focuses on “Efficiency”.
1. When will be the efficiency of a transformer maximum?
a) Copper losses = hysteresis losses
b) Hysteresis losses = eddy current losses
c) Eddy current losses = copper losses
d) Copper losses = iron losses
Answer: d
Explanation: When the variable copper losses of a transformer becomes equal to the fixed iron losses of a transformer then we will get maximum efficiency. From these losses we’ll get the value of current required.
2. Efficiency of a power transformer is near to the ___________
a) 100 per cent
b) 98 per cent
c) 50 per cent
d) 25 per cent
Answer: b
Explanation: The efficiency of the transformer obtained from various experiments conducted on various loads showed the efficiency greater than 90% always. Transformer thus, can be said highly efficient device.
3. On which factors transformer routine efficiency depends upon?
a) Supply frequency
b) Load current
c) Power factor of load
d) Load current and power factor of load
Answer: d
Explanation: Efficiency of the transformer can be calculated by the output power divided by input power. Both of these powers include power factor in their calculations while load current and load voltage is also required in calculations.
4. Normal transformers are designed to have maximum efficiency at ___________
a) Nearly full load
b) 70% full load
c) 50% full load
d) No load
Answer: a
Explanation: Every device is manufactured to get maximum efficiency at the rated loads, i.e. full load. Thus, transformer will give the maximum efficiency at nearly full load. Internal losses are so adjusted to get maximum efficiency.
5. At which load condition maximum efficiency of a distribution transformer will be achieved?
a) At no load
b) At 60% full load
c) At 80% full load
d) At full load
Answer: b
Explanation: The main difference between power transformer and distribution transformer is distribution transformer is designed for maximum efficiency at 60% to 70% load as these transformers normally doesn’t operate at full load all the time.
6. Power transformers other than distribution transformers are generally designed to have maximum efficiency around ______
a) No-load
b) Half-load
c) Near full-load
d) 10% overload
Answer: c
Explanation: Similar to normal transformers power transformers are also designed to get maximum efficiency at load which is near to the full load of a transformer specified. Only in the case distribution transformer maximum efficiency is achieved at 60% of full load.
7. For a transformer given, operating at constant load current, maximum efficiency will occur at ______
a) 0.8 leading power factor
b) 0.8 lagging power factor
c) Zero power factor
d) Unity power factor
Answer: d
Explanation: Maximum efficiency for a transformer will be achieved at full load. While in the case of power factor also every device is set to get maximum efficiency at unity power factor. Thus, one will have maximum efficiency if load is nearly equal to full load and at unity power factor.
8. Why efficiency of a transformer, under heavy loads, is comparatively low?
a) Copper loss becomes high in proportion to the output
b) Iron loss is increased considerably
c) Voltage drop both in primary and secondary becomes large
d) Secondary output is much less as compared to primary input
Answer: a
Explanation: At heavy loads current drawn by the transformer circuit increases, as we know, variable copper losses are proportional to the square of the current. Thus, we’ll get higher copper loss in proportion to the output.
9. The efficiencies of transformers compared to electric motors of the same power are ___________
a) About the same
b) Much smaller
c) Much higher
d) Can’t comment
Answer: c
Explanation: Transformer is a highly efficient device compare to all other electrical instruments. In motor we need to add windage and friction losses along with the copper losses and iron losses thus, we’ll get lee efficiency for motor compare to transformer.
10. A transformer having maximum efficiency at 75% full load will have ratio of iron loss and full load copper loss equal to ___________
a) 4/3
b) 3/4
c) 9/16
d) 16/9
Answer: c
Explanation: Condition for maximum efficiency is, Copper loss= Iron loss, i.e. Pc= I2 R = Pi. transformer can be operated at any load but maximum efficiency occurs at a particular load condition only. Let x be that load factor corresponds to maximum efficiency. Given that, maximum efficiency will occur at 3/4 load. The load factor= 2.
11. What is the correct formula of efficiency of a device?
a) Input /output
b) Output/losses
c) 1- )
d) Cannot be determined
Answer: c
Explanation: Efficiency of any device is equal to the ratio of output power to the input power. Here, one can write input power is equal to the addition of output power with losses. Thus, expressing all these terms mathematically will give the answer.
12. A 500 kVA transformer is having efficiency of 95% at full load and also at 60% of full load; both at unity power factor. Then P i is ___________
a) 16.45 kW
b) 9.87 kW
c) 14.57 kW
d) Can’t be calculated
Answer: b
Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + P i + k 2 *PC)]. Thus, η= 500*1/ (500 + P i +P C ) = 0.95. also from the second condition given η= 500*0.6/ = 0.95. Thus, solving simultaneously we get 9.87 kW.
13. A 500 kVA transformer is having efficiency of 95% at full load and also at 60% of full load; both at unity power factor. Then Pc is ___________
a) 16.45 kW
b) 9.87 kW
c) 14.57 kW
d) Can’t be calculated
Answer: a
Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + P i + k 2 *PC)]. Thus, η= 500*1/ (500 + P i +P C ) = 0.95. also from the second condition given η= 500*0.6/ (500*0.6 + P i +0.6 2 *PC) = 0.95. Thus, solving simultaneously we get an answer 16.45 kW.
14. For a power transformer operating at full load it draws voltage and current equal to 200 V and 100 A respectively at 0.8 pf. Iron and copper losses are equal to 120 kW and 300kW. What is the efficiency?
a) 86.44%
b) 96.44%
c) 97.44%
d) 99.12%
Answer: c
Explanation: Power output= VI cosθ= 200*100*0.8 = 16000 W . While total losses are equal to iron loss+ k 2 *copper losses =120+ 300= 420 W. Efficiency is equal to 1- 420/= 97.44%.
This set of Transformers Multiple Choice Questions & Answers focuses on “Voltage Regulation-1”.
1. The highest voltage for transmitting electrical power in India is _______
a) 33 kV
b) 66 kV
c) 132 kV
d) 00 kV
Answer: d
Explanation: Transmission voltage in power transfer in India is 750KV AC and these lines are erected by Power Grid Corporation for interstate connections throughout India. However, work on 800KV is in the progress. DC transmission voltage in India is 600KV.
2. A transformer can have zero voltage regulation at _______
a) Leading power factor
b) Lagging power factor
c) Unity power factor
d) Zero power factor
Answer: a
Explanation: At leading power factor the voltage regulation is given by I*. Thus, at a particular condition of angle φ we may get zero voltage regulation. While in lagging power factor case we have + sign in the above formula.
3. What will happen to a given transformer if it made to run at its rated voltage but reduced frequency?
a) Flux density remains unaffected
b) Iron losses are reduced
c) Core flux density is reduced
d) Core flux density is increased
Answer: d
Explanation: E=4.44fNAB is the emf equation for a transformer, now as E is kept constant we can say frequency is inversely proportional to the B value. Thus, as frequency increases we will get less core flux density and vice-versa.
4. In an actual transformer the iron loss remains practically constant from no load to full load because ___________
a) Value of transformation ratio remains constant
b) Permeability of transformer core remains constant
c) Core flux remains practically constant
d) Primary voltage remains constant
Answer: c
Explanation: The reason behind core-iron loss being constant is that hysteresis loss and eddy current loss both are dependent on the magnetic properties of the material which is used in the construction and design of the core of the transformer.
5. Negative voltage regulation indicates ___________
a) Capacitive loading only
b) Inductive loading only
c) Inductive or resistive loading
d) Cannot be determined
Answer: a
Explanation: The sign -ve arises in the voltage regulation calculations when, the load connected to the transformer is leading in the nature. The only condition when we’ll get negative voltage regulation when second term is higher than first term.
6. When will a transformer have regulation closer to zero?
a) On full-load
b) On overload
c) On leading power factor
d) On zero power factor
Answer: c
Explanation: Since voltage regulation of a transformer in the leading loading condition is not additive in nature, at particular power factor in leading we can get zero voltage regulation. While, in lagging condition we’ll get ultimately non-zero VR.
7. A good voltage regulation of a transformer indicates ______________
a) output voltage fluctuation from no load to full load is least
b) output voltage fluctuation with power factor is least
c) difference between primary and secondary voltage is least
d) difference between primary and secondary voltage is maximum
Answer: a
Explanation: Voltage regulation is defined as rise in the voltage when the transformer is thrown off from full load condition to no-load condition. Thus, least voltage regulation means output fluctuations depending on the load are very less.
8. Which of the following acts as a protection against high voltage surges due to lightning and switching?
a) Horn gaps
b) Thermal overload relays
c) Breather
d) Conservator
Answer: a
Explanation: Arcing horns in a transformer form a spark gap across the insulator with a lower breakdown voltage than the air path along the insulator surface, so an overvoltage it will cause the air to break down and the arc to form between the arcing horns, diverting it away from the surface of the insulator.
9. Minimum voltage regulation occurs when the power factor of the load is ______________
a) Unity
b) Lagging
c) Leading
d) Zero
Answer: c
Explanation: When the leading load is connected to the transformer difference of Rcosφ and Xsinφ is multiplied with the current, thus we may get -ve, zero voltage regulations at this condition. That is minimum voltage regulation.
This set of Transformers Questions and Answers for Freshers focuses on “Voltage Regulation – 2”.
1. Voltage regulation of transformer is given by _____________
a) E 2 -V 2 /V 2
b) E 2 -V 2 /E 2
c) V 2 -E 2 /E 2
d) V 2 -E 2 /V 2
Answer: b
Explanation: Voltage regulation is defined as change in the voltage or rise in voltage when transformer is load is thrown off. Thus, it is the difference of the no load voltage with the full load voltage divide by full load voltage to get % increase.
2. On which load power factor zero voltage regulation will be achieved?
a) 0
b) 1
c) Leading
d) Lagging
Answer: c
Explanation: At leading power factor the voltage regulation can be negative or zero. This can be found from this equation % regulation = εxcosθ – εrsinθ. Bu substituting the appropriate value of angle one can check this mathematically.
3. A transformer has resistance and reactance in per unit as 0.01 and 0.04 pu respectively. What will be its voltage regulation for 0.8 power factor lagging and leading?
a) 3.2% and 1.6%
b) 3.2% and -1.6%
c) 1.6% and -3.2%
d) Can’t be defined
Answer: b
Explanation: Voltage regulation for lagging power factor = × 100, Voltage regulation for 0.8 lagging power factor = × 100 = 3.2%. Voltage regulation for leading power factor = × 100, Voltage regulation for 0.8 leading power factor = × 100 = -1.6%.
4. At which power factor one will get maximum voltage regulation?
a) 0
b) 1
c) Leading
d) Lagging
Answer: d
Explanation: At lagging power factor the voltage regulation is given by I*. Thus, at a particular condition of angle φ we will get maximum voltage regulation. While in leading power factor case we have – sign in the above formula.
5. Which is the correct phasor equation indicating the transformer voltages lagging?
a) V 1 = V 2 + I*
b) V 2 = V 1 + I*
c) V 1 = V 2 + I*
d) V 1 = V 2 + I*
Answer: a
Explanation: According to the phasor diagram drawn for lagging current, we will have positive sign in the voltage regulation formula thus, V 1 = V 2 + I* gives the correct relation, while V 1 indicates the primary voltage.
6. Which is the correct phasor equation indicating the transformer voltages leading?
a) V 1 = V 2 + I*
b) V 2 = V 1 + I*
c) V 1 = V 2 + I*
d) V 1 = V 2 + I*
Answer: a
Explanation: According to the phasor diagram drawn for leading current, we will have negaitive sign in the voltage regulation formula thus, V 1 = V 2 + I* gives the correct relation, while V1 indicates the primary voltage.
7. What is the correct formula to get power factor angle in leading condition?
a) tan φ= X/R
b) tan φ= R/X
c) cos φ = R/√
d) cos φ= R/X
Answer: b
Explanation: For leading condition derivative of voltage regulation with respect to φ is obtained and solved for the power factor angle calculations we’ll get tan φ = R/X for leading condition, for lagging condition we’ll get tan φ= X/R.
8. What is the correct formula to get power factor angle in lagging condition?
a) sin φ= X/R
b) tan φ= R/X
c) cos φ = R/√
d) cos φ= R/X
Answer: c
Explanation: For lagging condition derivative of voltage regulation with respect to φ is obtained and solved for the power factor angle calculations we’ll get tan φ = R/X for leading condition, for lagging condition we’ll get tan φ= X/R. In terms of cosine function, we’ll get cos φ = R/√.
9. Zero voltage regulation of a transformer is achieved at 1 pf leading.
a) True
b) False
Answer: b
Explanation: Though zero voltage regulation occurs at leading power factor condition, it is not occurring at unity power factor leading. As at unity power factor leading, cos term will be equal to 1. Hence, we’ll get some non-zero VR at unity power factor.
This set of Transformers online quiz focuses on “Excitation Phenomenon”.
1. No-load current in the transformer is _______
a) Straight DC
b) Steps
c) Sinusoidal
d) Sinusoidal distorted
Answer: d
Explanation: The no-load current in a transformer is non-sinusoidal. The basic cause for this phenomenon, which lies in hysteresis and saturation non-linearities of the core material, will now be investigated; this can only be accomplished graphically.
2. The main reason for generation of harmonics in a transformer is ____________
a) Fluctuating load
b) Poor insulation
c) Mechanical vibrations
d) Saturation of core
Answer: d
Explanation: The iron core which is used in transformer is subjected to saturation effect. Thus, according to the hysteresis loop, the generation of harmonics at particular saturation level can be identified.
3. The magnetising current of a transformer is usually small because it has _______
a) Small air gap
b) Large leakage flux
c) Laminated silicon steel core
d) Fewer rotating parts
Answer: a
Explanation: Air Gap increases the reluctance in the magnetic path of magnetic lines of force. When there is an air gap in the magnetic circuit the reluctance is high owing to the permeability of air which is much lower as compared to ferromagnetic materials. The mmf required to overcome that is more.
4. Harmonics in transformer result in ___________
a) Increased core losses
b) Increased I 2 R losses
c) Magnetic interference with communication circuits
d) Increased core lossed, Increased I2R losses and magnetic interference
Answer: d
Explanation: harmonics are produced in an particular circuit as a result of magnetization of core material. Thus, magnetic losses are increased that is, iron losses are more so to maintain constant more current will be drawn giving rise to I 2 losses.
5. For given applied voltage, with the increase in frequency of the applied voltage ___________
a) Eddy current loss will decrease
b) Eddy current loss will increase
c) Eddy current loss will remain unchanged
d) Cannot be determined
Answer: b
Explanation: Eddy current losses are directly proportional to the square of the frequency of input applied rated voltage. This loss is independent of voltage applied to the transformer. Thus, eddy current loss will increase with frequency.
6. Which of the following is the most likely source of harmonics in a transformer?
a) Poor insulation
b) Overload
c) Loose connections
d) Core saturation
Answer: d
Explanation: Core material used in a transformer is iron, which undergoes saturation when it is excited by some external supply, which causes some unlikely effects like generation of harmonics which leads in losses for a particular transformer.
7. Third harmonic currents can be as large as ______________
a) 100% of fundamental
b) 10% of fundamental
c) 70% of fundamental
d) 40% of fundamental
Answer: d
Explanation: While odd symmetry is preserved and the current and flux maximas occur simultaneously, the current zeros are advanced in time with respect to the flux wave shape. As a consequence, the current has fundamental and odd harmonics, the strongest being the third harmonic which can be as large as 40% of the fundamental.
