Ques .31. An eight-pole wound rotor induction motor operating on a 60 Hz supply is driven at 1800 rpm by a prime mover in the opposite direction of the revolving magnetic field. The frequency of rotor current is (SSC-2015)
120 Hz
180 Hz
60 Hz
200 Hz
Answer.2. 180 Hz
The speed of the induction motor is
N = 120f/P
f = PN/120
f = (8 x 1800)/120
= 120
Frequency of rotor current is
f’ = f + 60 = 120 + 60
F’ = 180
Second Method
Ques.32. The rotor slots, in an induction motor, are usually not quite parallel to the shaft because of it; (SSC-2014)
Improve the power factor
Improves the efficiency
Help the rotor teeth to remain under the stator
Help in reducing the tendency of the rotor teeth to remain under the stator teeth
Answer.4. Help in reducing the tendency of the rotor teeth to remain under the stator teeth
Explanation:-
The slots are not made parallel to each other but are bit skewed as the skewing prevents magnetic locking of stator and rotor teeth and makes the working of the motor more smooth and quieter.
Ques.33. The synchronous speed of a three-phase induction motor having 20 poles and connected to a 50 Hz source is (SSC-2014)
1200 rpm
300 rpm
600 rpm
1000 rpm
Answer.2. 300 rpm
Explanation:-
Synchronous speed of synchronous motor is given by
Ns = 120f/P
=120 x 50/20 = 300 RPM
Ques.34. When the rotor of a three-phase induction motor is blocked, the slip is (SSC-2014)
1
0
0.1
0.5
Answer.1. 1
Explanation:-
When the rotor of a three-phase induction motor is blocked then the rotor speed “Nr” = 0
and
s = Ns – Nr/Ns
= Ns – 0/Ns = 1
Ques.35. A 3-phase 4 pole induction motor works on 3-phase 50 Hz supply. If the slip of the motor is 4%. The actual speed will be (SSC-2014)
720 RPM
1550 RPM
1460 RPM
1440 RPM
Answer.4. 1440 RPM
Explanation:-
Synchronous speed
Ns =120f/P = 120 x 50/4 = 1500 RPM
% S = (Ns – Nr) x 100/Ns
4 =(1500 – Nr) x100/1500
60 = 1500 – Nr
Nr = 1500 – 60 = 1440 RPM
Ques .36. Low-frequency operation of A.C series motor in traction application (SSC-2014)
Improves its commutation but starting current increases
Improves its commutation property but pf and η reduces
Improves its commutation, pf, and efficiency
Adversely affects commutation but pf and η improves
Answer.3. Improves its commutation, pf, and efficiency
Explanation:-
AC series motors are used to produce the propulsion of the vehicle. This system uses AC voltages from 15-25KV at a frequency of 16.7 (i.e., 16 2/3) or 25 Hz. This low frequency leads to giving better performance and more efficient operation by the series motor.
The low-frequency operation of the overhead line reduces communication interferences. Also, the reactance of the line is low at the lower frequency and hence the voltage drop in the line is reduced. Thus increasing efficiency and P.f
Ques.37. The speed of a p-pole synchronous machine in rpm is given by (SSC-2014)
120 fp
120f/P
120P/f
120f/2p
Answer.2. 120f/P
Explanation:-
The speed of p-pole synchronous machine is given as 120f/p
Ques.38. In a 3- phase induction motor crawling happens at (SSC-2014)
Any speed
No-load speed
Odd multiples of fundamental
Even multiples of fundamental
Answer.3. Odd multiples of fundamental
Explanation:-
An induction motor is a single excited machine. The crawling word it self-suggest crawl means moving with low speed. This characteristic is the result of improper functioning of the motor that means either motor is running at very slow speed or it is not taking the load.
The resultant speed is nearly 1/7th of its synchronous speed. This action is due to the harmonics fluxes produced in the gap of the stator winding of odd harmonics like 3rd, 5th, 7th, etc.
Ques.39. A 4-pole, 3-phase induction motor runs at 1440 rpm on a 50 Hz supply. Find the slip speed (SSC-2014)
2940 rpm
1500 rpm
1440 rpm
60 rpm
Answer.4. 60 rpm
Explanation:-
Synchronous speed Ns is
Ns = 120f/P
Ns = 120 x 50/4 = 1500 rpm
Slip speed = Ns – Nr
= 1500 – 1440
= 60 rpm
Ques.40. How many watt-seconds are supplied by a motor developing 2 hp (British) for 5 hours? (SSC-2013)
2.68452×10 7 watt-seconds
4.476×10 5 watt-seconds
2.646×10 7 watt-seconds
6.3943×10 6 watt-seconds
Answer.1. 2.68452×10 7 watt-seconds
Explanation:
1 HP=745.7 Watt of Power
2 HP = 1491.4 Watt of Power
So, 2 HP motor working for 5 hours = 1491.4 x 5= 7457 Watt-hour
To convert Watt-Hour into Watt-sec multiply it by 3600
