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# Multiple choice question for engineering

## Set 1

1. A 3-phase induction motor is fed at the stator at a frequency f. If the rotor speed is Nr and synchronous speed is Ns,
then the absolute speed of the resultant flux in space is
a) Ns
b) (Ns+Nr) if the flux rotates in the same same direction as the rotor
c) (Ns-Nr) if the flux rotates in the same same direction as the rotor
d) Nr

Answer: a [Reason:] The flux speed in space will be Ns irrespective of any rotor speed.

2. I. Leakage reactance is dependent on supply current.
II. Magnetizing reactance is dependent on the air gap flux.
Which of the following is correct?
a) I,II
b) II
c) I
d) None of the mentioned

Answer: a [Reason:] Leakage flux depends on the load current and the magnetizing reactance depends on the air gap flux.

3. I. Rotor resistance is dependent on the speed.
II. Core loss does not depend on the supply frequency and voltage.
a) I
b) II
c) I,II
d) None of the mentioned

Answer: a [Reason:] Core loss does depend on the supply ‘V’ and ‘f’.

4. The stator of a 3-phase, six pole induction motor’s stator is connected to 50 Hz supply while its rotor is fed from a 20 Hz supply.
The possible speeds at which the machine will run will be
(i) 1600rpm (ii) 600rpm (iii) 1400rpm (iv) 400 rpm
a) (ii),(iii)
b) (ii),(iv)
c) (i),(iii)
d) (i),(iv)

Answer: a [Reason:] Stator speed = 120*50/6 = 1000 rpm Rotor speed = 120*20/6 = 400 rpm Hence the possible speeds could be 1400rpm or 600rpm.

5. For a 3-phase 4-pole induction motor has open circuited slip rings. The frequency across the slip rings is 75 Hz. Then we can conclude that the motor is run at a speed of
(i) 750rpm (ii) 1500rpm (iii) 2250rpm (iv) 3750 rpm
a) (i),(iv)
b) (i),(iii)
c) (ii),(iii)
d) (ii),(iv)

Answer: a [Reason:] Stator speed = 120*50/4 = 1500 rpm Rotor speed = 120*50/4 = 2250 rpm Hence the possible speeds could be 750rpm or 3750rpm.

6. A 3-phase slip ring IM is has its copper bars replaced by the aluminium bars in the machine. With the above change,
a) starting torque increases, maximum torque remains same, smT increase
b) starting torque increases, maximum torque remains same, smT remains same
c) starting torque increases, maximum torque remains increases, smT increase
d) starting torque decreases, maximum torque remains same, smT decrease

Answer: a [Reason:] As the resistance has been changed as the aluminium has replaced copper, it will not affect the maximum torque.

7. A 3-phase slip ring IM is has its copper bars changed by the aluminium bars in the machine. With the above change,
a) starting torque remains same
b) starting torque increase
c) starting torque decrease
d) maximum torque increase

Answer: a [Reason:] Resistance increase, so the starting torque increase.

8. A center ammeter connected to the rotor end circuit of a 6-pole,50 Hz, induction motor makes 45 complete oscillations in a minute.
Then the rotor speed and the speed of stator field w.r.t. rotor is
a) 985, 15
b) 970, 1000
c) 985, 985
d) 985, 970

Answer: a [Reason:] Rotor frequency = s*f1 = 45/60 = 0.75 Hz. s = 0.75/50 = 0.015 rotor speed = 1000(1-0.015) = 985 rpm. Speed of stator field w.r.t. rotor = Ns-Nr = 1000-985=15 rpm.

9. Choose from the below which can be obtained by the equivalent circuit of an electrical machine?
a) Complete performance characteristics of the machine
b) Temperature coefficients of the machine components
c) Type of protection to be used in the machine
d) Design parameters of the windings

Answer: a [Reason:] Equivalent circuit can help to give the complete performance characteristics of an electrical machine.

