Electrical Drives And Controls Set 7 MCQs

Q1 | To get the speed of DC motor below the normal speed without wastage of electrical energy we use

ward leonard control
rheostatic control
any of the ward leonard or rheostatic method can be used
not possible

Answer: ward leonard control

Explaination:
ward leonard method of speed control is most efficient method of speed control in all aspects. we can get constant torque operation and constant power operations as well, with this method.

Q2 | can give uniform speed variation

in both directions
in one direction
below normal speed only
above normal speed only.

Answer: in both directions

Explaination:
speed control by ward leonard method, gives uniform speed variation in both the directions and above and below of normal speed as well. speed reversal is carried out smoothly by this control method.

Q3 | control method is

high initial cost
high maintenance cost
low efficiency at high loads
high cost, high maintenance and low efficiency

Answer: high cost, high maintenance and low efficiency

Explaination:
ward leonard speed control method requires large number of building blocks like generators, motors. thus, installing cost and maintenance cost of the whole unit is very high. apart from cost, it gives low efficiency at very high loads.

Q4 | TOPIC 5.1 SPEED CONTROL OF THREE PHASE INDUCTION MOTOR

slip frequency
supply frequency
the frequency corresponding to rotor speed
zero

Answer: slip frequency

Explaination:
the relative speed between rotor magnetic field and stator conductors is sip speed and hence the frequency of induced

Q5 | An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 720 rpm. The frequency of the rotor current of the motor in Hz is

2
4
3
1

Answer: 2

Explaination:
given a number of poles = 8. supply frequency is 50 hz. rotor speed is 720 rpm. ns = 120×f÷p=120×50÷8 = 750 rpm. s=ns-nr÷ns = 750 – 720÷750 = .04.

Q6 | Calculate the moment of inertia of the thin spherical shell having a mass of 73 kg and diameter of 36 cm.

1.56 kgm2
1.47 kgm2
1.38 kgm2
1.48 kgm2

Answer: 1.56 kgm2

Explaination:
the moment of inertia of the thin spherical shell can be calculated using the formula i=mr2×.66. the mass of the thin spherical shell and diameter is given. i=

Q7 | Calculate the phase angle of the sinusoidal waveform z(t)=78sin(456πt+2π÷78).

π÷39
2π÷5
π÷74
2π÷4

Answer: π÷39

Explaination:
sinusoidal waveform is generally expressed in the form of v=vmsin(ωt+α) where vm represents peak value, ω represents angular frequency, α represents a phase difference.

Q8 | A 50 Hz, 4poles, a single phase induction motor is rotating in the clockwise direction at a speed of 1425 rpm. The slip of motor in the direction of rotation & opposite direction of the motor will be respectively.

0.05, 0.95
0.04, 1.96
0.05, 1.95
0.05, 0.02

Answer: 0.05, 1.95

Explaination:
synchronous speed, ns=120×50÷4=1500 rpm. given a number of poles = 4. supply frequency is 50 hz. rotor speed is 1425 rpm. s=ns-nr÷ns=1500- 1425÷1500=.05. sb=2-s=1.95.

Q9 | The frame of an induction motor is made of

aluminum
silicon steel
cast iron
stainless steel

Answer: cast iron

Explaination:
the frame of an induction motor is made of cast iron. the power factor of an induction motor depends upon the air gap between stator and rotor.

Q10 | The slope of the V-I curve is 5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

.3254 Ω
.3608 Ω
.3543 Ω
.3443 Ω

Answer: .3443 Ω

Explaination:
the slope of the v-i curve is resistance. the slope given is 5° so r=tan(5°)=.3443 ω. the slope of the i-v curve is reciprocal of resistance.

Q11 | In an induction motor, when the number of stator slots is equal to an integral number of rotor slots

there may be a discontinuity in torque slip characteristics
a high starting torque will be available
the maximum torque will be high
the machine may fail to start

Answer: the machine may fail to start

Explaination:
when the number of stator slots is an integral multiple of a number of rotor slots the machine fails to start and this phenomenon is called cogging.

