Two Marks Questions and Answers

Two Marks Questions and Answers

Q. 1 Define leakage inductance or winding inductance. AU: Nov.-02, May-16

Ans. : Inductance offered by the winding due to the leakage flux associated with it is called leakage inductance. It is the ratio of leakage flux linkages with the winding to the current passing through the winding.

Q. 2 What material is used in the transformer core?AU: Nov.-02

Ans. High grade silicon steel laminations are used for the construction of the core.

Q. 3 What is meant by leakage flux ? AU: May-03, 04, 11

Ans. : Part of the primary flux as well as the secondary flux completes the path through air and links with the respecting winding only. Such a flux is called leakage flux. Leakage fluxes link with the respective windings only and not to both the windings.

Q. 4 Why transformers rated in kVA instead of kW ?AU: May-03, 08, 09,17, Dec.-06, 11, 15

Ans. : The copper loss (I²R) in the transformer depends on the current 'I' through the winding while the iron or core loss depends on the voltage 'V' as frequency of operation is constant. None of these losses depend on the power factor (cos ϕ) of the load. Hence losses decide the temperature rise and hence the rating of the transformer. As losses depend on V and I only, the rating of the transformer is specified as a product of these two parameters V × L Thus the transformer rating is in kVA and not in kW.

Q. 5 Why is the efficiency of transformers more than that of other rotating machines ? AU: May-03

Ans. : There are no moving parts in transformer hence the friction and mechanical losses are absent in transformer. Hence efficiency of the transformer is more than of other rotating machines.

Q. 6 Define voltage regulation of a transformer. AU: Dec.-03,13,16,18

Ans. : The regulation is defined as change in the magnitude of the secondary terminal voltage, when full load i.e. rated load of specified power factor supplied at rated voltage is reduced to no load, with primary voltage maintained constant expressed as the percentage of the rated terminal voltage.

Let E₂ =  Secondary terminal voltage on no load

V₂ = Secondary terminal voltage on given load

then mathematically voltage regulation at given load can be expressed as,

 % Voltage regulation = E₂-V₂  / V₂ × 100

Q. 7 Draw the typical equivalent circuit of a single phase transformer.

[Refer section 6.11] AU: Dec.-03

Q. 8 How can the eddy current loss be reduced? AU: May-04, Dec.-14

Ans. : The eddy current losses are reduced using the laminated construction. The path of the eddy currents is broken due to the insulating sheets present between the laminations. This prevents the flow of eddy currents through the core and reduces the eddy current losses.

Q. 9 What is an ideal transformer ? AU: May-04, Dec.-04

Ans. : A transformer is said to be ideal if it satisfies following properties:

i) It has no losses.

ii) Its windings have zero resistance.

iii) Leakage flux is zero i.e. 100 % flux produced by primary links with the secondary.

iv) Permeability of core is so high that negligible current is required to establish the flux in it.

Q. 10 State the different losses which Occur in a transformer AU: Dec.-04, May-08, 13

Ans. : The loses taking place in a transformer are: i) Hysteresis losses ii) Eddy current losses iii) Copper losses.

Q. 11 What is the condition for maximum efficiency and retno zero regulation of a transformer ? AU: May-05, Dec.-18

Ans. : The maximum efficiency occurs when

P_i = I₂² R_2e = P_cu

i.e. Copper losses = Iron losses

The leading power factor at which a transformer gives zero voltage regulation is, 

Q. 12 What is the application of equivalent circuit of a single phase transformer ? AU: Dec.-01, 05

Ans. : The equivalent circuit is the electrical model of the transformer. Once the equivalent circuit parameters are obtained then the transformer regulation and efficiency at any power factor and load conditions can be obtained without actually loading the transformer.

Q. 13 Mention the properties of oil used in transformer. AU: Dec.-05

Ans. : The properties of oil used in transformer are,

1. Good conductor of heat.

2. High coefficient of volume expansion.

3. High insulating strength.

4. Free from impurities like alkalies, sulphur etc.

5. Free from moisture content.

Q. 14 What is staggering in the construction of a transformers? Explain with a sketch. [Refer section 6.3.1] AU: Dec.-05

Q. 15 Define regulation up and regulation down for a transformer. AU: Dec.-05

Ans. : The net change in voltage when transformer is loaded is E₂ - V₂ where E₂ is no load voltage while V₂ is load voltage. This change if divided by E₂, it is called down regulation. This change if divided by V₂, it is called up regulation.