10. If one of the supply fuses blows off and gets open circuited, then for the connected induction machine continues to run with increased slip.
a) True
b) False

Answer: a [Reason:] It is true because as one of the phase is missing, it will lead to decrease in speed assuming machine was providing the rated load.

11. Due to the line to ground fault, phase ‘a’ in the induction motor phase supply gets cut, then machine
a) continues to run with excessive supply current
b) continues to run with reduced slip
c) stalls
d) all of the mentioned

Answer: a [Reason:] It continues to run at lowered speed but with increased current.

a) power factor improves
b) power factor remains same
c) power factor varies linearly
d) power factor increases till 40% load and then decreases

Answer: a [Reason:] Overall during the operation power factor improves as the reactance reduces and the power factor angle reduces.

13.For 3-phase induction motor, as load increases from no load towards the full load, torque _______ slip.
a) increases in proportion to
b) decreases in proportion to
c) remains constant to
d) increases hyperbolically to

Answer: a [Reason:] For the lower slip, torque will increase proportionally to the slip as verified by the torque slip characteristic.

A. power factor improves
B. torque increases
C. air gap flux falls steeply
a) A, B
b) B, C
c) A, B, C
d) A,C

15. Torque-slip characteristic of an induction motor is linear in the smaller slip values, because
a) effective rotor resistance is very large compared to reactance
b) rotor resistance is equal to stator resistance
c) rotor resistance is equal to rotor reactance
d) rotor reactance almost equal to stator reactance

Answer: a [Reason:] It is because effective resistance value(r2/s) is far more than reactance and so the torque is linear in this region.

16. If a 3-phase induction motor is fed from the rotor short circuiting the stator terminals, frequency of the current flowing in the short-circuited stator is
a) slip frequency
b) supply frequency
c) zero
d) frequency corresponding to the rotor speed

Answer: a [Reason:] In the stator the flux developed will revolve with a speed of slip frequency.

17. A constant torque variable power drive of induction machine has
a) constant air gap flux.
b) constant slip
c) variable air gap flux, constant slip
d) constant slip, constant air gap flux

Answer: d [Reason:] A constant torque drive should have constant flux and hence constant air gap flux. Also it should have constant slip.

18. If an emf at slip frequency is injected at an angle α with existing rotor induced emf, then
a) power factor only improves
b) speed changes
c) power factor improves and speed changes
d) power factor worsens and speed changes

Answer: a [Reason:] Power factor increases as the net angle decreases.

## Set 2

1. A 315V, 3-phase 400 MVA alternator is running at no load at constant excitations. The most appropriate representation in space domain is
a)
b)
c)
d)

Answer: a [Reason:] At no load Ia = 0. So Ef = Vt.

2. A 315V, 3-phase 400 MVA alternator is running at lagging power factor at constant excitations. The most appropriate representation in space domain is
a)
b)
c)
d)

Answer: a [Reason:] Ia should lag Ef.

3. A 315V, 3-phase 400 MVA alternator is running at leading power factor load at constant excitations. The most appropriate representation in space domain is
a)
b)
c)
d)

4. A 315V, 3-phase 400 MVA alternator is running at unity power factor at constant excitations. The most appropriate representation in space domain is
a)
b)
c)
d)

Answer: a [Reason:] Ia in phase with excitation voltage.

5. A 315V, 3-phase 400 MVA alternator is running at zero power factor lagging at constant excitations. The most appropriate representation in space domain is
a)
b)
c)
d)

Answer: a [Reason:] Ia has to be lagging by 90°.

6. A 315V, 3-phase 400 MVA alternator is running at zero power factor leading at constant excitations. The most appropriate representaion in space domian is
a)
b)
c)
d) none of the mentioned

7. With the alternator running at rated terminal voltage Vt, at rated current Ia = 1 per unit. What happens if load current is reduced to zero, if it was operating at lagging power factor?
a)
b)
c)
d)

Answer: a [Reason:] Check for the Ia lagging Ef and Ef > Vt.

