Previous Year Questions- Three Phase Induction Machines - 3 | Electrical Machines - Electrical Engineering (EE) PDF Download

Q36: A three phase squirrel cage induction motor has a starting current of seven times the full load current and full load slip of 5%
If a starting torque of 0.5 per unit is required then the per unit starting current should be      (2007)
(a) 4.65
(b) 3.75
(c) 3.16
(d) 2.13
Ans: (c)
Sol: T_s = 0.5 T_fl
and s_fl = 0.05(given)
In the case of direction line starting,
Per unit starting current
I_sc = 3.16pu

Q37: A three phase squirrel cage induction motor has a starting current of seven times the full load current and full load slip of 5%
If a star-delta starter is used to start this induction motor, the per unit starting torque will be       (2007)
(a) 0.607
(b) 0.816
(c) 1.225
(d) 1.616
Ans: (b)
Sol: In case of star-delta starter, the ratio of starting torque to full load torque
Per unit starting torque,
T_s = 0.816 pu

Q38: A three phase squirrel cage induction motor has a starting current of seven times the full load current and full load slip of 5%
If an auto transformer is used for reduced voltage starting to provide 1.5 per unit starting torque, the auto transformer ratio(%) should be      (2007)
(a) 57.77%
(b) 72.56%
(c) 78.25%
(d) 81.33%
Ans: (c)
Sol: In case of auto transformer, the starting torque can be expressed as ratio of full load
where,
 x = auto transformer ratio
Substituting the above values in equation (i)
Transformer ratio x = 0.7825 or 78.25

Q39: A three-phase squirrel cage induction motor has a starting torque of 150% and a maximum torque of 300% with respect to rated torque at rated voltage and rated frequency. Neglect the stator resistance and rotational losses. The value of slip for maximum torque is      (2007)
(a) 13.48%
(b) 16.42%
(c) 18.92%
(d) 26.79%
Ans: (d)
Sol: Assuming rated torque = T_rated
Starting torque
T_s = 150% of T_rated = 1.5T_rated
Maximum torque,
T_m = 300% of T_rated = 3 .0T_rated
as we know,
as we know, maximum torque occurs at
s_m = 3.73 is rejected, as for this value of slip motor works in braking mode. Therefore, for motoring mode s_m = 0.2679 or 26.79%

Q40: The electromagnetic torque T_e of a drive and its connected load torque T_L are as shown below. Out of the operating points A, B, C and D, the stable ones are          (2007 )
(a) A, C, D
(b) B, C
(c) A, D
(d) B, C, D
Ans: (a)
Sol: Accelerating torque = T_a = T_e − T_L
At point A, C, D
T_L=T_e
T_a = T_e − T_L = 0
if due to some disturbance speed decreases
T_e > T_L
⇒ T_a = T_e − T_L > 0
So, rotor accelerates and speed increases.
If speed increases due to some disturbances  
T_e < T_L
⇒ T_a = T_e − T_L < 0
So, rotor decelerates and speed decreases.
So, point A, C, D are still stable
At point B,
T_e = T_L
T_a = T_e − T_L = 0
If speed decreases due to some disturbance
T_e < T_L
T_a = T_e − T_L < 0
So rotor decelerates, and rotor speed keeps on decreasing.
If speed increases,
T_e > T_L
T_a = T_e − T_l > 0
Rotor accelerates, and rotor speed keeps on increasing. So, point B is unstable.

Q41: A 3-phase, 10 kW, 400 V, 4-pole, 50Hz, star connected induction motor draws 20 A on full load. Its no load and blocked rotor test data are given below. 
Neglecting copper loss in no load test and core loss in blocked rotor test, estimate motor's full load efficiency      (2006)
(a) 76%
(b) 81%
(c) 82.40%
(d) 85%
Ans: (b)
Sol: No-load losses, P₁ = 1002W = 1.002kW (does not depends on load)
Copper loss at 15A
P_cu = 762W
Full load current = 20A
So, copper loss on full load
Power available at shaft or output power on full load
P₀ = 10kW
Input power = Output power + losses

