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# Multiple choice question for engineering

## Set 1

1. Below is the phasor diagram of synchronous generator.

If the machine is made to run as synchronous motor, what will be the phasor diagram changes then?
a) Reverse Ia
b) Reverse Ef
c) Reverse Vt and Ia
d) Reverse Vt

Answer: a [Reason:] Only the direction of the armature current changes in the motor as it absorbs the electrical energy.

2. Below is the phasor diagram of synchronous generator.

If the machine is made to run as synchronous motor, what will be the new phasor diagram?
a)
b)
c)
d)

Answer: a [Reason:] Only the direction of the armature current changes in the motor as it absorbs the electrical energy. It has a component in phase opposite to Ef.

3. The voltage equation of synchronous motor is
a) Ef = Vt + Ia*(ra+jXa)
b) Ef = Vt – Ia*(ra+jXa)
c) Ef = Vt + Ia*(ra-jXa)
d) Ef = Vt – Ia*(ra-jXa)

Answer: a [Reason:] The voltage equation of synchronous motor is Ef = Vt + Ia*(ra+jXa).

4. Synchronous motor delivers lagging power at
b) lagging pf
c) zero pf
d) unity pf

Answer: a [Reason:] As Q < 0 for synchronous motor at leading pf.

5. A 420 V synchronous motor is working at 30° lagging load. What can be the expected phasor?
a)
b)
c)
d) None of the mentioned

Answer: a [Reason:] The Ia should lag Ef by 30°.

6. The synchronous motor is operating at upf load. The most suitable phasor will be?
a)
b)
c)
d)

Answer: a [Reason:] Ia must be in phase with the Ef.

7. The zero power factor of an alternator can be obtained by _________ at rated Ia.
a) over exciting
b) conducting short circuit of secondary terminals
c) under excitation
d) running as reluctance motor

Answer: a [Reason:] By over exciting an alternator gives the additional flux required to make it operate at zero power factor.

8. From the figure shown below for the OCC and SCC of a three phase alternator, the point ‘L’ corresponds to _______ and the point ‘A’ can be obtained by _____ the alternator.

a) terminal voltage, over exciting
b) excitation voltage, over exciting
c) terminal voltage, under exciting
d) excitation voltage, under exciting

Answer: a [Reason:] The point ‘L’ corresponds to Vt and the point ‘A’ can be obtained by over exciting the alternator.

9. In the figure below, the point ‘F’ corresponds to _______ which can be obtained by ________

a) field current required to circulate short circuit current, SCC
b) field current required to circulate full load current, SCC
c) mmf required to cancel the hysteresis losses, SCC
d) mmf required to compensate the leakage reactances, SCC

Answer: a [Reason:] Point ‘F’ corresponds to the short circuit current and it is obtained by SCC plot.

10. The operating point for an over excited alternator working near zpf lagging power factor gives ‘A’ as the operating point. Then if the same machine is working as an over excited synchronous motor will give same operating point at
b) zpf lagging
c) upf
d) there will never be the same operating point for the machine as motor as well as generator.

Answer: a [Reason:] Yes, by making it from lag to leading power factor, a synchronous machine can have same operating point for the alternator as well as synchronous motor.

11. Identify the over excited alternator operating at zpf.
a)
b)
c)
d)

Answer: a [Reason:] Ef>Vt, makes it work as over excited.

12. Identify the over excited synchronous motor operating at zpf.
a)
b)
c)
d)None of the mentioned

Answer: a [Reason:] Ef>Vt, and the Ia is in opposite to that of alternator.

13.
For the above graph depicting the performance of a synchronous motor, coordinates ‘x’ and ‘y’ are respectively?
In the ‘V’ curve shown in the above figure for a synchronous motor, the parameter of x and y coordinates are, respectively
a) Armature current and field current
b) Power factor and field current
c) Armature current and torque
d) Torque and field current

Answer: a [Reason:] V curve is between the armature current and the field current.

