Previous Year Questions- Three Phase Induction Machines - 2 | Electrical Machines - Electrical Engineering (EE) PDF Download

Q21: A single phase induction motor draws 12 MW power at 0.6 lagging power. Acapacitor is connected in parallel to the motor to improve the power factor of the combination of motor and capacitor to 0.8 lagging. Assuming that the real and reactive power drawn by the motor remains same as before, the reactive power delivered by the capacitor in MVAR is ______.      (SET-2 (2014))
(a) 6
(b) 7
(c) 8
(d) 9
Ans: (b)
Sol: Given , motor draws 12 MW at 0.6 pf lag
For improving the power factor to 0.8 lag, Q should be reduced i.e. ϕ reduces
ϕ′ = (16 − Q)MV AR
Where, Q is the reactive power delivered by the capacitor.
Now, power factor,
from power traingle,
Reactive power delivered by the capacitor, Q = 7.07 MVAR

Q22: A three-phase, 4-pole, self excited induction generator is feeding power to a load at a frequency f₁. If the load is partially removed, the frequency becomes f₂. If the speed of the generator is maintained at 1500 rpm in both the cases, then       (SET-2 (2014))
(a) f₁, f₂ > 50 Hz and f₁ > f₂ 
(b) $𝑓1<$f₁ < 50 Hz and f₂ > 50 Hz
(c) f₁, f₂ < 50 Hz and f₂ > f₁
(d) $𝑓1>$f₁ > 50 Hz and f₂ < 50 Hz
Ans: (c)
Sol: An induction generator always operates at negative slip,
As the generator speed is maintained at a constant speed of 1500 rpm. Therefore, the synchronous speed will be less that 1500 rpm in both the cases (i.e. when at full load as well as reduced load).
N_s = 1500 rpm for P = 4 and f = 50Hz.
Also, when load decreases then, frequency increases.
Hence,  f₂ > f₁
Therefore, option (C) is correct.  

Q23: A 3 phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1 Ω and reactance of 0.92 Ω . Neglect the voltage drop in stator and assume that the rotor resistance is constant. Given that the full load slip is 3%, the ratio of maximum torque to full load torque is       (SET-1 (2014))
(a) 1.567
(b) 1.712
(c) 1.948
(d) 2.134
Ans: (c)
Sol: Given,
Slip at maximum torque,
We know that,
Therefore, ratio of maximum torque to full-load torque

Q24: An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 700 rpm. The frequency of the rotor current of the motor in Hz is ______.      (SET-1 (2014))
(a) 7.56
(b) 1.85
(c) 2.15
(d) 3.33
Ans: (d)
Sol: Given, P = 8, f = 50Hz, N_r = 700rpm
Synchronous speed,
Therefore, frequency of rotor current,
Hence, answer is 3.33Hz.

Q25: A 4-pole induction motor, supplied by a slightly unbalanced three-phase 50 Hz source, is rotating at 1440 rpm. The electrical frequency in Hz of the induced negative sequence current in the rotor is      (2013)
(a) 100
(b) 98
(c) 52
(d) 48
Ans: (b)
Sol: For negative sequence component,

Q26: Leakage flux in an induction motor is     (2013 )
(a) flux that leaks through the machine
(b) flux that links both stator and rotor windings
(c) flux that links none of the windings
(d) flux that links the stator winding or the rotor winding but not both
Ans: (d)

Q27: The locked rotor current in a 3-phase, star connected 15 kW, 4 pole, 230 V, 50 Hz induction motor at rated conditions is 50 A. Neglecting losses and magnetizing current, the approximate locked rotor line current drawn when the motor is connected to a 236 V, 57 Hz supply is       (2012)
(a) 58.5 A
(b) 45.0 A
(c) 42.7 A
(d) 55.6 A
Ans: (b)
Sol: At standstill, the rotor current is
As losses are zero

Q28:  The slip of an induction motor normally does not depend on      (2012)
(a) rotor speed
(b) synchronous speed
(c) shaft torque
(d) core-loss component
Ans: (d)
Sol: From the above formula slip depends upon:
(1) Synchronous speed (N_s)
(2) Rotor speed (N_r)
as the shaft torque depends upon rotor speed therefore the slip also depends on shaft torque. And core-losses are independent of slip.

