100 Transformer MCQ With Full Explanation [Free PDF]

1. Which of the following is minimized by laminating the core of a transformer?

(A) Hysteresis loss

(B) Eddy current 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

(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

(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.

Eddy current loss in the transformer is given by,

Pe = Kf2B2mt2V

Hysteresis loss occurring in a material is

Ph = ηfB1.6mV

Where,

K - Co-efficient of eddy current. Its value depends upon the nature of magnetic material

Bm - Maximum value of flux density in Wb/m2

t - Thickness of lamination in meters

f - Frequency of reversal of magnetic field in Hz

V - Volume of magnetic material in m3

η is Steinmetz's constant (or) hysteresis constant

4. Under no load conditions, which of the following loss is negligible?

(A) Hysteresis loss

(B) Eddy current loss

(D) All losses have the same magnitude.

Ans: (C) Copper loss

Explanation:

Electrical or Copper losses (I²R 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.

Copper Loss = I2R

At no load value of current is negligible, so that copper losses are also negligible.

The copper loss at any load also given by

Pcu = x2 Pcf

Where, x = Fraction of load, Pcf = Full load copper loss,

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

(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 doesnÂ’t contribute to the copper loss. Core loss is calculated b 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

(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

(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:

HopkinsonÂ’s tes t 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

(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.

Copper loss = i2R

Iron loss:

Iron losses = Eddy current loss + hysteresis loss

Eddy current loss = Ke t2 f2 B2

Hysteresis loss = Kh f B^η

9. Eddy current losses in a transformer core can be reduced by

(A) reducing the thickness of laminations

(B) increasing the thickness of laminations

(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 o Electromagnetic induction.
Since the magnetic material is a conducting material, these EMF’sThese 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.

Mathematically, the eddy current loss is given by:

Eddy current loss in the transformer is given by:

Pe = Ke Bm2. t2. f2. V Watts

Where;

K - coefficient of eddy current. Its value depends upon the nature of magnetic material

Bm - Maximum value of flux density in Wb/m2

t - Thickness of lamination in meters

f - Frequency of reversal of the magnetic field in Hz

V - Volume of magnetic material in m3

From the above formula, we conclude that the Eddy current loss is proportional to the square of the frequency.

Observation:

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

(D) eddy current losses will reduce.

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

Explanation:

ϕ = B.A

where,

ϕ = flux (wb)

B= flux density of core (wb/m²)

A= area of core (m²)

EXPLANATION:

The area of the core is inversely proportional to flux density at constant ϕ.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

(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.
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

(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

(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

(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

W_h ∝ V₁^1.6 f^0.6 ; as frequency increases, the hysteresis loss will decreases.

W_e ∝ 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

(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

(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 substances vapor pressure. At a given temperature, a substances 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 fluids 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

(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

(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

(D) Circular type

Ans: (C) 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

(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

(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

(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

(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 los 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

(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

(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

(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 turn 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

(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:

1. Same voltage ratio & turns ratio (both primary and secondary voltage rating is same).

2. Same percentage impedance and X/R ratio.

3. Identical position of tap changer.

4. Same KVA ratings.

5. Same phase angle shift (vector group are same).

6. Same frequency rating.

7. Same polarity.

8. 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

(D) 100.

Answer: (A) 200

Explanation:

N_s/N_p = V_s/V_p

Here N_s = ? N_p = 100Â V_s = 400 V V_p = 200 V

N_s/100 = 400/200

N_s =2 X 100

N_s= 200

31. Flash point of transformer insulating oil should be more than

(A) 100°

(B) >140°

(D) Below 75°

Answer:(B) >140°

Explanation:

Transformer oil or insulating oil is an oil that is stable at high temperatures and has excellent electrical insulating properties.

As per standard rules, the flash point of transformer oil should be greater than or equal to 140 degrees.

32. In a transformer, the magnetic coupling between the primary and secondary circuit can be increased by

(A) increasing the number of turns

(B) using soft material for windings

(D) using transformer oil of better quality.

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

Explanation:

Magnetic coupling between two windings depends upon flux linkage between them. Flux linkage can be maximized between two coils by placing a magnetic core of low reluctance in between them.

Key Points

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".

33. If flux density in the core of a transformer is increased

(A) frequency on secondary winding will change

(B) wave shape on secondary side will be distorted

(D) eddy current losses will reduce.

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

Explanation:

Flux density and frequency is inversely proportional to the size of the transformer. KVA is directly proportional to the size of the transformer

34. Which loss in a transformer varies significantly with load?

(A) Hysteresis loss

(B) Eddy current loss

(D) Core loss.

Ans: (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.
The copper loss for the primary winding is I12 — R1 and for secondary winding is I22 — R2.
Where I1 and I2 are current in primary and secondary winding respectively.
R1 and R2 are the resistances of primary and secondary winding respectively.
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.

35. Voltage remaining constant, if the frequency is increased then

(A) eddy current losses will decrease

(B) Hysteresis losses will decrease

(D) Hysteresis losses remain unchanged.

Ans: (C) eddy Current losses will remain unchanged

Explanation:

If the frequency is increased, while V remains constant, Bm will decrease; however, ƒ Bm remain constant. Thus, eddy current loss Pe remains unchanged. Hence, as frequency increases.

36. The power factor in a transformer

(A) is always unity

(B) is always leading

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

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

Explanation:

Why is a transformer’s power factor not fixed?

Power factor depends on the types of load used by consumer.

In trnasformer various load can be connected to its secondary terminal. Like motor, light, fan etc..

So power factor will vary when load changes.

That's why power factor doesn't indicated in transformer nameplate's. Ratting is given in KVA Or MVA unit.

Power factor doesn't depends on transformer, it depends on connected load of transformer.

37. At no load the current taken by a transformer

(A) lags behind the applied voltage by 75°

(B) lags behind the applied voltage by 50°

(D) leads the applied voltage by 80°.

