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Latest Transformers MCQs – Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) MCQs

Latest Transformers MCQs – Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) MCQs

Latest Transformers MCQs

By practicing these MCQs of Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) MCQs – Latest Competitive MCQs , an individual for exams performs better than before. This post comprising of objective questions and answers related to Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) Mcqs “. As wise people believe “Perfect Practice make a Man Perfect”. It is therefore practice these mcqs of Transformers to approach the success. Tab this page to check ” Excitation Phenomenon & Variable Frequency Transformers ( Transformers )” for the preparation of competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs and others.

Transformers MCQs – Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) MCQs

The most occurred mcqs of Excitation Phenomenon & Variable Frequency Transformers ( ) in past papers. Past papers of Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) Mcqs. Past papers of Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) Mcqs. The Important series of Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) Mcqs are given below:

Excitation Phenomenon

1. No-load current in the transformer is _______
a) Straight DC
b) Steps
c) Sinusoidal
d) Sinusoidal distorted
Answer: d
Explanation: The no-load current in a transformer is non-sinusoidal. The basic cause for this phenomenon, which lies in hysteresis and saturation non-linearities of the core material, will now be investigated; this can only be accomplished graphically.

2. The main reason for generation of harmonics in a transformer is ____________
a) Fluctuating load
b) Poor insulation
c) Mechanical vibrations
d) Saturation of core
Answer: d
Explanation: The iron core which is used in transformer is subjected to saturation effect. Thus, according to the hysteresis loop, the generation of harmonics at particular saturation level can be identified.

3. The magnetising current of a transformer is usually small because it has _______
a) Small air gap
b) Large leakage flux
c) Laminated silicon steel core
d) Fewer rotating parts
Answer: a
Explanation: Air Gap increases the reluctance in the magnetic path of magnetic lines of force. When there is an air gap in the magnetic circuit the reluctance is high owing to the permeability of air which is much lower as compared to ferromagnetic materials. The mmf required to overcome that (maintaining flux density in the air gap) is more.

4. Harmonics in transformer result in ___________
a) Increased core losses
b) Increased I2R losses
c) Magnetic interference with communication circuits
d) Increased core losses, Increased I2R losses and magnetic interference
Answer: d
Explanation: harmonics are produced in an particular circuit as a result of magnetization of core material. Thus, magnetic losses are increased that is, iron losses are more so to maintain constant more current will be drawn giving rise to I2 losses.

5. For given applied voltage, with the increase in frequency of the applied voltage ___________
a) Eddy current loss will decrease
b) Eddy current loss will increase
c) Eddy current loss will remain unchanged
d) Cannot be determined
Answer: b
Explanation: Eddy current losses are directly proportional to the square of the frequency of input applied rated voltage. This loss is independent of voltage applied to the transformer. Thus, eddy current loss will increase with frequency.

6. Which of the following is the most likely source of harmonics in a transformer?
a) Poor insulation
b) Overload
c) Loose connections
d) Core saturation
Answer: d
Explanation: Core material used in a transformer is iron, which undergoes saturation when it is excited by some external supply, which causes some unlikely effects like generation of harmonics which leads in losses for a particular transformer.

7. Third harmonic currents can be as large as ______________
a) 100% of fundamental
b) 10% of fundamental
c) 70% of fundamental
d) 40% of fundamental
Answer: d
Explanation: While odd symmetry is preserved and the current and flux maximas occur simultaneously, the current zeros are advanced in time with respect to the flux wave shape. As a consequence, the current has fundamental and odd harmonics, the strongest being the third harmonic which can be as large as 40% of the fundamental.

8. Flux transient goes till the maximum value due to ___________
a) Doubling effect
b) Saturation effect
c) Hysteresis effect
d) Modulation effect
Answer: a
Explanation: A transient flux component (off-set flux) ft = fm originates so that the resultant flux is (ft + fss) which has zero value at the instant of switching. The transient component ft will decay according to the circuit time constant (L/R) which is generally low in a transformer. If the circuit dissipation (core-loss) is assumed negligible, the flux transient will go through a maximum value of 2fm.

