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# Multiple choice question for engineering

## Set 1

1. I. Voltages – Direct ratio
II. Current – Inverse ratio
III. Impedance – Direct ratio squared
Iv. Power – Inverse ratio squared
Mark the statements which are marked correct for an ideal transformer.
a) I,II,III,IV
b) III,IV
c) I,II,III
d) I,IV

Answer: c [Reason:] The power and the KVA of an ideal transformer always remains same assuming negligible impedances.

2. The non linear magnetization curve of a practical transformer will introduce
a) heating
b) higher order harmonics
c) magnetostriction
d) all of the mentioned

Answer: d [Reason:] If the magnetization is non linear in nature then it will cause a saturation in the core and harmonics will be introduced to cause humming sounds.

3. Transformer action requires a
a) constant magnetic flux
b) increasing magnetic flux
c) alternating magnetic flux
d) alternating electric flux

Answer: c [Reason:] As per the Faraday’s laws, the emf will be induced when flux is time varying as in transformer, there is no moving part.

4. If a transformer is fed from a 220V and dc supply rather than a 1-phase ac supply, then the transformer will
a) burn its windings
b) operate normal
c) will not operate
d) will give very small leakage flux

Answer: a [Reason:] Dc is nothing but ac at zero frequency. So the reactance offered will be zero at dc and the current will be limited only by the small resistance of the winding which will produce very high amount of current to flow through the windings, so burning them up.

5. In an ideal transformer, the impedance can be transformed from one side to the other
a) in direct proportion to square of turns-ratio
b) in direct proportion to turns-ratio
c) in inverse proportion to square turns-ratio
d) in inverse proportion to turns-ratio

Answer: a [Reason:] Impedance is transformed in square of the turns-ratio.

6. A transformer has sometimes more than two ratings depending upon the use of
a) the cooling application
b) type of windings
c) type of core
d) type of insulation to be given

Answer: a [Reason:] Yes, with different types of the cooling methods, the losses can be varied and so the current and the voltages.

7. Considering a transformer at no load is excited at rated voltage.
A small recognizable gap is made in the yoke of the limbs. With this altercation, the transformer core flux
a) will decrease and magnetizing current will increase
b) will remain constant and magnetizing current will increase
c) as well as magnetizing current will increase
d) as well as magnetizing current both will decrease

Answer: b [Reason:] The reluctance of the path increases here after making the cut. so the magnetizing current will increase but the core flux will be same as it depends on the applied voltage not the reluctance.

8. In an oil filled transformer, the application of oil is for
a) cooling
b) insulation
c) both cooling and insulation
d) preventing the accumulation of dust

Answer: c [Reason:] Oil can be used not only for cooling but also for insulation from the metallic parts.

9. We laminate transformer core to reduce
a) eddy current loss
b) hysteresis loss
c) both eddy current and hysteresis loss
d) ohmic loss

Answer: a [Reason:] Laminations provide larger area so that the current path increases and current reduces.

10. I. Core flux – Depends on applied voltage
II. Leakage flux – Depends on winding current
Which of the above two are matched correctly?
a) I
b) II
c) I,II
d) none of the mentioned

Answer: a [Reason:] Both the statements are correct, as the leakage flux depends on the current flowing in the respective winding and the core flux depends on the voltage.

11. In case of a power transformer, the no load current in terms of rated correct is
a) 10-20%
b) 2-6%
c) 15-20%
d) 30-50%

Answer: b [Reason:] The air gap is very less in the transformer, so the magnetizing current needed is around 2-6% only.

12. Energizing the transformer primary from a triangular wave voltage source makes the output voltage as
a) zero
b) a sine wave
c) a triangular wave
d) a pulsed wave

Answer: c [Reason:] Output will follow the shape of input wave. As the primary input is triangular in shape, its secondary will also be same.

13. In a single phase transformer,the no-load current lags the applied voltage by
a) 90°
c) 0

Answer: b [Reason:] Check the phasor diagram. 14. A 400/200V transformer has total resistance of 0.04 p.u on its L.V side. This resistance when referred to H.V side would be
a) 0.04
b) 0.02
c) 0.01
d) 0.08

Answer: a [Reason:] The per unit value does not change from the winding side of a transformer.