Q42: A 3-phase, 4-pole, 400 V, 50 Hz, star connected induction motor has following circuit parameters
The starting torque when the motor is started direct-on-line is (use approximate equivalent circuit model)         (2006)
(a) 63.6Nm
(b) 74.3Nm
(c) 190.8Nm
(d) 222.9Nm
Ans: (a)
Sol: rated line to line voltage, V_l−l = 400V
Phase to neutral voltage,  
Synchronous speed at 50 Hz

Q43: The speed of a 4-pole induction motor is controlled by varying the supply frequency while maintaining the ratio of supply voltage to supply frequency (V/f) constant. At rated frequency of 50 Hz and rated voltage of 400 V its speed is 1440 rpm. Find the speed at 30 Hz, if the load torque is constant      (2006)
(a) 882rpm
(b) 864rpm
(c) 840rpm
(d) 828rpm
Ans: (c)
Sol: Torque developed in induction motor
In stable region, slip is very low, so
So, T can be approximated as
as load torque is constant, T₁ = T₂

Q44: Determine the correctness or otherwise of the following assertion[A] and the reason[R].
Assertion [A] : Under V/f control of induction motor, the maximum value of the developed torque remains constant over a wide range of speed in the subsynchronous region.
Reason [R] : The magnetic flux is maintained almost constant at the rated value by keeping the ration V/f constant over the considered speed range.      (2005)
(a) Both [A] and [R] are true and [R] is the correct reason for [A]
(b) Both [A] and [R] are true and but [R] is not the correct reason for [A]
(c) Both [A] and [R] are false
(d) [A] is true but [R] is false
Ans: (a)
Sol: The resultant air-gap flux per pole is given by
Therefore, in order to avoid saturation in stator and rotor cores which would cause sharp increase in magnetization current, the flux f_p must be kept constant. To achieve this, V/f ratio must be kept constant.
Let, V₀ = Nominal voltage
f₀ = Nominal frequency
X'₂₀ = Nominal rotor standstill reactance reffered to stator then at any frequency f.
Then at any frequency f,
= constant, independent of f

Q45: A three-phase cage induction motor is started by direct-on-line (DOL) switching at the rated voltage. If the starting current drawn is 6 times the full load current, and the full load slip is 4%, then ratio of the starting developed torque to the full load torque is approximately equal to      (2005)
(a) 0.24
(b) 1.44
(c) 2.4
(d) 6
Ans: (b)
Sol: Torque in induction motor is given by,

Q46: Under no load condition, if the applied voltage to an induction motor is reduced from the rated voltage to half the rated value,      (2005)
(a) the speed decreases and the stator current increases
(b) both the speed and the stator current decreases
(c) the speed and the stator current remain practically constant
(d) there is negligible change in the speed but the stator current decreases
Ans: (d)
Sol: Let no-load current = I₀
Equivalent-circuit model of induction motor at no-load is drawn.  
If voltage is reduced, no-load current (I₀) reduces. Slip is extremely low (of the order 0.001) at no-load, even if voltage is reduced, there is negligible change in the speed.

Q47: For an induction motor, operation at a slip s, the ration of gross power output to air gap power is equal to       (2005)
(a) (1 − s)²
(b) (1 − s)
(c)
(d)
Ans: (b)
Sol: Let, P_g = Air gap power
P_cu = Rotor copper loss = sP_g
P_m = Gross power output = Air gap power - Rotor copper loss

Q48: On the torque/speed curve of the induction motor shown in the figure, four points of operation are marked as W, X, Y and Z. Which one of them represents the operation at a slip greater than 1 ?       (2005)
(a) W
(b) X
(c) Y
(d) Z
Ans: (a)
Sol: In braking mode, the motor runs in opposite direction to rotating field i.e. N_f (rotor speed) is negative, 
Q49: A 400 V, 15 kW, 4-pole, 50Hz, Y-connected induction motor has full load slip of 4%. The output torque of the machine at full load is     (2004)
(a) 1.66 Nm
(b) 95.50 Nm
(c) 99.47 Nm
(d) 624.73 Nm
Ans: (c)
Sol: N_s (Syn. speed) = (120f)/P
Torque developed

