Transformers MCQs With Explanatory Answers

Electrical Technology

Transformers (MCQs With Explanatory Answers)

1. A Transformer is designed to be operated on both 50 & 60 Hz frequency.For the Same rating, which one will give more out put; when,

  1.  Operates on 50 Hz
  2.  Operates on 60 Hz

Show Explanatory Answer

Answer:      1.   operates on 50 Hz  
Suppose,
When Transformer operates on 50 Hz Frequency
Transformer = 100kVA, R=700Ω, L=1.2 H, f= 50 Hz.
XL = 2πfL = 2 x 3.1415 x 50 x 1.2 = 377 Ω
impedance Z = √ (R2+XL2) = √ (7002 + 3772) = 795 Ω
Power factor Cos θ = R/Z = 700/795 =0.88
Transformer Output (Real Power)
kVA x Cos θ = 100kVA x 0.88
 88000 W = 88kW
Now,
When Transformer operates on 60 Hz Frequency
Transformer =100kVA, R=700Ω, L=1.2 H, f= 60 Hz.
XL = 2πfL = 2 x 3.1415 x 60 x1.2 = 452.4 Ω
impedance Z = √ (R2+XL2) = √ (7002 + 452.4 2) = 833.5 Ω
Power factor = Cos θ = R/Z = 700/833.5 =0.839
Transformer Output (Real Power)
kVA x Cos θ 100kVA x 0.839
=83900W = 83.9kW Output
Now see the difference (real power i.e., in Watts)
88kW- 83.9kW = 4100 W = 4.1kW
If we do the same (As above) for the power transformer i.e, for 500kVA Transformer, the result may be huge, as below.
(Suppose everything is same, without frequency)
Power Transformer Output (When operates on 50 Hz)
500kVA x 0.88 = 44000 = 440kW
Power Transformer Output (When operates on 60 Hz)
500kVA x 0.839 = 419500 = 419.5kW
Difference in Real power i.e. in Watts
440kW – 419.5kW = 20500 = 20kVA

2. In a Transformer , The primary flux is always _________ the secondary ( flux).

  1. Greater then
  2. Smaller then
  3. Equal
  4. Equal in both step up and Step down Transformer

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Answer:    4.    Equal in both step up and Step down Transformer
Flux in Primary and Secondary Winding is always equal.
Explanation:
Given Data;
Primary Number of Turns N1 = 524,
Secondary Number of Turns N2 = 70
Primary Input Voltage V1= 3300 Volts.
Secondary Current I2= 250 A.
Find/Calculate?
Secondary Voltage V2 =?
Primary Current I1=?
 Φm 1 = Φm2
We Know that,
N2/N1 = V2/V1 ====> V2 = (N2 x V1)/N1
Putting the Values
V2 = (70 x 3300)/525 = 440 Volts Ans.
Now if Neglecting Losses,
V1I1= V2I2 ====> I1/I2 = V2/ V1 …..Or…..I1 = (V2 x I2) / V1
Putting the Values,
I1 = 440 x 250/3300 = 33.3 Amp Ans.
Now turn around the Transformer equation.
E1 = 4.44 f N1 φm1
φm1 = E1 / 4.44 f N1
Putting the Values
Φm 1 = 3300 / (4.44 x 50 x 525) = 0.0283 Weber’s
Φm 1 = 28.3mWeber’s = Flux in Primary Windings
Same is on the other side,
E2 = 4.44 f N2 φm2
Φm2 = E2 / 4.44 f N2
Putting the values,
Φm2 = 440 / (4.44 x 50 x 70) = 0.0283 Weber’s
Φm2 = 28.3mWeber’s = Flux in secondary Windings
So You can see the flux (Φm) produced in Both Primary and Secondary Winding is same.

3. What would happen if we operate a 60 Hz Transformer on 50 Hz Source of Supply.(and how can we do that?

  1. Current will decrease (so increase the current)
  2. Current will increase ( so decrease the current)
  3. Current will be same in both cases.
  4. No Effect ( We can do that without changing anything)
  5.  We can’t perform such an operation.