8. Flux transient goes till the maximum value due to ___________
a) Doubling effect
b) Saturation effect
c) Hysteresis effect
d) Modulation effect
Answer: a
Explanation: A transient flux component ft = fm originates so that the resultant flux is which has zero value at the instant of switching. The transient component ft will decay according to the circuit time constant which is generally low in a transformer. If the circuit dissipation is assumed negligible, the flux transient will go through a maximum value of 2fm.
9. The transformer switching transient is referred as _____________
a) Inrush current
b) Outrush current
c) Middlerush current
d) Harmonic current
Answer: a
Explanation: In subsequent half-periods ft gradually decays till it vanishes and the core flux acquires the steady-state value, Because of the low time constant of the transformer circuit, distortion effects of the transient may last several seconds. The transformer switching transient is referred to as the inrush current.
10. Which of the following is not the method of reducing harmonics?
a) Adding filters
b) Adding capacitors
c) Adding inductors
d) Adding resistance
Answer: d
Explanation: Reduce the harmonic currents produced by the load. Add filters to siphon the harmonic currents off the system, or block the currents from entering the system, or supply the harmonic currents locally. Modify the frequency response of the system by filters, inductors, or capacitor.
11. It is possible to reduce harmonic currents completely.
a) True
b) False
Answer: b
Explanation: It is not possible to prevent harmonic currents completely. But they can be prevented from flowing through the main system by providing a separate low-impedance path for them. This is done by the use of rated series tuned circuits consisting of a reactor and capacitor, which have equal impedance at a specific harmonic frequency.
This set of Transformers Multiple Choice Questions & Answers focuses on “Autotransformer”.
1. Which of the following is the main advantage of an auto-transformer over a two-winding transformer?
a) Hysteresis losses are reduced
b) Saving in winding material
c) Copper losses are negligible
d) Eddy losses are totally eliminated
Answer: b
Explanation: Auto transformer is a special type of transformer which has primary and secondary winding both located on same winding. Thus, winding material required for a transformer is very less in the case of autotransformer.
2. Auto-transformer makes effective saving on copper and copper losses, when its transformation ratio is
a) Approximately equal to one
b) Less than one
c) Great than one
d) Cannot be found
Answer: a
Explanation: Copper In auto transformer /copper in two-winding transformer = 1- T2/T1. This means that an auto transformer requires the use of lesser quantity of copper given by the ratio of turns. Hence, if the transformation ratio is approximately equal to one, then the copper saving is good and the copper loss is less.
3. Total windings present in a autotransformer are __________
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: Autotransformer is the special transformer for which the single winding acts as a primary and secondary both. Thus, by taking the appropriate winding into consideration a variable secondary voltage is obtained.
4. Autotransformers are particularly economical when _________
a) Voltage ratio is less than 2
b) Voltage ratio is very high
c) Voltage ratio is higher than 2 in smaller range
d) Can be used anywhere
Answer: a
Explanation: Autotransformer is economical where the voltage ratio is less than 2 in which case electrical isolation of the two windings is not essential. The major applications are induction motor starters, interconnection of HV systems at voltage levels with ratio less than 2, and in obtaining variable voltage power supplies .
5. Which of the following is not true regarding the autotransformer compare to two-winding transformer?
a) Lower reactance
b) Lower losses
c) Higher exciting current
d) Better voltage regulation
Answer: c
Explanation: Autotransformer is the advance version of normal transformer. It is having better voltage regulation, higher efficiency due to lower losses, lower reactance and thus it also requires very small exciting current.
6. Two-winding transformer of a given VA rating if connected as an autotransformer can handle ___________
a) Higher VA
b) Lower VA
c) Same VA
d) Cannot be found
Answer: a
Explanation: A two-winding transformer of a given VA rating when connected as an autotransformer can handle higher VA. This is because in the autotransformer connection part of the VA is transferred conductively.
7. When auto-transformation ratio becomes equal to 1, which of the following statement is true?
a) VA rating of the autotransformer becomes far greater than VA rating of two winding transformer
b) VA rating of the autotransformer becomes far lower than VA rating of two winding transformer
c) VA rating of the autotransformer becomes equal to VA rating of two winding transformer
d) Can’t comment
Answer: a
Explanation: VA rating of autotransformer is = [1/1-a] * VA of two-winding transformer, thus, when a i.e. transformation ratio of autotransformer becomes closer to 1 one gets very high value of VA rating of an autotransformer.
8. An autotransformer compared to its two-winding counterpart has a higher operating efficiency.
a) True
b) False
Answer: a
Explanation: The losses are less in autotransformer compare to two-winding transformer. Thus, for the given same input to autotransformer as that of two-winding transformer more output will be available to secondary side.
9. Ratio of winding material needed for autotransformer to thr two winding transformer is ______
a) 1- V 2 /V 1
b) 1- N 2 /N 1
c) 1- V 2 /V 1 and 1- N 2 /N 1
d) 1- V 1 /V 2
Answer: c
Explanation: G auto /G TW = 1- V 2 /V 1 , as voltage is directly proportional to the number of turns we can say the value is also equal to the 1- N 2 /N 1 . Thus, one can write G TW -G auto = 1/a’ * GTW = saving of the conductor material using autotransformer.
10. For an auto-transformation ratio tending to the unity value, saving of the conductor material will be ___________
a) Tend towards 90% or more
b) Tend towards 0%
c) Can’t say
d) Wil remain fix
Answer: a
Explanation: G TW -G auto = 1/a’ * GTW = saving of the conductor material using autotransformer. So, if a’=10, saving is only 10% but for a’=1.1, saving is as high as 90%. Hence it is more economical when the turn-ratio is closer to unity.
11. What are the modes in which power can be transferred in an autotransformer?
a) Conduction
b) Induction
c) Conduction and Induction
d) Cannot be said
Answer: c
Explanation: In two winding transformer there is no electrical connection between primary and secondary. So, the power is transferred through induction. But in auto-transformer there is a common electrical path between primary and secondary. So, power is transferred through both conduction and induction processes.
12. A 100/10, 50 VA double winding transformer is converted to 100/110 V auto transformer. tWhat will be the rating of auto transformer?
a) 500 VA
b) 550 VA
c) 100 VA
d) 110 VA
Answer: b
Explanation: Secondary current of the two-winding transformer at rated voltage supply = 50/10= 5 A
Thus, autotransformer will also carry the same rated current in secondary giving the power output as 5*110= 550 VA.
13. For a 20kVA transformer with turn ratio 0.4 what amount of total power is transferred inductively?
a) 8kVA
b) 12kVA
c) 10kVA
d) 50kVA
Answer: b
Explanation: For an auto transformer power is transferred partially inductively and partially conductively. Thus, out of total power, power transferred inductively is given by *total power= 0.6*20= 12kVA.
14. For a 20kVA transformer with turn ratio 0.4 what amount of total power is transferred conductively?
a) 8kVA
b) 12kVA
c) 10kVA
d) 50kVA
Answer: a
Explanation: For an auto transformer power is transferred partially inductively and partially conductively. Thus, out of total power, power transferred conductively is given by *total power= 0.4*20= 8kVA.
This set of Transformers Multiple Choice Questions & Answers focuses on “Variable Frequency Transformer”.
1. Output transformers are also called for __________
a) Maximum efficiency
b) Maximum power
c) Maximum output
d) Maximum efficiency, power and output
Answer: b
Explanation: In electronic circuit applications maximum power is given more preference than maximum efficiency unlike in power system operations. So, all the transformers working for maximum power are called as output transformers.
2. The output transformers are also named as _________
a) Audio transformers
b) Power transformers
c) Industrial transformers
d) Electric transformers
Answer: a
Explanation: in electronic circuit applications the performance criterion is the maximum power unlike the maximum efficiency in power system applications. Such transformers are known as output transformers while in audio applications these are known as audio transformers.
3. Which of the following is the major requirement for the transformers used for electronic purposes?
a) Perfect DC isolation
b) Maximum efficiency
c) Constant amplitude voltage gain
d) Perfect DC isolation , maximum efficiency and constant voltage gain
Answer: c
Explanation: An important requirement of these transformers is that the amplitude voltage gain should remain almost constant over the range of frequencies of the signal.
4. Why frequency scale is logarithmic?
a) Frequency range is very small for these transformers
b) Frequency range is very large for these transformers
c) Frequency used is negligible
d) Can’t say
Answer: b
Explanation: When we investigate the gain and phase frequency characteristics of the transformer. This would include the effect of the output impedance of the electronic circuit output stage. In these characteristics as the frequency range is quite large the frequency scale used is logarithmic.
5. When series leakage inductances are ignored then the region is called as _____________
a) Low-frequency region
b) High-frequency region
c) Mid-frequency region
d) Ultra high-frequency region
Answer: c
Explanation: In this region the series leakage inductances can be ignored and the shunt inductance can be considered as open circuit. With these approximations the equivalent circuit as seen on the primary side can be drawn and gain is then calculated.
6. What is the gain of the transformer in mid-band region?
a) [N 2 /N 1 ] [R L ‘/(R+R L ’)].
b) [N 2 /N1] [RL ‘/(R+R L ’+jωL)].
c) AH=A0/ [1+ 2 ] 1/2
d) AL=A0/ [1+ 2 ] 1/2
Answer: a
Explanation: It immediately follows from the circuit analysis of transformer in mid-frequency region that VL and VS are in phase, the circuit being resistive only. As for the amplitude gain, it is given as
V L ’=V S [R L ‘/(R+R L ’)] [N 2 /N 1 ]V L =VS [R L ‘/(R+R L ’)] A 0 = V L /V S = [N 2 /N 1 ] [R L ‘/(R+R L ’)]
7. When series leakage inductances are not ignored but shunt inductance is an effective open circuit then the region is called as _____________
a) Low-frequency region
b) High-frequency region
c) Mid-frequency region
d) Ultra high-frequency region
Answer: b
Explanation: In this region the series inductances must be taken into account but the shunt inductance is an effective open circuit yielding the approximate equivalent circuit. Amplitude and phase angle as function of frequency are derived below.
8. What is the gain of the transformer in high-frequency region?
a) [N 2 /N 1 ] [R L ‘/(R+R L ’)].
b) [N 2 /N 1 ] [R L ‘/(R+R L ’+jωL)].
c) A H =A 0 / [1+ 2 ].
d) A L =A 0 / [1+ 2 ] 1/2
Answer: b
Explanation: It immediately follows from the circuit analysis of transformer in high-frequency region that AH is equal to [N 2 /N 1 ] [R L ‘/(R+R L ’+jωL)]. If recognised the term [N 2 /N 1 ] [R L ‘/(R+R L ’)] as the gain of transformer in mid-band region then A H =A 0 / [1+ 2 ] 1/2 .
9. When series leakage inductances are not ignored but shunt inductances are considered in parallel then the region is called as _____________
a) low-frequency region
b) high-frequency region
c) mid-frequency region
d) ultra high-frequency region
Answer: a
Explanation: In this region the series effect of leakage inductances is of no consequence but the low reactance shunting effect must be accounted for giving the approximate equivalent circuit and in the calculations of gain of a transformer.
10. What is the gain of the transformer in low-frequency region?
a) [N 2 /N 1 ] [R L ‘/(R+R L ’)].
b) [N 2 /N 1 ] [R L ‘/(R+R L ’+jωL)].
c) A H =A 0 / [1+ 2 ].
d) A L =A 0 / [1+ 2 ] 1/2
Answer: d
Explanation: The corner frequency ωL of this circuit is obtained by considering the voltage source as short circuit. This circuit is Lm in parallel with R||R￠L. Thus, ωL = [R||R L ’/Lm].
Complex gain can be expressed as A L =A 0 / [1+ 2 ] 1/2 .
11. Relative voltage ratio curve for an output transformer is _______________
a) Circular
b) Straight line
c) Bell shaped
d) Irregular curve
Answer: c
Explanation: The complete amplitude and phase response of the transformer on log frequency scale. At high frequencies, the interturn and other stray capacitances of the transformer windings need to be considered. The capacitance-inductance combination causes parallel resonance effect thus, amplitude peak shows up in the high-frequency region of the frequency response.
12. Phase angle characteristic of an output transformer is ____________
a) Curve increasing towards high-frequency region
b) Curve increasing towards low-frequency region
c) Curve increasing towards mid-frequency region then decreasing
d) Can’t define
Answer: a
Explanation: The phase-angle characteristics of an output transformer starts from below of relative voltage ratio characteristics drawn on the same axis. Further it goes on increasing towards high-frequency region non-linearly.
This set of Transformers Multiple Choice Questions & Answers focuses on “Three Phase Transformer Construction”.
1. Which type of winding is used in 3-phase shell-type transformer?
a) Circular type
b) Sandwich type
c) Cylindrical type
d) Rectangular type
Answer: b
Explanation: In core type of the transformer, winding is done by normal method, while in the shell type transformer, winding is sandwiched between corresponding shells of core material. Hence, it is sandwich type.
2. 3-phase transformers compare to 1-phase transformers ________
a) More economical
b) Easy in construction
c) Easy to construct
d) Easy to handle
Answer: a
Explanation: 3-phase transformers are also used to power large motors and other heavy loads. A three-wire three-phase circuit is more economical than an equivalent two-wire single-phase circuit at the same line to ground voltage because it uses less conductor material to transmit a given amount of electrical power.
3. How 3-phase transformers are constructed?
a) A bank of 3 single phase transformers
b) A single 3-phase transformer with the primary and secondary of each phase wound on three legs of a common core
c) Single 3-phase transformer or a bank of 3 1-phase transformers
d) By different method
Answer: c
Explanation: #-phase transformers can be either constructed by 2 methods. One is a bank of 3 single phase transformers and another method is a single 3-phase transformer with the primary and secondary of each phase wound on three legs of a common core.
4. Three phase transformer compare to a bank of 3 single phase transformers is ____________
a) Cheaper
b) Costlier
c) More in space
d) Can’t be determined
Answer: a
Explanation: 3-phase transformers can be manufactured by 2 methods. The three-phase transformer unit costs about 15% less than that of a bank and furthermore, the single unit occupies less space. There is little difference in reliability.
5. In mines we use _______________
a) a single unit of 3-phase transformer
b) a bank of 3 single phase transformers
c) use of 3-phase transformer is avoided
d) a single unit or a bank
Answer: b
Explanation: It is cheaper to carry spare stock of a single-phase rather than a three-phase transformer. In underground use a bank of single-phase units may be preferred as it is easier to transport these units. The bank also offers the advantage of a de-rated open-delta operation when one single-phase unit becomes inoperative.
6. In three phase bank phases are ___________________
a) the phases are electrically not connected and magnetically independent
b) the phases are electrically not connected and magnetically dependent
c) the phases are electrically connected and magnetically independent
d) the phases are electrically connected and magnetically dependent
Answer: c
Explanation: The 3-phase load bank is constructed by so that all three phases are made to be connected with together electrically. By the same time these banks form three magnetic circuits which are independent.
7. Where the tappings are provided in a transformer?
a) At the phase end of LV side
b) At the phase end of HV side
c) At the neutral side end of the HV side
d) At the middle of HV side
Answer: d
Explanation: Tappings are provided in order to maintain the terminal voltage across the loads within the permissible specified limits with respect to load variations. Tappings can be provided at phase end or neutral end of HV side. But to maintain magnetic symmetry tappings are preferred in the middle of the winding only.
8. Tappings are on _____________
a) LV side of a transformer
b) HV side of transformer
c) Not on any side
d) On both sides
Answer: b
Explanation: Generally, tappings are preferred on HV side, because the HV side is low current-high voltage side, tap changer involves low current interruption and large variations or small steps are possible due to more no of turns.
