Transformer Objective Question with Answers

1. Which of the following is minimized by laminating the core of a transformer?
(A) Hysteresis loss
(B) Eddy current loss
(C) Heat loss
(D) All of the above.

Ans: (B) Eddy current loss

Explanation:
  • In a transformer, the eddy current loss is proportional to the square of the diameter of the core.
  • Larger the diameter, the more the eddy current loss.
  • Hence the transformer core is laminated so that the net effective diameter of the transformer core reduces and thus eddy current loss can be minimized.
  • Eddy current loss. Copper loss. Stray loss. By laminating the core of transformer, we make the eddy current circulating path narrow, means increasing the resistance of eddy current and proportionately reducing the eddy current.

What is eddy current loss?
Eddy current loss is conductive I2R loss produced by circulating currents induced in response to AC flux linkage, flowing against the internal resistance of the core.

What is eddy current?
Eddy currents (also called Foucault's currents) are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday's law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field.


2. The function of the breather in a transformer is
(A) To provide oxygen to the cooling oil
(B) To provide cooling air
(C) To arrest flow of moisture when outside air enters the transformer
(D) To filter the transformer oil.

Ans:(C) To arrest flow of moisture when outside air enters the transformer

Explanation:
The function of the breather in the transformer:
  • Silica gel is used to absorb moisture and prevent entering the oil tank while breathing. It is used in an oil transformer breather.
  • The Colour of fresh silica gel is blue. The moist silica gel became pink in color.
  • A transformer consists of Breather, Conservator and Buchholz relay, etc.
  • The breather is used in the transformer to filter out the moisture from the air. Breather consists of silica gel which absorbs moisture from the air.
  • Conservator tank present at the top of the transformer which allows adequate space for expansion of oil. Therefore, during an overloading condition, the oil moves to the conservator tank.
  • Buchholz relay is used for the protection of transformers from the faults occurring inside the transformer.
  • Whenever the transformer is loaded, the temperature of the transformer insulating oil increases. Consequently, the volume of oil is increased.
  • As the volume of the oil is increased, the air above the oil level in the conservator will come out. At low oil temperature, the volume of the oil is decreased, causes air to enter into a conservator tank.
  • The air consists of moisture in it and this moisture can be mixed up with oil. Hence to filter the air from moisture silica gel breather is used.

3. Iron loss in a transformer occurs in
(A) core
(B) winding
(C) insulating oil
(D) main body.

Ans: (A) core

Explanation:
  • Iron losses are caused by the alternating flux in the core of the transformer as this loss occurs in the core it is also known as core loss.
  • Iron loss is further divided into hysteresis and eddy current loss.

4. Under no load conditions, which of the following loss is negligible
(A) Hysteresis loss
(B) Eddy current loss
(C) Copper loss
(D) All losses have the same magnitude.

Ans: (C) Copper loss

Explanation:
Electrical or Copper losses :
  • The copper losses are the winding losses taking place during the current flowing through the winding. These losses occur due to the resistance in the winding.
  • At no load value of current is negligible, so that copper losses are also negligible.
  • Hence at no load copper loss is negligible.
  • These losses are determined with the help of a short-circuit test.
  • The short circuit test is performed on the high voltage winding of the transformer or electrical machine while the low voltage side is short-circuited i.e. performed at rated current.
  • The equivalent resistance, impedance, and leakage reactance are also calculated by the short circuit test.
Important Points
  • The no-load test gives information regarding no-load losses also known as constant losses such as core loss, friction loss, and windage loss.
  • The no-load test is performed when the rotor rotates with synchronous speed and there is no load torque.
  • Rotor copper loss at no load is very less than its value is negligible.
  • In this test, a small current is required to produce adequate torque.
  • This test is used to evaluate the resistance and impedance of the magnetizing path of induction motor or shunt branch parameters.
  • During the no-load test of a 3 phase electrical machine, the machine draws power for core loss and windage-friction loss.

5. Open circuit test of a transformer gives
(A) hysteresis loss
(B) eddy current loss
(C) the sum of hysteresis and eddy current loss
(D) copper loss.