15. Energising the transformer primary from a triangular wave flux makes the output voltage as
a) square wave shifted by 90°
b) a sine wave
c) a triangular wave
d) a square wave

Answer: d [Reason:] E = -N*dϕ/dt. Differentiation of triangular wave results in square wave.

16. The output voltage seen at the CRO connected at the secondary terminals is square wave. Then the flux density used for energizing the primary is
a) triangular
b) sinusoidal
c) saw tooth wave
d) square wave

Answer: a [Reason:] E = -N*dϕ/dt. Integration of square wave will be triangular in nature.

## Set 2

1. In salient-pole machines, the armature mmf cannot be accounted for by introducing one equivalent reactance due to
a) non-uniform airgap
b) variable resistance
c) non-uniform air gap
d) any of the mentioned

Answer: a [Reason:] Due to saliency of the of the rotor poles, the air gap varies and the so one reactance is not sufficient to express the armature mmf.

2. If the internal power factor angle of synchronous motor is 30 degree. Then the direct-axis component of the armature current will be
a) 0.5 pu
b) 0.866 pu
c) 1.73 pu
d) 0

Answer: a [Reason:] Id = Ia*sinψ = 1*sin30 = 0.5 pu.

3. If the internal pf angle of a synchronous motor is 30 degree. Then the quadrature axis component of the armature is
a) 0.866 pu
b) 0.5 pu
c) 1.73 pu
d) 0

Answer: a [Reason:] Id = Ia*cosψ = 1*cos(3*0).

4. A salient pole synchronous generator has following per-unit parameter Xd = 1.2 pu, Xq = 0.8pu, ra = 0.25 ohms.
Then the armature current at 0.8 lagging is
a) 1,-36.9°
b) 1, 36.9°
c) 1, 73°
d) 1, 27°

Answer: a [Reason:] In per unit, armature current will be lagging by 36.9°.

5. The internal power factor angle is given for a lagging load of a 3- phase alternator.
a) ψ = δ – θ
b) ψ = δ + θ
c) ψ = -δ -θ
d) ψ = δ -θ

Answer: b [Reason:] 6. Mark the correct expression for alternator working at a lagging power factor
a) tanψ = (IaXq – Vtsinθ)/(Vtcosθ – Ia*ra)
b) tanψ = (IaXq + Vtsinθ)/(Vtcosθ – Ia*ra)
c) tanψ = (IaXq + Vtsinθ)/(Vtcosθ + Ia*ra)
d) tanψ = (IaXq – Vtsecθ)/(Vtcosecθ – Ia*ra)

Answer: a [Reason:] The internal power factor angle is given as tanψ = (IaXq – Vtsinθ)/(Vtcosθ – Iara) for an alternator.

7. For a 400 V, 3 phase alternator gives an open circuit voltage of 380V and armature current of 38A at a field current of 20A. Then the synchronous reactance of the machine is
a) 10 ohm
b) 38/√3
c) 19 ohm
d) 1.9 ohm

Answer: a [Reason:] Z = Vt/Ia at the same field current = 380/38 = 38 ohms.

8. While finding the parameters for an alternator, its voltage regulation is found to be negative. Then the load connected to it must be
a) capacitive nature
b) inductive nature
c) resistive nature
d) can not be judged based on voltage regulation

Answer: a [Reason:] The given voltage regulation is mentioned as negative. So it must have a load which is a source of reactive power and hence it must be a capacitor.

9. for a synchronous motor, its power factor
a) improves with increase in excitation and may even become leading at higher excitations
b) is independent of its excitation
c) decreases with increase in excitation
d) increases with loading for a given excitation

Answer: a [Reason:] A per the inverted V- curves it can be easily concluded that as the excitation increases synchronous motor becomes leading.

10. For obtaining maximum current when we conduct the ‘Slip Test’ on a synchronous machine, its armature field will align along
a) d-axis
b) q-axis
c) 45° to d-axis
d) 45° to q-axis

Answer: b [Reason:] The maximum current is observed at low reluctance which is along q-axis.