Q12 | rotor?

30 revolution per minute
40 revolution per minute
60 revolution per minute
50 revolution per minute

Answer: 50 revolution per minute

Explaination:
supply frequency=50 hz. no- load speed of motor = 1000 rpm. the full load speed of motor=950 rpm. since the no- load speed of the motor is almost 1000 rpm, hence synchronous speed near to 1000 rpm.

Q13 | Calculate the active power in a 487 H inductor.

2482 w
1545 w
4565 w
0 w

Answer: 0 w

Explaination:
the inductor is a linear element. it only absorbs reactive power and stores it in the form of oscillating energy. the

Q14 | Calculate the active power in a 788 ω resistor with 178 A current flowing through it.

24.96 mw
24.44 mw
24.12 mw
26.18 mw

Answer: 24.96 mw

Explaination:
the resistor is a linear element. it only absorbs real power and dissipates it in the form of heat. the voltage and current are in the same phase in case of the resistor so the angle between v & i is 90°.

Q15 | an external voltage source is in phase with the main voltage then speed will

increase
decrease
remain unchanged
first increases then decrease

Answer: increase

Explaination:
in the rotor injection method, when an external voltage is in phase with the main voltage net voltage increases and the value of slip decreases and the value of rotor speed increases.

Q16 | R.M.S value of the periodic square waveform of amplitude 72 V.

72 v
56 v
12 v
33 v

Answer: 72 v

Explaination:
r.m.s value of the periodic square waveform is vm and r.m.s value of the trapezoidal waveform is vm÷3½. the peak value of the periodic square waveform is vm.

Q17 | frequency of the rotor current of the motor in Hz is 2.

9.98
9.71
9.12
9.37

Answer: 9.37

Explaination:
given a number of poles = 2. rotor speed is 550 rpm.

Q18 | Calculate the average value of the sinusoidal waveform x(t)=848sin(1.65πt+2π÷0.68).

0
78 v
15 v
85 v

Answer: 0

Explaination:
sinusoidal waveform is

Q19 | phase with the main voltage then speed will

increase
decrease
remain unchanged
first increases then decrease

Answer: decrease

Explaination:
in the rotor injection method, when an external voltage is in opposite phase with the main voltage net voltage decreases and the value of slip increases and the value of rotor speed decreases.

Q20 | The rotor injection method is a part of the slip changing technique.

true
false

Answer: true

Explaination:
rotor injection method comes under slip changing technique. it uses an external voltage source to change the slip value. the load torque remains constant here.

Q21 | synchronous speed changing technique.

true
false

Answer: true

Explaination:
slip recovery scheme comes under the synchronous speed changing technique. it uses an external induction machine to change the frequency value. the load torque remains constant here.

Q22 | relationship between voltage and current is a straight line.

.16 Ω
.26 Ω
.25 Ω
.44 Ω

Answer: .16 Ω

Explaination:
the slope of the v-i curve is resistance. the slope given is 9.1° so r=tan(9.1°)=.16 Ω. the slope of the i-v curve is reciprocal of resistance.

Q23 | If induction motor air gap power is 1.8 KW and gross developed power is .1 KW, then

s
s2
s3
s4

Answer: s

Explaination:
at low values of slip, the electromagnetic torque is directly proportional to slip value. due to heavy loading slip value decreases which increases the ratio of r2÷s.

Q24 | KW.

8 sec
4 sec
7 sec
3 sec

Answer: 4 sec

Explaination:
the fundamental time period of

Q25 | Calculate the total heat dissipated in a resistor of 44 Ω when 0 A current flows through it.

0 w
2 w
1.5 w
.3 w

Answer: 0 w

Explaination:
the resistor is a linear element. it only absorbs real power and dissipates it in the form of heat. the voltage and current are in the same phase in case of the resistor so the angle between v & i is 90°.