Ans: (C) the sum of hysteresis and eddy current loss

Explanation:
Open circuit test are used for finding core losses or no load losses and short circuit test are used for finding copper loss in the winding.

Open circuit test:
  • It is done by keeping one of the windings open (without load, usually high voltage winding is open) and applying rated voltage to other winding (usually low voltage winding because it is easier to apply rated voltage).
  • The current drawn from this terminal is the no-load current corresponding to core loss component. Since the no-load current is very small it doesn’t contribute to the copper loss. Core loss is calculated by multiplying the applied voltage and no-load current.

It is used to find
  • The core/iron loss of the transformer
  • The no-load current
  • Equivalent resistance referred to the metering side

Short circuit test:
  • It is done by shorting one of the winding terminals (usually low voltage terminal) and applying a small voltage across the other winding terminals (high voltage terminal because the current in HV terminal will be less and easy to handle) and using a wattmeter to measure the power dissipated in the LV terminal. The wattmeter will indicate the full load copper loss.

A short-circuit test is done to find
  • The full load copper loss
  • Short circuit current

6. Copper loss in a transformer occurs in
(A) core
(B) winding
(C) main body
(D) bushings.

Ans: (B) winding

Explanation:
  • These losses occur due to ohmic resistance of the transformer windings.
  • Hence at no load copper loss is negligible.
  • These losses are determined with the help of the short-circuit test.
  • The short circuit test is performed on the high voltage winding of the transformer or electrical machine while the low voltage side is short-circuited i.e. performed at rated current.
  • The equivalent resistance, impedance, and leakage reactance are also calculated by the short circuit test.

7. Which test on a transformer provides information about regulation, efficiency and Heating under load conditions?
(A) Open circuit test
(B) Back-to-back test
(C) Hopkinson test
(D) Short circuit test.

Ans: (B) Back-to-back test

Explanation:
  • Sumpner's or back to back test on a transfonner provides information about regulation, efficiency and Heating under load conditions
  • By this test we know about temperature rise in the transformer.
  • Back-to-back test on transformer is a method for determining transformer efficiency, voltage regulation and heating under loaded conditions. In this method, two identical transformers are the connected back-to-back such that their primaries are in parallel across the same voltage source and the secondaries in series so that one transformer is loaded on the other.
  • Short circuit and open circuit tests on transformer can give us parameters of equivalent circuit of transformer but they cannot help us in finding the heating information.
  • This transformer's test is also known as regenerative test or Sumpner's test.

Note:
  • Hopkinson’s test is also known as Regenerative test, Back-to-Back test and Heat Run test.
  • Hopkinson's test is used in DC machines.

8. For transformer, the condition for maximum efficiency is
(A) hysteresis loss = eddy current loss
(B) core loss = hysteresis loss
(C) copper loss = iron loss
(D) total loss = 2/3 x copper loss.

Ans: (C) copper loss = iron loss

Explanation:
The transformer will give the maximum efficiency when their copper loss is equal to the iron loss.

Copper loss:
A loss in a transformer that takes place in the winding resistance of a transformer is known as the copper loss.

Iron loss:
Iron losses = Eddy current loss + hysteresis loss

9. Eddy current losses in a transformer core can be reduced by
(A) reducing the thickness of laminations
(B) increasing the thickness of laminations
(C) increasing the airgap in the magnetic circuit
(D) reducing the airgap in the magnetic circuit.

Ans: (A) reducing the thickness of laminations

Explanation:
Eddy current losses:
  • When an alternating magnetic field is applied to a magnetic material, an emf is induced in the material itself according to Faraday’s law of Electromagnetic induction.
  • Since the magnetic material is a conducting material, these EMF’s circulates current within the body of the material. These circulating currents are called Eddy currents. They are produced when the conductor experiences a changing magnetic field.
  • The process of lamination involves dividing the core into thin layers held together by insulating materials.
  • Due to lamination effective cross-section area of each layer reduces and hence the effective resistance increases.
  • As effective resistance increases, the eddy current losses will get decrease.
  • The Eddy current loss is proportional to the square of the frequency.
  • Since the Eddy current loss is proportional to the square of the thickness of the lamination.
  • The eddy current loss in a transformer can be reduced by decreasing the thickness of the laminations.