## Set 3

1. When the load on an alternator is varied, its terminal voltage also changes due to
a) armature resistance
b) leakage reactance
c) armature reaction
d) all of the mentioned

Answer: a [Reason:] Armature reactance, leakage reactance and armature reaction cause the terminal voltage to differ from the excitation voltage.

2. Voltage regulation for the alternator operating at leading power factor is negative.
a) True
b) False

Answer: a [Reason:] Voltage regulation for an alternator operating at leading power factor is negative due to magnetizing nature of armature reaction.

3. Voltage regulation for an alternator operating at leading power factor is negative due to
a) magnetizing nature of armature reaction
b) demagnetizing nature of armature reaction
c) cross-magnetizing nature of armature reaction
d) all of the mentioned

Answer: a [Reason:] Field flux gets aided by the armature reaction leading to higher Vt.

4. Alternator on infinite bus bar has constant
a) terminal voltage and frequency
b) frequency
c) power factor
d) power factor and terminal voltage

Answer: a [Reason:] Infinite bus bar has constant voltage and frequency.

5. The emf method of the voltage regulation is applicable only to cylindrical rotor alternator due to
a) resultant air gap flux is not affected by angular position of rotor
b) uniform angular position of rotor
c) non uniform angular position of rotor
d) saliency of the poles is a trouble while estimating the emf

Answer: a [Reason:] It is so because resultant air gap is unaffected by rotor position.

6. Emf method is also known as
a) pessimistic method
b) optimistic method
c) zero power factor method
d) none of the mentioned

Answer: a [Reason:] Unsaturated impedance is greater than saturated impedance, so the drop accounted will be less than the actual making the voltage regulation more than actual.

7. In mmf method
a) all the emf is scaled to mmf
b) only the mmf values are considered neglecting impedance drop
c) all the emf are taken zero
d) emf is converted to saturated impedance drops

Answer: a [Reason:] In mmf method all the emf is scaled to mmf

8. Mmf method of voltage regulation is called _______ while the emf method is __________
a) optimistic, pessimistic
b) pessimistic, pessimistic
c) optimistic, optimistic
d) pessimistic, optimistic

Answer: a [Reason:] In the mmf method the calculated value is lesser than the actual value so it is less misleading than emf method which predicts a greater value.

9. The preferred order of the calculating the voltage regulation is
a) ZPF > ASA > MMF > EMF
b) ZPF > MMF > ASA > EMF
c) ASA > MMF > ASA > EMF
d) EMF > ASA > ZPF > MMF

Answer: a [Reason:] Emf method is least preferred over the other methods and zpf method gives the accurate results as it does not alter the values.

10. In a calculation, the actual voltage regulation is 33.1% while the calculated value of the voltage regulation is 56.8%. This infers that the chosen method has been
a) emf
b) mmf
c) asa
d) zpf

Answer: a [Reason:] As the given value is more than actual, so it must be emf method.

11. In a calculation, the actual voltage regulation is 33.1% while the calculated value of the voltage regulation is 25%. This infers that the chosen method has been
a) emf
b) mmf
c) asa
d) zpf

Answer: b [Reason:] As the given value is less than actual, so it must be mmf method.

12. In a calculation, the actual voltage regulation is 33.1% while the calculated value of the voltage regulation is 32.9%. This infers that the chosen method has been
a) zpf or asa
b) emf
c) mmf
d) mmf or zpf

Answer: a [Reason:] As the given value is very close to actual, so it must be zpf or asa method of regulation.

13. If the emf from the air gap line if the 3-phase alternator is 440V per phase and armature current is 110 A. The synchronous reactance is
a) 4 ohms
b) 2 ohms
c) 6.92 ohms
d) 2.32 ohms

Answer: a [Reason:] Z = V/I = 440/110 = 4 ohms.

14. If the machine size increases, the three phase alternator resistance _______ and synchronous reactance _______
a) increases, decreases
b) decreases, increases
c) remains constant, decreases
d) decreases, decreases

Answer: a [Reason:] If the machine size increases, the copper amount will increase but the air gap will also increase having more reluctance so the reactance will reduce.