14. We can increase the torque of a reluctance synchronous motor if we
a) Increase reluctance of the magnetic circuit along the direct axis
b) Decrease the reluctance of the magnetic circuit along the quadrature axis
c) Increase the ratio of the quadrature axis reluctance to direct axis reluctance
d) Decrease the ratio of quadrature axis reluctance to direct axis reluctance

Answer: c [Reason:] Te = -1/2(φ^2)(Rlq-Rld) sin2δ.

15. Switched reluctance motors are basically
a) salient pole synchronous motor but without excitation winding
b) stepper motor with salient poles
c) synchronous motor with salient poles on stator and rotor
d) stepper motor with closed loop control and with rotor position sensor

Answer: d [Reason:] A switched reluctance motor has stepper motor.

## Set 2

1. A V-V connected transformer can be connected in parallel to delta-delta connected transformer but not to
a) delta-star
b) star-delta
c) star-V
d) all of the mentioned

Answer: a [Reason:] The V-V connected transformer and D-D connected transformers have same phase displacement, so they only can be connected in parallel to each other.

2. Three units of single phase transformers and one single three-phase transformer rating
a) will be same for one rating
b) can never be made same
c) may be same
d) none of the mentioned

Answer: a [Reason:] Three single phase transformers and one single unit of three phase transformer will be same only. this is done to reduce the cost and spacing.

3. The transformer which is more feasible to use in the distribution ends should be
a) star-delta
b) delta-star
c) scott
d) delta-delta

Answer: a [Reason:] Star-delta will have lower voltage at delta end.

4. For the given circuit below, the voltage across the terminal ‘a’ and ‘b’ is

a) 1.73*V
b) V/1.73
c) V
d) 3*V

Answer: a [Reason:] From delta to star, the voltage between the lines will increase by 1.73.

5. The voltage across the terminal ‘r’ and ‘y’ is

a) 1.73*V
b) V/1.73
c) V
d) 3*V

Answer: b [Reason:] From star to delta, the line voltage will reduce by 1.73.

6. The delta-delta connections are used in applications of
a) large l.v. transformers
b) small h.v. transformers
c) large h.v. transformers
d) small l.v. transformers

Answer: a [Reason:] Delta connected winding handle line voltages so it needs more turns in the winding but thin wires.

7. Open delta transformers can be obtained from
a) delta-delta
b) star-delta
c) delta-star
d) any of the mentioned

Answer: a [Reason:] If one of the transformers is removed from the bank of only delta-delta, then it behaves with 58% power delivery.

8. If one of the transformers is removed from the bank of only delta-delta, then it behaves with 58% power delivery.
a) True
b) False

Answer: a [Reason:] It is true as the circuit will still be closed and the transformer will operate will lesser operating point.

9. Below is the circuit diagram for delta-delta transformer have a transformation ratio ‘k’. The current flowing in the windings of phase r-y in the below diagrams is

a) k*I/1.73
b) I/k*1.73
c) 1.73*k*I
d) 3*k*I

Answer: a [Reason:] The phase current will be reduced by 1/1.73.

10. If para magnetic core is used in the place of the ferromagnetic core of the transformer, then magnetostriction will
a) be vanished
b) reduce
c) increase
d) not be affected

Answer: a [Reason:] If the core is not ferromagnetic then the transformer will not operate at all so no humming sound.

11. When we magnetize the ferromagnetic core of the transformer, core length_______ and it ______ when demagnetized.
a) decreases, increases
b) decreases, decreases
c) increases,increases
d) none of the mentioned

Answer: a [Reason:] Due to aligned magnetic dipoles in the material, the size will reduce by a small margin and vice versa while demagnetization.

12. A 400 V, 10 KVA transformer at 50 Hz, is operated at the frequency of 40 Hz, then the humming
a) increases
b) decreases
c) remains same
d) increases to very high

Answer: a [Reason:] If the frequency is reduced, the core flux density increases, so the noise also will increase.

13. While using three phase transformers, it is taken care about the third harmonics current in the system.
This is present when transformer set is delta on either side.
a) True
b) False

Answer: a [Reason:] Third harmonic current are co-phase in nature and require closed path which is present in delta connected only.