10. In a transformer, the magnetic coupling between the primary and secondary circuits can be increased by
(A) increasing the number of turns
(B) using soft material for windings
(C) using the magnetic core of low reluctance
(D) using transformer oil of better quality.

Answer: (C) using the magnetic core of low reluctance

Explanation:
Using transformer oil of better quality. Magnetic coupling between two windings depends upon flux linkage between them. Flux linkage can be maximized between two coils by placing magnetic core of low reluctance in between them.

Transformer
  • It is an electrical device that converts the high voltage into low or low voltage into high.
  • A transformer is an electrical device that, by the principles of electromagnetic induction, transfers electrical energy from one electric circuit to another, without changing the frequency.
  • The energy transfer usually takes place with a change of voltage and current.
  • Transformers do not generate electrical power, they transfer electrical power from one AC circuit to another using magnetic coupling.
  • The core of the transformer is used to provide a controlled path for the magnetic flux generated in the transformer by the current flowing through the windings, which are also known as coils.
  • When an input voltage is applied to the primary winding, alternating current starts to flow in the primary winding. As the current flows, a changing magnetic field is set up in the transformer core. As this magnetic field cuts across the secondary winding, an alternating voltage is produced in the secondary winding.
  • The ratio between the number of actual turns of wire in each coil is the key in determining the type of transformer and what the output voltage will be.
  • The ratio between the output voltage and the input voltage is the same as the ratio of the number of turns between the two windings.
  • The output voltage is stepped up, and considered to be a "step-up transformer".
  • If the secondary winding has fewer turns than the primary winding, the output voltage is lower. This is a "step-down transformer".

11. If flux density in the core of a transformer is increased
(A) frequency on secondary winding will change
(B) wave shape on the secondary side will be distorted
(C) size of the transformer can be reduced
(D) eddy current losses will reduce.

Answer: (C) size of the transformer can be reduced

Explanation:
Ï• = B.A

where,
Ï• = flux (wb)
B= flux density of core (wb/m2)
A= area of core (m2)

Notes:
The area of the core is inversely proportional to flux density at constant Ï•.The more flux density means the less area of the core and hence reduces the weight per KVA. If flux density increases, voltage per turn will increase due to which number of turns required is decreased thus overall turns and conductor requirements will be reduced and weight of transformer is also reduced.

12. Which loss in a transformer varies significantly with load?
(A) Hysteresis loss
(B) Eddy's current loss
(C) Copper loss
(D) Core loss.

Answer: (C) Copper loss

Explanation:
  • Copper loss is directly proportional to the load current. so if load current increase, then the copper loss also increases.
  • Copper loss is due to ohmic resistance of the transformer windings.
  • It is clear that Cu loss is proportional to the square of the current, and current depends on the load. Hence copper loss in the transformer varies with the load.

13. Natural air cooling is generally restricted for transformer upto
(A) 1.5 MVA
(B) 5 MVA
(C) 15 MVA
(D) 20 MVA

Ans: (A) 1.5 MVA

Explanation:
  • The natural air cooling method is also known as the self-cooled method.
  • In this method, the heat generated by the transformer is cooled by the circulation of natural air.
  • This method of cooling is effective for smaller output transformer upto 1.5 MVA.
  • Beyond that level, natural air cooling is not very effective and the transformer can be heated upto a dangerous level which can cause serious damage to the transformer.

14. The size of the transformer core mainly depends on
(A) Frequency
(B) Area of core
(C) Flux density of core
(D) Both frequency and area of core

Ans:(D) Both frequency and area of core

Explanation:
From emf equation of transformer

E=4.44fNAB

Where
E= Voltage
f= frequency
A= Area of the core
N= number of turns
B =magnetic flux density

In general, we can say
A=E/(4.44fNB)

For the constant value of E,N,B if we increase F, then Area of the core will decreases means the size of the transformer will reduce.
Higher frequency implies faster MMF variations with time hence higher emf inducted on coils, then for same voltage lower core area is needed or lower number of turns, in any case, lower volume.

