Transformer objective Questions Part-1

1. A three phase Transformer , transformer load current is 140A and secondary voltage is 440 V ratting of transformer is ?
A.100 KVA
B. 150KVA
C. 200KVA
D. 250KVA

Answer: A. 100KVA

Explanation:
For Three phase Transformer formula 

Power P = √3 * Voltage * Current 

Here we take 

Voltage V = 440V 

Current I = 140 A 

Here √3 = 1.73

than power P = 1.73 * 440 * 140 

                      = 106568 VA 

                      = 107KVA  is equal to 100 KVA 

Transformer ratting is 100KVA 

Now we Note Transformer related points which is useful for exams 
  • Transformer is static device which has no any rotating parts 
  • Transformer never run on Dc supply If we give Dc supply to transformer Transformer Primary will be burn 
  • Transformer convert only Voltage and current only and frequency remain constant for all load.
  • Transformer having breather and silica gel filled inside the breather which is used for taking air during breathing work.
  • Silica gel having Blue color when not absorb moisture And become pink when fully absorb moisture. 
  • Pressure relief valve used for protection of transformer internal fault.
  • Bucholz relay used for protection of transformer internal fault. this relay give alarm during fault generated if fault is not cleared and become high than transformer tripped.
  • Isolation transformer is used for isolation purpose only and primary to secondary turn ratio is 1:1.
  • Isolation transformer is used in sensitive equipment's for protection for electrical shocks and used in medical equipment's.
  • Transformer step down or step up AC voltage & Current.
  • In an Autotransformer primary and secondary winding are connected magnetically As well electrically.
  • impedance ratio of transformer is equal to square of turn ratio.
2.Secondary winding of current transformer(CT) should be ?
A. Open 
B. Closed 
C. Closed or Open 
D. None Of above 

Ans: B. Closed 

Explanation: 
  • If keep open Secondary winding of current Transformer Heavy voltages Generated In secondary winding and insulation of secondary winding may damage.
  • 66Kv or 11Kv or other voltage classes having very higher voltages and and very higher currents which can not be directly measured.
  • For measurement of this higher current , Current Transformer is required which step down this higher current to measurable limit.
  • Current transformer is step down transformer which is step down the current from higher level to lower level.
  • Current transformer having Primary Winding and Secondary winding.
  • In current transformer primary Winding having only one number of turns and secondary winding having large number of turns.
  • Primary winding of current transformer having very large cross section area secondary of current transformer having very small cross section area.
  • Current transformer used for measurement of current , protection of Transformer and lines etc.
  • Current transformer secondary side current are o.1 A , 1A or 5A which is very lower current and which can easily measured by measuring instruments.
  • Current transformer primary side having very higher current 10A to 3000A or more.
Good To know :
A 300 : 10 , current Transformer is used along with an ammeter . if ammeter reading is 3 A, estimate line current or primary current 

Here Current transformer ratio = 300 / 10 
where formula for CT ratio = I1 / I2 

here ratio = 300 / 10 given and also ammeter current I2= 3 A which is CT secondary given 

I1 / I2 = 300 / 1o
I1 / 2.7 = 300 / 10
I1 /3 =30
I1 = 30 * 3
I1 = 90 A 

so here line current or CT primary current is 90 A 

3. Primary Winding of Potential transformer having 
A. Large Number of Turn 
B. Small Number of Turn 
C. No Turn
D. None of above 