14. A. Positive sequence components – 7th harmonic component
B. Negative sequence components – 5th harmonic component
C. Zero sequence components – 3rd harmonic component
Above are the matching made for the harmonics and the associated harmonic component, then the correct marking is
a) A,B,C
b) C
c) A,B
d) B,C

Answer: a [Reason:] 3m-1 phases are displaced 120° in opposite and (3m+1) in the same direction as the fundamental. Here ‘m’ is even integer.

15. A recording of the output of the emf induced for star and delta are recorded.
Then shape of the emf induced in Y-connected 3-phase transformer is non sinusoidal in nature due to
a) 3rd harmonic component of currents is absent
b) 3rd harmonic component of currents is present
c) negative sequence component of current is present
d) none of the mentioned

Answer: a [Reason:] In a star connected transformer, closed path to generate 3rd harmonic is not there. So the emf shape will be peaky in nature.

16. Shape of emf generated by delta connected transformer is not always sinusoidal.
a) True
b) False

Answer: b [Reason:] No, for a delta transformer , closed path is present to generate 3rd harmonics current and so the emf will be sinusoidal always.

## Set 3

1. In power transformers, it is found that the iron or core losses practically varies very less, this is because
a) constant core flux
b) constant leakage flux
c) constant leakage and core flux

Answer: a [Reason:] Due to the constant core flux, the iron losses also remain constant.

2. It is advised to coat the laminations of the core with some enamel, to ensure
a) insulation
c) reduction in humming sound
d) all of the mentioned

Answer: a [Reason:] It is done to provide insulation.

3. Primary and the secondary winding of the transformer are interlaced mainly for
a) reducing leakage flux
b) reducing cost
c) reducing heating
d) saving the copper of the winding

Answer: a [Reason:] Interlacing is done to provide less air gaps and improved efficiency by reduction of leakage fluxes around the core.

4. Transients currents are maximum in the transformer when it is switched on with secondary ______ and when input voltage wave is passing through ________
a) open, zero
b) open, peak
c) closed, zero
d) closed, peak

Answer: a [Reason:] At t=0, the current flowing will be very high as emf is very high due to t = 0.

5. I. Silicon steel – Power transformer
II. Ferrite – High frequency transformers
III. Alnico – permanent magnet
Identify the matching which are not matched correctly.
a) I,II,III
b) I,III
c) II,I
d) None of the mentioned

Answer: d [Reason:] All are matched correct. High frequency transformer uses ferrites as the magnetic domains reverse very fastly.

6. I. Silicon steel – Power transformer
II. Ferrite – High frequency transformers
III. Alnico – permanent magnet
Mark the matching which are not matched correctly.
a) I,II,III
b) I,III
c) II,I
d) I

Answer: a [Reason:] All are matched correct. High frequency transformer uses ferrites as the magnetic domains reverse very fastly.

7. I. Silicon steel – High frequency transformers
II. Ferrite – Power transformer
III. Alnico – permanent magnet
Choose the matching which are not matched correctly.
a) I,II,III
b) I,III
c) II,I
d) III

Answer: c [Reason:] High frequency transformer uses ferrites as the magnetic domains reverse very fastly. And power transformer uses CRGO steel.

8. I. OC test – Copper and iron losses
II. SC test – Copper losses
III. Load Test – Total losses
IV. Sumpner’s Test – Iron losses
Which of the matches made above are matched correctly?
a) II,III
b) II,III,IV
c) I,IV
d) II,IV

Answer: a [Reason:] OC test gives iron losses. Sumpner’s test gives iron and core losses.

9. In a transformer,eddy-current loss is 1000 watts which is half of the total core loss.
If thickness of lamination and frequency are increased by 10%, the new core-losses are
a) 3200W
b) 3300W
c) 2200W
d) 3100W

Answer: a [Reason:] Pe = 1000W, Pi = 2000W Here, V/f = constant Pe will be same 1000W, Pi = 200+2000 = 2200W Total core losses = 1000+2200 = 3200W.

10. The voltage conversion ratio of the transformer is inverse of the current transformation.
a) True
b) False

Answer: a [Reason:] It is true because, V1/V2 = N2/N1; I2/I1=1/N.