15. Voltage remaining constant, if the frequency is increased then
(A) eddy, current losses will decrease
(B) hysteresis losses will decrease
(C) eddy current losses will remain unchanged
(D) Hysteresis losses will remain unchanged.

Answer: (B) hysteresis losses will decrease & (C) eddy current losses will remain unchanged

Explanation:
As V/f ratio is not equal
Wh ∝ V11.6 f0.6 ; as frequency increases, the hysteresis loss will decreases.
We ∝ V12 ; which is independent of frequency. Hence eddy’s current loss will be constant.

16. The power factor in a transformer
(A) is always unity
(B) is always leading
(C) is always lagging
(D) depends on the power factor of the load.

Answer: (D) depends on the power factor of the load.

Explanation:
  • As the load on a transformer increases, the reactance decreases, and the power factor increases. At full load, the power factor approaches 1. Loads with a low power factor draw considerably more current than loads with a power factor near unity.
  • In transformer, routine efficiency is depends on load current and power factor of load.
  • Efficiency curve with respect to load current is double valued function, so that same efficiency is possible at two different load conditions.
  • By keeping load current constant, the maximum efficiency will be obtained exactly at UPF. Efficiency is same for lagging and leading power factor and it is dependent only on magnitude of power factor and independent on type of power factor of load.

17. The transformer oil should have __ Volatility and __ Viscosity.
(A) Low & High
(B) High & High
(C) Low & Low
(D) High & Low

Ans:(C) Low & Low

Explanation:
  • Transformer oil or insulating oil is an oil that is stable at high temperatures and has excellent electrical insulating properties.
  • Volatility is quantified by the tendency of a substance to vaporize. Volatility is directly related to a substance’s vapor pressure. At a given temperature, a substance with higher vapor pressure vaporizes more readily than a substance with lower vapor pressure.
  • Therefore the transformer should have low volatility.
  • Viscosity is a measure of a fluid’s resistance to flow.
  • Low viscosity substance moves quickly.
  • Therefore transformer Oil having low viscosity i.e greater fluidity will cool transformers at a much better rate.

18. During the open circuit test of a transformer
(A) Primary is supplied rated voltage
(B) Primary is supplied current at reduce the voltage
(C) Primary is supplied rated KVA
(D) Primary is supplied full load current

Ans: (A) Primary is supplied rated voltage

Explanation:
  • An open circuit test is performed to determine the iron loss in the transformer.
  • In this method, the secondary of the transformer is left open-circuited.
  • A wattmeter is connected to the primary.
  • An ammeter is connected in series with the primary winding. A voltmeter is optional since the applied voltage is the same as the voltmeter reading.
  • Rated voltage is applied at primary.
  • If the applied voltage is normal voltage then normal flux will be set up. Since iron loss is a function of applied voltage, the normal iron loss will occur. Hence the iron loss is maximum at rated voltage. This maximum iron loss is measured using the wattmeter.

19. At no load the current taken by a transformer
(A) lags behind the applied voltage by 80°
(B) lags behind the applied voltage by 50'
(C) leads the applied voltage by 50'
(D) leads the applied voltage by 80°.

Answer: (A) lags behind the applied voltage by 80°

Explanation:
  • When the transformer is operating at no load, the secondary winding is open-circuited, which means there is no load on the secondary side of the transformer and, therefore, current in the secondary will be zero.
  • While primary winding carries a small current I0 called no-load current which is 2 to 10% of the rated current.
  • the Power factor of a transformer on no load is poor due to the magnetizing reactance of the transformer.
  • So no-load current of a transformer has a small magnitude and low power factor.
  • Ideally, a transformer draws the magnetizing current and lags the primary applied voltage by 90°.
  • But the transformer also has a core loss current component which will be in phase with applied voltage.
  • No-load current is nothing but the vector summation of these two currents.
  • Hence, the no-load current will not lag behind applied voltage by exactly 90° but it lags somewhat less than 90°.
  • It is in practice generally about 75°.