Ans: (A) lags behind the applied voltage by 75°

Explanation:

Ideally, a transformer draws the magnetizing current, lags primary applied voltage by 90° But the transformer also has core loss current component which will be in phase with applied voltage. And 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 tha 90°. It is in practice generally about 75°.

38. The efficiency of a transformer does not depend on

(A) current

(B) load

(D) all of the above.

Ans: (C) power factor

Explanation:

Efficiency of the transformer is defined as the ratio of output power to input power at any load.

Power factor doesn't depends on transformer, it depends on connected load of transformer.

What are the factors that affect the efficiency of the transformer?

1. The heating effect of current in a coil

Power is lost as heat I^2R whereby I is the current flowing through the coil and R is the resistance of the coil.
We can increase the efficiency by using thick copper wires of low resistance. Also use of coolant to decrease the temperature of the transformer.

2. Heating effect of induced eddy currents

In the iron core. When the magnetic field in the iron core fluctuates, eddy currents are generated in the iron core.
We can increase the efficiency by using a laminated iron core whereby each layer is insulated with enamel paint to prevent the flow of eddy currents. The high resistance between layers of the iron core decrease the prevalence of eddy currents and heat.

3. Magnetization of the Iron Core.

The energy used in the magnetization and de-magnetization of the iron core each time current changes its direction is known as hysterisis. This energy is lost as heat which subsequently heats up the iron core.
We can increase the efficiency by using a soft iron core that is easily magnetized and de-magnetized.

4. Flux leakage.

Some of the induced magnetic flux from the primary coil is not transmitted to the secondary coil, therefore the e.m.f in the secondary coil is decreased.
The secondary coil(windings) are intertwined tightly with the primary coils. The iron core should form a closed loop.
We can increase the efficiency if the secondary coil (windings) is intertwined tightly with the primary coils. The iron core should form a closed loop.

39. Which of the following transformer will be smallest in size ?

(A) 10 kVA, 50 Hz

(B) 10 kVA, 60 Hz

(D) 10 kVA, 200 Hz.

Ans: (D) 10 kVA, 200 Hz.

Explanation:

The transformers which handle high power have their size more or less proportional to the frequency as the power lost in iron increases with the frequency and therefore the Xmer gets heated up more rapidly. Therefore for its efficient cooling the surface area has to be increased which demands for larger Xmers.

40. A short circuit test on a transformer gives

(A) copper losses at full load

(B) copper losses at half load

(D) sum of iron losses and copper losses.

Ans: (A) copper losses at full load

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.

41. Leakage fluxes in a transformer may be minimized by

(A) sectionalizing and interleaving the primary and secondary windings

(B) constantly cooling the core

(D) reducing the reluctance of the iron corc to the minimum.

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

Explanation:

Flux leakage in a transformer can be minimized by winding the primary and secondary coils one over the other.

Core of the transformer is made of soft iron.

42. Essential condition for the parallel operation of transformer is the

(A) they must have equal kVA ratings

(B) their voltage ratings must be in proportion to the load shared

(D) their ratio of transformation must be in proportion to the load shared.

Ans: (B) their voltage ratings must be in proportion to the load shared

Explanation:

1.Same voltage ratio and turns ratio (both primary and secondary voltage rating is same)

2. Same percentage impedance and X/R ratio.

3.Identical position of tap changer

4.Same KVA ratings

5.Same phase angle shift (vector group are same)

6.Same frequency rating

7.Same polarity

8.Same phase sequence.

43. Which of the following transformers will be largest in size?

(A) 1 kVA, 25 Hz

(B) 1 kVA, 50 Hz

(D) 1KVA, 100 Hz.

Ans:(A) 1 kVA, 25 Hz

Explanation:

44. A step-up transformer increases

(A) power

(B) power factor

(D) frequency

Ans: (C) voltage

Explanation:

Step transformer increase voltage and reduce the current
Step down transformer decrease the voltage and increase the current.
In step up transformer, The number of turns on the secondary of the transformer is greater than that of the primary.
the primary winding of the step-up transformer is made up of thick insulated copper wire because the high magnitude current flows through it and secondary winding made from thin copper wire beacuse low magnitude current flowing through it.
In step Down transformer, The number of turns on the primary of the transformer is greater than the turn on the secondary of the transformer.
The primary winding of the step-down transformer is made up of thin insulated copper wire because the low magnitude current flows through it and secondary winding made from thick copper wire because high magnitude current flowing through it.

45. In a step-down transformer

(A) secondary turns are less than primary turns

(B) Secondary power is less than primary power

(D) secondary current is always more than primary current.

Ans: (A) secondary turns are less than primary turns

Explanation:

In step Down transformer, The number of turns on the primary of the transformer is greater than the turn on the secondary of the transformer.
The primary winding of the step-down transformer is made up of thin insulated copper wire because the low magnitude current flows through it and secondary winding made from thick copper wire because high magnitude current flowing through it.

46. Power transformers are usually designed to have maximum efficiency

(A) near full load

(B) at 75% of full load

(D) between 50% and 75% of full load.

Ans:(A) near full load

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 50-60% of full load.

47. Distribution transformers are usually designed to have maximum efficiency

(A) near full load

(B) near 75% of full load

(D) near no load.

Ans: (C) near 50% of full load

Explanation:

Power transformers are operated on full load most of the time. The loading of distribution transformers depends on the demand and is loaded about 50 to 60% on an average.

Hence for maximum efficiency, Power transformers are designed for maximum efficiency at full load and distribution transformers are designed for maximum efficiency at some point of intermediate loading (Depends on the loads) and normally it is 50 % of full load.

48. Iron losses of a transformer can be approximately calculated if

(A) the gauge of sheet used in core is known

(B) the material of core and yoke is known

(D) the shape of core and yoke is known.