Q29: A three-phase 440 V, 6 pole, 50 Hz, squirrel cage induction motor is running at a slip of 5%. The speed of stator magnetic field to rotor magnetic field and speed of rotor with respect of stator magnetic field are       (2011)
(a) zero, -50 rpm
(b) zero, 955 rpm
(c) 1000 rpm, -5 rpm
(d) 1000 rpm, 955 rpm
Ans: (a)
Sol: Stator and rotor magnetic field rotates at same speed. So difference in speed is zero. Speed of stator magnetic field is N_s ( synchronous speed)
So, speed of rotor with respect to stator magnetic field = N_r - N_s
= 950 - 1000 = -50 rpm

Q30: A separately excited dc machine is coupled to a 50 Hz, three-phase, 4-pole induction machine as shown in figure. The dc machine is energized first and the machines rotate at 1600 rpm. Subsequently the induction machine is also connected to a 50 Hz, three-phase source, the phase sequence being consistent with the direction of rotation. In steady state       (2010)
(a) both machines act as generator
(b) the dc machine acts as a generator, and the induction machine acts as a motor
(c) the dc machine acts as a motor, and the induction machine acts as a generator
(d) both machines act as motors
Ans: (c)
Sol: As both the machines are coupled, rotor of both the machines rotate at same speed.
Rotor speed of DC machine = 1600 rpm
Rotor spedd of induction machine =
N_r rotor speed of DC machine
N_r = 1600 rpm
Speed of stator field in induction machine = N_s
Slip of rotor wrt stator field in induction machine
As slip is negative, induction machine acts as a generator and the dc machine acts as a motor.

Q31: A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetizing reactance are denoted by r_s, r_r, X_s, X_r and X_m, respectively. The magnitude of the starting current of the motor       (2010)
(a)
(b)
(c)
(d)
Ans: (a)
Sol: Starting current in induction motor can be as large as 5 to 6 times the full load current. As compared to starting current (I_st), exciting current (I_m) is very small  (I_m < ltI_st)
Therefore, shunt branch can be neglected in the circuit model of the motor.

Q32: A 3-phase, 440 V, 50 Hz, 4-pole slip ring induction motor is feed from the rotor side through an auto-transformer and the stator is connected to a variable resistance as shown in the figure.
The motor is coupled to a 220 V, separately excited d.c generator feeding power to fixed resistance of 10 Ω. Two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such the motor runs at 1410 rpm and the following readings were recorded W₁ = 1800W, W₂ = −200W.
Neglecting all losses of both the machines, the dc generator power output and the current through resistance (R_ex) will respectively be        (2008)
(a) 96 W, 3.10 A
(b) 120 W, 3.46 A
(c) 1504 W, 12.26 A
(d) 1880 W, 13.71 A
Ans: (c)
Sol: Slip = 90/1500 = 0.06
Total power input to induction motor within = 1800 - 200 = 1600 W
Power output of induction motor
Since all the losses are neglected.
Induction motor power output = d.c. generator input = d.c. generator output = 1504 W

Q33: A 3-phase, 440 V, 50 Hz, 4-pole slip ring induction motor is feed from the rotor side through an auto-transformer and the stator is connected to a variable resistance as shown in the figure.
The motor is coupled to a 220 V, separately excited d.c generator feeding power to fixed resistance of 10 Ω. Two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such the motor runs at 1410 rpm and the following readings were recorded
W₁ = 1800W, W₂ = −200W.
The speed of rotation of stator magnetic field with respect to rotor structure will be       (2008)
(a) 90 rpm in the direction of rotation
(b) 90 rpm in the opposite direction of rotation
(c) 1500 rpm in the direction of rotation
(d) 1500 rpm in the opposite direction of rotation
Ans: (d)
Sol: If the three phase supply is given to the rotor winding through an autotransformer, the three phase rotor current will generate a rotating field in the air-gap, rotating at the synchronous speed woth respect to rotor. If the rotor is kept stationary, this rotating field will also rotate in the gap at the synchronous speed voltage and current will be induced in the stator winding and a torque will be developed. If rotor is allowed to move, it will rotate as per the Lenz's law, opposing the rotation of the rotating field decreasing the induced voltage in the stator winding. Thus at a particular speed, the frequency of the stator circuit will correspond to the slip speed.
Synchronous speed,
As stator magnetic field rotates 90 rpm in the opposite direction of the rotation of rotor, therefore, speed of the stator field with respect to roto= 1410 + 90 = 1500rpm

Q34: A 400 V, 50 Hz, 4-pole, 1400 rpm, star connected squirrel cage induction motor has the following parameters referred to the stator:
Neglect stator resistance and core and rotational losses of the motor. The motor is controlled from a 3-phase voltage source inverter with constant V/f control. The stator line-to-line voltage(rms) and frequency to obtain the maximum torque at starting will be :        (2008)
(a) 20.6 V, 2.7 Hz
(b) 133.3 V, 16.7 Hz
(c) 266.6 V, 33.3 Hz
(d) 323.3 V, 40.3 Hz
Ans: (b)
Sol: For max. torque slip
For starting torque, s_m = 1
In const V/f control method,

Q35: A 400 V, 50 Hz 30 hp, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be       (2008)
(a) 23.06 kW
(b) 24.11 kW
(c) 25.01 kW
(d) 26.21 kW
Ans: (c)
Sol: Line to line supply voltage = V_l−l = 400V
Current drawn by the motor, I_L= 50A at 0.8 pf
Input power,
Air-gap power (P_g) in the power that is transferred from the stator to the rotor via the air-gap magnetic field.
 P_g = Input power - Core loss - Copper loss in stator
= 27.71 - 1.5 - 1.2 = 25.01 kW