20. The efficiency of a transformer does not depend on
(A) current
(B) load
(C) power factor
(D) all of the above.

Answer: (C) power factor

Explanation:
  • In transformer, routine efficiency is depends on load current and power factor of load.
  • Efficiency curve with respect to load current is double valued function, so that same efficiency is possible at two different load conditions.
  • By keeping load current constant, the maximum efficiency will be obtained exactly at UPF. Efficiency is same for lagging and leading power factor and it is dependent only on magnitude of power factor and independent on type of power factor of load.

21. Which type of winding is used in a 3-phase shell type transformer?
(A) Rectangular Type
(B) Cylindrical Type
(C) Sandwich Type
(D) Circular type

Ans: Sandwich Type

Explanation:
Sandwich Type Winding
  • LV winding is placed close to the core which is at ground potential.
  • HV section lies between two LV sections.
  • Sandwich winding provide control over the short circuit impedance of the transformer.
  • In sandwich coils, leakage can be controlled easily by bringing HV and LV coils close on the same magnetic axis.
  • Reactance can be reduced by increasing the number of sandwich coil hence mutual flux is increased.

22. A transformer has negative voltage regulation when its load power factor is
(A) Lagging
(B) Leading
(C) Unity
(D) Any of the above

Answer: (B) Leading

Explanation:
  • For leading power factor load, the secondary voltage increases slightly with an increase in the load current. Thus for leading power factor loads, the regulation is negative (raise in voltage as load current increases)
  • When load is of the capacitive type, V2 > E2 & hence, regulation becomes negative.

23. If the secondary of a 1: 10 steps up transformer is connected to the primary of a 1 : 5 step-up transformers, the total transformation ratio will be
(A) 15
(B) 30
(C) 50
(D) 2500.

Answer: (C) 50

Explanation:
Transformation Ratio of transformer 1
V2/V1 = T2/T1 = 10

Transformation ratio of transformer 2
V2/V1 = T2/T1 = 5

Total Transformation Ratio
= Transformation Ratio of TR1 X Transformation Ratio of TR2
= 10 X 5
=50

24. A 1600 kVA, 200 Hz transformer is operated at 50 Hz. its kVA rating should be restricted to
(A) 800 kVA
(B) 400 kVA
(C) 200 kVA
(D) 100 kVA.

Answer: (B) 400 kVA

Explanation:
  • If frequency of the transformer is decreased (assuming its applied voltage is same), magnetizing current in the primary of the transformer increases as it is inversely proportional to frequency and directly proportional to the applied voltage.
  • This magnetizing current is used to set up flux in core of the transformer. If magnetizing current goes beyond certain limit, the transformer core may saturate.
  • Therefore, it is required to reduce applied voltage along with frequency in same proportion to keep the magnetizing current same.
V/f ratio must be constant.

Calculation:
Frequency is reduced form 200 Hz to 50 Hz i.e. frequency is reduced by 4 times.
To keep the V/f ratio constant, the supply voltage also must be reduced by 4 times and hence kVA rating will also be reduced by 4 times.
Required rating of the transformer = 400 kVA

25. The main purpose of performing short circuit test in a transformer is to measure its
(A) Copper loss
(B) Core loss
(C) Insulation Resistance
(D) Total loss

Answer:(A) Copper loss

Explanation:
  • Short circuit test is conducted to find the copper loss.
  • It is calculated under the assumption that core loss is neglected.
  • When SC test is conducted on the LV side it would require a larger voltage to get the rated current.
  • Hence core loss cannot be neglected in this case and wattmeter doesn’t give the copper loss alone.
  • Therefore to get accurate results that test is done on the HV side.

26. A short circuit test on a transformer gives
(A) copper losses at full load
(B) copper losses at half load
(C) iron losses at any load
(D) the sum of iron losses and copper losses.

Answer: (A) copper losses at full load

Explanation:
  • Short-circuit test is carried out at rated current to determine the full load copper loss.
  • Since the transformer winding resistance is not affected much by frequency in the power frequency range this test need not be necessarily conducted at the rated frequency.
  • This test is carried out with the instruments placed on the high voltage side while the low voltage side is short circuited by a thick conductor.
  • This is because the rated current on the high voltage side is lower than at low voltage side and therefore economic instruments may be used.