Ans: (C) the weight of core and yoke is known

Explanation:

Iron losses of transformer are measured by conducting open circuit test. During open circuit test transformer works at very low power factor.

The low power factor wattmeter is the instrument that measures lower values of power factor accurately.

Hence low pf wattmeter is used to measure iron losses of transformer.

Note:

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 at low power factor corresponding to core loss component. Since the no-load current is very small it doesn't contribute to the copper loss. Core loss i calculated by multiplying the applied voltage and no-load current.
As the secondary side is open, the entire coil will be purely inductive in nature. So, the power will be lagging due to inductive property of the circuit. So LPF (Low Power Factor) Wattmeter is used in open circuit test of transformer.

It is used to find

The core loss or iron loss of the transformer
The no-load current
Equivalent resistance referred to 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. Wattmeter will indicate the full load copper loss.
In short circuit test the secondary winding of transformer is short circuited. As secondary side is short circuited the entire coil will be purely resistive in nature. So, the power factor will be High or unity.

Short circuit test is done to find

The full load copper loss
Short circuit current

49. In high-frequency transformers,

(A) carbon cores are used

(B) wooden cores are used

(D) aluminum cores are used.

Ans:(C) ferrite cores are used

Explanation:

Ferrite core power transformers are widely used in switched-mode power supplies (SMPSs). The powder core enables high-frequency operation, and hence much smaller size-to-power ratio than laminated-iron transformers.

50. In a transformer it is difficult to measure the efficiency by output-input measurement methods because

(A) the output is sinusoidal and hence cannot be measured

(B) losses are abnormally high

(D) output is out of phase with respect input.

Ans: (C) the efficiency of a transformer is usually high and hence accurate measurement will be necessary.

Explanation:

In a transformer, it is difficult to measure the efficiency by output-input measurement methods because the efficiency of a transformer is usually high and hence accurate measurement will be necessary.
The transformer will give the maximum efficiency when its copper loss is equal to the iron loss.
The maximum efficiency occurs at the unity power factor.
The all-day efficiency of a transformer is related to the distribution transformer.
It is computed on the basis of energy consumed during a period of 24 hours.
The load varies throughout the day.
Distribution Transformers are those transformers that change the voltage level to another voltage level suitable for utilization purposes at the consumer’s premises. A distribution transformer must have it's secondary directly connected to the consumer’s terminals.
The iron loss takes place in the core of the transformer. Thus, the iron or core loss occurs for the whole day in the distribution transformer. The second type of loss known as a copper loss takes place in the windings of the transformer also known as a variable loss
It occurs only when the transformers are in the loaded condition. Hence, the performance of such transformers cannot be judged by commercial or ordinary efficiency, but the efficiency is calculated or judged by A
All Day Efficiency knew as operational efficiency or energy efficiency. It can be defined as the ratio of output to input in kWh for 24 hours.
The maximum efficiency in a transformer that occurs at copper losses is equal to iron losses

51. The transformer laminations are insulated from each other by

(A) Mica strip

(B) Paper

(D) Any of the above

Ans: (C) Thin coating of Varnish

Explanation:

The purpose of providing a coating of varnish in windings are

To keep moisture out of the windings to preserve insulation.
To protect the windings from humming or vibrating when magnetized.
To increase the electrical insulation and dielectric strength.
Provide heat dissipation from coil.

52. Two transformers are connected in parallel. These transformers have different percentage impedance. It is likely to result in

(A) Loading in the transformer is not proportional to their KVA rating.

(B) Short circuit in secondary

(D) Power factor of one of the transformer is leading while that of the other is lagging.

Ans: (A) Loading in the transformer is not proportional to their KVA rating.

Explanation:

If percentage impedance is unequal means one of the transformers has high impedance as compared to other transformers.
In this case, the transformer having high impedance will have a smaller amount of current flowing through the windings.
While Low impedance transformer offers a large amount of current to flow in the windings.
Therefore high percentage information will be lightly loaded while subjected to heavy load.
Whereas low percentage impedance will be overloaded for the same load.

53. Which of the following is the main advantage of autotransformer over a two winding transformer?

(A) Reduces hysteresis losses

(B) Reduce eddy current losses

(D) Saving of copper material

Ans: (D) Saving of copper material

Explanation:

An autotransformer is built with only a single conductor, which serves both as the primary and secondary coil. Thus, it saves the cost of extra wire for the secondary winding.

54. Which winding of the transformer has less cross-section area?

(A) Primary winding

(B) Secondary Winding

(D) Low Voltage Winding

Ans: (C) High Voltage Winding

Explanation:

We know that current density is defined as the ratio of current to the perpendicular cross-section area through which current is crossing.
Value of current density of HV winding is more in comparison to LV winding because of better cooling of HV winding
Since HV winding is placed far from the core in Comparison to LV winding which is placed near to the core.
Current density is inversely proportional to the area of the core so the thin wire is used for HV winding.
Therefore HV winding has Low cross-sectional area.

55. The core used in the high-frequency transformer is usually

(A) Copper Core

(B) Iron Core

(D) Air core

Ans: (D) Air core

Explanation:

An air-core transformer is designed to be used at a high frequency, used in radio circuits.
The currents are usually small but the voltages can vary.
They can be used to change voltages, match one stage to other, for matching antennas to the radio circuit.
They have many uses within a radio.

56. Open circuit test on a transformer yields

(A) core losses

(B) copper losses

(D) leakage of reactance.

Ans:(A) core losses

Explanation:

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 at low power factor corresponding to the core loss component. Since the no-load current is very small it doesn’t contribute to the copper loss. Core loss i calculated by multiplying the applied voltage and no-load current.
As the secondary side is open, the entire coil will be purely inductive in nature. So, the power will be lagging due to the inductive property of the circuit. So LPF (Low Power Factor) Wattmeter is used in the open circuit test of the transformer.