15. Ideally the voltage regulation of an alternator should be
a) zero
b) infinite
c) 50%
d) 100%

Answer: a [Reason:] Ideally the terminal voltage and the excitation voltage should be same. So V.R. = E-V/V = 0%.

## Set 4

1. A 10 kVA, 400/200 V, 1-phase transformer with 2% resistance and 2% leakage reactance. It draws steady short circuit current at angle of
a) 45°
b) 75°
c) 135°
d) 0°

Answer: [Reason:] The power factor angle will be atan(x/r)=45°.

2. While conducting open circuit test and short circuit test on a transformer, status of low-voltage and high-voltage windings will be such that in
a) OC test – h.v. open, SC test-l.v. short-circuited
b) OC test – l.v. open, SC test-h.v. short-circuited
c) OC test – l.v. open, SC test-l.v. short-circuited
d) OC test – h.v. open, SC test-h.v. short-circuited

Answer: a [Reason:] In conducting short circuit test, l.v. winding is short circuited. In OC test h.v. is open circuited.

3. While conducting testing on the single phase transformer, one of the student tries to measure the resistance by putting an ammeter across one terminal of primary and other to secondary, the reading obtained will be
a) infinite
b) zero
c) finite
d) negative finite

Answer: a [Reason:] As the primary and secondary are physically isolated, the impedance will be infinite for not electrically connected circuit.

4. If the per unit leakage impedance for the primary of a transformer is ‘x’ on the given rated base value. If the voltage and volt-amperes are doubled, then the changed per unit impedance will be
a) 0.5x
b) 2x
c) 4x
d) x

Answer: a [Reason:] pu(new base)=(x)*(MVA(new)/MVA(old))*(kV(old)/kV(new))^2 =x*2*(1/4) =0.5x.

5. If the per unit leakage impedance for the primary of a transformer is ‘x’ on the given rated base value. If the voltage and volt-amperes are halved, then the changed per unit impedance will be
a) 0.5x
b) 2x
c) 4x
d) x

Answer: b [Reason:] pu(new base)=(x)*(MVA(new)/MVA(old))*(kV(old)/kV(new))^2 =x*0.5*(4/) =2x.

6. The voltage regulation for transformer is given by
a) (E2-V2)/E2
b) (E2-V2)/V2
c) (V2-E2)/E2
d) (V2-E2)/V2

Answer: a [Reason:] Voltage regulation is the change in secondary voltage with secondary rated voltage.

7. While estimating voltage regulation of a transformer, keeping
a) primary voltage constant
b) secondary voltage constant
c) voltage changes constant at primary
d) all of the mentioned

Answer: a [Reason:] V.R. is calculated keeping the primary constant because then the core flux will change and the change of secondary voltage can not be fixed.

8. Identify the phasor diagram for the negative voltage regulation from the below diagrams.
a) b) c) d) Answer: a [Reason:] Negative V.R. is achieved at leading power factor.

9. Identify the phasor diagram for the zero voltage regulation from the below diagrams.
a) b) c) d) Answer: a [Reason:] When E2=V2, then V.R. will be zero.

10. Identify the phasor diagram for the maximum voltage regulation from the below diagrams.
a) b) c) d) Answer: a [Reason:] Maximum voltage regulation occurs when load power factor angle and leakage impedance angle are equal.

11. The transformer phasor diagram under the short circuit can be identified as
a) b) c) d) Answer: a [Reason:] For the short-circuit condition of a transformer, voltage across the secondary will be voltage drop across winding only.

12. With the reference of the diagram below, Choose the most appropriate
A: B: a) A-lagging pf, B-leading pf
b) B-lagging pf, A-leading pf
c) A-lagging pf, B-lagging pf

Answer: a [Reason:] Maximum voltage regulation occurs at lagging pf while zero or minimum V.R. occurs at leading pf.

13. A 200/400 V single phase transformer has leakage impedance z= r+jx. Then we can expect zero voltage regulation at power factor of
b) x/r lagging
d) r/x lagging

Answer: a [Reason:] ZVR occurs at the leading pf of load at x/r.