9. The transformer switching transient is referred as _____________
a) Inrush current
b) Outrush current
c) Middlerush current
d) Harmonic current
Answer: a
Explanation: In subsequent half-periods ft gradually decays till it vanishes and the core flux acquires the steady-state value, Because of the low time constant of the transformer circuit, distortion effects of the transient may last several seconds. The transformer switching transient is referred to as the inrush current.

10. Which of the following is not the method of reducing harmonics?
a) Adding filters
b) Adding capacitors
c) Adding inductors
d) Adding resistance
Answer: d
Explanation: Reduce the harmonic currents produced by the load. Add filters to siphon the harmonic currents off the system, or block the currents from entering the system, or supply the harmonic currents locally. Modify the frequency response of the system by filters, inductors, or capacitor.

11. It is possible to reduce harmonic currents completely.
a) True
b) False
Answer: b
Explanation: It is not possible to prevent harmonic currents completely. But they can be prevented from flowing through the main system by providing a separate low-impedance path for them. This is done by the use of rated series tuned circuits consisting of a reactor and capacitor, which have equal impedance at a specific harmonic frequency.

Autotransformer

1. Which of the following is the main advantage of an auto-transformer over a two-winding transformer?
a) Hysteresis losses are reduced
b) Saving in winding material
c) Copper losses are negligible
d) Eddy losses are totally eliminated
Answer: b
Explanation: Auto transformer is a special type of transformer which has primary and secondary winding both located on same winding. Thus, winding material required for a transformer is very less in the case of autotransformer.

2. Auto-transformer makes effective saving on copper and copper losses, when its transformation ratio is
a) Approximately equal to one
b) Less than one
c) Great than one
d) Cannot be found
Answer: a
Explanation: Copper In auto transformer /copper in two-winding transformer = 1- T2/T1. This means that an auto transformer requires the use of lesser quantity of copper given by the ratio of turns. Hence, if the transformation ratio is approximately equal to one, then the copper saving is good and the copper loss is less.

3. Total windings present in a autotransformer are __________
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: Autotransformer is the special transformer for which the single winding acts as a primary and secondary both. Thus, by taking the appropriate winding into consideration a variable secondary voltage is obtained.

4. Autotransformers are particularly economical when _________
a) Voltage ratio is less than 2
b) Voltage ratio is very high
c) Voltage ratio is higher than 2 in smaller range
d) Can be used anywhere
Answer: a
Explanation: Autotransformer is economical where the voltage ratio is less than 2 in which case electrical isolation of the two windings is not essential. The major applications are induction motor starters, interconnection of HV systems at voltage levels with ratio less than 2, and in obtaining variable voltage power supplies (low voltage and current levels).

5. Which of the following is not true regarding the autotransformer compare to two-winding transformer?
a) Lower reactance
b) Lower losses
c) Higher exciting current
d) Better voltage regulation
Answer: c
Explanation: Autotransformer is the advance version of normal transformer. It is having better voltage regulation, higher efficiency due to lower losses, lower reactance and thus it also requires very small exciting current.

6. Two-winding transformer of a given VA rating if connected as an autotransformer can handle ___________
a) Higher VA
b) Lower VA
c) Same VA
d) Cannot be found
Answer: a
Explanation: A two-winding transformer of a given VA rating when connected as an autotransformer can handle higher VA. This is because in the autotransformer connection part of the VA is transferred conductively.

 

Efficiency And Voltage Regulation MCQs

 

7. When auto-transformation ratio becomes equal to 1, which of the following statement is true?
a) VA rating of the autotransformer becomes far greater than VA rating of two winding transformer
b) VA rating of the autotransformer becomes far lower than VA rating of two winding transformer
c) VA rating of the autotransformer becomes equal to VA rating of two winding transformer
d) Can’t comment
Answer: a
Explanation: VA rating of autotransformer is = [1/1-a] * VA of two-winding transformer, thus, when a i.e. transformation ratio of autotransformer becomes closer to 1 one gets very high value of VA rating of an autotransformer.

8. An autotransformer compared to its two-winding counterpart has a higher operating efficiency.
a) True
b) False
Answer: a
Explanation: The losses are less in autotransformer compare to two-winding transformer. Thus, for the given same input to autotransformer as that of two-winding transformer more output will be available to secondary side.