It is used to find

The core loss (iron losses or constant losses) of the transformer
The no-load current
Equivalent resistance referred to metering side
Shunt branch parameters i.e. magnetizing impedance

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. Wattmeter will indicate the full load copper loss.
In the short circuit test, the secondary winding of the transformer is short-circuited. As the secondary side is short-circuited the entire coil will be purely resistive in nature. So, the power factor will be High or unity.

Short circuit test is done to find

The full load copper loss or ohmic loss
Short circuit current

57. The leakage flux in a transformer depends upon

(A) load current

(B) load current and voltage

(D) load current, voltage frequency and power factor.

Ans: (A) load current

Explanation:

In a Transformer, Core flux is the difference of primary flux and Secondary flux which are opposite to each other in direction.
So, some of the primary flux passes through the core and remaining becomes leakage flux (Because Secondary flux forces it to get out of the core).
The same is the case with Secondary flux. Now, flux is directly proportional to Voltage and Current.
When Current increases due to increased load (and voltage remains the same).Then both primary and secondary flux increase. Because both of them increases, so their difference remains the same. And all remaining flux is forced out. Hence leakage flux increases with current, but Core flux remains constant.

58. Which of the following is not a routine test on transformers?

(A) Polarity test

(B) Radio interference test

(D) Impedance test

Ans:(B) Radio interference test

Explanation:

Core insulating voltage test: It is used before installations in such tests.
Impedance voltage test: It is done on a transformer in order to check net impedance offered by a transformer circuit at rated supply.
Polarity test: It is also done before SC and OC test on the transformer. The relative polarities of primary and secondary terminals at any instant can be determined using this test.
Radio interference test is not a routine test on transformers.

59. A transformer will have the highest efficiency near

(A) 25% of rated load

(B) 50% of rated load

(D) 93% of rated load.

Ans:(D) 93% of rated load.

Explanation:

Transformer have higher efficiency near full load.

The efficiency of the transformer is obtained from various experiments conducted on different loads. The efficiency of a power transformer is always greater than 90%. Thus, it can be said highly efficient device.

Power Transformer is maintained at rated conditions to get maximum efficiencies and If the load on a power transformer is low then the load on the power transformer is shared among other power transformers which are in parallel so that they will have high efficiency and the initial transformer is shut down.

Power transformers are operated on full load hence power transformers are designed to have maximum efficiency at full load.

When the load is not connected across the secondary of the transformer, the output power of the transformer is zero. As efficiency is the ratio of output power by input power, the efficiency will be zero at the no-load condition.

Power transformers are operated on full load most of the time. While the loading of distribution transformers depends on the demand and is loaded about 50 to 60% on an average.
Hence for maximum efficiency, Power transformers are designed for maximum efficiency at full load and distribution transformers are designed for maximum efficiency at some point of intermediate loading (Depends on the loads) and normally it is 50 % of full load.

Note:

Transformer efficiency is defined as the ratio of its output power to input power
At light load, hysteresis losses (magnetization and demagnetization of the core of the transformer) and eddy current losses are high compared to output power; Hence, the efficiency is low at this condition.
When the load on the transformer is further increased, its efficiency increases
At heavy loads, copper losses are high compared to the output power; Hence the efficiency is low at this condition
Efficiency is maximum at both copper losses are equal to iron losses
Transformer efficiency is the ratio of its output power to input power and it depends on both load power-factor and load magnitude
At low load, a transformer has low efficiency as hysteresis losses (magnetization and demagnetization of a core of transformer) and eddy current losses dominate over output power

59. A 1 KVA transformer, at full load, will have

(A) copper losses greater than iron losses

(B) iron losses greater than copper loses

(D) windage losses less than copper losses.

Ans: (A) copper losses greater than iron losses

Explanation:

The efficiency of a transformer will be maximum when copper losses are equal to iron losses. Iron losses include both hysteresis and eddy current losses.

Copper-loss is the power loss due to the resistance of the transformer windings. It varies with the load on the transformer.

60. In a transformer if peak voltage is fed to the primary

(A) the iron losses will be less

(B) the iron losses will be more

Ans:(A) the iron losses will be less

Explanation:

When peak voltage is fed to the primary of a transformer, the following effects can be observed:

1. Flux density: The flux density in the core of the transformer will be maximum when peak voltage is applied to the primary. This is because the voltage applied to the primary is proportional to the flux produced in the core. Therefore, the higher the voltage applied, the higher the flux density.

2. Iron loss: Iron loss is the power dissipated in the core due to hysteresis and eddy currents. When peak voltage is applied to the primary, the iron losses will be less. This is because the magnetic field in the core will be established quickly due to the high flux density, and the hysteresis and eddy current losses will be reduced.

3. Copper loss: Copper loss is the power dissipated in the windings due to the resistance of the wire. When peak voltage is applied to the primary, the copper losses will be higher. This is because the current in the windings will be higher due to the higher voltage applied.

4. Windage loss: Windage loss is the power dissipated in the transformer due to the friction and turbulence of the cooling air. When peak voltage is applied to the primary, the windage losses will be more or less constant. This is because the windage losses are not affected by the voltage applied to the primary.

61. The phase difference between the primary and secondary voltage of a transformer is

(A) 0

(B) 90°

(D) Between 30' and 60°

Ans:(C) 180Â°

Explanation:

Transformer Phase Difference

The phase difference between the primary and secondary voltage of a transformer is determined by the transformer's construction and the way it is connected.

A transformer is an electrical device that is used to transfer electrical energy from one circuit to another. It works on the principle of electromagnetic induction, which states that when a magnetic field changes in strength or direction, an electromotive force (EMF) is induced in a nearby conductor.

When an AC voltage is applied to the primary winding of a transformer, it creates a changing magnetic field inside the core. This magnetic field induces a voltage in the secondary winding, which is connected to a load. The voltage induced in the secondary winding is proportional to the turns ratio of the transformer.