Answer: A. Large Number of Turn 

Explanation:
  • Potential Transformer Having Large Number of turn in primary Side and very less number of turn on secondary side.
  • In case of Current transformer Primary Side having very less number of turns and Secondary side having very large number of turn.
  • Potential transformer also used for measurement of very large voltage . Potential transformer stepping down very large voltage to 110V which can be measure easily by voltmeter.
  • This Transformer design is same as power transformer but accuracy is main consideration for Potential transformer.
  • Potential transformer having larger core and conductor size compared to power transformer.
  • There are two types of potential transformer are used 
  •  A Shell type Transformer used for Low voltage measurement and core type used for high voltage measurement. 
  • The oil immersed potential transformer used for the voltage level above 7KV.
  • This transformer Treated at parallel transformer under open circuit secondary .
  • In case of Current Transformer in can be treated as series transformer under short circuit conditions. 
  • In case of Current transformer secondary should be always short circuited and in case of potential transformer secondary is nearly under open circuited conditions.
  • In case of current transformer very small voltage exist across its terminals because its connected in series.
  • In case of Voltage transformer full line voltage appear across its terminal because its connected in parallel.
  • In case of current transformer winding carries full line current and in case of potential transformer winding is impressed with full line voltage. 
  • In case of current transformer Primary current is independent of the secondary circuit conditions and in case of potential transformer primary current depends on the secondary circuit conditions.
Good To Know 
A 66000V : 110v potential Transformer is used along with voltmeter reading 90 V , estimate the value of the line voltage or Primary Voltage.

Here is given that 
Potential Transformer ratio = 66000 / 110 v 
so V1 / V2 = 66000 /110 V 
here secondary voltage V2= 90V 

so 

V1 / V2 = 66000 /110 
V1 /90 = 600
V1 = 600 * 90 
V1 = 54000 V 
V1 = 54KV 

Here primary  voltage is 54 Kv 


4. Sumpner's test is conducted on transformers to determine
A. temperature
B. stray losses
C. all-day efficiency
D. none of the above

Answer: 

Explanation:
  • Sumpner’s test is conducted on transformers to determine temperature effects.
  • Sumpner‟s test also know as back to back test is conducted by connecting two transformers in parallel across a single rated voltage supply. 
  • In Sumpner’s test the two secondary windings are connected in series with opposing polarity.
  • Generally Sumpner’s test is employed for large transformers because for large sized transformers the ordinary full-load tests are way too expensive from the viewpoint of electrical energy consumption alongside difficulties in finding a suitable load. 
  • The thermal tests require operation under full load for many hours. This difficulty is resolved using Sumpner’s test.
  • It is a non-loading heat run test on transformers.
  • The maximum temperature rise in a large transformer is determined by the full load test by measuring the temperature of their oil after every particular interval at a time.
  • The full load test on a small transformer is very convenient, but on the large transformer, it is very difficult.
  • This test is called, back-to-back test, regenerative test. Two identical transformers are used for the back to back test.
  • While performing back to back test, the amount of power consumed is equal to iron and copper losses of two transformer at full load.

5. A transformer cannot raise or lower the  voltage of a D.C. supply because
A. there is no need to change the D.C. voltage
B. a D.C. circuit has more losses 
C. Faraday's laws of electromagnetic induction are not valid since the rate of change of flux is zero
D. none of the above

Answer: C. Faraday's laws of electromagnetic induction are not valid since the rate of change of flux is zero

Explanation:
  • The transformer works on the principle of mutual induction, for which current in one coil must change uniformly
  • If dc supply is given, the current will not change due to constant supply and transformer will not work
  • Faraday's laws of electromagnetic induction are not valid since the rate of change of flux is zero
  • Practically winding resistance is very small
  • For dc, the inductive reactance is zero as dc has no frequency
  • So total impedance of winding is very low for dc
  • Thus, winding will draw very high current if dc supply is given to it
  • This may cause the burning of windings due to extra heat generated and may cause permanent damage to the transformer

6. Primary winding of a transformer
A. is always a low voltage winding 
B. is always a high voltage winding
C. could either be a low voltage or high voltage winding
D. none of the above

Answer: C. could either be a low voltage or high voltage winding

Explanation:
  • A transformer is a static device that transfers power from one circuit to another circuit without a change in the frequency.
  • Generally, the winding which is connected to the supply is called as primary winding and the winding on which load connected is called as secondary winding.
  • In the transformer, two windings are electrically isolated but magnetically coupled. Hence the transformer is a coupled circuit.
  • The transformer is a singly excited device since it requires only one external voltage source to energize any number of windings placed on its core.
  • The transformer can be treated as a phase-shifting device since it offers a displacement of approx. 1800 between two circuits.
  • As the amount of flux in the transformer core is constant irrespective of power transfer, it can be treated as a "Constant flux device".
  • The transformer is a negative feedback circuit because it satisfies 'Lenz's law'.
  • Transformer: An electrical device that is used to transfer electrical energy from one electrical circuit to another is called a transformer.
  • In a transformer, there are two coils- The primary coil (P) and secondary coil (S).
  • A Transformer is used to convert low voltage (or high current) to high voltage (or low current) and high voltage to low voltage.