% down regulation = E₂-V₂ /  E₂ × 100

% up regulation = E₂-V₂  / V₂ × 100

Q. 16 How are sledging in transformer oil caused?AU June-06

Ans. : While heating if the oil is exposed to oxygen then the reaction takes place due to which large deposits of dark and heavy material are formed in the oil. This is called sledging of oil. Such deposits can clog the cooling ducts. The sledging is possible due to decomposition of oil with long term and continuous use.

Q. 17 Give the expression for the load current when the transformer operates at its maximum efficiency. AU: Dec.-06

Ans. : The load current at nmax in terms of full load current is,

Q. 18 Under what value of power factor a transformer gives zero voltage regulation? AU: May-07

Ans. : The leading power factor at which a transformer gives zero voltage regulation is,

Q. 19 Draw the no-load vector diagram of a transformer.AU : Nov.-08

[Refer section 6.8]

Q. 20 Distinguish between core and shell type transformers. [Refer section 6.4]

AU: Nov.-07

Q. 21 What are the conditions of parallel operation of transformers. [Refer section 6.23] AU: May-10, Dec.-10

Q. 22 Give the principle of transformer. AU: May-10

Ans. : The principle of mutual induction states that when two coils are inductively coupled and if current in one coil is changed uniformly then an e.m.f. gets induced in the other coil. This e.m.f. can drive a current, when a closed path is provided to it. The transformer works on the same principle.

Q. 23 What is transformer?

Ans. : The transformer is a static piece of apparatus by means of which an electrical power is transformed from one alternating current circuit to another with the desired change in voltage and current, without any change in the frequency.

Q. 24 What are the advantages of staggering in transformer?

Ans. :

1. It avoids continuous air gap.

2. The reluctance of magnetic circuit gets reduced.

 3. The continuous air gap reduces the mechanical strength of the core. The staggering helps to increase the mechanical strength of the core.

Q. 25 State the e.m.f. equation of transformer.

Ans. : The e.m.f. equation of transformer is,

E₁ = 4.44 f ϕ_m N₁ volts and

E₂ =4.44 f ϕ_m N₂ volts

Q. 26 How do you reduce leakage flux in a transformer ? AU Dec.-17

Ans. :

1. By interleaving and sectionalizing the windings.

2. By reducing number of primary and secondary turns.

3. By reducing the magnetizing current.

4. By reducing the reluctance of the core.

5. By making transformer window long and narrow.

6. By arranging the primary and secondary windings concentrically.

Q. 27 What are no load losses in a two winding ase transformer and state the reasons for such losses. AU: Dec.-10

Ans. : The primary current in the transformer under no load condition has to supply iron losses which includes hysteresis and eddy current losses and small primary copper losses due to the primary winding resistance. The hysteresis losses are due to the cyclic magnetisation of the core due to alternating current through it while the eddy current loses are due to induced e.m.f. in the core material.

Q. 28 The e.m.f. per turn for a single phase 2200/220 V, 50 Hz transformer is 11 V. Calculate the number of primary and secondary turns. AU: May-11

Ans. :

Q. 29 What are the functions of transformer oil?

Ans. :

i) It is good conductor of heat hence transfers the heat from the core effectively.

ii) Due to high coefficient of volume expansion adequate circulation can be obtained.

iii) To provide additional insulating strength as it is an insulating medium.

Q. 30 What is Steinmetz's constant ? Give its value.

Ans. : The hysteresis loss is proportional to the n power of maximum flux density i.e. (Bm)ⁿ. The constant 'n' is called Steinmetz's constant. Its value is 1.67 for an alternating supply. It varies from 1.06 to 1.88 for the various frequency ranges.

Q. 31 List the advantages of stepped core arrangement in a transformer ?

Ans. :

1. To reduce the space effectively.

2. To obtain reduced length of mean turn of the winding

3. To reduce copper loss.

Q. 32 Why are breathers used in transformers ?

 Ans. : Breathers are used to entrap the atmospheric moisture and thereby not allowing it to pass on to the transformer oil. Also to permit the oil inside the tank to expand and contract as its temperature increases and decreases.

Q. 33 What happens if DC supply is applied to the transformer? AU: May-12, Dec.-14, 15

Ans. : If d.c. supply is given, the current will not change due to constant supply hence mutual induction is not possible and transformer will not work. The resistance of primary winding is very small and inductive reactance is zero for d.c. Hence primary will draw very high current for d.c. supply which may cause damage to the transformer due to extra heat generated. This may cause saturation of the core. Hence d.c. supply is not applied to the transformers.

Q. 34 Differentiate between α core and shell type transformer. [Refer section 6.4.4] AU: May-14

Q. 35 What is Inrush current in a transformer ? AU: May-16

Ans. : When primary voltage is applied to the transformer, the flux goes through transient period before achieving steady state. During transient period the total flux achieves a maximum value of 2 ϕ_m at a particular instant. This is called doubling effect. Due to this, transformer draws very high exciting current as core of transformer goes into deep saturation. This large current drawn during the transients is called current inrush in a transformer.