9. In core type 3-phase transformer flux path chooses how many paths to return?
a) 2
b) Single
c) 3
d) Many
Answer: a
Explanation: Flux of each limb in core type uses the other two limbs for its return path with the three-magnetic flux’s in the core generated by the line voltages differing in time-phase by 120 degrees. Thus, the flux in the core remains nearly sinusoidal, producing a sinusoidal secondary supply voltage.
10. Why shell type 3-phase transformer is used in large power transforming applications?
a) Can be made with more height
b) Can be made with less height
c) More height and less height flexibility
d) Due to other reasons
Answer: b
Explanation: The shell-type five-limb type three-phase transformer construction is heavier and more expensive to build than the core-type. Five-limb cores are generally used for very large power transformers as they can be made with reduced height.
11. A three-phase transformer generally has the three magnetic circuits interlaced.
a) True
b) False
Answer: b
Explanation: A three-phase transformer generally has the three magnetic circuits which are interlaced to give a uniform distribution of the dielectric flux between the high and low voltage windings. The exception is a three-phase shell type transformer. In the shell type of construction, even though the three cores are together, they are non-interlaced.
This set of Transformers Multiple Choice Questions & Answers focuses on “Three Phase Transformer Connections”.
1. When does star/star transformers work satisfactorily?
a) Load is unbalanced only
b) Load is balanced only
c) On balanced as well as unbalanced loads
d) Independent of load type
Answer: b
Explanation: With the unbalanced load connected to the neutral, the neutral point shifts thereby making the three line-to-neutral voltages unequal. The effect of unbalanced loads can be illustrated by placing a single load between phase a and the neutral on the secondary side.
2. When does delta/star transformer work satisfactorily?
a) Load is balanced only
b) Load is unbalanced only
c) On balanced as well as unbalanced loads
d) Independent of load type
Answer: c
Explanation: Large unbalanced/balanced loads can be handled satisfactory. The Y-D connection has no problem with third harmonic components due to circulating currents in D. It is also more stable to unbalanced loads since the D partially redistributes any imbalance that occurs.
3. Scott connections are used in _______________
a) three-phase to single phase transformation
b) three-phase to two-phase transformation
c) single phase to three-phase transformation
d) all phase transformations
Answer: b
Explanation: Scott connections are used to convert three-phase to two-phase conversion, to start two phase motors and two phase furnaces. It requires two single phase transformers with adjustable tappings, one transformer is main transformer which is centre tapped through teaser transformer primary with a suitable number of turns to get a balanced two-phase supply.
4. In a three-phase star – delta transformer, what is the angle difference between primary and secondary phase voltages?
a) Delta side leads by 300
b) Delta side lags by 300
c) Star side leads by 300
d) Star side lags by 300
Answer: a
Explanation: This is a vector group and has + 30° displacement. Therefore, delta side leads by + 30°. So, it can be stated that delta side is having lead of 300 over star side because of the connections made.
5. Which can be also called as 00 /1800 connection?
a) Star/star
b) Direct star
c) Delta/star
d) Star/delta
Answer: a
Explanation: Star connection is formed on each side by connecting together phase winding terminals. The voltages of the corresponding phases are in phase. This is known as the 0°-connection. If the winding terminals on secondary side are reversed, the 180°-connection is obtained.
6. What is the ratio of transformation of star/star connection?
a) Phase transformation x:1, line transformation x:1
b) Phase transformation x:1, line transformation 2x:1
c) Phase transformation x:1, line transformation x/3:1
d) Can’t say
Answer: a
Explanation: The phase transformation ratio is given as x:1, where x simply denotes the turns ratio of a transformer given, thus, in the star/star connection we will get the same ratios, as at a particular point, voltage reading on primary will be proportional to secondary with x.
7. Delta/delta connection is also called as ____________
a) 00-connection
b) 900-connection
c) 1800-connection
d) 00/1800-connection
Answer: d
Explanation: Delta/delta connection is also called as 00-connection as seen from the phasor diagram that primary and secondary line voltages are in phase with each other. By reversing connection, we can get 1800 phase shift.
8. What is the ratio of transformation of delta/delta connection?
a) Phase transformation x:1, line transformation x:1
b) Phase transformation x:1, line transformation 2x:1
c) Phase transformation x:1, line transformation x/3:1
d) Can’t determine
Answer: a
Explanation: The phase transformation ratio is given as x:1, where x simply denotes the turns ratio of a transformer given, thus, in the delta/delta connection also similar to the star/star, we will get the same ratios, as at a particular point, voltage reading on primary will be proportional to secondary with x.
9. Open delta connection has VA rating of _______________
a) √3 times delta/delta VA rating
b) 1/√3 times delta/delta VA rating
c) 3 times delta/delta VA rating
d) 1/3 times delta/delta VA rating
Answer: b
Explanation: When one of the transformer in delta/delta connection is removed we get open delta connection. This connection can handle the power of √3VI. While on the similar line delta/delta connection can handle the power of 3VI.
10. Star/delta connection is also called as ____________
a) 300-connection
b) 00-connection
c) -300-connection
d) 300/-300-connection
Answer: d
Explanation: Star/delta connection is also called as +/-300-connection as seen from the phasor diagram that primary and secondary line voltages are either ahead or below by 300 phases with each other. By reversing connection, we can get another condition phase shift.
11. What is the ratio of transformation of star/delta connection?
a) Phase transformation x:1, line transformation x:1
b) Phase transformation x:1, line transformation √3x:1
c) Phase transformation x:1, line transformation 3x:1
d) Can’t determine with information available
Answer: a
Explanation: The phase transformation ratio is given as x:1, where x simply denotes the turns ratio of a transformer given, thus, in the star/delta connection we will get the √3 factor in ratios, as at a particular point, voltage reading on primary will be proportional to secondary with √3x.
12. x/√3:1 ratio is obtained in ______________
a) Star/delta
b) Delta/star
c) Delta/delta
d) Star/star
Answer: b
Explanation: The phase transformation ratio is given as x:1, where x simply denotes the turns ratio of a transformer given, thus, in the star/delta connection we will get the 1/√3 factor in ratios, as at a particular point, voltage reading on primary will be proportional to secondary with 1/√3x.
13. Which both connections have the same line transformation ratios?
a) Star/star and delta/delta
b) Star/delta and delta/star
c) Star/zig-zag star and delta/zig-zag star
d) Star/star, delta/delta and star/delta, delta/star
Answer: d
Explanation: Star/star and delta/delta both connections have phase transfer ratio of x:1 and line transfer ratio also equal to x:1, while star/zig-zag star and delta/zig-zag star connections have line transformation ratio equal to 2/√3x:1.
This set of Transformers Interview Questions and Answers for freshers focuses on “Three Phase Bank of Single Phase Transformers”.
1. For very high voltage transformers which connection is cheaper on primary side?
a) Star
b) Delta
c) Open delta
d) Can be star/ delta/ open delta
Answer: a
Explanation: In star connection with earthed neutral, the maximum voltage of the phase winding to ground is 1/√3 or 58% of the line voltage, while in delta connection this is equal to the line voltage in case of earthing of one of the lines during a fault. Thus, star on HV side is cheaper.
2. In which of the circuit given positive and negative sequence currents will flow in primary?
a) Star/delta
b) Star/star
c) Open delta/delta
d) Delta/delta
Answer: d
Explanation: The only positive and negative sequence currents flow in the lines on the delta side. This could also be achieved by star-connected primary provided the primary and secondary star points are grounded. But this is not recommended on account of flow of ground current for unbalanced secondary loads.
3. When star/star connection is used?
a) Small HV
b) Small LV
c) High HV
d) High LV
Answer: a
Explanation: This case is economical for small HV transformers as it minimizes the turns/phase and winding insulation. A neutral connection is possible. However, the Y /Y connection is rarely used because of difficulties associated with the exciting current.
4. When delta/delta connection is used?
a) Small HV
b) Small LV
c) High HV
d) High LV
Answer: d
Explanation: This suits large LV transformers as it needs more turns/phase of smaller section. A large load unbalance can be tolerated. The absence of a star point may be a disadvantage. This connection can operate at 58% normal rating as open-delta when one of the transformers of the bank is removed for repairs or maintenance.
5. For using as a step-up transformer which connection is used?
a) Star/star
b) Delta/delta
c) Delta/star
d) Star/delta
Answer: c
Explanation: This is the most commonly used connection for power systems. At transmission levels star connection is on the HV side, i.e. D/Y for step-up and Y/D for step-down. The neutral thus available is used for grounding on the HV side.
6. At distribution level transformer with which connection is used?
a) Star/star
b) Delta/delta
c) Delta/star
d) Star/delta
Answer: c
Explanation: At the distribution level the D/Y transformer is used with star on the LV side which allows mixed 3-phase and 1-phase loads, while delta allows the flow of circulating current to compensate for neutral current on the star side.
7. Third-harmonic currents have phase difference of _____________
a) 0 0
b) 90 0
c) 180 0
d) 27 0
Answer: a
Explanation: The phase difference in third-harmonic currents and voltages on a 3-phase system is 3 * 120° = 360° or 0° which means that these are cophasal. Therefore, third-harmonic currents and voltages cannot be present on the lines of a 3-phase system as these do not add up to zero.
8. Core flux in the transformer is _______________
a) sinusoidal
b) flat-topped
c) square wave
d) triangular
Answer: b
Explanation: The supply voltage provides the input current to the transformer primary, here primary current is only sinusoidal magnetizing current. Thus, this sinusoidal magnetic current will produce core flux, flat-topped.
9. In delta/delta connection flux is almost _________________
a) sinusoidal
b) flat-topped
c) triangular
d) square wave
Answer: a
Explanation: The supply voltage provides only sinusoidal magnetizing current so that core flux is flat-topped; but the third-harmonic emfs induced cause circulating currents in deltas restoring the flux to almost sinusoidal.
10. Apart from connection which of the following is different in star/delta or delta/star compare to delta/delta?
a) Flux is flat-topped
b) Impedance offered to third-harmonic currents in delta is less
c) Impedance offered to third-harmonic currents in delta is constant
d) Impedance offered to third-harmonic currents in delta is more
Answer: d
Explanation: Because of one delta connection the same conditions are obtained as in D/D connection except that the impedance offered to the flow of third-harmonic currents in delta is now larger and so are third-harmonic voltages.
11. In star/star connection the voltage can be correctly expressed as ____________
a) e aN = e a sin 2ωt + e a3 sin 3ωt
b) e aN = e a sin ωt + e a3 sin 3ωt
c) e aN = e a sin 3ωt + e a3 sin 3ωt
d) e aN = e a sin 6ωt + e a3 sin 3ωt
Answer: b
Explanation: In the case of isolated neutrals, third-harmonic voltages are present in each phase as explained earlier. Further, since these voltages are cophasal, no third-harmonic voltages are present between lines. The voltage of phase a to neutral can now be expressed as shown in option b.
12. Rate of change of voltage in star/star connection is ______________
a) ω
b) 2 ω
c) 3 ω
d) Can’t determine
Answer: b
Explanation: While fundamental frequency voltages in the three phases have a relative phase difference of 120°, the third-harmonic voltages in them are cophasal , but their phase with respect to the fundamental frequency .
13. Voltage at the neutral point oscillates at frequency of 2ω, this phenomenon is called as _________
a) oscillating neutral
b) doubling voltage
c) doubling current
d) doubling neutral
Answer: a
Explanation: Voltage of the neutral point oscillates at frequency 2 ω, phenomenon is known as oscillating neutral and is highly undesirable because of which the star/star connection with isolated neutrals is not used in practice. If the neutrals are connected, it effectively separates the three transformers. Third-harmonic currents can now flow via the neutrals.
This set of Transformers Multiple Choice Questions & Answers focuses on “Tests on Three Phase Transformers”.
1. In three-phase transformer, the harmonic fluxes are suppressed because of ____________
a) high reluctance path
b) low reluctance path
c) any reluctance paths
d) independent on reluctance path
Answer: a
Explanation: In core type transformer, the third-harmonic fluxes in all the three limbs are simultaneously directed upwards or downwards so that this flux must return through air . The high reluctance path tends to suppress the third-harmonic flux.
2. Suppressing of harmonic fluxes becomes more prominent in _______
a) fifth harmonic currents
b) third harmonic currents
c) fourth harmonic currents
d) second harmonic currents
Answer: a
Explanation: The phenomenon gets more complex now and at core densities exceeding 1.5 T, the total harmonic content is very marked in the magnetizing current .
3. To reduce effect of 5th harmonic current ____________
a) separate path must be provided
b) no need of separate path
c) add a resistor in series
d) add a resistor in parallel
Answer: a
Explanation: To reduce the strong fifth harmonic in the magnetizing current for the star/star connection with isolated neutral, a path must be provided through iron for the third-harmonic flux. Hence, the use of a 5-limb core is adopted.
4. For performing back to back test on 3-phase transformer, transformers should be ___________
a) non-identical
b) identical
c) they can be identical or non-identical
d) they should not be identical nor non-identical
Answer: b
Explanation: According to connection of back to back test, arrangement of three-phase transformers should be done by keeping the fact into consideration that both of these transformers should be identical.
5. In back to back test two secondaries are connected ___________
a) in proper phase sequence
b) in phase opposition
c) in proper phase sequence and with phase opposition
d) in opposite phase sequence
Answer: c
Explanation: According to the connection arrangement for the back-to-back test on two identical 3-phase transformers. Then, two secondaries must be connected in phase opposition and in proper phase sequence.
6. Auxiliary transformer is not needed in the back to back test.
a) True
b) False
Answer: b
Explanation: Auxiliary transformer is needed in the bak to back test as, we need to circulate the full-load current. This transformer is connected at secondaries or at primaries of the three-phase transformers.
7. Where the auxiliary transformers are connected in back to back test of 3-phase transformer?
a) Primaries
b) Secondaries
c) In the middle
d) Can be connected to primaries or secondaries as well
Answer: d
Explanation: The auxiliary transformer for circulating full-load current is included in the circuit of the two secondaries; it could also be included in the circuit of the primaries. Thus, with only losses supplied from the mains, a “heat run” test could be conducted on the transformers.
8. Auxiliary transformer connected to delta/delta transformer is of the type _______________
a) single phase transformer
b) three-phase transformer
c) two-phase transformers
d) can be of any type
Answer: a
Explanation: The primaries are normally excited from the mains. Each secondary delta is opened at one junction and a single-phase transformer can be employed to circulate full-load current in both the deltas.
9. If one of the transformers is removed from the bank of only delta-delta, then it will behave as power delivery transformer of ________
a) 58%
b) 78%
c) 45%
d) 100%
Answer: a
Explanation: It is true as the circuit will still be closed and the transformer will operate will lesser operating point. This new circuit so formed is also called as open delta circuit, which will deliver 58% of output.
10. Three units of single phase transformers and one single three-phase transformer_____________
a) will be same for one rating
b) can never be made same
c) may be same
d) depends on other factors
Answer: a
Explanation: Three single phase transformers and one single unit of three phase transformer will be same only, thus, will have same rating. This is done to reduce the cost and spacing, and to gain various other advantages.
11. A V-V connected transformer can be connected in parallel to delta-delta connected transformer
but not to _________
a) delta-star
b) star-delta
c) star-V
d) star-delta and star-V both
Answer: a
Explanation: The V-V transformer can be obtained from D-D transformer. The V-V connected transformer and D-D connected transformers have same phase displacement, so they only can be connected in parallel to each other.
This set of Transformers Questions and Answers for Experienced people focuses on “Additional Testing for Important Transformers”.
1. Which of the following is not a routine test for transformers?
a) Impedance test
b) Core insulation voltage test
c) Radio interference test
d) Polarity test
Answer: c
Explanation: Impedance test is done on a transformer in order to check net impedance offered by a transformer circuit at rated supply. Polarity test is also done before SC and OC test on transformer. Core insulation is also checked before installations in such tests.