11. The voltage regulation of a transformer at full-load 0.8 p.f lagging is 2%. Its voltage regulation at full load 0.8 p.f leading
a) will be positive
b) will be negative
c) may be positive
d) may be negative

12. The Transformer equivalent circuit referred to primary with converting two winding transformer in impedance forms is
a)
b)
c)
d)

Answer: a [Reason:] When referred to primary, secondary inductance winding will be turns ratio squared.

13. Consider a 10 KVA, 2000/200 V, single phase transformer having below following parameters:
r1=10Ω, r2=0.1Ω, l1= 40mH, l2=0.4 mH, M = 10H. Then the expected primary self inductance will be
a) 100.04H
b) 99.06H
c) 1.0004H
d) 10.04H

Answer: a [Reason:] Primary self inductance = L1 = Lm1 + l1 = (N1/N2)*M + l1 = 2000/200*10 + 0.04 =100.04 H.

14. There is a 10 KVA, 2000/200 V, single phase transformer having below following parameters:
r1=10Ω, r2=0.1Ω, l1= 40mH, l2=0.4 mH, M = 10H. Then the expected secondary self inductance will be
a) 100.04H
b) 99.06H
c) 1.0004H
d) 10.04H

Answer: c [Reason:] Secondary self inductance = L2 = Lm2 + l2 = (N2/N1)*M + l2 = 200/2000*10 + 0.0004H = 0.0004 H.

15. There is a 10 KVA, 2000/200 V, single phase transformer operating at 50 Hz. Maximum flux linkages in secondary will be
a) 0.9 AT
b) 9 AT
c) 1 AT
d) 0.09 AT

Answer: a [Reason:] Flux linkages on secondary = N*φ = E/(4.44*f) = 200/(4.44*50) = 0.9 AT

16. A 10 KVA, 2000/200 V, single phase transformer operating at 50 Hz with a current of 50 A flowing through the l.v. winding. Then the Self inductance of l.v. winding is
a) 12.7mH
b) 1.27 mH
c) 12.7H
d) 0.127 H

Answer: a [Reason:] Flux linkages on l.v. = N*φ = E/(4.44*f) = 200/(4.44*50) = 0.9 AT Self inductance = mutual flux linkage/peak current = 0.9/(1.414*50) = 0.0127 H.

## Set 4

1. The primary and secondary of a transformer are ________ coupled but _______ connected.
a) magnetically, not electrically
b) electrically, not magnetically
c) magnetically, also magnetically
d) electrically, also electrically

Answer: a [Reason:] Transformer is the machine which has physical spacing and has magnetic circuit to exchange the voltage.

2. We can employ transformers for a power range of
a) lower and higer values
b) lower values
c) higher values
d) medium values

Answer: a [Reason:] A transformer can be put in use upto a varying range of the power and it is usually available in readily in market.

3. A transformer has comparitively much higher efficiency than a similar induction machine due to
a) small air gaps
b) no moving parts
c) strong coupling
d) all of the mentioned

Answer: d [Reason:] Transformer does not has any moving components so losses are anyway reduced and also the coupling is very strong between two sides.

4. It was needed that to isolate dc noise coming from the transmitted signal, to attain the same which machine can be used without suffering significant loss
a) transformer
b) dc machine
c) induction machine
d) stepper motor

Answer: a [Reason:] We should use transformer to achive the isolation of dc.

5. The most widely used material in the core of the transformer is
a) cold rolled grain oriented sheet steel
b) cold rolled grain steel
c) soft iron
d) steel

Answer: a [Reason:] CRGO has magnetization in the rolling direction and low core losses and very high permeability than present materials.

6. The leakage flux is the flux in side the transformer which
a) links either of the windings
b) links both of the windings
c) yoke of the core
d) windows of the core

Answer: a [Reason:] Leakage flux is meant to to be loss as it does not link two windings.

7. It is advised that staggering of the butt joints
a) reduces reluctance of the path
b) increases air gap
c) increases mechanical strength
d) all of the mentioned

Answer: c [Reason:] Staggering is done for the steel butt joints of the transformer to gain more mechanical strength as the continuous air gap reduces the same.

