Important Multiple choice Question Of Transformer 2022

Ques.71. A 10 kVA auto-transformer, turn ratio is 0.4. Find the power transferred inductively

  1. 10 kVA
  2. 6 kVA
  3. 4 kVA
  4. 0 kVA

Answer.3. 4 kVA

Explanation:-

Here turn ratio = 0.4

Inductively transferred power = (1 – K) × input power = 10 (1- 0.4) = 6 kVA

Conductively transferred power = K × input power = 10 × 0.4 = 4 kVA

 

Ques.72. Which of the following is not a type of construction used for a single-phase transformer?

  1. Core type
  2. Shell type
  3. Berry type
  4. Roof type

Answer.4. Roof type

Explanation:-

According to the core construction and the manner in which the primary and secondary are placed around it, the transformers are named as follows:

  1. Core-type transformers
  2. Shell-type transformers
  3. Berry-type transformers

Core-type transformers:- In the core type, one-half of the primary winding and one-half of the secondary winding are placed around each limb. This arrangement reduces the leakage of flux. The flux produced by one winding but not linking with the other winding is called the Leakage Flux. Obviously, the leakage flux can be reduced by winding the coils to have the same axis and to be in very close proximity.

Core type transformer

Shell-type transformers:- In the shell type, both the windings are placed around a central limb, the two outer limbs acting simply as low reluctance paths for the flux. The core type has a longer mean length of the magnetic flux but a shorter mean length for the winding coils and so it is better suited for very high voltage transformers which require a large number of turns in their windings and also, it offers better scope for insulation and its inspection. On the other hand, the shell type offers better mechanical support and protection for the windings. However, the shell type requires specialized fabrication facilities.

Shell type

Berry-type transformers:- In the Berry type construction, the core is placed around the windings. It is essentially a shell-type construction but with the magnetic paths distributed evenly around the windings. This is more difficult to manufacture and hence not as commonly used as the other two.

berry type transformer

 

Ques.73. The primary winding of a 230/6 V, 50 Hz transformer is energized from 110 V, 50 Hz supply. The secondary output voltage will be ______.

  1. 6.0 V
  2. 2.86 V
  3. 3.0 V
  4. 12 V

Answer.2. 2.86 V

Explanation:-

In a transformer, the relation between the number of turns, current, and voltages is given by:

N1/N2 = V1/V2 = I2/I1

N1 and N2 = number of turns in the primary and secondary windings respectively

V1 and I1 = Voltage and current respectively at the primary end

V2 and I2 = Voltage and current respectively at the secondary end

Calculation:

Given that, V1 = 230 V

V2 = 6 V

The ratio of voltage will be the same

V1/V2 = V’1/V’2

230/6 = 110/V’2

V’2  = 2.86 V

 

Ques.74. A transformer works on the principle of:

  1. Faraday’s law of electrolysis
  2. Mutual induction between inductively coupled circuits
  3. Self-induction of coils
  4. Conduction of current between primary and secondary

Answer.2. Mutual induction between inductively coupled circuits

Explanation

A transformer is a static piece of apparatus used for transferring power from one circuit to another without a change in frequency. It can raise or lower the voltage with a corresponding decrease or increase in current. In its simplest form, a transformer consists of two conducting coils having a mutual inductance. The primary is the winding that receives electric power, and the secondary is the one that may deliver it. The coils are wound on a laminated core of magnetic material.

The physical basis of a transformer is mutual inductance between two circuits linked by a common magnetic flux through a path of low reluctance. The two coils possess high mutual inductance.

If one coil is connected to a source of alternating voltage, an alternating flux is set up in the laminated core, most of which is linked up with the other coil in which it produces mutually induced emf (electromotive force) according to Faraday’s laws electromagnetic induction, i.e.

$$e = M\frac{di}{dt}$$

Where

e = induced EMF

M = mutual inductance

 

Ques.75. Which of the following is not fitted on the transformer?

  1. Commutator
  2. Breather
  3. Conservator
  4. Buchholz Relay

Answer.1. Commutator

Explanation:-

The accessories with the transformer tank are thermometer pockets, drain cock, rollers, or wheels for moving the transformer from one place to another, breather, bushings, and Buchholz relay.

The conservator, breather, and Buchholz relay provide protection to the transformer. For cooling, cooling tubes are welded with the tank, but for radiators, separate radiators are individually welded and then bolted onto the transformer tank.

Transformers with a voltage rating of 6 kV and an output rating of 25 kVA and more should be provided with an oil conservator.

Conservator:- The conservator is of the airtight cylindrical metal drum which is supported horizontally on a neighboring wall. When the transformer is working its temperature rises, rise being excessive in case of overloads. Due to these changes in temperature the oil in the tank of the transformer undergoes the process of expansion and contraction.

Conservators are provided over the transformer tanks to absorb this expansion and contraction of oil, without allowing the oil to come in contact with the air. Usually, the conservator capacity should be 10 to 12 percent of the oil volume in the tank.

Breather:- A breather mounted on the transformer tank contains calcium chloride or silica gel, which extracts the moisture from the air. Due to changes in the oil volume, the displacement of air due to these changes of oil volume takes place through the breather, which can extract the moisture from the air. The silica gel extracts the moisture from the air.