The phase difference between the primary and secondary voltage of a transformer is determined by the relative position of the windings and the way they are connected. In general, the phase difference between the primary and secondary voltage is 180 degrees.

When the primary voltage is at its maximum, the secondary voltage is at its minimum, and vice versa. This means that the primary and secondary voltages are out of phase by 180 degrees.

However, in some cases, the phase difference between the primary and secondary voltage can be between 30 and 60 degrees. This can happen if the windings are not perfectly aligned or if there is a phase shift in the load.

Conclusion:

In conclusion, the phase difference between the primary and secondary voltage of a transformer is generally 180 degrees. However, it can be between 30 and 60 degrees in some cases. The phase difference is determined by the transformer's construction and the way it is connected.

62. In a transformer, the copper loss at half load as compared to that at full load will be

(A) 4 times

(B) 2 times

(D) One-fourth.

Ans:(D) One-fourth.

Explanation:

The copper losses of the transformer at X times the full load is proportional to X^2 times the losses at full load.

63. Harmonics in transformer result in

(A) Increase core losses

(B) Increases I2R Losses

(D) All of the above

Ans: (D) All of the above

Explanation:

We know that

Inductive reactance XL = ωL = 2πfL.

So as the frequency changes it also changes XL ( Inductive Reactance).

And I= V/XL

So if XL changes current also gets affected.

Whenever high current flows and if the frequency is higher than normal than the transformer gets Overheated.

Therefore Eddy’s current and Hysteresis Loss will increase in the transformer .

64. Oil is provided in an oil-filled transformer for

(A) Lubrication

(B) Insulation

(D) both cooling and insulation

Ans: (D) both cooling and insulation

Explanation:

There are two main functions of the transformer oil

Coolant: The copper coil in transformer carries a very high current and soon becomes hot. So transformer oil reduces the temperature of the transformer hence prevent burning of coil.

Insulator: Transformer oil has a great dielectric strength so it can withstand with a quite high voltage, so is used as a insulator in the transformer.

65. Iron loss in a transformer can be determined by

(A) Open circuit test

(B) Short Circuit test

(D) None of the above

Ans: (A) Open circuit test

Explanation:

Open circuit test is also called a no-load test.

It is used to determine the no-load losses (core loses in a transformer

Since the no-load current is low therefore the copper loss is neglected in this case.

66. Which of the following is not a part of transformer?

(A) Conservator

(B) breather

(D) Buchholz relay

Ans: (C) Exciter

Explanation:

Exciter is generally used for starting of motor. Since the transformer is a static device and there is no moving part as in the motor.

Parts of the Transformer:

1.) Windings:

Transformers have two windings: the primary winding and the secondary winding.
The primary winding is the coil that draws power from the source.
The secondary winding is the coil that delivers the energy at the transformed or changed voltage to the load.

2.) Core:

The transformer core is used to provide a controlled path for the magnetic flux generated in the transformer.
It is used to transmit power from one source to another through electromagnetic induction.

3.) Conservator tank:

The function of a conservator is to take up the contraction and expansion of oil without allowing it to come in contact with outside air.

4.) Breather:

Silica gel is used in breathers for controlling the level of moisture and prevents it from entering the equipment.
They are mainly useful in protecting the transformer oil from the damaging effects of moisture.

5.) Radiator:

The radiator accelerates the cooling rate of the transformer.
Thus, it plays a vital role in increasing the loading capacity of an electrical transformer.

67. The path of the magnetic flux in transformer should have

(A) low reluctance

(B) low resistance

(D) high resistance.

Ans: (A) low reluctance

Explanation:

1. If there is a low resistance, then there will be a smoother flow of magnetic flux through the core material.

2. Easily magnetized materials have low resistance and high permeability, whereas non-magnetic materials have high reticence and low permeability.

3. The formula of reluctance is r = l/μA. The higher the value of μ, the more will be the flux flows and the lower the flux, the lesser will be the reluctance value.

Magnetic reluctance, or magnetic resistance, is a concept used in the analysis of magnetic circuits. It is analogous to resistance in an electrical circuit, but rather than dissipating electric energy it stores magnetic energy.

Low the reluctance, less the opposition to flux therefore more flux can pass through the transformer core.

68. The desirable properties of transformer core material are

(A) low permeability and low hysteresis loss

(B) high permeability and high hysteresis loss

(D) low permeability and high hysteresis loss.

Ans: (C) high permeability and low hysteresis loss

Explanation:

Desirable properties of the material for transformer core:

1. Low hysteresis loss

2. Low reluctance

3. High permeability

4. Adequate mechanical strength

69. Which of the following test is performed to determine the leakage reactance

(A) Short circuit test

(B) Open circuit test

(D) Test by an Impedance bridge

Ans: (B) Short circuit test

Explanation:

Short circuit or Impedance test

Short circuit test or Impedance test is performed to determine

⇒Copper loss at full load

⇒Equivalent impedance (Zo1 or Zo2)

⇒Leakage reactance (Xo1 or Xo2)

In this test low voltage winding is short-circuited by a thick conductor.

Short circuit test is performed on HV side of the transformer.

Low voltage (5 to 10%) is applied to the primary and slowly increase till full load current are flowing both in primary and secondary.

The ammeter reading gives full load current IL.

Since applied voltage is small so flux (Φ) is also small therefore core loss is small hence core loss can be neglected.

70. During SC test the Power Input to a transformer comprises predominately

(A) Core loss

(B) Copper loss

(D) Eddy current loss

Ans: (B) Copper loss

Explanation:

Low voltage (5 to 10%) is applied to the primary and slowly increase till full load current are flowing both in primary and secondary.

The ammeter reading gives full load current IL.