2. Which of the following is not required for transformer testing to get an accurate result?
a) Class 0.1 current transformer
b) Class 0.1 voltage transformer
c) High power factor wattmeter
d) Voltmeters
Answer: c
Explanation: To obtain accurate results it is essential that low power factor wattmeter, precision grade ammeters, voltmeters, and class 0.1 current and voltage transformers are used.
3. All the instruments needed for transformer testing should be tested within span of ____________
a) a month
b) 3 months
c) 6 months
d) 12 months
Answer: d
Explanation: These instruments should be checked at intervals not exceeding 12 months to ensure that the requisite accuracy is maintained. All these instruments are highly sensitive in their operation, thus to maintain accuracy testing is essential.
4. Epstein square method is used in power transformer installations, to check ___________
a) Core frame insulation
b) Core-plate checks
c) Core-loss measurement
d) Winding copper checks
Answer: a
Explanation: Incoming core plate is checked for thickness and quality of insulation coating. A sample of the material is cut and built up into a small loop known as an Epstein Square from which a measurement of specific loss is made. Such a procedure is described in BS 6404.
5. Generally Core frame insulation test is done at _________
a) 10 kV DC
b) 10 kV AC
c) 2KV RMS
d) 2KV DC
Answer: c
Explanation: This is checked by Megger and by application of a 2 kV RMS or 3 kV DC test voltage on completion of erection of the core. These checks are repeated following replacement of the top yoke after fitting the windings.
6. Why tanks are tested?
a) Stiffness
b) Vacuum withstand capacity
c) Stiffness and vacuum withstand capacity
d) Any other purpose
Answer: c
Explanation: The first tank of any new design should be checked for stiffness and vacuum-withstand capability. For 275 and 400 kV transformers, a vacuum equivalent to 25 mbar absolute pressure should be applied. This need only be held long enough to take the necessary readings and verify that the vacuum is indeed being held, which might take up to 2 hours for a large tank.
7. How much pressure is applied for transformers rated 132 kV and below ones?
a) 330 mbar
b) 300 mbar
c) 400 mbar
d) 500 mbar
Answer: a
Explanation: For transformers rated 132 kV and below a more modest vacuum test equivalent to 330 mbar absolute pressure should be applied. The permissible permanent deflections following this test should be similar to those allowed for 275 and 400 kV transformer tanks reduced pro-rata for smaller tanks.
8. Which of the following parameters don’t define that transformer is built correctly?
a) Losses
b) Polarity
c) Tap changing operation
d) Ratio
Answer: a
Explanation: Tests to prove that the transformer has been built correctly: include ratio, polarity, resistance, and tap change operation. While losses prove the that transformer is guaranteed in its operation.
9. Which of the following parameters don’t define that transformer is built correctly?
a) Losses
b) Temperature rise
c) Noise level
d) Resistance
Answer: d
Explanation: Tests to prove guarantees; these are losses, impedance, temperature rise, noise level. While resistance, tap changing, ratio, polarity are the parts of tests that prove that transformer has been built correctly.
10. Which of the following tests don’t indicate that transformer will work satisfactorily for at least 30 Years?
a) Temperature rise
b) Dielectric test
c) Overvoltage test
d) Load current runs
Answer: a
Explanation: Tests to prove that the transformer will be satisfactory in service for at least 30 years. The tests in this category are the most important and the most difficult to frame: they include all the dielectric or overvoltage tests, and load current runs.
11. Which of the following is not in the category of special test for a transformer?
a) Short-circuit test
b) Harmonics on the no-load current
c) Power taken by fan and oil-pump motors
d) Open-circuit test
Answer: d
Explanation: Special tests are tests, other than routine or type tests, agreed between manufacturer and purchaser, for example: test with lightning impulse chopped on the tail, zero-sequence impedance on three-phase transformers, and other tests included in options.
12. How temperature rise of an oil-immersed transformer is not found out?
a) Short circuit equivalent test
b) Delta/delta test
c) Back-to-back test
d) Normal SC test
Answer: d
Explanation: When a test for temperature rise is specified it is necessary to measure the temperature rise of the oil and the windings at continuous full load, and the various methods of conducting this test are as follows: short-circuit equivalent test, back-to-back test, delta/delta test, open-circuit test.
This set of Transformers Multiple Choice Questions & Answers focuses on “Transformer Installations”.
1. Transformers are generally designed for ___________
a) on-site problem solving
b) one-time use
c) off-site problem solving
d) short-time use
Answer: a
Explanation: Every transformer is designed for use of it for multiple years, thus transformers are designed to handle the problems on site itself, because it not only saves time but also makes repairing work easy.
2. When all additional clamping is removed from a transformer?
a) When transformer is made start
b) After start of transformer function
c) Instant before start of transformer function
d) At the time of installations
Answer: d
Explanation: When the tank is correctly positioned on the plinth it must then be carefully examined for any signs of damage or any other indication that it might have been mishandled during transport. Any special provisions by way of protection applied during transport must be removed. If additional clamping has been applied to the core and windings for transport, this must be released or removed according to the instruction manual.
3. While transformer is transported, its valves are _______________
a) kept open
b) closed with blanking plates
c) can be kept open or can be closed according to type
d) valves are absent
Answer: b
Explanation: Removal of blanking plates giving access to the tank. Such opening of the tank must be kept to a minimum time, to reduce the possibility of moisture entering the tank; to assist in this, manufacturers of large high-voltage transformers provide equipment to blow dry air into the tank and thus maintain a positive internal pressure.
4. If any transformer is not in operation for some months, still it is advisable to keep it filled with _______________
a) nitrogen
b) oil
c) normal air
d) hydrogen helium mixture
Answer: b
Explanation: Even if the transformer is not required for service for some months, it is desirable that it should be filled with oil as soon as possible and certainly within three months of the original date of draining the oil in the factory. If it is being kept in storage for a period longer than three months at some location other than its final position, it should similarly be filled with oil.
5. When oil filling procedure has to be started?
a) After fitting of all bushings
b) After replacing access covers
c) After erection of conservator and Buchholz pipework
d) After fitting bushings, replacing covers, erecting pipework
Answer: d
Explanation: When all bushings have been fitted, access covers replaced, and conservator and Buchholz pipework erected, any cooler bank erected and associated pipework installed or tank-mounted radiators fitted, preparations can begin for filling with oil.
6. Materials used in construction of transformer are stressed ______________
a) electrically
b) mechanically
c) both electrically and mechanically
d) magnetically
Answer: c
Explanation: The materials used in their construction are highly stressed both electrically and mechanically, and to achieve satisfactory operation extensive precautions are taken in manufacture, particularly in respect of insulation quality.
7. After site erection vacuum is made with pressure range of ___________
a) 1 mbar
b) 3 mbar
c) 7 mbar
d) 15 mbar
Answer: c
Explanation: After completion of site erection, a vacuum pump is applied to the tank and the air exhausted until a vacuum equivalent to between 5 and 10 mbar can be maintained. If this work is carried out by the transformer manufacturer, or his appointed subcontractor, there will be no doubt as to the ability of the tank to withstand the applied vacuum.
8. For 400 kV transformer, moisture content required is _______________
a) below 0.1%
b) below 0.5%
c) above 0.5%
d) below 1%
Answer: b
Explanation: When a new 400 kV transformer is processed in the factory, the aim is to obtain a moisture content in the cellulose insulation of less than 0.5%. When an oiled cellulose insulation is exposed to atmosphere, the rate of absorption of moisture depends on the relative humidity of the atmosphere.
9. Which readings should be taken while drying out process of transformer?
a) Insulation reading between HV and LV
b) Temperature
c) Time
d) Insulation, temperature, time all readings
Answer: d
Explanation: During the drying-out process the following readings should be taken at frequent regular intervals: Insulation resistance between high-voltage and low-voltage windings and between each winding and earth, temperature and time.
10. Short circuit method is used for _____________
a) drying out the transformer and oil simultaneously in the transformer tank
b) drying the transformer only, out of its tank.
c) for any type of drying
d) never used
Answer: c
Explanation: Short-circuit method is also used for: Drying out the transformer and oil simultaneously in the transformer tank, drying the transformer only, out of its tank. During the process, the low-voltage winding is short-circuited, a low single-phase or three-phase voltage being applied to the high-voltage windings.
This set of Transformers Multiple Choice Questions & Answers focuses on “Parallel Operation of Transformers”.
1. For two transformers connected in parallel, not having unequal percentage impedances, which statement is correct?
a) Short-circuiting of the secondaries
b) Power factor of one of the transformers is leading while that of the other lagging
c) Transformers having higher copper losses will have negligible core losses
d) Loading of the transformers not in proportion to their kVA ratings
Answer: d
Explanation: In parallel operation of a transformer, loading gets divided between the whole set in proportion of their impedances. Thus, if impedances are not same then, loading of transformers will not be in the ratio of their kVA ratings.
2. For the parallel operation of two single phase transformers it is necessary that they should have ________
a) same efficiency
b) same polarity
c) same kVA rating
d) same number of turns on the secondary side
Answer: b
Explanation: Polarity of the two transformers must be same if the are supposed to be operated in parallel mode of operation. It is okay if they have unequal efficiencies, or unequal kVA ratings, as parallel connection is still possible.
3. Transformers operating in parallel mode of operation will share the load depending upon their ___________
a) leakage reactance
b) per unit impedance
c) efficiencies
d) ratings
Answer: b
Explanation: In parallel operation of a transformer, loading of the transformers gets divided between the whole set in proportion of their per unit impedances. Thus, if per unit impedances are not same then, loading will not be in the ratio of their kVA ratings.
4. What will happen if the transformers working in parallel are not connected with regard to polarity?
a) The power factor of the two trans-formers will be different from the power factor of common load
b) Incorrect polarity will result in dead short circuit
c) The transformers will not share load in proportion to their kVA ratings
d) Cannot be determined
Answer: b
Explanation: As far polarity is concerned, the transformers are connected in the same sequence when they are operated in parallel combination. An incorrect polarity connection of these transformers will result in dead short circuit.
5. If the percentage impedances of the two transformers working in parallel are different, then ______________________
a) transformers will be overheated
b) power factors of both the transformers will be same
c) parallel operation will be not possible
d) parallel operation will still be possible, but the power factors at which the two transformers operate will be different from the power factor of the common load
Answer: d
Explanation: If the percentage impedances of the two transformers which are working in the parallel are different, then parallel operation is still possible, but load sharing will not be in the ratio of their kVA loads.
6. A delta/star transformer is connected in parallel to a star/delta transformer. The turn ratio former is x times latter, the x is _________
a) 3
b) 1/3
c) √3
d) 1/√3
Answer: a
Explanation: Turn ratio of star1/ Turn ratio of delta 1= Vph/Vph/√3= √3
Ture ratio of star 2 / Turn ratio of delta 2= Vph/√3/Vph= 1/√3
Thus, turn ratio of star1/delta1 = 3 * turn ratio of star 2/ delta 2.
7. While connecting two transformers in parallel voltage around the local loop _________
a) positive
b) negative
c) equals zero
d) insufficient information
Answer: c
Explanation: The transformers must be connected properly as far as their polarities are concerned so that the net voltage around the local loop is zero. A wrong polarity connection results in a dead short circuit.
8. For three phase power transformers relative phase displacement should be ____________
a) 00
b) 300
c) 900
d) 1800
Answer: a
Explanation: Three-phase transformers must have zero relative phase displacement on the secondary sides and must be connected in a proper phase sequence. Only the transformers of the same phase group can be paralleled.
9. Y/Y and Y/D transformers can be paralleled.
a) True
b) False
Answer: b
Explanation: For example, Y/Y and Y/D transformers cannot be paralleled as their secondary voltages will have a phase difference of 30°. Transformers with +30° and –30° phase shift can, however, be paralleled by reversing the phase-sequence of one of them.
10. Why transformers connected in parallel should have same voltage ratio?
a) To avoid full load circulating current
b) To avoid no-load circulating current
c) To avoid other losses
d) To avoid all type of currents
Answer: b
Explanation: The transformers must have the same voltage-ratio to avoid no-load circulating current when transformers are in parallel on both primary and secondary sides. Since the leakage impedance is low, even a small voltage difference can give rise to considerable no-load circulating current and extra I 2 R loss.
11. Why transformers are paralleled?
a) Economical than replacing by one single large unit
b) Less maintaining cost
c) To handle more load
d) Economical factors, more load capacity, less maintenance
Answer: d
Explanation: When the load outgrows the capacity of an existing transformer, it may be economical to install another one in parallel with it rather than replacing it with a single larger unit. Also, sometimes in a new installation, two units in parallel, though more expensive, may be preferred over a single unit for reasons of reliability—half the load can be supplied with one unit out. Further, the cost of maintaining a spare is less with two units in parallel.
This set of Transformers Interview Questions and Answers for Experienced people focuses on “Load Division Between Transformers in Parallel”.
1. If the primary voltages at two transformers V1 and V2 are not equal then on load, we’ll get ____________
a) V 1 -V 2 at secondary
b) E 1 -E 2 at secondary
c) V 1 +V 2 at secondary
d) E 1 +E 2 at secondary
Answer: b
Explanation: When two transformers paralleled on both sides with proper polarities but on no-load. The primary voltages V 1 and V 2 are obviously equal. If the voltage-ratio of the two transformers are not identical, the secondary induced emfs, E1 1 and E 2 though in phase will not be equal in magnitude and the difference (E 1 -E 2 ) will appear across the switch S.
2. If two transformers’ secondaries are connected to each other with unequal primary voltage ratio then, __________
a) no circulating current will flow
b) very high short circuit current will flow
c) small circulating current will flow
d) insufficient information
Answer: c
Explanation: When secondaries are paralleled by closing the switch, a circulating current appears even though the secondaries are not supplying any load, as a result of difference in their voltage ratios.
3. The circulating current flowing through the circuit at no load condition depends on ________________
a) total leakage impedance of the two transformers
b) difference in their voltage ratios
c) difference in voltage ratios, leakage impedance of 2 transformers
d) other parameters
Answer: c
Explanation: The circulating current flowing at no-load condition depend upon the total leakage impedance of the two transformers and the difference in their voltage ratios. Only a small difference in the voltage-ratios can be tolerated.
4. If the transformers have equal voltage ratio then, ____________
a) exciting current can be neglected
b) summation of two transformer currents is not equal to the net load current
c) difference of two transformer currents is equal to the net load current
d) current will not flow
Answer: a
Explanation: When the transformers have equal voltage ratio, E1 = E2, the equivalent circuit of the two transformers would then be simple because of the assumption that the exciting current can be neglected in comparison to the load current.
5. Which is the correct formula for current flowing through the transformer 1, when they’re having equal voltage ratio?
a) I 1 = Z 2 /(Z 1 +Z 2 ) *I L
b) I 2 = Z 2 /(Z 1 +Z 2 ) *I L
c) I 1 = Z 1 /(Z 1 +Z 2 ) *I L
d) I 1 = Z 2 /(Z 1 -Z 2 ) *I L
Answer: a
Explanation: Since both the transformers are having same number of turns and thus equal voltage ratios, it is easier for analysis of current as division in two branches will be according to the Ohm’s law, which is the answer.
6. Individual currents are in two loaded transformers ______________
a) inversely proportional to the respective leakage impedances
b) inversely proportional to the net leakage impedances
c) inversely proportional to another leakage impedance
d) directly proportional to the respective leakage impedances
Answer: a
Explanation: the individual currents are inversely proportional to the respective leakage impedances. Thus, if the transformers are to divide the total load in proportion to their kVA ratings, it is necessary that the leakage impedances be inversely proportional to the respective kVA ratings.