Show Explanatory Answer

Answer:     2.     Current will increase ( so decrease the current)
Explanation:                Suppose this is a 60 Hz transformerTransformer: Transformers MCQs question answersTransformer: Transformers MCQs question answers

4. A Step-Up Transformer which has 110/220 turns.What would happen if we replace it with 10/20 turns? (because Turns ratio would be same in both cases) 

  1. induced E.M.F wold be same
  2. Induced E.M.F would be decreased

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Ans:    2.   Induced E.M.F would be decreased
Explanation:
Click image to enlargeTransformers MCQs With Explanatory Answers

5. The rating of transformer may be expressed in ____________.

  1. kW
  2. kVAR
  3. kVA
  4. Horse power.

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Answer:    3.   kVA
Explanation:
There are two type of losses in a transformer;
1. Copper Losses
2. Iron Losses or Core Losses or  Insulation Losses
Copper losses ( I²R )depends on Current which passing through transformer winding while Iron Losses or Core Losses or  Insulation Losses depends on Voltage.
That’s why the rating of Transformer is in kVA,Not in kW.

6. What will happen if the primary of a transformer is connected to D.C supply?

  1. Transformer will operate with low efficiency
  2. Transformer will operate with high efficiency
  3.  No effect
  4. Transformer may start to smoke and burn

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Answer:    4.   Transformer may start to smoke and burn.
Explanation:What will happen if the primary of a transformer is connected to D.C supply?

7. What would happen if a power transformer designed for operation on 50 Hz (frequency) were connected to a 500 Hz (frequency) source of the same voltage?

  1.  Current will be too much high
  2.  Transformer may start to smoke and burn
  3.  Eddy Current and Hysteresis loss will be excessive
  4.  No effect

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Answer:    3.    Eddy Current and Hysteresis loss will be excessive
Explanation;power transformer designed for operation on 50 Hz

8. What would happen if a power transformer designed for operation on 50 Hz (frequency) were connected to a 5 Hz (frequency) source of the same voltage? 

  1.  Current will be too much low
  2.  Transformer may start to smoke
  3.  Eddy Current and Hysteresis loss will be excessive
  4.  No effect

Show Explanatory Answer

Answer:    2.    Transformer may start to smock
Explanation:power transformer designed for operation on 50 Hz 
9. A Step Up transformer _____________. 
  1. Step Up the level of Voltage
  2. Step down the level of current
  3. Step up level the power
  4. Step up the level of Frequency
  5. 1 and 2 only

Show Explanatory Answer

Answer:   5.   1 and 2 only.
Explanation:
A Step up transformer only step up the level of voltage and step down the level of current.
because the input power is same.
So according to P=VI→ I = P/V…. We can see that, when Voltage increases, current decreases.
So in Step up transformer, input power is same, therefore, when voltage increases, then current decreases.

10. Under what condition is D.C supply applied safely to the primary of a transformer? 

  1. We can connect directly to DC. No condition required
  2. We can’t connect to DC Supply
  3. A High resistance should be connect in series with primary, but circuit will be useless.
  4.  The above statement is wrong

Show Explanatory Answer

Answer:    3.    A High resistance should be connect in series with primary, but circuit will be useless.
Explanation:Under what condition is D.C supply applied safely to the primary of a transformer?