Important points:
  • Short circuit test is used to determine the copper losses and open circuit test is used to determine the core losses.
  • Open circuit test is carried out on the low voltage side while the high voltage side is open circuited.

27. The short circuit test in a transformer is performed on
(A) Low voltage side
(B) High voltage side
(C) Either 1 & 2
(D) Both 1 & 2

Answer: (B) High voltage side

Explanation:
  • Short circuit test in the transformer is conducted on the high voltage side and low voltage side is short circuited.
  • In this test, we need to increase the applied voltage gradually towards rated voltage up to rated load current passing in the secondary short-circuited winding by keeping frequency constant.
  • In HV winding, rated current is less than LV winding. As rated current is less on HV side, it is convenient to conduct this test on HV side by short circuiting the LV terminals.
  • This ensures that the low range of meters can be used for conducting this test as wattmeter is connected to the primary side (High voltage side).
  • Therefore, the primary side is generally chosen as primary side.
  • Short circuit test can be performed on either side but generally performed on HV side.

Important Points:
  • Open circuit test in the transformer is conducted on the low voltage side and high voltage side is open circuited.
  • Open circuit test is used to find shunt branch parameters of equivalent circuit of transformer and the constant losses or no-load losses in transformer.
  • Short circuit test is used to find the full load copper losses or variable losses of the transformer.

28. Leakage fluxes in a transformer may be minimized by
(A) sectionalizing and interleaving the primary and secondary windings
(B) constantly cooling the core
(C) underrating the transformer
(D) reducing the reluctance of the iron core to the minimum.

Answer: (A) sectionalizing and interleaving the primary and secondary windings

Explanation:
  • Transformer insulation is provided between core and LV winding, LV winding, and HV winding.
  • Due to high voltage surges, every piece of electrical equipment must be provided with proper insulation. Otherwise, the equipment may get damaged.
  • Generally, both HV and LV windings are placed on the same limb of the transformer to avoid short-circuits.
Leakage flux in a transformer is minimized by interleaving the primary and secondary windings.

Additional Information
We can minimize the leakage flux in a transformer by following methods
1) By reducing the magnetizing current to the minimum
2) By reducing the reluctance of the iron core to the minimum
3) By reducing the number of primary and secondary trn to the minimum
4) By sectionalizing and interleaving the primary and secondary windings

29. Essential condition for the parallel operation of the transformer is
(A) they must have equal kVA ratings
(B) their voltage ratings must be in proportion to the load shared
(C) they must operate at the same frequency
(D) their ratio of transformation must be in proportion to the load shared.

Answer: (C) they must operate at the same frequency

Explanation:
  • For parallel connection of single phase transformers, primary windings of the transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars.
  • Various conditions that must be fulfilled for the successful parallel operation of single phase transformers:
  • Same voltage ratio & turns ratio (both primary and secondary voltage rating is same).
  • Same percentage impedance and X/R ratio.
  • Identical position of tap changer.
  • Same KVA ratings.
  • Same phase angle shift (vector group are same).
  • Same frequency rating.
  • Same polarity.
  • Same phase sequence.

Some of these conditions are convenient and some are mandatory.
  • The convenient are: Same voltage ratio & turns ratio, same percentage impedance, same KVA rating, same position of tap changer.
  • The mandatory conditions are: same phase angle shift, same polarity, same phase sequence and same frequency.
  • When the convenient conditions are not met paralleled operation is possible but not optimal.
  • For the parallel operation of three phase transformers same conditions are applicable with the additional condition that the transformers should be belong to same vector group.
30. A transformer having 100 turns on the primary side is applied with 200 V ac. In order to get 400 V ac on the secondary side the number of turns on the secondary side must be
(A) 200
(B) 800
(C) 50
(D) 100.

Answer: (A) 200

Explanation:

Ns/Np = Vs/Vp

Here Ns = ? Np = 100 Vs = 400 V Vp = 200 V

Ns/100 = 400/200

Ns =2 X 100

Ns = 200
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