7. Which of the following is the correct ratio, for transformers having equal voltage ratios?
a) Z 1 /Z 2 = S 2 / S 1
b) Z 1 /Z 2 = S 1 / S 2
c) Z 1 /Z 2 = I 1 / I 2
d) Depends upon the type of connection
Answer: a
Explanation: If the transformers are to divide the total load in proportion to their kVA ratings, it is necessary that the leakage impedances be inversely proportional to the respective kVA ratings. Thus, Z 1 /Z 2 = S 2 / S 1 = V L I 2 / V L I 1 . Hence, Z 1 /Z 2 = I 2 / I 1 .
8. Which of the following is the correct statement?
a) S L = S 2 Z 1 +Z 2 /Z 2
b) S L = S1 Z 1 +Z 2 /Z 1
c) S L = S 2 Z 1 +Z 2 /Z 1
d) S L = S 1 +S 2
Answer: c
Explanation: We can define maximum load in kVA relating with rated kVA of transformer 1 as, S 2 = (Z 1 /Z 2 +Z1) *S L . Thus, by rearranging terms we get S L = S 2 Z 1 +Z 2 /Z 1 and SL= S 1 Z1+Z 2 /Z 2 .
9. Maximum load kVA is _____________________
a) greater than addition of individual rated kVAs
b) lesser than addition of individual rated kVAs
c) equal to addition of individual rated kVAs
d) depends on the loading condition
Answer: b
Explanation: As S L = S2 Z 1 +Z 2 /Z 1 and S L = S 1 Z 1 +Z 2 /Z 2 , because of individual leakage impedances are inversely proportional to the respective kVA ratings. Important thing is in either of the above cases maximum kVA loading is lesser than the addition of both rated kVAs.
10. Which is the correct formula of current flowing through one of the transformers having unequal ratios?
a) I 2 = E 2 Z 1 -(E 1 -E 2 ) Z L / (Z 1 Z 2 + ZL(Z 1 +Z 2 ))
b) I 1 = E 2 Z 1 +(E 1 +E 2 ) Z L / (Z 1 Z 2 + ZL(Z 1 +Z 2 ))
c) I 1 = E 2 Z 1 -(E 1 -E 2 ) Z L / (Z 1 Z 2 – ZL(Z 1 -Z 2 ))
d) I 2 = E 2 Z 1 +(E 1 -E 2 ) Z L / (Z 1 Z 2 + ZL(Z 1 +Z 2 ))
Answer: a
Explanation: We know that a small difference in voltage ratios can be tolerated in the parallel operation of transformers. Thus, in the unequal voltage ratio condition current will flow from depending on both of the options stated above.
11. A 600-kVA, single-phase transformer with 0.012 pu resistance and 0.06 pu reactance is connected in parallel with a 300-kVA transformer with 0.014 pu resistance and 0.045 pu reactance to share a load of 800 kVA at 0.8 pf lagging. Find how they share the load when both the secondary voltages are 440 V.
a) S 1 = 377+j305.2
b) S 2 = 377-j305.2
c) S 1 = 264+ j171.6
d) S 1 = 377-j305.2
Answer: d
Explanation: Z 1 = 0.012+j 0.06
Z 2 = 2
Z 1 +Z2= 0.04+ j0.15
The load is SL = 800
Thus, S 1 = Z 2 /Z 1 +Z 2 S L , we get S 1 = 377- j305.2
S 2 = Z 1 /Z 1 +Z 2 S L , we get S 2 = 264-j171.6.
This set of Transformers Multiple Choice Questions & Answers focuses on “Three Winding Transformers”.
1. Three winding and two winding transformers work on same working principle.
a) True
b) False
Answer: a
Explanation: Both the transformers work on the same principle of electromagnetic induction. Transformers may be built with a third winding in addition to the primary and secondary winding called the tertiary. There are various purposes which dictate the use of a tertiary winding.
2. Why tertiary winding is used?
a) To make circuit stable
b) To provide voltage supply to the substations different from the primary and secondary voltage levels
c) To connect static capacitors for reactive power injection
d) Stable circuit, different voltage output, static capacitor connection
Answer: d
Explanation: Tertiary winding can be used to supply the substation auxiliaries at a voltage different from those of the primary and secondary windings. Static capacitors or synchronous condensers may be connected to the tertiary winding of a transformer for reactive power injection into the system for voltage control.
3. Tertiary winding is generally _________
a) star connected
b) delta connected
c) open-delta connected
d) star or open-delta connected
Answer: b
Explanation: Tertiary winding of the three-winding transformer is generally delta connected which is helpful in reducing the impedance offered to the zero sequence currents thus, it allows a larger earth-fault current to flow for proper operation of protective equipment.
4. Tertiary winding allows the flow of _______________
a) third harmonic currents
b) fifth harmonic currents
c) seventh harmonic currents
d) eleventh harmonic currents
Answer: a
Explanation: Delta-connected tertiary winding limits voltage imbalance when the load is unbalanced. It also permits the third harmonic current to flow which ultimately reduces third-harmonic voltage. Thus, performs as a stabilizing winding to the transformer.
5. Tertiary winding can be used for ____________
a) to maximize the core area
b) to measure voltage during HV testing of transformer
c) to fill up the space between primary and secondary windings
d) to measure voltage during open circuit testing of transformer
Answer: b
Explanation: There are many uses of tertiary winding, such as, three windings may be used for interconnecting three transmission lines at different voltages. Tertiary winding can serve the purpose of measuring voltage of an HV testing transformer.
6. Any unbalanced load can be divided into _______________
a) positive and negative sequence components
b) positive, negative and zero sequence components
c) only in positive and zero sequence components
d) only in negative and zero sequence components
Answer: b
Explanation: Any unbalanced load can be divided into three 3-phase sets . The zero-sequence component caused by a line-to-neutral load on the secondary side cannot be balanced by primary currents as the zero-sequence currents cannot flow in the isolated neutral star connected primary.
7. Which currents flow through the delta-connected tertiary winding?
a) Positive sequence currents
b) Negative sequence currents
c) Zero sequence currents
d) All sequence currents
Answer: c
Explanation: Iron path is available for the zero-sequence flux in a bank of single-phase units and in the 5-limb core and as a consequence the impedance offered to the zero-sequence currents is very high inhibiting the flow of these currents. The provision of a delta-connected tertiary permits the circulation of zero-sequence currents in it, thereby considerably reducing the zero-sequence impedance.
8. Simple phase equivalent circuit of three winding transformer can be shown as ____________
a) simple star network
b) simple parallel network
c) simple series network
d) can be shown accordingly
Answer: a
Explanation: Particular winding can be represented as equivalent resistor and reactance. For simplicity, the effect of the exciting current can be ignored in the equivalent circuit. It may be noted that the load division between the secondary and tertiary is completely arbitrary.
9. How impedance of particular winding can be calculated?
a) Z 1 = Z 12 +Z 13 +Z 23
b) Z 1 = 1/2½ (Z 12 +Z 13 -Z 23 )
c) Z 1 = Z 12 -Z 13 +Z 23
d) Z 1 = Z 12 +Z 13 -Z 23
Answer: c
Explanation: The single-phase equivalent of three-winding transformer can be shown. According to the short circuit test one can easily find the short circuit impedance between winding 1 and 2 as Z 12 and thus, formula for primary impedance can be calculated.
This set of Transformers Multiple Choice Questions & Answers focuses on “Phase Conversion”.
1. Only Scott connection is used for _________
a) Converting three-phase to two-phase conversion
b) Converting three-phase to single-phase conversion
c) Converting single-phase to two-phase conversion
d) None of the mentioned
Answer: a
Explanation: Scott connection is used for obtaining two-phase supply as it is needed for various special purposes like supplying two-phase electric arc furnaces and can easily obtained from three-phase supply by this method.
2. In Scott connection, according to the vector diagram two windings are placed at ______________
a) At 1200 to each other
b) Perpendicular with respect to each other
c) At 600 to each other
d) Can’t say
Answer: b
Explanation: For a transformer vector diagram, if one winding is kept along the perpendicular axis of another winding then one can obtain two-phase supply from three-phase supply by using Scott connection. Physically this can be obtained by appropriate turns ratio.
3. What is the ratio of voltage/turn of two windings of a transformer in operated in Scott connection?
a) ½
b) 1/√2
c) 1
d) 1/√3
Answer: c
Explanation: The transformer primaries must have √3 N 1 /2 and N1 turns; this would mean equal voltage/turn in each transformer. A balanced 2-phase supply could then be easily obtained by having both secondaries with equal number of turns, N 2 .
4. The primaries of two transformers in Scott connection are in turns ratio of ___________
a) √3/2: 1
b) 2/√3: √2
c) 1:1
d) Can’t say
Answer: a
Explanation: A 2-phase supply could thus be obtained by means of transformers; one connected perpendicular according to the vector diagram is called the teaser transformer and the other is connected across, with turns ratio √3/2: 1 respectively.
5. The secondaries of two transformers are in ratio ____
a) 1:1
b) √3/2: 1
c) √3: 1
d) √2: 1
Answer: a
Explanation: Though transformer primaries are in the ratio of √3/2: 1 with respect to each other secondary windings of these transformers are in ratio of 1:1. This simply means that both of these transformers have equal voltage per turns.
6. In Scott connection neutral point is located at _______________
a) Three phase side at teaser
b) Two-phase side
c) Three phase side secondary
d) Anywhere
Answer: a
Explanation: Neutral point if required can be located on three phase side of a transformer at teaser in Scott connection which is used to obtain two-phase supply from three-phase supply. Neutral point is located at teaser winding dividing the winding in the ratio 1:2.
7. For single-phase to three-phase change in transformer which of the following connection is most suitable?
a) Scott connection
b) Scott connection with resistor network
c) Scott connection with some energy storing device
d) Any of the mentioned
Answer: c
Explanation: A single-phase power pulsates at twice the frequency, while the total power drawn by a balanced 3-phase load is constant. Thus a 1-phase load can never be transferred to a 3-phase system as a balanced load without employing some energy-storing device .
8. For 3/6-phase converter transformers __________
a) Scott connection is used
b) Each secondary winding is divided
c) Each primary winding is divided
d) Can’t be done
Answer: b
Explanation: For converting 3-phase to 6-phase, three-phase delta connection is used on primary side. Each secondary phase is divided into two equal halves with appropriate polarity and corresponding 6-phase voltage is obtained.
9. In 3/6-phase connection of transformers _______________
a) Secondaries are two stars in phase sequence
b) Secondaries are two deltas in phase sequence
c) Secondaries are two stars in phase opposition
d) Secondaries are two deltas in phase opposition
Answer: c
Explanation: Six-phase voltages are obtained by means of two stars in phase opposition, each star being formed from three respective half-windings. This is done without Scott connection.
10. 3/6-phase circuit is employed for providing DC path in some circuits.
a) True
b) False
Answer: a
Explanation: In certain applications like thyristors and rectifiers six-phase supply is required for providing path for the DC current. Therefore, it becomes necessary to convert three-phase AC supply into six-phase.
11. Two single-phase furnaces let’s say, A and B are supplied at 100 V by means of a Scott-connected transformer combination from a 3-phase 6600 V system. The voltage of furnace A is leading. Calculate on of the line current on the 3-phase side, when the furnace A takes 400 kW at 0.707 pf lagging and B takes 800 kW at unity pf.
a) 110 A
b) 99 A
c) 250 A
d) 149 A
Answer: b
Explanation: N 1 /N 2 = 6600/100= 60
√3/2 (N 1 /N 2 ) = 57.16
Furnace currents are= I a = 400*1000/ = 5658 A
Similarly, I b = 8000 A
On 3-phase side, I A = 5658/57.16 = 99 A.
This set of Transformers Multiple Choice Questions & Answers focuses on “Tap Changing Transformers-1”.
1. Why tapping is necessary?
a) To provide a neutral point
b) To vary secondary voltage
c) To control real and reactive power flow in the network
d) For neutral point, variation in secondary voltage, controlling power flow
Answer: d
Explanation: Voltage variation in power systems is a normal phenomenon owing to the rapid growth of industries and distribution network. System voltage control is therefore essential for many purposes as listed above.
2. Common range pf tap changing is __________
a) 5%
b) 7%
c) 15%
d) 10%
Answer: a
Explanation: Adjustment is normally carried out by off-circuit tap changing, the common range being 5% in 2.5% steps. Daily and short-time control or adjustment is carried out by means of on-load tap changing gear, which makes this application easier.
3. How tap changing is achieved?
a) Voltage variation with constant flux and constant voltage turn
b) With varying flux
c) Either by voltage variation or by flux variation
d) Can’t be done by variation in voltage or flux
Answer: c
Explanation: Tap changing may be achieved in one of the three conditions, viz.
voltage variation with constant flux and constant voltage turn,
with varying flux,
a mix of and .
4. Negative tapping means ________________
a) less turns than positive tap
b) less turns than principal tap
c) less turns than zero tap
d) more turns than principal tap
Answer: b
Explanation: The principal tapping is one to which the rating of the winding is related. A positive tapping means more turns and a negative tapping implies less turns than those of the principal tap. Tap changing may be achieved by three conditions.
5. Tapping is done at _________________
a) primary only
b) secondary only
c) primary or secondary side
d) on both sides
Answer: c
Explanation: Tapping can be done to either primary or secondary sides. The taps may be placed on the primary or secondary side which depends on construction. Tapping is thus so useful for various applications.
6. Bushing insulators are required when ____________
a) tapping is done near the neutral end
b) tapping is done at the line ends
c) tapping is done
d) they are not required
Answer: b
Explanation: The taps may be placed on the primary or secondary side which partly depends on construction. If tapping is near the line ends, fewer bushings insulators are required. If the tappings are placed near the neutral ends, the phase-to-phase insulation conditions are eased.
7. For achieving large voltage regulation tapping should be done at _____________
a) at the upper ends
b) at the lower ends
c) at the centre
d) anywhere
Answer: c
Explanation: For achieving large voltage variation, tappings should be placed near the centres of the phase windings to reduce magnetic asymmetry. It is generally done on that winding which is placed outside.
8. In case of large voltage variations transformers are tapped at _____________
a) LV
b) HV
c) Both LV and HV
d) Can’t be tapped
Answer: b
Explanation: Centre tapped arrangement cannot be put on LV windings placed next to the core because of accessibility and insulation considerations. The HV winding placed outside the LV winding is easily accessible and can, thus, be tapped easily.
9. Tapping of HV winding is advantageous for _____________
a) step-up transformer
b) step-down transformer
c) both type of transformers
d) never advantageous
Answer: b
Explanation: It is not possible to tap other than an integral number of turns and this may not be feasible with LV side tappings. An example can illustrate this, 250 V phase winding with 15 V/turn cannot be tapped closer than 5%. It is therefore essential to tap the HV windings which is advantageous in a step-down transformer.
10. Tap changing do not causes change in losses of a transformer.
a) True
b) False
Answer: b
Explanation: Axial mmf unbalance is minimized by thinning out the LV winding or by arranging parts of the winding more symmetrically. For very large tapping range, a special tapping coil can be employed. Tap changing causes changes in leakage reactance, core loss, I2R loss and perhaps some problems in parallel operation of dissimilar transformers.
11. Off-circuit tap changer is used when transformer is ________________
a) deenergized
b) energized
c) can be used in either energy conditions
d) never used
Answer: a
Explanation: The cheapest method of changing the turn ratio of a transformer is the use of off-circuit tap changer. As the name off-circuit tap changer indicates, it is required to deenergize the transformer before changing the tap.
