Discussion
A 6-pole, 3-phase alternator running at 1000 r.p.m. supplies to an 8-pole, 3-phase induction motor has a rotor current of frequency 2Hz. The speed at which the motor operates is -
Answer: B
The motor runs at 720 r.p.m.
The supply frequency comes from the alternator, the synchronous speed of the motor from that frequency, and the slip from the rotor frequency.
\[ \begin{aligned} f &= \frac{PN}{120} = \frac{6 \times 1000}{120} = 50\ \text{Hz} \\ N_{s} &= \frac{120 \times 50}{8} = 750\ \text{r.p.m.} \\ s &= \frac{f_{r}}{f} = \frac{2}{50} = 0.04 \\ N &= 750(1 - 0.04) = 720\ \text{r.p.m.} \end{aligned} \]
750 r.p.m. is the synchronous speed, which an induction motor never quite reaches, and 960 r.p.m. applies the 4% slip to the alternator's 1000 r.p.m. instead of the motor's synchronous speed.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development