11. An Auto-transformer (which has only one winding) may be used as a ______?

  1. Step-Up Transformer
  2. Step-Down Transformer
  3. Both Step-Up and Step-Down transformer
  4. None of the above

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Answer:      3.    Both Step-Up and Step-Down transformer
Explanation:An Auto-transformer (which has only one winding) may be used

12.  E.M.F Equation of the Transformer is _________.

  1. E1 = 4.44 f N1 Øm      ,    E2=4.44 f N2 Øm
  2. E1= 4.44 f N1 Bm A   ,   E2 = 4.44 f N2 BmA
  3. E1=  4.44 N1 Øm/T     ,    E2=4.44 N2  Øm/T
  4. All of the above
  5. None of the above

Show Explanatory Answer

Answer:             4.       All of the above
Explanation:
Take the basic Equation of the transformer (Option 1) E1 = 4.44 f N1 Øm      ,    E2=4.44 f N2 Øm ,
and then, first put the value of  Øm = Bm A. So the equation becomes as in Option 2.
Now put the value of Frequency ( f = 1/T ) in Equation on Option (1). So the equation becomes as in Option 3.

13. The friction losses in  Real Transformers are _________?

  1. 0%
  2. 5%
  3. 25%
  4. 50%

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Answer:          1.        0%
Explanation: Transformer is a Static Devise. So, no rotation, No Friction losses.

14.  In Three Phase Transformer, The load Current is 139.1A, and Secondary Voltage is 415V. The Rating of the Transformer would be ___________.

  1. 50kVA
  2. 57.72kVA
  3. 100kVA
  4. 173kVA

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Answer:        3.      100kVA
Explanation:
Rating of a Three Phase Transformer:
P = √3. V x I
Rating of a Three phase transformer in kVA
kVA = (√3. V x I) /1000
Now
P = √3 x V x I (Secondary voltages x Secondary Current)
P= √3 x 415V x 139.1A = 1.732 x 415V x 139.1A= 99,985 VA = 99.98kVA=100kVA
For more Detail
How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)?

15  In Single Phase Transformer, The Primary Current and Primary Voltage is 4.55 and 11kV respectively. The Rating of the transformer would be________?

  1. 50kVA
  2. 86kVA
  3. 100kVA
  4. 150kVA

Show Explanatory Answer

Answer:        1.      50kVA
Explanation:
Rating of a Single Phase Transformer:
P =  V x I
Rating of a Single phase transformer in kVA
kVA = (V x I) /1000
Now
P =  V x I (Primary voltages x Primary Current)
P =  11000V x 4.55A  = 50,050VA = 50 kVA
For more Detail .. Read the rating of transformer post in MCQs No 14 explanatory section titled as
“How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)”?

16.  An Isolation Transformer Has Primary to Secondary turns ratio of __________.

  1. 1 : 2  
  2. 2 : 1 
  3. 1 : 1 
  4. Can be any ratio

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 Answer:   3.    1:1
Explanation: Isolation Transformer is used for isolation purpose only. Isolation transformer transfer electrical power from the source circuit to another circuit with connecting electrically (but magnetically) for preventing electric shock and also used in sensitive devices (like medical equipment etc). Thus, isolation between two electrical circuit can be done by Isolation transformer with turns ratio of 1:1.

17.   In an Auto Transformer, The Primary and Secondary are__________Coupled.

  1. Only Magnetically
  2. Only Electrically
  3. Magnetically as well as Electrically
  4. None of the above

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Answer:     3.    Magnetically as well as Electrically
Explanation: As we know that in a Transformer, Primary and Secondary winding are magnetically coupled. But in case of Auto transformer, there is only one winding (which is used both as a Primary and Secondary). Thus, in an In an Auto Transformer, The Primary and Secondary are Magnetically as well as Electrically Coupled.
for More detail: Check MCQs No 11 with diagram.

18.   A Transformer______________.

  1. Changes ac to DC
  2. Changes dc to AC
  3. Steps up or down DC Voltages & Current
  4. Steps up or down AC Voltages & Current

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Answer:       4.     Step up or Step down AC Voltage & Current
Explanation: A Transformer does not work on DC and operates only and only on AC, therefore it Step up of Step down the level of AC Voltage or Current.
For More detail: Check MCQs No 9

19.   Transformer is a device which:________________.

  1. Transfer Electrical power from one electrical circuit to another Electrical circuit
  2. It’s working without changing the frequency
  3. Work through on electric induction.
  4. When, both circuits take effect of mutual induction
  5. Can step up or step down the level of voltage.
  6. Its Working without  changing the Power.
  7. All of the above

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Answer:       7.     All of the above
Explanation: Transformer

20. For a transformer with number of secondary windings more than the number of primary windings, the secondary current will be _____________ ?