8. Consider two transformers X and Y having exact ratings, but have flux densities of 1.5T and 2T respectively. The weight of the transformer A per KVA will be
a) more than that of B
b) lesser than that of B
c) equal to that of B
d) can not be said from the given data

Answer: a [Reason:] Flux density = flux/Area. Hence A has more area than B and so weight of A will be more than that of B.

9. Consider two transformers X and Y having identical ratings, but have flux densities of 1.5T and 2T respectively. The weight of the transformer B per KVA will be
a) more than that of A
b) lesser than that of A
c) equal to that of A
d) can not be said from the given data

Answer: b [Reason:] Flux density = flux/Area. Hence A has more area than B and so weight of B will be less than that of A.

10. Transformers do not require any type of cooling as it well ventilated.
a) True
b) False

Answer: b [Reason:] It is wrong to say that transformers do not need any cooling. It does need various types of cooling like oil cooling, air cooled etc.

11. A coupling magnetic field inside a rotating machine or static machine like transformers must involve with
a) mechanical parts
b) electrical parts
c) both electrical and mechanical
d) either of the electrical or mechanical parts

Answer: a [Reason:] The coupling field should interact with both the electrical as well as mechanical parts in order to achieve electromechanical energy conversion.

12. A coupling magnetic field inside a rotating machine or static machine like transformers must involve with
I. electrical system to extract energy from electrical system.
II. mechanical system to extract energy from electrical system.
a) Only I is true
b) Only II is true
c) I and II are true
d) I and II are false

Answer: b [Reason:] The energy conversion which involves both electrical as well as mechanical systems must have mutual field.

13. I. Pulse transformers use soft ferrites.
II. The transformers used in the radio recievers use air core.
a) Only I is true
b) Only II is true
c) I and II are true
d) I and II are false

Answer: c [Reason:] Pulse transformers are used in isolation transformer and need high permeablity.

14. Stepping of the core is implemented in the core to
a) reduce conductor material and copper losses
b) reduce core loss
c) to provide mechanical strength
d) to reduce magnetizing current

Answer: a [Reason:] Stepping of the core is done so that the copper material reduces.

15. Core type transformers have, LV and HV windings are arranged such that
a) Half LV near the core and half HV outside LV on each limb
b) LV one one limb and HV on the other
c) Half LV outside the core and half HV inside LV on each limb
d) LV and HV windings are sandwiched

Answer: a [Reason:] Core windng has first of all both the windings on the same limb and HV is placed outside to have tappings to the transformer.

16. The shell type transformers have, LV and HV windings are arranged such that
a) Half LV near the core and half HV outside LV on each limb
b) LV one one limb and HV on the other
c) Half LV outside the core and half HV inside LV on each limb
d) LV and HV windings are sandwiched

Answer: d [Reason:] Shell type windng has mix of LV and HV winding sandwiched over each other.

## Set 5

1. I. Maximum voltage regulation occurs at the leading p.f.
II. Maximum voltage regulation occurs when load p.f. angle and impedance angle of the leakage impedance are same.
III. V.R. at zero p.f. is always zero.
IV. V.R. of a transformer may be negative at leading p.f.
Which of the statements made here are correct regarding the transformer?
a) I,III
b) II,IV
c) 1,III
d) I,IV,III

Answer: b [Reason:] V.R. is always negative at leading p.f. and the load p.f. angle and impedance angle of the leakage impedance should be same for maximum V.R.

2. I. Maximum voltage regulation occurs at the leading p.f.
II. Maximum voltage regulation occurs when load p.f. angle and impedance angle of the leakage impedance are same.
III. V.R. at zero p.f. is always zero.
IV. V.R. of a transformer may be negative at leading p.f.
Which of the statements made here are correct regarding the transformer?
a) I,II,III
b) II,IV
c) 1,III
d) I,IV,III

3. I. Maximum voltage regulation occurs at the leading p.f.
II. Maximum voltage regulation occurs when load p.f. angle and impedance angle of the leakage impedance are same.
III. V.R. at zero p.f. is always zero.
IV. V.R. of a transformer may be negative at leading p.f.
Which of the statements made here are incorrect regarding the transformer?
a) I,III
b) II,IV
c) 1,III
d) I,IV,III

Answer: a [Reason:] V.R. is always negative at leading p.f. and the load p.f. angle and impedance angle of the leakage impedance should be same for maximum V.R.