Since applied voltage is small so flux (Φ) is also small therefore core loss is small hence core loss can be neglected.

71. The maximum load that a power transformer can carry is limited by its

(A) Voltage ratio

(B) Temperature Rise

(D) Dielectric strength of coil

Ans:(A) Voltage ratio

Explanation:

The amount of power a transformer can transfer depends primarily on the magnetic properties and the volume of its core.

This power is frequency-dependent, and it scales essentially with P ∝ V ∝ F.

For a given core size and core material (which also depends on the application), the number of turns then determines the inductance and the capacitance of the primary and secondary coils.

In power applications, the number of turns in a winding is usually chosen in such a way that the magnetic flux density in the core at the rated voltage is of the same order of magnitude as the saturation flux density of the core material.

Decreasing the number of turns would lead to saturation of the transformer core, losses would rapidly increase, and finally, the transformer would stop operating as a transformer altogether.

When increasing the number of turns you will not fully utilize the core anymore (or, alternatively, you can make the core slimmer), but at the same time, you need more conductor material for the winding (remember that the thickness of the wire was determined by the current rating!). So, in effect, the transformer will become bulkier and more expensive.

Smaller turn number means larger core but smaller winding volume, larger turn number means smaller core but larger winding volume.

72. A shell-type transformer has

(A) High eddy current losses

(B) Reduce magnetic leakage

(D) All of the above

Ans: (B) Reduce magnetic leakage

Explanation:

In shell-type transformer both the primary and secondary coils are turned in the same core, i.e in the middle leg.

Shell type transformer has a double magnetic circuit and three limbs.

At first, either the primary or the secondary winding are wound around the middle leg, and then above it the other one is wound. In this case, there is least possible leakage.

When the primary side is excited, it would produce the flux that must cut the secondary coil.

Here at the time of production of flux it simultaneously cuts the secondary coil with the least amount of leakage and produces the necessary output voltage.

73. A transformer can have zero voltage regulation closer to zero

(A) On full load

(B) On overload

(D) On zero power factor

Ans:(C) On leading power factor

Explanation:

Voltage regulation is defined as the ratio of the difference between no-load voltage and full load voltage to no-load voltage. Voltage regulation depends on the nature of the load.

For lagging and unity power factor V2 > E2.

where,

V2 = Secondary terminal voltage on given load.

E2 = Secondary terminal voltage on no load.

As the load increase, it tends to become capacitive and V2 start increasing.

Now at a certain point of leading power factor, V2 = E2, and regulation become zero.

In leading, the power factor condition load is just able to supply the reactive power for the active power flow.

74.Core losses in a transformer

(A) vary from 1% to 3% between no load and full load

(B) vary from 10% to 50% between no load and full load

(D) depend only on the supply voltage.

Ans: (A) varies from 1% to 3% between no load and full load

Explanation:

Losses in the transformer:

1.) Core loss:

It is a loss that happens in the core of a transformer when it is subjected to the change in magnetic flux.
Hence, core loss depends upon supply voltage and frequency.
Core losses remain constant with an increase or decrease in the load.
Core loss is the sum of eddy current loss and the hysteresis loss.
The induced emf produces a current known as eddy current. The losses due to this current are known as eddy current losses.
The losses in a transformer that are due to magnetization saturation in the core are known as hysteresis losses.

2.) Copper loss:

The losses produced due to heat produced by electrical currents in the transformer windings are known as the copper losses.
These losses increase with an increase in loading conditions.

3.) Stray magnetic losses:

The losses occurring in the transformer due to the presence of external magnetic field nearby transformer are known as stray magnetic losses.

75. The function of transformer oil in a transformer is

(A) to provide insulation and cooling

(B) to provide protection against lightning

(D) to lubricate the moving parts.

Ans:(A) to provide insulation and cooling

Explanation:

Function of transformer oil :

1. The construction of the transformer should be such that the heat generated at the core and at the windings should be removed efficiently.

2. Moreover, in order to avoid the insulation deterioration, the moisture should not be allowed to creep into the insulation

3. Both these objectives can be achieved by immersing the built-up transformer in a closed tank filled with noninflammable insulting oil called transformer oil.

4. In order to increase the cooling surface exposed to ambient, tubes or fins are provided on the

outside of tank walls.

76. Buchholz relay is used on

(A) Air-cooled transformers

(B) Oil cooled transformers

(D) Fumace transformers

Ans:(B) Oil cooled transformers

Explanation:

Detailed Solution

Buchholz relay is a protection device which is normally used in large oil absorbed transformers; It is a kind of oil and gas actuated relay.
Whenever a small fault happens within the electrical device, heat is generated by the fault currents.
The generated heat causes decomposition of electrical device oil and gas bubbles are generated.
These gas bubbles run in the upward direction and obtain collected within the Buchholz relay.
The collected gas relocates the oil in Buchholz relay and therefore the displacement is similar to the amount of gas collected.
The dislocation of oil causes the higher float to shut the higher mercury switch to connect an alarm circuit
Hence, once a small fault happens, then the alarm will be activated; The collected quantity of gas specifies the level of fault occurred.
Different types of faults have different types of oil flow and thus a fault can be recognized easily, and proper action can be taken.

77. Buchholz relay is generally not provided on transformers below

(A) 1 kVA

(B) 5 KVA

D) 500 KVA

Ans: (D)500 KVA

Explanation:

Buchholz relay is not provided in relays having rating below 500 kVA from the point of view of economic considerations. Buchholz relay detects incipient faults in a transformer.

78. Operating time of the Buchholz relay is of the order of

(A) 0.1 microsecond

(B) 0.1 milli second

(D) 1 second.

Ans:(C) 0.1 second

Explanation:

The Buchholz relay protects the transformer from internal faults. It is the gas actuated relay. The Buchholz relay is placed between the main tank and the conservator. Such type of relay is used in the transformer having the rating higher than 500KVA. It is not used in small transformer because of economic consideration. The minimum operating time of the relay is 0.1 seconds.