9. Ratio of winding material needed for autotransformer to thr two winding transformer is ______
a) 1- V2/V1
b) 1- N2/N1
c) 1- V2/V1 and 1- N2/N1
d) 1- V1/V2
Answer: c
Explanation: Gauto/GTW= 1- V2/V1, as voltage is directly proportional to the number of turns we can say the value is also equal to the 1- N2/N1. Thus, one can write GTW-Gauto = 1/a’ * GTW = saving of the conductor material using autotransformer.

10. For an auto-transformation ratio tending to the unity value, saving of the conductor material will be ___________
a) Tend towards 90% or more
b) Tend towards 0%
c) Can’t say
d) Wil remain fix
Answer: a
Explanation: GTW-Gauto = 1/a’ * GTW = saving of the conductor material using autotransformer. So, if a’=10, saving is only 10% but for a’=1.1, saving is as high as 90%. Hence it is more economical when the turn-ratio is closer to unity.

11. What are the modes in which power can be transferred in an autotransformer?
a) Conduction
b) Induction
c) Conduction and Induction
d) Cannot be said
Answer: c
Explanation: In two winding transformer there is no electrical connection between primary and secondary. So, the power is transferred through induction. But in auto-transformer there is a common electrical path between primary and secondary. So, power is transferred through both conduction and induction processes.

12. A 100/10, 50 VA double winding transformer is converted to 100/110 V auto transformer. What will be the rating of auto transformer?
a) 500 VA
b) 550 VA
c) 100 VA
d) 110 VA
Answer: b
Explanation: Secondary current of the two-winding transformer at rated voltage supply = 50/10= 5 A
Thus, autotransformer will also carry the same rated current in secondary giving the power output as 5*110= 550 VA.

13. For a 20kVA transformer with turn ratio 0.4 what amount of total power is transferred inductively?
a) 8kVA
b) 12kVA
c) 10kVA
d) 50kVA
Answer: b
Explanation: For an auto transformer power is transferred partially inductively and partially conductively. Thus, out of total power, power transferred inductively is given by (1-k)*total power= 0.6*20= 12kVA.

14. For a 20kVA transformer with turn ratio 0.4 what amount of total power is transferred conductively?
a) 8kVA
b) 12kVA
c) 10kVA
d) 50kVA
Answer: a
Explanation: For an auto transformer power is transferred partially inductively and partially conductively. Thus, out of total power, power transferred conductively is given by (k)*total power= 0.4*20= 8kVA.

Variable Frequency Transformer

1. Output transformers are also called for __________
a) Maximum efficiency
b) Maximum power
c) Maximum output
d) Maximum efficiency, power and output
Answer: b
Explanation: In electronic circuit applications maximum power is given more preference than maximum efficiency unlike in power system operations. So, all the transformers working for maximum power are called as output transformers.

2. The output transformers are also named as _________
a) Audio transformers
b) Power transformers
c) Industrial transformers
d) Electric transformers
Answer: a
Explanation: in electronic circuit applications the performance criterion is the maximum power unlike the maximum efficiency in power system applications. Such transformers are known as output transformers while in audio applications these are known as audio transformers.

3. Which of the following is the major requirement for the transformers used for electronic purposes?
a) Perfect DC isolation
b) Maximum efficiency
c) Constant amplitude voltage gain
d) Perfect DC isolation , maximum efficiency and constant voltage gain
Answer: c
Explanation: An important requirement of these transformers is that the amplitude voltage gain (ratio of output/input voltage amplitude) should remain almost constant over the range of frequencies (bandwidth) of the signal.

4. Why frequency scale is logarithmic?
a) Frequency range is very small for these transformers
b) Frequency range is very large for these transformers
c) Frequency used is negligible
d) Can’t say
Answer: b
Explanation: When we investigate the gain and phase frequency characteristics of the transformer. This would include the effect of the output impedance (resistance) of the electronic circuit output stage. In these characteristics as the frequency range is quite large the frequency scale used is logarithmic.