This set of Transformers test focuses on “Tap Changing Transformers – 2”.
1. Face plate carrying suitable studs is mounted on _____________
a) lower yoke
b) upper yoke
c) away from tapped positions
d) anywhere
Answer: b
Explanation: The face plate carrying the suitable studs can be mounted at a convenient place on the transformer such as upper yoke or located near the tapped positions on the windings. The movable contact arm A may be rotated by handwheel mounted externally on the tank.
2. Why microswitch is provided to open the circuit breaker?
a) To prevent inadvertent operation
b) To make circuit energized again
c) To avoid unauthorised operation
d) For various other purposes
Answer: a
Explanation: To prevent inadvertent operation, an electromagnetic latching device or microswitch is provided to open the circuit breaker so as to deenergize the transformer as soon as the tap changer handle is moved; well before the contact of the arm with the stud opens.
3. With the introduction of on-load tap changers, operating efficiency ___________
a) improves
b) decreases
c) depends on the application
d) stays constant
Answer: a
Explanation: On-load tap changers are used to change the turn ratio of transformer to regulate system voltage while the transformer is delivering load. With the introduction of on-load tap changer, the operating efficiency of electrical system gets considerably improved.
4. During the operation of on-load tap changer main circuit ______________
a) can be opened
b) always closed
c) depends on application type
d) depends on the construction
Answer: b
Explanation: During the operation of an on-load tap changer the main circuit should not be opened to prevent sparking and no part of the tapped winding should get short-circuited. All forms of on-load tap changing circuits are provided with an impedance, which is introduced to limit short circuit current during the tap changing operation.
5. On-load tap changers are generally _______________
a) resistor type
b) reactor type
c) can be resistor type or reactor type
d) can’t say
Answer: c
Explanation: All forms of on-load tap changing circuits are provided with an impedance, which is introduced to limit short circuit current during the tap changing operation. The impedance can either be a resistor type or centre-tapped reactor can also be used. The on-load tap changers can in general be classified as resistor or reactor type.
6. In modern type, all tap-changers are made of _____________
a) resistor type
b) reactor type
c) can be either resistor or reactor type
d) can’t say
Answer: a
Explanation: The impedance can either be a resistor or centre-tapped reactor. The on-load tap changers can in general be classified as two types- resistor or reactor type. In modern designs the current limiting is almost invariably carried out by a pair of resistors.
7. Why an energy storage device is provided on on-load tap changing transformers?
a) To ensure transition is started
b) To store additional energy
c) To ensure that transition once started is completed
d) To check the extra energy produced due to transition
Answer: c
Explanation: To ensure that the transition once started gets completed, an energy storage is provided which acts even if the auxiliary power supply happens to fail. It thus, serves an important role.
8. Highest tap changers available as on today are ___________
a) 1475 kV impulse
b) 2000 kV impulse
c) 575 kV impulse
d) 955 kV impulse
Answer: a
Explanation: At present tap changers are available for the highest insulation level of 1475 kV impulse and 630 kV power frequency voltage. Efforts are underway to develop tap changers suitable for still higher insulation levels.
9. Which of the following is/are part/s of tap changing circuit?
a) Automatic voltage regulator
b) A time delay relay
c) Compounding elements
d) AVR, time delay relay, compounding elements
Answer: d
Explanation: The main components are an automatic voltage regulator, a time delay relay, and compounding elements. The time delay here, prevents unwanted initiation of a tap change by a small transient voltage fluctuation. It may be set for a delay upto 1 min.
10. Compact tap changers are made of _____________
a) thyristorized tap changers
b) vacuum switches in diverter switch
c) metallic tap changers
d) rubbered tap changers
Answer: b
Explanation: More compact tap changers with high reliability and performance are being made by employing vacuum switches in the diverter switch. Also, now thyristorized tap changers are available widely in market for special applications where a large number of operations are desired.
This set of Transformers Multiple Choice Questions & Answers focuses on “Voltage and Current Transformers”.
1. What will happen if secondary of a current transformer is open-circuited?
a) hot because of heavy iron losses
b) hot because primary will carry heavy current
c) cool as there is no secondary current
d) depends on other parameters
Answer: a
Explanation: If secondary of current transformer is made open-circuited the transformer temperature will rise to higher value because of heavy iron losses taking place in the circuit due to high flux density.
2. The secondary winding of which of the following listed transformers is always kept closed?
a) Step-up transformer
b) Step-down transformer
c) Potential transformer
d) Current transformer
Answer: d
Explanation: Current transformer works on the principle of shorted secondary. It simply means that burden on the system Z b is equal to 0. This transformer produces a current in its secondary which is proportional to the current in its primary.
3. Current transformers are used for _____________________
a) to provide to measure voltages
b) to measure high value of currents
c) to short-circuit the unwanted instruments
d) to measure low value of currents
Answer: b
Explanation: It is the current ratio transformer meant for measuring large currents and provide a step-down current to current measuring instruments like an ammeter. Such instruments present a short-circuit to the CT secondary.
4. The secondary of a current transformer is always kept short-circuited while operating because it _______________________
a) avoids core saturation and high voltage induction
b) is safe to human beings
c) protects the primary circuit
d) to keep temperature within limits
Answer: a
Explanation: Secondary side of current transformer is always kept short circuited in order to avoid core saturation and provide high voltage induction, so that current transformer can be used to measure high values of currents.
5. In CT deep saturation will cause when _________________
a) if circuit is short-circuited
b) if circuit is open-circuited
c) in both OC and SC conditions
d) if operated at very high supply
Answer: b
Explanation: Most important precaution in use of a CT is that in no case should it be open circuited . As the primary current is independent of the secondary current, all of it acts as a magnetizing current when the secondary is opened. This results in deep saturation of the core which cannot be returned to the normal state and so the CT is no longer usable.
6. Current transformers are __________________
a) series connected type of instrument transformers
b) parallel connected type of instrument transformers
c) series-parallel connected type of instrument transformers
d) parallel connected normal transformers
Answer: a
Explanation: Current transformer is a series connected type of instrument transformer. They are designed to present negligible load to the supply which is being measured and also have an accurate current ratio and phase relationship to enable accurate secondary connected metering.
7. Voltage transformers are designed to have _____________
a) high leakage reactance
b) high magnetizing current
c) high magnetizing reactance
d) low magnetizing reactance
Answer: c
Explanation: Voltage transformers often known as potential transformers are designed with minimum errors and to achieve this they are constructed with low leakage reactance, low loss and high magnetizing reactance.
8. Which of the following can be considered as error for Potential transformer?
a) Magnitude errors
b) Phase errors
c) Unit errors
d) Magnitude and phase errors
Answer: d
Explanation: V 2 /V 1 differs from the desired value (N 1 /N 2 ) in magnitude and phase resulting in magnitude and phase errors. Such errors are required to be kept within the limit defined by the precision required. In order to achieve this a PT-potential transformer is designed specially.
This set of Transformers Multiple Choice Questions & Answers focuses on “Audio Frequency Transformer”.
1. Audio frequency transformers are used ________________
a) Input stage of audio frequency electronic amplifier
b) Intermediate stage of audio frequency electronic amplifier
c) Output stage of audio frequency electronic amplifier
d) Anywhere in the amplifier circuitry
Answer: c
Explanation: Audio frequency transformers are used at the output stage of audio frequency electronic amplifier for matching the load to the output impedance of the power amplifier stage. Here the load is fixed but the frequency is variable over a band.
2. Which of the following is the most desirable frequency response for audio frequency transformers?
a) Rising
b) Flat-topped
c) Decreasing
d) Depends on the application
Answer: b
Explanation: The response being the ratio V2/V1. A flat frequency response over the frequency band of interest is most desirable in these cases. The corresponding phase angle is called phase response. A small angle is acceptable.
3. Stray capacitance in the IF region acts as __________
a) closed circuit
b) open circuit
c) first close then open circuit
d) first open then close circuit
Answer: b
Explanation: Shunting effect of transformer windings stray capacitance CS. In the intermediate frequency range the shunt branch acts like an open circuit and series impedance drop is also negligibly small such that V2/V1 remains fixed .
4. Magnetizing susceptance is low in __________________
a) IF region
b) LF region
c) HF region
d) In all regions
Answer: b
Explanation: In the LF region the magnetizing susceptance is low and draws a large current with a consequent large voltage drop in . As a result, V2/V1 drops sharply to zero as Bm = 0.
5. Stray capacitance susceptance has a strong shunting effect in _________________
a) IF region
b) LF region
c) HF region
d) In all regions
Answer: c
Explanation: In the HF region BS = 1/ω CS has a strong shunting effect and V 2 /V 1 drops off, can be shown by the complete frequency response of a transformer on logarithmic frequency scale as decreasing.
6. How low reactance condition is achieved in potential transformer?
a) By using minimum number of turns
b) By varying thickness of wire
c) By interlacing both windings on same limb
d) By other methods
Answer: c
Explanation: Low reactance is achieved by interlacing primary and secondary both on core limb. High magnetizing reactance according to physics, requires minimum iron path and high permeability steel. Low loss requires low-loss steel and very thin laminations.
7. Which of the following is most important way to reduce errors in PT?
a) Low magnetizing current
b) High magnetizing reactance
c) Low loss
d) High burden
Answer: d
Explanation: There are various ways to minimize the errors in potential transformer, like low leakage reactance, low loss and high magnetizing reactance, etc. But the most important thing for low PT errors is to make the burden (Z b ) as high as feasible.
8. Potential transformers are ____________
a) step up transformers
b) step down transformers
c) can be of any type according to the application
d) step up and step down transformer
Answer: b
Explanation: Potential transformer or voltage transformer are used in electrical power system for stepping down the system voltage to a different safe value which can be fed to low ratings meters and relays.
This set of Transformers Multiple Choice Questions & Answers focuses on “Grounding and Welding Transformer”.
1. When grounding transformer is employed?
a) When neutral is available
b) When neutral is not available
c) When transformer of type Y/D is used
d) Always
Answer: b
Explanation: In case the neutral of the power transformer is not available for grounding , a special Y-D transformer is employed only for neutral grounding, such a transformer is called a grounding transformer.
2. Grounding transformer is ______________
a) Step-up transformer
b) Step-down transformer
c) Autotransformer
d) Any transformer can be grounding transformer
Answer: b
Explanation: Grounding transformer is special type of transformer which is employed when neutral is not available in power transformers. It is constructed in Y-D winding format and always a step-down transformer.
3. The secondaries of grounding transformer are generally ___________________
a) Star only
b) Star with neutral taken out
c) Delta
d) Open-delta
Answer: c
Explanation: The star connected primaries are connected to the system and its neutral is grounded. The secondaries are in delta and generally do not supply any load but provide a closed path for harmonic currents to circulate in them.
4. Which harmonic currents flow through the grounding transformer secondary?
a) Triple
b) Fifth
c) Eleventh
d) Seventh
Answer: a
Explanation: The secondaries of grounding transformer are in delta and do not supply any load but provide a closed path for triple n harmonic currents to circulate in them. Under balanced conditions the current in a grounding transformer is its own exciting current.
5. Under fault conditions current flowing through the grounding transformer is _____________
a) very large
b) zero
c) some small non-zero value
d) depends on fault
Answer: a
Explanation: Under balanced conditions the current in a grounding transformer is its own exciting current. Under fault conditions large current may flow in it. Hence a grounding transformer should be of sufficient rating to withstand the effects of LG faults.
6. Welding transformer is _______________________
a) step-up transformer
b) step-down transformer
c) auto transformer
d) one-one transformer
Answer: b
Explanation: Welding transformer is special type of transformer which is basically a step-down transformer. Welding transformer has high reactance both in primary winding and secondary winding.
7. V-I characteristics of welding transformer are _____________________
a) increasing slowly
b) increasing steeply
c) decreasing slowly
d) decreasing steeply
Answer: d
Explanation: Welding transformer has primary and secondary winding are placed in separate limbs or in the same limbs but spaced distance apart. This high reactance causes steeply drooping V-I characteristics.
8. For welding transformer as current increases induced emf ______________________
a) stays constant
b) increases
c) decreases
d) depends on application
Answer: c
Explanation: Welding transformers have steeply drooping characteristics. with increase in current, the leakage flux increase and the induced emf will come down. This is why the increase in primary or secondary current increases the reactance voltage drop across the respective windings.
9. Welding transformers work on principle that weld is actually __________________
a) open circuit
b) short circuit
c) circuit with finite resistance
d) circuit with finite reactance
Answer: b
Explanation: The increase in primary or secondary current of welding transformer increases the reactance voltage drop across the respective windings, which is essential to limit the welding current as the weld is practically a short circuit.
This set of Transformers Quiz focuses on “Transformers as Magnetically Coupled Circuits”.
1. Considering transformer as a magnetically coupled circuit, the voltage at the primary of the transformer is _________________
a) R 1 i 1 + L 1 *di 1 /dt – M*di 2 /dt
b) R 1 i 1 + L 1 *di 1 /dt + M*di 2 /dt
c) R 1 i 1 – L 1 *di 1 /dt + M*di 2 /dt
d) R 1 i 1 – L 1 *di 1 /dt – M*di 2 /dt
Answer: a
Explanation: If transformer is considered as a magnetically coupled circuit, then by fundamental Kirchoff’s law one can find equations for applied voltage, current and inductance of each coil of a transformer.
2. Considering transformer as a magnetically coupled circuit, the voltage at the secondary of the transformer is _________________
a) R 2 i 2 + L 2 *di 2 /dt – M*di 1 /dt
b) R 2 i 2 + L 2 *di 2 /dt + M*di 1 /dt
c) -R 2 i 2 + L 2 *di 2 /dt – M*di 1 /dt
d) -R 2 i 2 – L 2 *di 2 /dt + M*di 1 /dt
Answer: d
Explanation: If transformer is considered as a magnetically coupled circuit, then by fundamental Kirchoff’s voltage law one can find equations for applied voltage, current and inductance of each coil of a transformer, considering the direction of current, which implies negative sign.
3. Core magnetizing current is __________________
a) i 1 -i 2 /a
b) i 1 +i 2 /a
c) i 1 /a-i 2 /a
d) i 1 /a+i 2 /a
Answer: a
Explanation: The current flowing through the primary is divided into two parts, one of which flows through series parameters of the circuit which is equal to i2/a and the remaining current flows through the parallel branch.
4. Which of the following ratio is equal to N 1 /N 2 ?
a) L 1 /L 2
b) L 2 /L 1
c) √ (L 1 /L 2 )
d) √ (L 2 /L 1 )
Answer: c
Explanation: When tight coupling is considered, k=1. Because, leakage is equal to 0. So, φC 1 = φ 1 and φC 2 = φ 2 . Thus, it follows from equations N 1 /N 2 (M/L 1 ) = N 2 /N 1 (M/L 2 ). By solving this equality, we get the answer.
5. A transformer has turn ratio of a = 10. The primary on application of 200 V draws 4 A with secondary open circuited which is found to have a voltage of 1950 V. Then, L1 is equal to ______________
a) 0.19 H
b) 0.159 H
c) 0.9 H
d) 0.259 H
Answer: b
Explanation: Xm = 200/4 = 50 Ω, we also know that Xm = 2πfL1.
Here frequency denoted by f is equal to 50 Hz. While Xm is calculated previously by taking ratio of voltage with currents.