79. In a distribution transformer, normally

(A) core losses are equal to copper losses

(B) core losses are more than copper losses

(D) core losses are half of the copper losses.

Ans:(C) core losses are less than copper losses

Explanation:

Distribution transformers operate mostly at light load or no-load conditions
For this transformer under open circuit conditions core losses are high when compared to copper losses
The primary of distribution transformers is energized for all twenty-four hours and core losses occur throughout the day whereas copper losses will occur only when secondary is supplying load
These are designed to keep core losses minimum and copper losses are relatively less important
A distribution transformer is a transformer that provides the final voltage transformation in the electric power distribution system.
The distribution transformer is a Step-down transformer.
These are used to stepping down the voltage used in the distribution lines to the level used by the end customer.
Distribution transformers have the rating of up to 200 kVA.
Power transformers are used in transmission network of voltages 200 MVA

A transformer has mainly two types of losses, these are, iron losses and copper losses. Iron loss, which is also referred as core loss, consists of hysteresis loss and eddy current loss. These two losses are constant when the transformer is charged. That means the amount of these losses does not depend upon the condition of secondary load of the transformer. In all loading condition, these are fixed. But the copper loss which is also referred as I²R loss entirely depends upon load current I. Distribution transformers have core losses less than full load copper losses.

80. The secondary winding of which of the following transformers is always kept closed?

(A) Current transformer

(B) Voltage transformer

(D) Step down transformer

Answer.(A) Current transformer

Explanation:

If the current transformer secondary is not shorted when unused and kept open then it can develop a very high voltage across secondary which may damage transformer insulation.

81. If the supply frequency of a transformer increases, the secondary output voltage of the transformer

(A) Increase

(B) Decrease

(D) Any of the above

Answer.(C) Remain the same

Explanation:

The transformer is a static device that changes Voltage from one side of its coil to the other at a constant frequency.

Frequency does not change because of the working principle of the transformer-based on Mutual induction which happens without any change in frequency.

82. Power transformers are designed to have maximum efficiency at

(A) Full load

(B) 50% load

(D) no load

Answer.(A) Full load

Explanation:

Power transformers are used for transmission as a step-up device hence they are not directly connected to consumers therefore, load fluctuation is very less. So the power transformer can operate on full load.

83. A Buchholz relay will operate in a transformer whenever there is

(A) large internal fault

(B) saturation of magnetic circuit

(D) any of the above.

Answer: (A) large internal fault

Explanation:

Buchholz relay is used for the protection of transformers from the faults occurring inside the transformer.

Short circuit faults such as inter-turn faults, incipient winding faults and core faults may occur due to the impulse breakdown of the insulating oil or simply the transformer oil.

Buchholz relay is a protection device which is normally used in large oil absorbed transformers; It is a kind of oil and gas activated relay
Whenever a small fault happens within the electrical device, heat is generated by the fault currents
The generated heat causes decomposition of electrical device oil and gas bubbles are generated
These gas bubbles run in the upward direction and obtain collected within the Buchholz relay
The collected gas relocates the oil in Buchholz relay and therefore the displacement is similar to the amount of gas collected
The dislocation of oil causes the higher float to shut the higher mercury switch to connect an alarm circuit
Hence, once a small fault happens, then the alarm will be activated; The collected quantity of gas specifies the level of fault occurred
Different types of faults have different types of oil flow and thus a fault can be recognised easily, and proper action can be taken

84. Which of the following protection is normally not provided on small distribution transformers?

(A) Over current protection

(B) Buchholz relay

(D) All of the above.

Ans:(B) Buchholz relay

Explanation:

Buchholz relay is not provided on small distribution transformers because it is gas actuated protection relay universally used on all oil-immersed transformers having a rating more than 500 kVA. It is not provided in relays having rating below 500 kVA from the point of view of economic considerations.

85. In a transformer

(A) open circuit and short circuit tests are conducted on low voltage side

(B) open circuit and short circuit tests are conducted on high voltage side

(D) open circuit test is conducted on low voltage side and short circuit test on high voltage side.

Ans:(D) open circuit test is conducted on low voltage side and short circuit test on high voltage side.

Explanation:

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).
No-load test (or) open-circuit test in a transform is preferred on the low voltage side because the low voltage is sufficient to obtain rated flux in the core and large on-load current for convenient reading.
The current drawn from this terminal is the no-load current at a low power factor corresponding to the core loss component.
Since the no-load current is very small it doesnÂ’t contribute to the coppe loss. Core loss is calculated by multiplying the applied voltage and no-load current.
As the secondary side is open, the entire coil will be purely inductive in nature. So, the power will be lagging due to the inductive property of the circuit. So LPF (Low Power Factor) Wattmeter is used in the open circuit test of the transformer.

It is used to find

The core loss (iron losses or constant losses) of the transformer
The no-load current
Equivalent resistance referred to the metering side
Shunt branch parameters i.e. magnetizing impedance

Short circuit test:

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 the frequency constant.
In HV winding, the 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 the wattmeter is connected to the primary side (High voltage side).
Therefore, the primary side is generally chosen as the primary side.
Short circuit test can be performed on either side but generally performed on the HV side.

Short circuit test is done to find

The full load copper loss or ohmic loss
Short circuit current

86. For a transformer, operating at constant load current, maximum efficiency will occur at

(A) Zero power factor

(B) Unity power factor unity

(D) 0.8 lagging power factor

Ans:(B) Unity power factor unity

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

87. The function of a breather in a transformer is

(A) to provide protection against overcurrents

(B) to suppress harmonics

(D) to control the level of oil in the tank.