Important Points:
Step-up transformer: 
  • The transformer which increases the potential is called a step-up transformer.
  • The number of turns in the secondary coil is more than that in the primary coil. The primary winding will be low voltage winding. 
Step-down transformer: 
  • The transformer which decreases the potential is called a step-down transformer.  
  • The number of turns in the secondary coil is less than that in the primary coil. The primary winding will be high voltage winding.

7. The efficiency of a transformer will be maximum when
A. copper losses = hysteresis losses
B. hysteresis losses = eddy current losses
C. current losses = copper losses
D. copper losses = iron losses

Answer: D. copper losses = iron losses

Explanation:
  • The transformer will give the maximum efficiency when their copper loss is equal to the iron loss.
  • The efficiency of a transformer will be maximum when the constant losses i.e. iron losses and variable losses i.e. copper losses are equal. Formula to calculate transformer efficiency is (input - losses) / input = 1 - (losses / input).

8. No-load current in a transformer
A. to lags behind the voltage by about 75°
B. leads the voltage by about 75°
C. lags behind the voltage by about 15°
D. leads the voltage by about 15°

Answer: A. to lags behind the voltage by about 75°

Explanation:
  • In the case of no-load, the secondary terminal of the transformer is open.
  • There is no path available for the current to flow on the secondary side.
  • Hence, the transformer does not draw current from the source.
  • A small magnitude of current flows through the primary transformer (no-load current I0) called excitation current (used for excitation of the core).​
  • No Load Power Factor, cos ϕ0 = (Iμ / I0) ≈ 0.2 to 0.25
  • Hence, 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, lags primary applied voltage by 90°. 
  • But the transformer also has 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°.

9. The purpose of providing an iron core in a transformer is to
A. provide support to windings
B. reduce hysteresis loss
C. decrease the reluctance of the magnetic path
D. reduce eddy current losses

Answer: 

Explanation: 
  • In transformers, the two coils are wound onto the same iron core. 
  • The purpose of the iron core is to channel the magnetic flux generated by the current flowing around the primary coil so that as much of it as possible also links the secondary coil. 
  • The iron core is also used to decrease the magnitude of magnetizing current.


Important Points
Transformer:
A Transformer is a static electrical machine that transfers AC electrical power from one circuit to the other circuit at a constant frequency, but the voltage level can be altered which means voltage can be increased or decreased according to the requirement.

Construction:
The transformer consists of
  • Magnetic circuit (consisting of core, limbs, yoke, and damping structure)
  • Electrical circuit (consisting of primary and secondary windings)
  • Dielectric circuit (consisting of insulations in different forms and used at different places)
  • Tanks and accessories (conservator, breather, bushings, cooling tubes, etc.)
Core:
  • The purpose of the iron core is to channel the magnetic flux generated in the primary coil.
  • The main problem with the transformer core is its hysteresis and eddy current losses.
  • Hysteresis loss in the transformer mainly depends upon its core materials.
  • A small quantity of Silicon alloyed with low carbon content steel produces material for the transformer core.
  • It has low hysteresis loss and high permeability. 
  • The core of the transformer is made up of soft iron it has high coercivity and low retentivity.
  • The primary coil and the secondary coil are wound on the soft iron core.
  • The iron core decreases the reluctance of the magnetic path to flow magnetic flux.
  • The soft iron core has high permeability and it provides a complete linkage of the magnetic flux of the primary coil with the secondary coil. 

10. Which of the following is not a part of transformer installation ?
A. Conservator
B. Breather
C. Buchholz relay 
D. Exciter

Answer: D. Exciter

Explanation:
The Exciter is not part of a transformer. It is used in DC machines.

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.

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