6. 5. A transformer has turn ratio of a = 10. The primary on application of 200 V draws 4 A with secondary open circuited which is found to have a voltage of 1950 V. Then, M is equal to ______________
a) 1.9 H
b) 1.55 H
c) 9 H
d) 2.59 H
Answer: b
Explanation: Xm = 200/4 = 50 Ω, we also know that Xm = 2πfL1. So, by calculations at f= 50 Hz we get, value of L1 equal to 0.159 H. now, 1950 = √2π N2 φmax =√2π ψmax.
ψmax = 1950/ √2π = 8.78 Wb-T.
M= ψmax/ i1= 8.78/ = 1.55 H.
7. What is the coupling factor if self-inductances and mutual inductance are 5.096 H, 0.05098 H and 0.5096 H respectively?
a) 1
b) 1.499
c) 1.2
d) 0.699
Answer: a
Explanation: Coupling factor is calculated by the ratio of mutual inductance with square root of product of respective self-inductances. That is, k = M/ √ = 0.5096 /√ = 0.9999 =1.
8. Total cross-sectional area of earthing conductor depends on _____________
a) Minimum fault current
b) Maximum fault current
c) System output voltage
d) System input voltage
Answer: b
Explanation: The chief points to be borne in mind when installing an earthing equipment are, that it must possess sufficient total cross-sectional area to carry the maximum fault current, and it must have a very low resistance in order to keep down to a safe value the potential gradient in the earth surrounding the plates, etc., under fault conditions.
This set of Basic Transformers Questions and Answers focuses on “Neutral Earthing in Transformers”.
1. Categories for a transformer earthing of a system neutral are ___
a) 2
b) 3
c) 4
d) Many
Answer: b
Explanation: Fortunately for transformer designers, earthing of a system neutral can only fall into one of three categories. These are: Neutral solidly earthed, Neutral earthed via an impedance, Neutral isolated.
2. Which of the following neutral earthing method is disadvantageous?
a) Neutral solidly earthed
b) Neutral earthed via an impedance
c) Neutral isolated
d) Neutrally
Answer: c
Explanation: Due to the problems and disadvantages of the third alternative that is neutral isolated, it is unlikely that it will be encountered in practice so that it is only necessary to be able to design for the first two.
3. Minimum voltage above which electrical system needs to be earthed is _____
a) 50 V
b) 5 V
c) 10 V
d) 20 V
Answer: d
Explanation: According to regulations and acts made in earlier years Every electrical system rated at greater than 50 V shall be connected to earth. Here minimum voltage is defined at the level of 50 V. Various other rules are defined according to the voltage ranges.
4. What is the low voltage range according to the earthing procedures?
a) 5 V – 100 V
b) 0 V – 100 V
c) 50 V – 1000 V
d) 0 V – 1000 V
Answer: c
Explanation: Low voltage is defined as exceeding 50 V but not exceeding 1000 V and is mainly referring to 415 V distribution networks. For low voltage systems the regulations say that ‘no impedance shall be inserted in any connection with earth.
5. Low voltage systems should be solidly earthed.
a) True
b) False
Answer: a
Explanation: Low-voltage systems must be solidly earthed. The system of protective multiple earthing, which can be advantageous on 415 V distribution networks in some situations, is permitted on low-voltage systems subject to certain other conditions but this still requires that the neutral should be solidly earthed ‘at or as near as is reasonably practicable to the source of voltage.
6. Earthed neutral allows rapid operation of protection to ___________
a) Line to line faults
b) Earthed faults
c) Other faults
d) For all faults
Answer: b
Explanation: An earthed neutral allows rapid operation of protection immediately an earth fault occurs on the system. The earthed neutral in conjunction with sensitive earth fault protection results in the faulty section being isolated at an early stage of the fault.
7. Which method will reduce the cost of insulation between earth and cables?
a) Neutral solidly earthed
b) Neutral earthed via an impedance
c) Neutral isolated
d) Neutrally
Answer: a
Explanation: If the neutral is solidly earthed, the voltage of any live conductor cannot exceed the voltage from line to neutral. As under these conditions the neutral point will be at zero potential, it is possible to effect appreciable reductions in the insulation to earth of cables and overhead lines, which produces a corresponding saving in cost.
8. For an un-earthed line voltage of line to earth can be ______________
a) More than line to neutral of earthed system
b) Less than line to neutral of earthed system
c) Equal to line to neutral of earthed system
d) Can’t specify
Answer: a
Explanation: On an unearthed system the voltage to earth of any line conductor may have any value up to the breakdown value of the insulation to earth, even though the normal voltage between lines and from line to neutral is maintained.
9. High voltage line is only disadvantageous in ______________
a) Small overhead lines
b) Small underground lines
c) Long overhead lines
d) Long underground lines
Answer: c
Explanation: The only disadvantage of connecting a high-voltage system to earth is that this introduces the first earth from the outset and it thus increases the susceptibility to earth faults. This can be inconvenient in the case of a long overhead line, particularly in areas of high lightning incidence.
10. Earth contact material resistance should be ____________
a) Less than 2 ohms
b) More than 2 ohms
c) Depends on the circuit or system
d) Very high
Answer: a
Explanation: It is not always appreciated that it is very difficult to obtain resistance values of less than about 2 ohms from a single earth plate, and often it is still more difficult to maintain the value after the earthing system has been installed for some time.
11. If parallel arrangement is done the minimum distance between two earthing conductors should be _______
a) 10 meters
b) 5 meters
c) More than or equal to 10 meters
d) 2 meters
Answer: c
Explanation: Where a parallel arrangement is employed, each plate, rod, etc., all such things should be installed outside the resistance area of any other. Strictly, this requires a separation of the order of 10 meters.
This set of Transformers Multiple Choice Questions & Answers focuses on “Noise in Transformers”.
1. If para magnetic core is used in the place of the ferromagnetic core of the transformer, then what will happen to magnetostriction?
a) be vanished
b) reduce
c) increase
d) not be affected
Answer: a
Explanation: If the core is not ferromagnetic then the transformer will not operate at all. Thus, humming sound which is produced generally, by a transformer of a ferromagnetic core will get vanished.
2. In which of the following category, noise level test of a transformer falls in?
a) Special test
b) Routine test
c) Type test
d) Different test
Answer: a
Explanation: Special Tests of transformer include Dielectric tests, Measurement of zero-sequence impedance of three-phase transformers, Short-circuit test, Measurement of acoustic noise level, Measurement of the harmonics of the no-load current.
3. The noise produced by a transformer is termed as _____________
a) zoom
b) hum
c) ringing
d) buzz
Answer: b
Explanation: The transformer when operated produces noise which is termed in an electric system as hum. Sound pressure and frequency are the objective characteristics measured by a sound level meter, it is possible to obtain a rating proportional to the loudness of a sound from the appropriate meter readings.
4. The hum in a transformer is in direct proportion with ______________
a) load changes
b) oil in the transformer
c) magnetostriction
d) mechanical vibration
Answer: c
Explanation: The basic cause of transformer noise is due to magnetostriction, which is an expansion and contraction of the iron core due to the magnetic effect of alternating current flowing through the transformer coils. This produces an audible sound called hum.
5. A 400 V, 10 KVA transformer at 60 Hz, is operated at the frequency of 40 Hz, then the humming __________
a) increases
b) decreases
c) remains same
d) increases to very high
Answer: a
Explanation: We know that frequency can be related with core flux density in the inverse proportion. Thus, if the frequency is reduced, the core flux density increases, which will also increase the sound produced by transformer.
6. How the A-weighted sound power level is calculated?
a) L WA = L pA – 10 log 10 S/S O
b) L WA = L pA + 10 log 10 S/S O
b) L WA = L pA + 10 log 10 S*S O
b) L WA = L pA – 10 log 10 S*S O
Answer: b
Explanation: The sound power level can be calculated using the sound pressure levels determined above by computing the effective area for the measurement surface according to the relevant method of measurement and relating this to the standard measurement surface, which is one square metre. The A-weighted sound power level is thus: L WA = L pA + 10 log 10 S*S O .
7. Noise emitted by the transformers when drawn against frequency shows the ______________
a) increasing curve
b) straight line increasing
c) straight line decreasing
d) decreasing curve
Answer: d
Explanation: Consideration of magnetostrictive strain in the transformer core reveals that magnetostriction can be expected to produce a longitudinal vibration in the laminations at just this measured frequency. Unfortunately, the magnetostrictive strain is not truly sinusoidal in character, which leads to the introduction of the harmonics.
8. Step-lap is _____________
a) special component used in transformer to control sound
b) special construction technique
c) special supply technique
d) graphical nature shown by transformer sound
Answer: b
Explanation: The other main source of noise from the transformer core is due to alternating attractive and repulsive forces between the laminations caused by flux transfer across the air gaps at the leg to yoke and inter-yoke joints. These forces can be reduced by special building and design techniques of which the best known and most widely used is the step-lap form of construction.
9. Main and basic design of the transformer is tested by ___________
a) routine tests
b) special tests
c) type tests
d) different tests
Answer: c
Explanation: To check that transformer meets customer’s specifications and design expectations, the transformer has to go through different tests in manufacturer premises. Some of these tests are carried out for confirming the basic design of that transformer. Type test of transformer confirms main and basic design criteria of transformer production lot.
10. The most carried method of attenuation of transformer is ___________
a) changing core material
b) increasing oil tank size
c) building walls around transformer
d) adding some another fluid in oil tank
Answer: c
Explanation: The most obvious method of attenuation is by the provision of a suitable barrier between the transformer and the listener. The simplest form of barrier is a screening wall, the effectiveness of which will vary with height and density as well as with the frequency of the noise.
11. Transformers are located ______________
a) anywhere
b) in the direction of down wind
c) in the direction of up wind
d) can be located in any wind direction according to the application
Answer: b
Explanation: Topographical features of the site should be exploited to the full in order to reduce noise. Where possible the transformer should be located in the prevailing down-wind direction from houses. Existing walls and mounds should, if possible, be kept between dwellings and the transformer.
This set of Transformers Multiple Choice Questions & Answers focuses on “Transformer Protection”.
1. Satisfactory parallel operation of transformers needs ______
a) 5 conditions
b) many conditions
c) 10 conditions
d) 7 conditions
Answer: b
Explanation: The satisfactory parallel operation of transformers is dependent upon five principal characteristics; that is, any two or more transformers which it is desired to operate in parallel should possess: The same inherent phase angle difference between primary and secondary Terminals, The same voltage ratio, The same percentage impedance, The same polarity, The same phase sequence.
2. Voltmeter connected across two similar terminals of parallel operated transformers should give _______
a) 0 reading
b) maximum reading
c) sum of individual reading
d) division of individual readings
Answer: a
Explanation: Polarity applied so as to indicate the directional relationship of primary and secondary terminal voltages of a single unit. Any two single-phase transformers will have the same polarity when their instantaneous terminal voltages will be in phase. With this condition a voltmeter connected across similar terminals will indicate zero.
3. Transformers are subjected to transients because _________________
a) open-circuit currents
b) short-circuit currents
c) inrush currents
d) both OC and SC currents
Answer: d
Explanation: The transients to which transformers are mainly subjected are: Impact of high-voltage and high-frequency waves arising from various causes, including switching in, system switching transients with slower wavefronts than the above, switching inrush currents, short-circuit currents.
4. Any transformer needs to be protected from __________
a) transformer faults
b) faults occurring on the transformer connected systems
c) faults within and on system
d) other faults
Answer: c
Explanation: The subject of transformer protection falls naturally under two main headings. These are the protection of the transformer against the subsequent events of effects of faults occurring on any part of the system. Protection of the system against the effects of faults arising in the transformer.
5. How many types of faults can occur in a system?
a) 2
b) 3
c) 5
d) Many
Answer: b
Explanation: Considering first the means to be adopted for protecting the transformer itself against the effects of system faults, three distinct types of disturbances have to be provided for. These are: Short-circuits, high-voltage, high-frequency disturbances including lightning, pure earth faults.
6. Ferroresonanace can be added in __________________
a) faults due to system
b) faults in the transformer
c) manual faults
d) other faults
Answer: a
Explanation: To this list could be added ferroresonance, which can occur under certain conditions in any system containing capacitance and inductance elements such as those associated respectively with cables and transformers.
7. System-short circuits may occur due to ________________
a) Line to line contacts
b) Line to neutral contacts if neutral is not earthed
c) Line to neutral contacts if neutral is earthed
d) LL and LG faults
Answer: d
Explanation: System-short circuits may occur across any two or even three lines, or, if the neutral point is solidly earthed, between any one line and the earthed neutral. Thus, mechanical stress gets created in the transformer circuit.
8. Mechanical stress produced in the circuit ______________
a) directly proportional to the square of the voltage
b) inversely proportional to the square of the voltage
c) directly proportional to the square of the currents
d) inversely proportional to the square of the currents
Answer: c
Explanation: The short-circuit currents produce very high mechanical stresses in the equipment through which they flow, these stresses being proportional to the square of the currents. The magnitude of these short-circuit currents can be limited by increasing the system impedance, usually incorporating this into the supply transformers.
9. High voltage, high frequency surges can occur in the system due to _________________
a) atmospheric disturbances
b) line faults
c) manual faults
d) line to neutral faults
Answer: a
Explanation: High-voltage, high-frequency surges may arise in the system due to lightning, external flashover on overhead lines, switching operations and to the effects of atmospheric disturbances. These surges principally take the form of travelling waves having high amplitudes and steep wavefronts.
10. Surge impedance can be calculated as _________
a) L/C
b) C/L
c) √
d) √
Answer: c
Explanation: The amplitudes of the waves in the overhead line and at the transformer terminals depend upon the respective values of their surge impedance, which is given by the formula: √; where Z=surge impedance in ohms, L=inductance in henrys, C=capacitance in farads, of the circuit concerned.
11. Surge protection can’t be implemented by the addition of rod gaps.
a) True
b) False
Answer: b
Explanation: Surge protection is implemented by the addition of rod gaps or surge arresters adjacent to the transformer to shunt the surges to earth. These attenuate the surge magnitudes in the view of the windings and their resulting insulation stresses to levels which can be withstood by suitably proportioned insulation distribution without causing resonant instability and dangerous oscillations within the windings.
This set of Transformers Multiple Choice Questions & Answers focuses on “Transformer Maintenance”.
1. In how many types transformers can be used with relay protection system?
a) 5
b) 10
c) 2
d) Many
Answer: c
Explanation: The relay is available in two forms: for use with line current transformers with ratios matched to the load current to give zero differential current under healthy conditions; with tapped interposing transformers for use with standard line current transformers of any ratio.
2. Ingress of air into the oil tank can be avoided by using ___________
a) relays
b) plastic coatings
c) metal coatings
d) fuses
Answer: a
Explanation: Relay are used to protect the transformer circuitry in core-bolt insulation failure, short-circuited core laminations, bad electrical contacts, local overheating, loss of oil due to leakage, ingress of air into the oil system.
3. When tripping of the transformer from the main circuit is required?
a) Local overheating
b) Short-circuited core laminations
c) Core-bolt insulation failure
d) Puncture of bushings
Answer: d
Explanations: For all other options removing transformer from the circuit is not necessary as alarms and consequent actions help to remove that fault. While in following four cases transformer must be removed from the main circuitry: short-circuit between phases, winding earth fault, winding short-circuit, puncture of bushings.
4. Transformers used generally don’t belong to type “construct and forget.”
a) True
b) False
Answer: b
Explanation: Transformers generally have no moving parts and there is nothing to wear out, so that there is little to be maintained. Indeed, many small transformers, particularly those installed in distribution networks, once commissioned remain in service for many years with minimal attention. These are examples of the ‘fit and forget’ methodology.