Buchholz relay:- Buchholz relay is designed to detect the faults within the transformer. So, it is a very important part of the transformer. Most of the internal fault within the transformer generates the gas. This relay can only be fitted to the transformer having conservators and is installed on the pipe between the transformer and its conservator.

This relay consists of two internal floats which operate mercury switches connected to the external alarm circuit and tripping circuit. Normally, the relay is full of oil. At the time of the fault, small bubbles of gas are generated. When these bubbles pass through the conservator, they are trapped in the relay hence oil level in the relay falls, which rotates the upper float and operates the mercury switch which triggers an alarm.

Note:- The commutator is a part of the D.C machine

 

Ques.76. If the iron core of a transformer is replaced by an air core, then the hysteresis losses in the transformer will

  1. Increase
  2. Decrease
  3. Remains unchanged
  4. Become zero

Answer.4. Become zero

Explanation:-

When the iron core of a transformer is replaced by an air core, then the hysteresis losses in the transformer will become zero.

Iron core transformer:- A transformer having a magnetic core usually is called an iron-core transformer, regardless of the specific composition of the core.

Air core transformer:- A transformer having a non-magnetic core usually is called an air-core transformer, although the core might be (at least in part) paper or plastic.

Transformer action demands only the existence of alternating flux linking the two windings. No doubt such action will be obtained if an air core is used but it will be obtained much more effectively if an iron core is used. It is because the flux is then substantially confined to a definite path (i.e., iron path) having a much higher permeability than air.

Hysteresis is the property exhibited by ferrous materials, particularly those having iron in its composition.

So naturally, air-cored transformers will not have any hysteresis losses. In fact, it is this reason air-cored transformers are used in electronic applications, particularly those involving higher frequencies, where iron losses due to hysteresis are high.

In an air-core transformer, the iron core is absent and the flux is linked with the windings through the air. So that hysteresis losses are eliminated.

In addition to the noise-free operation, an air-core transformer is quite lightweight due to the absence of a heavy-weight iron core. That is why this air-core transformer is most suitable for portable, lightweight electronic devices and high-frequency devices.

Air core transformers are generally used in radio transmitters and communication devices etc.

 

Ques.77. A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 95 V when loaded. The regulation of the transformer is

  1. 4.5%
  2. 9%
  3. 5%
  4. 1%

Answer.3. 5%

Explanation:-

% voltage regulation = (No-load voltage − Full Load voltage)/No voltage voltage

Given that, no-load voltage = 100 V

full load voltage = 95 V

% V.R = (100 − 95) × 100/100

= 5%

 

Ques.78. Voltage regulation of transformer is given by

  1. (V0 – V) / V0
  2. (V0 – V) / V
  3. (V – V0) / V0
  4. (V – V0) / V

Answer.1. (V0 – V) / V0

Explanation:-

A transformer voltage regulation is defined as ‘the change in secondary voltage, from no load to full load, at a specified power factor, expressed as a percentage of its no-load voltage, with the applied primary supply voltage being held constant.

Voltage regulation = (no voltage regulation) − (full load voltage)/(no load voltage)

Voltage regulation = (Vo − V)/Vo

 

Ques.79. A transformer operates most efficiently at 3/4th full load. Its iron loss (Pi) and full load copper loss (Pc) are related as

  1. Pi/Pc = 16/9
  2. Pi/Pc = 3/4
  3. Pi/Pc = 9/16
  4. Pi/Pc = 4/3

Answer.3. Pi/Pc = 9/16

Explanation:-

The maximum efficiency in any machine occurs, when the Copper losses (Pcu) is equal to the iron losses/constant losses (Pcu).

Copper losses (Pcu) = iron losses (Pi) at the maximum efficiency

Then, Total losses = 2 × iron losses (Pi)

Copper losses are proportional to the square load current (i.e. Square of X% full load)

∴ x2 (full load Cu losses) = iron losses

Condition for the maximum efficiency at x of full load.

x2 of full load =  Pi/Pcu

(3/4)2 = Pi/Pcu

9/16 = Pi/Pcu

 

 

Ques.80. In T-T connection, the percentage tapping of main transformer and teaser transformer respectively are

  1. 50%, 5%
  2. 50%, 86.6%
  3. 86.6%, 86.6%
  4. 86.6%, 50%

Answer.2. 50%, 86.6%

Scott (T-T) connection:

The Scott connection is used to convert three-phase power into two-phase power using two single-phase transformers. The Scott connection is very similar to the T connection in that one transformer, called the main transformer, must have a center or 50% tap, and the second or teaser transformer must have an 86.6% tap on the primary side. The difference between the Scott and T connections lies in the connection of the secondary windings.

Scott connection 1

In the Scott connection, the secondary windings of each transformer provide the phases of a two-phase system. The voltages of the secondary windings are 90° out of phase with each other. The Scott connection is generally used to provide two-phase power for the operation of two-phase motors.

Scott-connected transformers are similar to T-connected ones in that the main transformer must have a 50% tap on both the primary and secondary, and the teaser transformer must have an 86.6% tap.

1. The Scott connection is used to produce two-phase power from a three-phase connection.

2. A two-phase system has voltages 90° out of phase with each other

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