8. With the alternator running at rated terminal voltage Vt, at rated current Ia = 1 per unit. What happens if load current is reduced to zero, if it was operating at leading power factor?
a)
b)
c)
d)

Answer: a [Reason:] Check for the Ia leading Ef and Ef < Vt.

9. If the machine was operating at upf, then the new excitation voltage will be
a) greater than 1 pu
b) less than 1 pu
c) 1 pu
d) zero

Answer: a [Reason:] The new excitation will be more than 1 pu to compensate for the flux required.

10. The variation of Vt vs Ia is for the alternator is given below.

Choose the appropriate

Answer: a [Reason:] A is zpf lead for the alternator and D will be zpf lag.

11. Mark the plot for terminal voltage vs armature current for the alternator.
a)
b)
c)
d)

Answer: a [Reason:] After the maximum excitation the alternator will have a fall out if the excitation is increased beyond synchronous speed.

12. Alternator compounding characteristic is obtained by
a) keeping Vt constant
b) keeping Ef constant
c) keeping Ia constant
d) varying Vt

Answer: a [Reason:] Alternator compounding characteristic is obtained by keeping Vt constant.

13. At zpf lag, the excitation should be increased in order to maintain the armature terminal voltage constant.
a) True
b) False

Answer: a [Reason:] Due to demagnetizing armature reaction at zpf lag in an alternator.

14. The rating of the alternator is decided by
a) losses
b) voltage
c) armature current
d) temperature

Answer: a [Reason:] The losses determine the rating to be used for alternator.

15. The core losses in synchronous machine is _______ dependent.
a) voltage
b) current
c) temperature
d) insulation

Answer: a [Reason:] Core losses depend on the voltage applied to the machine.

## Set 3

1. In the design of a single mode step index optical fiber close to upper cut off, the single mode operation is NOT preserved if
a) Radius as well as operating wavelength are halved
b) Radius as well as operating wavelength are doubled
c) Radius is halved and operating wavelength is doubled
d) Radius is doubled and operating wavelength is halved

Answer: d [Reason:] The cut off frequency is given by f = 2πa sin α/wavelength, where a is the radius of core. For a single mode step index fiber, f must lie between 0 and 2.405. Thus to get f below 2.405, the radius must be doubled and wavelength must be halved.

2. Which parameters cannot be computed from the Smith chart?
a) Impedance
c) Reflection coefficient and VSWR
d) Intrinsic impedance

Answer: d [Reason:] From the Smith chart, the parameters impedance, admittance, reflection coefficient and VSWR can be computed directly. The intrinsic impedance cannot be calculated.

3. Consider a transmission line of characteristic impedance 50 ohm. Let it be terminated at one end by +j50 ohm. The VSWR produced by it in the transmission line will be
a) 1
b) 0
c) Infinity
d) +j

Answer: c [Reason:] The reflection coefficient is given by R = j50 – 50/j50 + 50 = j – 1/j + 1. Thus the reflection coefficient is given by 1. The VSWR = 1 + R/1 – R = 2/0 = infinity.

4. A transmission line whose characteristic impedance is a pure resistance
a) Must be a lossless line
b) Must be a distortionless line
c) May not be a lossless line
d) May be a lossless and may not be a distortionless line

Answer: d [Reason:] When the characteristic line is a pure resistance, the line must be a distortionless line. But it may not be a lossless line.

5. A two wire transmission line of characteristic impedance Z is connected to a load impedance ZL(ZL and Zo are not equal). Impedance matching cannot be achieved with
a) A quarter wavelength transformer
b) A half wavelength transformer
c) An open circuited parallel stub
d) A short circuited parallel stub

Answer: b [Reason:] When the characteristic and load impedance are not same, a half wavelength transformer cannot be used for impedance matching because when l = wavelength/2, we get Zin = ZL.