Ans: (C) to arrest the flow of moisture from atmospheric air

Explanation:

Transformer Breather Helps to Prevent Atmospheric Moisture and Reduce Maintenance Costs. During the breathing cycle of a transformer, it is crucial to prevent atmospheric moisture from entering the transformer, which can contaminate the oil

88. A good voltage regulation of a transformer means

(A) difference between primary and secondary voltage is least

(B) difference between primary and secondary voltage is maximum

(D) output voltage fluctuation from no load to full load is least.

Ans:(D) output voltage fluctuation from no load to full load is least.

Explanation:

Voltage regulation is the measure of how well a power transformer can maintain constant secondary voltage given a constant primary voltage and wide variance in load current. The lower the percentage (closer to zero), the more stable the secondary voltage and the better the regulation it will provide.

89. A transformer can have zero voltage regulation at

(A) zero power factor

(B) lagging power factor

(D) unity power factor.

Ans:(C) leading power factor

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.

90. The colour of dry silica gel is

(A) pale yellow

(B) pale pink

(D) blue

Ans:(D) blue

Explanation:

Ready to use silica gel is blue in colour. When the silica gel has soaked up a lot of moisture, the silica gel turns to pink. Once the silica gel turns pink it cannot adsorb any more moisture. It needs to be regenerated.

91. The colour of moist silica gel is

(A) red

(B) Pink

(D) yellow.

Ans: (B) Pink

Explanation:

92. The noise of the transformer mainly due to

(A) Cooling fan

(B) magnetostriction in an iron core

(D) All of the above

Ans: (B) magnetostriction in iron core

Explanation:

Most transformers use laminated iron cores. The lamination may not all be tightly glued together so that one or more may be free to slightly move or vibrate in response to the force of the ac magnetic field. In other cases, the core material, ferrite or lamination, may exhibit a slight magneto strictive effect.

93. In a transformer the primary flux is ___ secondary flux.

(A) Greater than

(B) Smaller than

(D) Equal to

Ans: (D) Equal to

Explanation:

The transformer is said to be a constant main flux device. It is due to the high permeability and greater mutual flux which maintain a constant value.

If you increase the load in the case of a transformer, the secondary will draw more current and hence demagnetizing the core.

This, in turn, draws more current from the primary to maintain the main flux at a constant value.

Similarly, for reducing the load, the secondary current decreases, and hence the primary current also reduces.

94. What would happen if a transformer is connected to a DC supply?

(A) No effect

(B) Operate with high efficiency

(D) Operate with low frequency

Ans: (D) Damage the transformer

Explanation:

A transformer works on the principle of mutual induction, in which you need a varying magnetic field in a winding to induce an EMF in the secondary winding.

In DC generally change in frequency with respect to time is zero.

So a dc source cannot provide varying magnetic field, hence mutual induction is not possible.

Since the primary winding has a low value of winding resistance, the high value of current flowing through it can damage the winding.

95. An autotransformer can be used as

(A) Step up device

(B) Step down device

(D) None of the above

Ans: (C) Both step up and step down

Explanation:

Normal transformers have two winding placed on two different sides i.e. primary and secondary.

In Auto Transformer, one single winding is used as primary winding as well as secondary winding i.e primary and secondary shares the common single winding.

The primary is electrically connected to the secondary, as well as magnetically coupled to it. Auto transformers are often used to step up or step down voltages upto 240 V range.

96. The friction loss in a transformer is

(A) 20%

(B) 0%

(D) more than 50%

Ans: (B) 0%

Explanation:

As we know that transformer is a static device so there is no rotating part hence no friction loss.

97. In an Auto Transformer, The Primary and Secondary are_____Coupled

(A) Electrically only

(B) Magnetically only

(D) None of the above

Ans: (C) Both electrically & magnetically

Explanation:

Normal transformers have two winding placed on two different sides i.e. primary and secondary.

In Auto Transformer, one single winding is used as primary winding as well as secondary winding i.e primary and secondary shares the common single winding.

Therefore the primary is electrically connected to the secondary, as well as magnetically coupled to it.

98. The essential condition for parallel operating of two single phase transformer is that they should have same

(A) efficiency

(B) capacity

(D) polarity.

Ans:(D) polarity.

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:

1. Same voltage ratio & turns ratio (both primary and secondary voltage rating is same).

2. Same percentage impedance and X/R ratio.

3. Identical position of tap changer.

4. Same KVA ratings.

5. Same phase angle shift (vector group are same).

6. Same frequency rating.

7. Same polarity.

8. 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.

99. In a transformer the resistance between the primary and secondary must be

(A) Zero

(B) 1 killo ohm

(D) Infinite

Ans:(D) Infinite

Explanation:

In a transformer, the coils are not electrically connected therefore the resistance is ideally infinite. BUT an auto transformer does the same using a single coil as primary with one or more taps for secondary in different parts of the coil. In this case, the resistance will ideally be ZERO, or a short-circuit if you will.

100. The secondary of a current transformer is always short circuited under operating conditions because

(A) it avoids core saturation and high voltage induction

(B) it avoids high voltage surges

(D) it is same for operation.

Ans: (A) it avoids core saturation and high voltage induction

Explanation:

The secondary side of the current transformer is always kept short-circuited in order to avoid core saturation and high voltage induction so that the current transformer can be used to measure high values of currents.

Current transformer works on the principle of shorted secondary
It means that the burden on the system Z_b is equal to 0
Thus, the current transformer produces a current in its secondary which is proportional to the current in its primary

Important Points:

Most important precaution in use of a CT is that in no case should it be open-circuited (even accidentally)
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
Again, due to large flux in the core, the flux linkage of secondary winding will be large which in turn will produce a large voltage across the secondary terminals of the CT
This large voltage across the secondary terminals will be very dangerous and will lead to insulation failure and there is a good chance that the person who is opening the CT secondary while the primary is energized will get a fatal shock.

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100 Transformer MCQ With Full Explanation [Free PDF]

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