5. When series leakage inductances are ignored then the region is called as _____________
a) Low-frequency region
b) High-frequency region
c) Mid-frequency region
d) Ultra high-frequency region
Answer: c
Explanation: In this region the series leakage inductances can be ignored (as these cause negligible voltage drops) and the shunt inductance (magnetizing inductance) can be considered as open circuit. With these approximations the equivalent circuit as seen on the primary side can be drawn and gain is then calculated.

6. What is the gain of the transformer in mid-band region?
a) [N2/N1] [RL ‘/(R+RL’)].
b) [N2/N1] [RL ‘/(R+RL’+jωL)].
c) AH=A0/ [1+(ω/ ωH)2]1/2
d) AL=A0/ [1+(ω/ ωL)2]1/2
Answer: a
Explanation: It immediately follows from the circuit analysis of transformer in mid-frequency region that VL and VS are in phase, the circuit being resistive only. As for the amplitude gain, it is given as
VL’=VS [RL ‘/(R+RL’)] [N2/N1]VL=VS [RL ‘/(R+RL’)] A0= VL/VS= [N2/N1] [RL ‘/(R+RL’)]

7. When series leakage inductances are not ignored but shunt inductance is an effective open circuit then the region is called as _____________
a) Low-frequency region
b) High-frequency region
c) Mid-frequency region
d) Ultra high-frequency region
Answer: b
Explanation: In this region the series inductances must be taken into account but the shunt inductance is an effective open circuit yielding the approximate equivalent circuit. Amplitude and phase angle as function of frequency are derived below.

8. What is the gain of the transformer in high-frequency region?
a) [N2/N1] [RL ‘/(R+RL’)].
b) [N2/N1] [RL ‘/(R+RL’+jωL)].
c) AH=A0/ [1+(ω/ ωH)2].
d) AL=A0/ [1+(ω/ ωL)2]1/2
Answer: b

Explanation: It immediately follows from the circuit analysis of transformer in high-frequency region that AH is equal to [N2/N1] [RL ‘/(R+RL’+jωL)]. If recognised the term [N2/N1] [RL ‘/(R+RL’)] as the gain of transformer in mid-band region then AH=A0/ [1+(ω/ ωH)2]1/2.

9. When series leakage inductances are not ignored but shunt inductances are considered in parallel then the region is called as _____________
a) low-frequency region
b) high-frequency region
c) mid-frequency region
d) ultra high-frequency region
Answer: a
Explanation: In this region the series effect of leakage inductances is of no consequence but the low reactance (ωLm) shunting effect must be accounted for giving the approximate equivalent circuit and in the calculations of gain of a transformer.

10. What is the gain of the transformer in low-frequency region?
a) [N2/N1] [RL ‘/(R+RL’)].
b) [N2/N1] [RL ‘/(R+RL’+jωL)].
c) AH=A0/ [1+(ω/ ωH)2].
d) AL=A0/ [1+(ω/ ωL)2]1/2
Answer: d
Explanation: The corner frequency ωL of this circuit is obtained by considering the voltage source as short circuit. This circuit is Lm in parallel with R||R¢L. Thus, ωL = [R||RL’/Lm].
Complex gain can be expressed as AL=A0/ [1+(ω/ ωL)2]1/2.

11. Relative voltage ratio curve for an output transformer is _______________
a) Circular
b) Straight line
c) Bell shaped
d) Irregular curve
Answer: c
Explanation: The complete amplitude and phase response of the transformer (with source) on log frequency scale. At high frequencies, the interturn and other stray capacitances of the transformer windings need to be considered. The capacitance-inductance combination causes parallel resonance effect thus, amplitude peak shows up in the high-frequency region of the frequency response.

12. Phase angle characteristic of an output transformer is ____________
a) Curve increasing towards high-frequency region
b) Curve increasing towards low-frequency region
c) Curve increasing towards mid-frequency region then decreasing
d) Can’t define
Answer: a
Explanation: The phase-angle characteristics of an output transformer starts from below of relative voltage ratio characteristics drawn on the same axis. Further it goes on increasing towards high-frequency region non-linearly.

Latest Transformers MCQs – Excitation Phenomenon & Variable Frequency Transformers ( Transformers ) MCQs