Q50: The synchronous speed for the seventh space harmonic mmf wave of a 3-phase, 8-pole, 50 Hz induction machine is      (2004)
(a) 107.14 rpm in forward direction
(b) 107.14 rpm in reverse direction
(c) 5250 rpm in forward direction
(d) 5250 rpm in reverse direction
Ans: (a)
Sol: Synchronous speed
mmf wave contain space harmonics e.g. fifth and seventh harmonics which correspond to poles five and seven time that of the fundamental.
Since the space-phase difference between fundamental poles of the winding phase is (0°, 120°, 240°), this difference is (0°, 240°, 120°) for fifth harmonic poles and (0°, 120°, 240°) for the seventh. Hence, the fifth harmonic poles rotate backwards with synchronous speed N_s/5 and the seventh harmonic poles rotate forward at 
Q51: The direction of rotation of a 3-phase induction motor is clockwise when it is supplied with 3-phase sinusoidal voltage having phase sequence A-B-C. For counter clockwise rotation of the motor, the phase sequence of the power supply should be      (2004)
(a) B-C-A
(b) C-A-B
(c) A-C-B
(d) B-C-A or C-A-B
Ans: (c)
Sol: To reverse the direction of rotation, phase sequence of the supply has to be reversed. For clockwise direction, the phase sequence was A-B-C. For counter-clockwise direction, the phase sequence has to be A-C-B.

Q52: An ac induction motor is used for a speed control application. It is driven from an inverter with a constant V/f control. The motor name-plate details are as follows
(no. of poles = 2)
V:415 V
Ph:3
f:50 Hz
N:2850 rpm
The motor runs with the inverter output frequency set at 40 Hz, and with half the rated slip. The running speed of the motor is     (2003)
(a) 2400 rpm
(b) 2280 rpm
(c) 2340 rpm
(d) 2790 rpm
Ans: (c)
Sol: where, by (V/f) control,
(synchronous speed at 40 Hz)
N₂ = New running speed of motor

Q53: A 3-phase induction motor is driving a constant torque load at rated voltage and frequency. If both voltage and frequency are halved, following statements relate to the new condition if stator resistance, leakage reactance and core loss are ignored
1. The difference between synchronous speed and actual speed remains same
2. The airgap flux remains same
3. The stator current remains same
4. The p.u. slip remains same
Among the above, correct statements are     (2003)
(a) All
(b) 1, 2 and 3
(c) 2, 3 and 4
(d) 1 and 4
Ans: (b)
Sol: So, air-gap flux remains same.
(X₂′)₁ = nominal rotor stanstill reactance referred to stator (at frequency f₁)
Torque developed in induction motor
For normal steady state oprating region, slip is very low
As the load torque is constant
Difference between synchronous speed and actual speed

Therefore, the difference between synchronous speed and actual speed remain same.
Stator current
Therefore, stator current remains same.

Q54: No-load test on a 3-phase induction motor was conducted at different supply voltage and a plot of input power versus voltage was drawn. This curve was extrapolated to intersect the y-axis. The intersection point yields       (2003)
(a) Core loss
(b) Stator copper loss
(c) Stray load loss
(d) Friction and windage loss
Ans: (d)
Sol: Power input at no-load (P₀) provided losses only as the shaft output is zero. These losses comprise P_i(Iron/core loss) and  P_wf(windage and friction loss).
As the voltage is reduced below the rated value, the core-loss decreases as the square of voltage. Since the slip does not increase significantly, the windage and friction loss remains almost constant. The plot of P₀ versus V is extrapolated at V = 0 which gives, P_wf as P_i = 0 art zero voltage.

Q55: If a 400V, 50 Hz, star connected, 3 phase squirrel cage induction motor is operated from a 400 V, 75 Hz supply, the torque that the motor can now provide while drawing rated current from the supply?       (2002)
(a) reduces
(b) increases
(c) remains the same
(d) increase or reduces depending upon the rotor resistance
Ans: (a)
Sol: As the motor is drawing rated current from the supply.
Hence as frequency increases, torque decreases.