Q. 36 List out the merits and demerits of core and shell type transformer. (Refer section 6.4.4) AU: Dec.-17

Q. 37 What is the important property of deltamax cores?  AU: Dec.-18

Ans. : The properties of deltamax cores are,

i) It has very square hysteresis loop.

ii) Saturation is of about 1.2 to 1.5 T.

The Fig. 6.29.1 shows B-H loop for deltamax core.

Q. 38 If a transformer has 50 turns in the primary winding and 10 turns in the secondary winding, what is the reflective resistance if the secondary load resistance in 250 Ω ? AU: May-19

Ans.:

Q. 39 A certain transformer has a turns ratio of 1 and a coupling coefficient of 0.85. When 2 V ac is applied to the primary, what is the secondary voltage ? AU: May-19

Ans. :

Q. 40  The full load copper loss in a transformer is 600 W and iron loss is 400 W. What will be the agil copper loss and iron loss at half-load ? AU: Dec.-19

Ans. :

Iron loss remains same at all loads i.e. 400 W.

Q. 41 Define all day efficiency of a transformer. AU May-04, 16, Dec.-05, 06, 12, 19

Ans. : In distribution transformers, the energy output is calculated in kilo watt hours (kWh) and energy spent in supplying the various losses is also determined in kilo watt hours (kWh) over its operation for a day. Then ratio of total energy output to total energy input (output + losses) over a day is calculated. Such ratio is called Energy Efficiency or All Day Efficiency of a transformer. All day efficiency is defined as,

% All day η = Output energy in kWh during a day / Input energy in kWh during a day × 100

= Output energy in kWh during a day / Output energy + Energy spent for total losses × 100

Q. 42 Explain why the wattmeter in O.C. test on transformers reads core loss and that in S.C. test reads copper loss at full load. AU: May-06, 17

Ans. : As secondary is open, I₂ = 0. Thus its reflected current on primary I '2 is also zero. So we have primary current I₁ = I₀. The transformer no load current is always very small, hardly 2 to 4 % of its full load value. As I₂ = 0, secondary copper losses are zero. And I1 = Io is very low hence copper losses on primary are also very very low. Thus the total copper losses in O.C. test are negligibly small. As against this the input voltage is rated at rated frequency hence flux density in the core is at its maximum value. Hence iron losses are at rated voltage. As output power is zero and copper losses are very low, the total input power is used to supply core losses. While in S.C. test the voltage applied is very low hence the core losses are very low while the current flowing through the winding is rated hence the copper losses are at rated current. Hence the entire losses are copper losses in S.C. test which are measured by the wattmeter.

Q. 43 Why is the short-circuit test on a transformer performed on HV (High voltage) side? AU: May-07, 09

Ans. : The high voltage side is a low current side hence it is convenient to connect high voltage side to the supply and short the low voltage side. The voltage on high voltage side can be easily adjusted to such a value that the high voltage side carries the rated current. As this is low current side, this current can be easily achieved without any danger and can be safely measured with the available laboratory ammeters. As the voltage necessary in short circuit test is very low, greater accuracy can be achieved using high voltage side for supplying the voltage.

Q. 44 Why polarity test has to be done in a transformer ? AU: Dec.-07

Ans. : On the primary side of a two winding transformer, one end is positive with respect to other at any instant. At the same instant, one end of secondary winding is positive with respect to other. These relative polarities of primary and secondary windings must be known for operating transformers in parallel or for using transformer in the polyphase circuits. To obtain these polarities, polarity test is done on the transformers.

Q. 45 Which equivalent circuit prameters can be determined from the open circuit test on a transformer? AU: May-11

Ans. : From the open circuit test the exciting branch parameters can be obtained which includes resistance R₀ indicating core loss component and no load inductance X₀ indicating magnetising reactance.

Q. 46 Why all day efficiency is lower than commercial efficiency? AU: May-12

Ans. : All day efficiency is calculated for those transformers which are used to supply power continuously to the consumer for 24 hours a day. The primary of such transformers is energised for 24 hours a day hence core losses take place for 24 hours a day whether the load is connected to it or not. Due to these losses, the all day efficiency is lower than the commercial efficiency.

Q. 47 Calculate the hourly loss of energy in kWh in a specimen of iron, the hysteresis loop of which is equivalent in area to 250J/m³. Frequency 50 Hz; specific gravity of iron 7.5; weight of specimen 10 kg. AU: May-19

Ans. :