5. Which of the following is the reason relating to the maintenance while operation?
a) To obtain the maximum practicable operating efficiency
b) To obtain optimum life
c) To minimise the risk of premature and unexpected failure
d) Maximum efficiency, life, minimum temperature
Answer: d
Explanation: The objects in maintaining any item of plant are: To obtain the maximum practicable operating efficiency, to obtain optimum life, to minimise the risk of premature and unexpected failure.
6. Stored oil should be checked continuously for _________________
a) impurities in the oil
b) oil levels
c) oil moisture content
d) oil reactions
Answer: c
Explanation: Bulk oil storage tanks will be fitted with a breather usually of the silica gel type. It is important that the desiccant in this should be frequently checked and maintained in a dry state. Oil stored in drums will also breathe via the bungs on the covers.
7. Dissolved gas contents are need to be observed because _____________________
a) dissolved oxygen
b) dissolved gases from outside air
c) dissolved gases from arcs occurring in the transformer
d) all dissolved gases
Answer: c
Explanation: The generation of gas in oil-filled equipment by disruptive discharges and severe overheating results from the chemical reactions which occur as a result of such faults. The resultant effect of the high thermal and disruptive discharge conditions are due to the severity of the fault and the presence of other materials such as solid insulation.
8. How many places of arcing can possible?
a) 4 places
b) 5 places
c) 8 places
d) 2 places
Answer: a
Explanation: Arcing of winding clamping-pressure adjusting screws, arcing of a connection to a winding stress shield, burning of core plates at their edges consistent with severe circulating currents, indication of overheating of core frames and adjacent core frame insulation.
9. What is the advantage of gas-monitor?
a) Time delay
b) Alarms at presence of a particular gas
c) Less efficiency
d) Requires high power
Answer: b
Explanation: The disadvantage of this device is that it simply alarms at the presence of a particular gas. As should be evident from the above, it is not so much the presence of gas, or gases, which are indicative of a fault so much as a sudden change in the status quo.
10. For high reliability of transformer action, how many fundamental checks should be done?
a) 2
b) 3
c) 4
d) Can’t tell
Answer: b
Explanation: User must regularly monitor three fundamental cases for good efficiency: breather systems must be adequately maintained so that water content is kept at the lowest practicable level, the transformer must be adequately cooled at all times, any overloading maintained within permitted limits and action taken on any indications of possible overheating, the transformer must not be subjected to excessive over voltages.
This set of Transformers online test focuses on “Operation under Abnormal Conditions”.
1. Increased transformer insulation stresses are due to _______________
a) third harmonic currents
b) third harmonic voltages
c) fifth harmonic currents
d) fifth harmonic voltages
Answer: b
Explanation: Following are the effects due to third-harmonic voltages: Increased transformer insulation stresses, electrostatic charging of adjacent lines and telephone cables, possible resonance at third-harmonic frequency of transformer windings and line capacitance.
2. Overheating of transformer windings and of load are due to ________
a) third harmonic currents
b) third harmonic voltages
c) fifth harmonic currents
d) fifth harmonic voltages
Answer: a
Explanation: Following are the effects due to third-harmonic currents: Overheating of transformer windings and of load, telephone and discriminative protective gear magnetic disturbances, increased iron loss in transformers.
3. Operation other than rated power will lead to ______________
a) increase in temperature
b) decrease in temperature
c) short circuit
d) open circuit
Answer: a
Explanation: Operation at other than rated load will result in hot-spot temperature rises differing from those corresponding to rated conditions and rated temperature rise is based on a hot-spot temperature of 98°C with a 20°C ambient.
4. Time required for an insulation to reach at end is given by __________________
a) Kirchhoff
b) Joule
c) Ampere
d) Arrhenius
Answer: d
Explanation: Expressed in quantitative terms the time required for insulation to reach its end of life condition is given by the Arrhenius law of chemical reaction rate. Within a limited range of temperatures this can be approximated to the simpler Montsinger relationship.
5. Arrhenius law is given by ___________________
a) L=αβT
b) L=αβ/T
c) L=α+βT
d) L=α+β/T
Answer: d
Explanation: L is defined as the time for the reaction to reach a given stage, but which might in this case be defined as end of life, while α and β are constants. T is absolute temperature. In limited range of temperatures, it can be approximated to Montsinger relationship.
6. Rate of aging is __________________
a) v= M e -pθ
b) v= M e pθ
c) v= M θe -pθ
d) v= M e -θ
Answer: b
Explanation: Montsinger relationship is defined as L= e -pθ , where p is a constant, θ is the temperature in degrees Celsius. Investigators have not always agreed on the criteria for which L is representative of end of life, but for the purposes of this evaluation this is not relevant and of more significance is the rate of ageing. This is the inverse of the lifetime.
7. Relative age rating is ___________________
a) 2 * 6
b) 2 – 6
c) 2 / 6
d) 2
Answer: c
Explanation: If 98°C is then taken as the temperature at which normal ageing rate occurs, then the relative ageing rate at any other temperature θ is given by the expression:
V= ageing rate at θ/ ageing rate at 98°C= 2 / 6
8. Hot-spot temperature is dependent on ____________
a) 4 categories
b) 6 categories
c) 8 categories
d) 10 categories
Answer: a
Explanation: Hot-spot temperature is made up of the following components: Ambient temperature, top oil temperature rise, average gradient, difference between average and maximum gradient of the windings.
9. In 4-wire star connection 3rd harmonic currents may flow through ______________
a) lines
b) phases
c) from fourth wire to neutral
d) all connections
Answer: d
Explanation: With a four-wire star connection, third-harmonic voltages from lines to neutral or earth are suppressed partially or completely according to the impedance of the third-harmonic circuit. With a four-wire star connection, third-harmonic currents may flow through the phases and through the line wires and fourth wire from the neutral.
This set of Transformers Multiple Choice Questions & Answers focuses on “Generator and Power Station Transformers”.
1. Generator transformers are ________________
a) Step-up transformers
b) Step-down transformers
c) Auto-transformers
d) One-one transformers
Answer: a
Explanation: Generator transformers are employed in generating stations to connect the power station to the transmission system. Generator transformers step up the generator output at low voltage to the voltage at which the transmission system operates.
2. In the CCGT, how many step-up transformers will require?
a) 3
b) 4
c) 5
d) 6
Answer: b
Explanation: In the case of combined cycle gas turbine plants this can mean having four step-up transformers, three associated with the gas turbines and one with the steam turbine on a single unit.
3. Which of the following is one of the criteria of selecting particular generator transformer?
a) Low HV voltage
b) Low LV currents
c) High impedance
d) On-load tap-changer
Answer: d
Explanation: An on-load tapchanger is required to allow for variation of the HV system volts and generator power factor. LV volts will generally remain within š5% under the control of the generator automatic voltage regulator .
4. Generator transformers can undergo sudden load-changes.
a) True
b) False
Answer: a
Explanation: Generator transformers may be subjected to sudden load rejection due to operation of the electrical protection on the generator. This can lead to the application of a sudden overvoltage to the terminals connected to the generator.
5. Station transformers are generally used for ____________
a) Providing generator voltage to transmission
b) Providing power to load from transmission
c) Isolating DC
d) To supply power section auxiliary
Answer: d
Explanation: The station transformer generally supplies the power station auxiliary system for starting up the boiler/turbine generator unit or gas turbine/generator and for supplying those loads which are not specifically associated with the generating unit, for example lighting supplies, cranes, workshops and other services.
6. Which of the following does not follow the criteria of station transformer?
a) LV at 11 kV
b) HV at 275-400 kV
c) Low impedance
d) On-load tap-changer required
Answer: c
Explanation: According to the UK standards, Impedance must be such that it can be paralleled with the unit transformer at 11 kV to allow changeover from station to unit supplies and vice versa without loss of continuity and without exceeding the permissible fault level for the unit and station switchgear this usually means that it is about 15%.
7. Operating load factor of station transformer must be _________
a) low
b) high
c) zero
d) infinite
Answer: a
Explanation: Operating load factor is low, i.e. for much of its life the station transformer will run at half-load or less. Load losses can therefore be relatively high, but fixed losses should be as low as possible.
8. For a unit transformer HV voltage must be _____________
a) 400 kV
b) 200 kV
c) 24 kV
d) 100 kV
Answer: c
Explanation: The HV voltage is relatively low, being equal to the generator output voltage, i.e. usually between 11 and 23.5 kV. The LV voltage is usually 11 kV nominal, although on some combined cycle gas turbine stations 6.6 kV is used to supply the unit auxiliaries.
9. What voltage of On-load tap-changer is required for unit transformer?
a) 11 kV
b) 23 kV
c) 400 kV
d) Not required
Answer: d
Explanation: Paralleling of unit and station transformers during changeover of station and unit supplies can result in a large circulating current between station and unit switchboard. This generally adds to the unit transformer load current, and subtracts from that of the station transformer.
10. On-load power factor for generator transformer is __________
a) high
b) low
c) can’t define
d) zero
Answer: a
Explanation: in the case of the generator transformer, operating load factor is high, so that load losses and no-load losses should both be as low as is economically practicable.
This set of Transformers MCQs focuses on “Transformers for HVDC Conversion”.
1. HVDC transformers are used widely _____________
a) for transmission purpose
b) for generation purpose
c) for supplying DC machines
d) for every use of transformer
Answer: a
Explanation: An HVDC converter converts electric power from high voltage alternating current to high-voltage direct current , or vice versa. HVDC is used as an alternative to AC for transmitting electrical energy over long distances or between AC power systems of different frequencies.
2. HVDC converter transformers are generally __________________
a) unidirectional
b) didirectional
c) nondirectional
d) multidirectional
Answer: b
Explanation: Almost all HVDC converters are bi-directional; they can convert either from AC to DC or from DC to AC . A complete HVDC system always includes one converter operating as a rectifier and another one operating as an inverter .
3. Insulation of HVDC transformers should be ___________
a) dry always
b) wet always
c) can be dry or wet
d) no need of insulation
Answer: a
Explanation: Because of the effect of moisture on the resistivity of insulation material, it is necessary to obtain and maintain a high level of dryness in the insulation of HVDC transformers. This is equally important in service as it is in the factory at the time of testing.
4. Tap winding is constructed so as to ____________________
a) maximize voltage regulation
b) minimize the voltage regulation
c) maximize the impedance regulation
d) minimize the impedance regulation
Answer: d
Explanation: The extent of the tap winding and its location such as to minimise impedance variation results in a high voltage being developed across it under impulse conditions, placing demands on the winding insulation design as well as the impulse withstand capability of the tapchanger itself.
5. Electronic converters for HVDC can be divided into _______
a) two types
b) three types
c) many types
d) only one type is possible
Answer: a
Explanation: Electronic converters for HVDC are classified into two main categories. Line-commutated converters- LCC are made with electronic switches that can only be turned on. Voltage-sourced converters- VSC are made with switching devices that can be turned both on and off.
6. Efficient LCC HVDC converters generally use __________
a) thyristor
b) mercury valves
c) diodes
d) mechanical switches
Answer: a
Explanation: Line-commutated converters-LCC use switching devices that are either uncontrolled or that can only be turned on by control action, such as thyristors. Though HVDC converters can be constructed from diodes, such converters can only be used in rectification mode and the lack of controllability of the DC voltage is a disadvantage.
7. HVDC LCCs have ______________
a) 1 degree of freedom
b) 2 degrees of freedom
c) 5 degrees of freedom
d) Many degrees of freedom
Answer: a
Explanation: As thyristors can only be turned on by control action, and rely on the external AC system to affect this turn-off process, the control system only has one degree of freedom – when to turn on the thyristor. This limits the usefulness of HVDC in some circumstances, some applications.
8. Which of the following is the problem with cast resin transformers?
a) Voids
b) Less efficiency
c) Low capacity
d) Unavailability
Answer: a
Expressions: The possibility of voids and of resin cracking is one problem which has been identified. One measure which can help to resist cracking is the incorporation into the resin of some reinforcement, such as, for example, glass fibre.
This set of Transformers Multiple Choice Questions & Answers focuses on “Distribution Transformers”.
1. Distribution transformers are generally designed for maximum efficiency around __________
a) 90% load
b) zero load
c) 25% load
d) 50% load
Answer: d
Explanation: The main difference between power and distribution transformer is distribution transformer is designed for maximum efficiency at 60% to 70% load as they normally don’t operate at full load all the time. Its load depends on distribution demand.
2. In a power or distribution transformer about 10 per cent end turns are heavily insulated _______________
a) to withstand the high voltage, drop due to line surge produced by the shunting capacitance of the end turns
b) to absorb the line surge voltage and save the winding of transformer from damage
c) to reflect the line surge and save the winding of a transformer from damage
d) insufficient information
Answer: a
Explanation: End turns produce shunt capacitance while circuit is made ON. Thus, to provide protection from line surge produced by the shunting capacitance of the end turns and to withstand high voltage they are heavily insulated.
3. Distribution transformers have _______________
a) smaller size than Power transformer
b) larger size than power transformers
c) size can be more or less depending on type
d) more weight than power transformers
Answer: a
Explanation: Distribution transformers are very likely to be made in a different factory from larger transformers. Being smaller and lighter they do not require the same specialised handling and lifting equipment as larger transformers. Impregnation under very high vacuum and vapour-phase drying equipment is not generally required.
4. Joints are used in distribution transformers because _____________
a) smaller in size
b) economical aspects
c) availability
d) less losses
Answer: d
Explanation: Joints form a greater proportion of the total iron circuit in the case of a small distribution transformer core compared to that of a large power transformer and so, measures to reduce losses at the joints will show a greater benefit.
5. How power losses and noise levels are lowered down?
a) By using another core material
b) By using step-lap construction
c) By using Different winding method
d) By using different oil
Answer: b
Explanation: The competitive nature of the industry, gives an incentive to provide low losses and noise levels in order to sustain in the market, both of which are improved by using the step-lap construction.
6. Foil windings are used for distribution transformers _________________________
a) low voltage winding
b) high voltage winding
c) for both windings
d) not used
Answer: d
Explanation: Foil windings are frequently used as low-voltage windings. In this form of construction, the winding turn, of copper or aluminium foil, occupies the full width of the layer. This is wound around a plain mandrel, with intermediate layers of paper insulation, to form the required total number of turns for the winding.
7. Foil winding provides ________________
a) less mechanical short circuit strength
b) low degree of electromagnetic balance
c) high degree of electromagnetic balance
d) can’t tell by just winding type
Answer: c
Explanation: Foil winding arrangement represents a very cost-effective method of manufacturing low-voltage windings and also enables a transformer to be built which has a high degree of electromagnetic balance and hence good mechanical short-circuit strength.
8. Voltage per turn in distribution transformers is ____________
a) very high
b) very low
c) depends on application
d) depends on other components of system
Answer: b
Explanation: Distribution transformers because of the small frame sizes resulting from low kVA ratings, the volts per turn is usually very low so that for a high-voltage winding a considerable number of turns will be required.
9. Which of the following can be used for HV winding in distribution transformers?
a) Coil winding
b) Foil winding
c) Cross-over coil
d) Single dash folding
Answer: c
Explanation: Another alternative for high-voltage windings is the use of ‘crossover’ coil which shows an individual coil. Each section of the winding, or coil, is itself a small multilayer spiral winding having a relatively short axial length.
10. Which of the following protective component is not provided on small distribution transformers?
a) Overfluxing protection
b) Buchholz relay
c) Overcurrent protection
d) Overcurrent and overvolt protection
Answer: b
Explanation: Class C dry-type transformers are those based on glass fibre-reinforced boards, aromatic polyamide paper conductor insulation and similar materials capable of operating at temperatures up to around 220°C. Generally small, lighter transformers are made of this type.