  1. More than the primary current
  2. Less than the primary current
  3. Equal to the primary current
  4. Zero

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Answer:       2.    Less than the primary current

Explanation: For a transformer, ratio of its primary voltage to secondary voltage is equal to ratio of primary turns to secondary turns. Thus, if number of secondary windings is more than number of primary windings, the secondary voltage will also be more than the primary voltage. For a constant impedance coil, current through the coil is inversely proportional to the voltage. Hence, as secondary voltage is higher, secondary current is lower and thus secondary current is less than the primary current.

21. DC power is never applied to transformer

  1. True
  2. False

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Answer: 1,    True

Explanatory Answer:

DC voltage applied to the primary of a transformer sets up a constant magnetic flux in its core. Due to absence of any rate of change of flux, there is no induced EMF in the primary to oppose the voltage. Thus, the resistance is very low and current is very high. This high current causes huge heat loss, causing the transformer core to burn and hence, DC power is never applied to transformer.

22. Impedance Ratio of a transformer is equal to _____________.

  1. Square of turns ratio
  2. Turns ratio
  3. 1
  4. Infinite

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Answer: 1 .Square of turns ratio

Explanation: For an ideal transformer with no power loss, the output power is equal to input power

i.e., VPIPcosφ = VSIScosφ

cosφ = Power Factor

Therefore, VP/VS = IS/IP = NP/NS

Now, ZP = Primary Impedance = VP/IP

ZS = Secondary Impedance = VS/IS

Thus, ZP/ZS = (VPxIS)/(VSxIP) = (NP/NS)2

 

23. For a single-phase transformer with 250 primary turns and 50 secondary turns, connected across a 1500 Volts, 50Hz supply, the maximum value of flux is—–

  1. 1 Wb
  2. 027Wb
  3. 04Wb
  4. 5Wb

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Answer: 2

For a single-phase transformer, Maximum flux,

Transformer

Now,

= Primary voltage = 1500 Volts

= Input Frequency = 50Hz

NP = Number of primary turns = 250

Substituting the values, we get, Maximum flux,

Transformer

24. For a single phase, 230/2300 Volts, 50Hz core type transformer of cross section 25 cm, if the maximum flux density is 1.12 wb/m2, the number of primary and secondary turns is———–

  1. 8, 148
  2. 16, 160
  3. 23, 230
  4. 14, 140

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Answer: 14

Given

Primary Voltage, VP = 230 Volts

Secondary Voltage, Vs = 2300 Volts

Maximum Flux Density, Bmax = 1.12 wb/m2

Area of cross section, A = 0.0625 m2

Therefore, Maximum flux, φmax = BmaxA = 0.07 wb

Also, φmax = VP/4.44xfxNP

Where, f = supply frequency = 50 Hz

Thus, NP = 14.8

EMF per turn, E = φmaxx4.44xf = 15.54 V/turn

Thus, number of secondary turns, NS = VS/E = 148

 

25. For a 300KVA, 11000 Volts/440 Volts, 50Hz single phase transformer, the values of primary and secondary currents are ————–

  1. 30 Amperes, 750 Amperes
  2. 2 Amperes, 681.8 Amperes
  3. 1 Amperes, 752.5 Amperes
  4. 5 Amperes, 637.5 Amperes

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Answer: 2

Explanation

Given, Transformer Rating = 300KVA

Primary Voltage = 11000 Volts

Primary Current = Transformer Rating x 1000/Primary Voltage = 27.2 Amperes

Secondary Voltage = 440 Volts

Secondary Current = Transformer Rating x 1000/Secondary Voltage = 681.8 Amperes

 

26. For a practical transformer at no-load, the input power is equal to the iron losses

  1. True
  2. False

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Answer: 1 True

Explanation

For a practical transformer, if the secondary winding is kept open circuited, it is said to be in no-load condition. As the secondary current is zero, there is no magnetic leakage in the primary and a small current is drawn to supply to the iron losses. Thus, the no load input power is equal to the iron losses.