14. A 200/400 V single phase transformer has leakage impedance z= r+jx. Then we can expect magnitude of load pf of ____ at zero voltage regulation.
a) cosθ = x/z
b) cosθ = r/z
c) cosθ = x/r
d) cosθ = r/x

Answer: a [Reason:] Cosθ = x/z for the transformer at zero V.R.

15. If the pu impedance of a single phase transformer is 0.01+j0.05, then its regulation at p.f. of 0.8 lagging will be
a) 3.8%
b) 2.2%
c) -3.8%
d) -2.2%

Answer: a [Reason:] V.R. = (r(pu)*cosθ+x(pu)*sinθ)*100 % = (0.01*0.8 + 0.05*0.6)*100 = 3.8%

16. If the pu impedance of a single phase transformer is 0.01+j0.05, then its regulation at p.f. of 0.8 leading will be
a) 3.8%
b) 2.2%
c) -3.8%
d) -2.2%

Answer: d [Reason:] V.R. = (r(pu)*cosθ-x(pu)*sinθ)*100 % = (0.01*0.8 – 0.05*0.6)*100 = -2.2%

## Set 5

1. Which is the most basic non-volatile memory?
a) Flash memory
b) PROM
c) EPROM
d) ROM

Answer: d [Reason:] The basic non-volatile memory is ROM or mask ROM, and the content of ROM is fixed in the chip which is useful in firmware programs for booting up the system.

2. Who has invented flash memory?
a) Dr.Fujio Masuoka
b) John Ellis
c) Josh Fisher
d) John Ruttenberg

Answer: a [Reason:] Flash memory is invented by Dr. Fujio Masuoka at Toshiba in the 1980s which are a non-volatile memory.

3. Which of the following is serial access memory?
a) RAM
b) Flash memory
c) Shifters
d) ROM

Answer: c [Reason:] The memory arrays are basically divided into three which are random access memory, serial access memory, and content address memory. Serial access memory is divided into two, theses are shifters and queues.

4. Which is the early form of non-volatile memory?
a) magnetic core memory
b) ferrimagnetic memory
c) anti-magnetic memory
d) anti-ferromagnetic

Answer: a [Reason:] The early form of non-volatile memory is known as magnetic core memory in which the ferromagnetic ring was magnetised to store data.

5. Which of the following memories has more speed in accessing data?
a) SRAM
b) DRAM
c) EPROM
d) EEPROM

Answer: a [Reason:] SRAM have more speed than DRAM because it has 4 to 6 transistors arranged as flip-flop logic gates, that is it can be flipped from one binary state to another but DRAM has a small capacitor as its storage element.

6. In which memory, the signals are multiplexed?
a) DRAM
b) SRAM
c) EPROM
d) EEPROM

Answer: a [Reason:] The signals in address bus are multiplexed with DRAM non-multiplexed with SRAM.

7. How many main signals are used with memory chips?
a) 2
b) 4
c) 6
d) 8

Answer: b [Reason:] The main signals associated with memory chips are four. These are the signals associated with address bus, data bus, chip select signals, and control signals for read and write operations.

8. What is the purpose of address bus?
a) to provide data to and from the chip
b) to select a specified chip
c) to select a location within the memory chip
d) to select a read/write cycle

Answer: c [Reason:] Address bus is used to choose a particular location in the memory chip. Data bus is used to provide data to and from the chip. Chip select signals are used to select a particular chip within the memory.

9. Which are the two main types of processor connection to the motherboard?
a) sockets and slots
b) sockets and pins
c) slots and pins
d) pins and ports

Answer: a [Reason:] The type of processor which connects to a socket on the bottom surface of the chip that connects to the motherboard by Zero Insertion Force Socket. Intel 486 is an example of this type of connection. The processor slot is one which is soldered into a card, which connects to a motherboard by a slot. Example for slot connection is Pentium 3.

10. Which of the following has programmable hardware?
a) microcontroller
b) microprocessor
c) coprocessor
d) FPGA