6. A transmission line of pure resistance characteristic impedance is terminated with an unknown load. The measured value of VSWR on the line is 2 and a voltage minimum point is found to be at the load. The load impedance is then
a) Complex
b) Purely capacitive
c) Purely resistive
d) Purely inductive

Answer: c [Reason:] Given, voltage minimum point is at load. If minimum voltage or maximum voltage occurs at the load for a lossless transmission line then the load impedance is purely resistive.

7. Which one of the following field patterns represents a TEM wave travelling in the positive x direction?
a) E = +8y, H = -4z
b) E = -2y, H = -3z
c) E = +2z, H = +2y
d) E = -3y, H = +4z

Answer: b [Reason:] To get wave travelling in the positive direction, the E component has to be in negative y direction and the H component has to be in negative z direction. Thus E = -2y and H = -3z is the right option.

8. The modes in a rectangular waveguide are denoted by TEmn / TMmn, where m and n are the Eigen values along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is true?
a) The TM10 mode does not exist
b) The TE10 mode does not exist
c) The TE10 and TM10 both exist and have same cut off frequencies
d) When m and n are increases, the cut off frequency decreases

Answer: a [Reason:] For TM mode, TM00, TM10, TM01 does not exist. For TE mode, TE00 does not exist but TE10 and TE01 exists.

9. In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz?
a) To increase the sensitivity of measurement
b) To transmit the signal to a far off place
c) To study amplitude modulation
d) Because crystal detector fails at microwave frequencies

Answer: d [Reason:] Since crystal detector fails at microwave frequencies, microwave signal amplitude is modulated at 1kHz.

10. Consider a lossless antenna with a directive gain of +6 decibel. If 1 milliwatt of power is fed to it, the total power radiated by the antenna will be (in milliwatt)
a) 4
b) 1
c) 7
d) 1/4

Answer: b [Reason:] For a lossless antenna, the input power and the radiated power will be same. The efficiency will be 100%. Thus the radiated power will also be 1 milliwatt.

## Set 4

1. The Smith chart is a polar chart which plots
a) R vs Z
b) R vs Znorm
c) T vs Z
d) T vs Znorm

Answer: b [Reason:] The Smith chart is a frequency domain plot. It is the polar chart of the reflection coefficient R with respect to the normalised impedance Znorm.

2. The Smith chart is graphical technique used in the scenario of transmission lines. State true/false.
a) True
b) False

Answer: a [Reason:] The Smith chart is used for calculating the reflection coefficient and standing wave ratio for normalised load impedance of a transmission line.

3. The Smith chart consists of the
a) Constant R and variable X circles
b) Variable R and constant X circles
c) Constant R and constant X circles
d) Variable R and variable X circles

Answer: c [Reason:] The Smith chart consists of the constant resistance circles and the constant reactance circles. The impedances are plotted using these circles. Also stub matching can be done using the Smith chart.

4. The circles in the Smith chart pass through which point?
a) (0,1)
b) (0,-1)
c) (-1,0)
d) (1,0)

Answer: d [Reason:] All the constant resistance and reactance circles in the Smith chart pass through the (1,0) point. This is the midpoint of the Smith Chart. The resistance is unity and reactance is zero at this point.

5. Moving towards the clockwise direction in the Smith chart implies moving
a) Towards generator
c) Towards stub
d) Towards waveguide

Answer: a [Reason:] On moving towards the clockwise direction in the Smith chart, we are traversing towards the generator. This is used to calculate the normalised load impedance.

6. The centre of the point having a normalised resistance of 1.2 ohm and reactance of 1.5 ohm is
a) (0.54,0)
b) (0.45,0)
c) (0.36,0)
d) (0.78,0)

Answer: a [Reason:] The centre of a point in Smith chart is given by C = (r/1+r, 0). On substituting for r = 1.2, we get centre as (1.2/1+1.2,0) = (0.54,0).