4. I. Maximum voltage regulation occurs at the leading p.f.
II. Maximum voltage regulation occurs when load p.f. angle and impedance angle of the leakage impedance are same.
III. V.R. at zero p.f. is always zero.
IV. V.R. of a transformer may be negative at leading p.f.
Which of the statements made here are incorrect regarding the transformer?
a) I,II,III
b) IV
c) 1,III
d) I,IV,III

5. Which of the below mentioned losses occur in a transformer?
a) Hysteresis losses ;Eddy current losses; Dielectric losses; Stray load losses
b) Hysteresis losses ;Eddy current losses;
c) Dielectric losses; Stray load losses
d) Hysteresis losses ;Eddy current losses; Stray load losses

Answer: a [Reason:] Hysteresis losses-due to magnetic material;Eddy current losses-on the core area; Dielectric losses- due to insulation material; Stray load losses- due to leakgae through the parts of the transformer.

a) decreases, increases
b) increases,increases
c) decreases, decreases
d) increases, decreases

Answer: a [Reason:] The mutual flux falls when the transformer is working at lagging p.f. and it increases at leading p.f.

7. It is possible to attain maximum efficiency in a transformer when the
a) core losses are equal to rated full load copper losses
b) core losses are more than rated full load copper losses
c) core losses and full load copper losses are constant
d) copper loss also becomes constant

Answer: a [Reason:] Maximum efficiency is achieved at the condition when fixed core losses and copper losses at rated condition are equal.

8. The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The the total losses at this condition is
a) 408W
b) 4.08kW
c) 204W
d) 2.04kW

Answer: a [Reason:] Efficiency = 1-(losses/input) = 1-(losses/output+losses) 0.98 = 1-(Losses/20000+losses) Losses = 408 W.

9. The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is
a) 0.01
b) 0.1
c) 1.0
d) 0.0196

Answer: a [Reason:] pu resistance = Ohmic losses/KVA = 200/20000 = 0.01 pu.

10. The load current at which maximum efficiency occurs is independent of the load power factor.
a) True
b) False

Answer: b [Reason:] Because Core losses and the effective resistance of the secondary are almost unaffected by load p.f.

11. The full load voltage drop in a 1-phase transformer is 2% and 4% respectively due to resistance and leakage reactance. Then the voltage drop is maximum at
a) 0.45 lagging
c) 0.9 lagging

Answer: a [Reason:] Z = 4.47; cosθ = percentage r/percentage z = 2/(4.47) = 0.45.

12. The full load voltage drop in a 1-phase transformer is 2% and 4% respectively due to resistance and leakage reactance. Then the voltage drop is zero at
a) 0.45 lagging
c) 0.9 lagging

Answer: c [Reason:] Z = 4.47; cosθ = percentage x/percentage z = 4/4.47 = 0.89 .

13. Core losses of a transformer as a function of frequency is best approximated by which of the below characteristics?
a)
b)
c)
d)

Answer: a [Reason:] We can represent the core losses per cycle eq. as Pc/f = A+B*f, where A and B are constants.

14. From the above graph of the core losses per cycle is, the hysteresis losses per phase for a transformer at a operating frequency of 40 Hz is

a) 20W
b) 30W
c) 80W
d) 32W

Answer: a [Reason:] Ph = k1*f = 0.5*40 = 20W.

15. From the above graph of the core losses per cycle is, the eddy losses per phase for a transformer at a operating frequency of 40 Hz is

a) 15 W
b) 20W
c) 35W
d) 32W

Answer: a [Reason:] Pe = k2*f2; Using the slope of the line, (k2)*50 =0.8-0.5 = 0.3 (k2) = 0.006 Pe = 50*50*0.006 = 15 W

16. The efficiency of a transformer can be calculated accurately from
a) Open circuit test, Short circuit test
b) Open circuit test, Short circuit test, Sumpner’s test
c) Sumpner’s test