 

27. For actual transformers with higher load, leakage flux is prominent

  1. True
  2. False

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Answer: 1

Explanation

In actual transformers, when the primary winding is energized with AC supply, an alternating flux is produced which not only links with the other winding, but also with that winding as well. This is known as leakage flux. At higher loads as primary and secondary currents are higher, the leakage flux is also higher.

 

28. Which among these losses does not depends on load and is constant?

  1. Eddy Current Losses
  2. Hysteresis losses
  3. A and C
  4. Copper losses

Show Explanatory Answer

Answer: 3

In an actual transformer, not all of input energy is transferred to output. Rather, a certain amount is lost in the core and as heat. There are two types of losses – Iron and Copper losses. Iron losses are of two types – Eddy Current losses and Hysteresis losses.

Hysteresis loss occurs due to magnetization of atoms in the magnetic material of the core, forming small magnetic domains. This leads to causes huge requirement of energy to power frequent orientation of the magnetic domains in the directions of alternating flux. Since the net hysteresis loss depends upon the maximum magnetic flux density, supply frequency and volume of the core material, it is independent of load and is constant at any load or no load.

Eddy current losses occur due induction of EMF in core laminations which produces circulating eddy currents. These losses also depend upon the magnetic flux density, supply frequency, thickness of laminations and volume of material and are independent of load.

 

29. For the transformer at no-load, the primary current is 4A at power factor 0.35 and connected across a power supply of 230 Volts, 50Hz, the core loss and magnetizing current are —————-respectively

  1. 400 W, 2A
  2. 322 W, 1.4A
  3. 450 W, 3A
  4. 433 W, 2.2A

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Answer: 2

Given:

No load primary voltage, VP = 230 Volts

Supply frequency = 50 Hz

No load primary current, Io = 4 A

Power Factor, cosφ = 0.35

Hence, Iron loss component of no load primary current, IW = Iocosφ = 1.4 A

Magnetizing current component,

Total Iron losses = VP x Iw = 322 W

 

30. For the circuit given below along with the parameters, the value of secondary voltage is———–

Note: Primary current = 25A

Input Power = 12kW

Transformer Rating = 10kVA, 600V/120V

Supply frequency = 50 Hz

  1. 100 Volts
  2. 105 Volts
  3. 18 Volts
  4. 8 Volts

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Answer:    ( 3 )

Given:

Primary Voltage, VP = 600V

Power Factor, cosφ = Transformer Rating/ (Primary Voltage x current) = 0.8

Primary reactance, XP = 100πx Inductance = 3.14 Ohms

Primary resistance, RP = 0.35 Ohms

Phase angle, φ = arccos (0.8) = 36.86

Sinφ = 0.6

Primary induced EMF, EP = VP – IPRP cosφ – IPXP sinφ = 600-7-47.1= 545.9Volts

Turns ratio, K = 120/600 = 0.2

Therefore, secondary induced EMF, ES = EP x K = 109.18 Volts

Secondary Voltage, VS = ES = 109.18 Volts

 

31.   For a transformer with ohmic loss of 2% with respect to output voltage and reactance drop of 6% with respect to voltage, the regulation at 0.75 lagging power factor is————

  1. 5%
  2. 4.5%
  3. 6%
  4. 8%

Show Explanatory Answer

Answer: 2

Given:

Percentage ohmic loss = 2

Percentage reactance drop = 6

Power Factor, cosφ = 0.75

Sinφ = 0.5

Percentage regulation = Percentage ohmic loss x cosφ + Percentage reactance drop x sinφ = 1.5+3 = 4.5%.

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