7. The normalised load impedance of the transmission line 50 ohm with a load of 30 ohm is
a) 30
b) 150
c) 5/3
d) 3/5

Answer: d [Reason:] The normalised impedance is calculated by dividing the impedance with the characteristic impedance. Given that the load impedance is 30 ohm, the normalised load impedance of the 50 ohm transmission line is 30/50 = 3/5 ohm.

8. The radius of the point having a normalised resistance of 1 ohm is
a) 1
b) 0.2
c) 0.5
d) 0.25

Answer: c [Reason:] The radius of the point with a radius r is given by R = 1/r+1. On substituting for r = 1, we get R = 1/1 + 1 = ½ = 0.5.

9. The best stub selection for the transmission line will be
a) Series open
b) Series short
c) Shunt open
d) Shunt short

Answer: d [Reason:] Normally series stubs are not preferred as modification of the stub parameters requires changing the whole stub setup. Shunt stubs enable modification with ease. Open circuited stubs are not preferred as it will radiate power like an antenna, which is undesirable. Hence shorted stubs are used.

10. The centre and radius of a line with normalised load impedance of 1 + 0.5j is
a) (1,2) and 2
b) (2,1) and 2
c) (1,2) and 1
d) (2,1) and 1

Answer: a [Reason:] The centre and radius of a line are (1, 1/x) and 1/x, where x is the reactance. Here x = 0.5, from the given data. Thus C = (1,2) and R = 2.

## Set 5

1. The source voltage of a 75ohm transmission line is given by 150V. Find the load current.
a) 0.5
b) 2
c) 4
d) 1

Answer: b [Reason:] The load current is given by IL = VS/Z0. On substituting for VS = 150 and Z0 = 75, we get IL = 150/75= 2A.

2. The guided terminations are used to
a) Increase reflection
b) Increase transmission
c) Eliminate reflection loss
d) Eliminate attenuation

Answer: c [Reason:] The guided termination refers to the waveguide shorted by conducting plates. This is done in order to eliminate the reflection losses.

3. Which type of wave does the resonant cavity produce?
a) Standing waves
b) Guided waves
c) Transmitted waves
d) Attenuated waves

Answer: a [Reason:] Resonant cavity is the waveguide shorted by a conducting plate. This is to reduce the reflection losses. Such arrangement leads to standing waves.

4. Which of the following parameter cannot be calculated from the standing waves?
a) Peak voltage and peak current
b) SWR
c) Reflection and transmission coefficients
d) Attenuation constant

Answer: d [Reason:] The peak voltage and current can be directly measured from the standing waves. The standing wave ratio, reflection coefficient and the transmission coefficient can also be calculated from it. Only the attenuation constant cannot be calculated directly.

5. For efficient transmission, the characteristic impedance of the transmission line has to be
a) 50 ohm
b) 75 ohm
c) Either 50 or 75 ohm
d) Neither 50 nor 75 ohm

Answer: c [Reason:] Generally, for ideal transmission lines, the characteristic impedance should be either 50 ohm or 75 ohm.

6. The cavity resonators are used in the klystron amplifiers for
a) Amplifying RF signals
b) Amplifying microwave signals
c) Attenuating RF signals
d) Attenuating microwave signals

Answer: b [Reason:] The cavity resonators are employed in the klystron amplifiers for amplifying the microwave signals.

7. The cavity resonators used in the reflex klystron oscillators are for
a) Generating RF signals
b) Generating microwave signals
c) Amplifying RF signals
d) Amplifying microwave signals

Answer: b [Reason:] Oscillators are devices that generate signal waveforms. The reflex klystron oscillator is used to generate microwave signals.

8. One of the applications of the cavity resonators is duplexer in RADAR systems. State true/false.
a) True
b) False

Answer: a [Reason:] Cavity resonator is used in duplexers of RADAR systems, as resonant cavity in transmit receive (TR) tubes and antitransmit receive (ART) tubes.

9. Cavity wave meters are used to measure which parameter of the wave?
a) Wavelength
b) Reflection factor
c) Phase
d) Frequency