Question
Download Solution PDFएक 50 Hz 4 ध्रुव एकल फेज मोटर सर्पी 3.4% के साथ चल रही है। मोटर की गति क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFधारणा:
तुल्यकालिक गति को निम्न रूप में दिया जाता है
\({N_s} = \frac{{120\;f}}{P}\)
सर्पी को निम्न रूप में दिया जाता है
\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}}\)
Nr = (1 - s) Ns
जहाँ
Nr = rpm में रोटर की गति
f = Hz में आपूर्ति आवृत्ति
P = ध्रुवों की संख्या
गणना:
दिया हुआ-
f = 50 Hz, P = 4,
s = 3.4% = 0.034
अब तुल्यकालिक गति की गणना इसप्रकार की जा सकती है
\({N_s} = \frac{{120 \times 50}}{4} = 1500\;rpm\)
इसलिए 1 -ϕ मोटर की गति है
Nr = (1 - 0.034) x 1500
Nr = 1449 rpm
Last updated on May 28, 2025
-> ISRO Technician recruitment notification 2025 has been released.
-> Candidates can apply for the ISRO recruitment 2025 for Technicians from June 2.
-> A total of 64 vacancies are announced for the recruitment of Technician.
-> The selection of the candidates is based on their performance in the Written Exam and Skill Test.
-> Candidates who want a successful selection must refer to the ISRO Technician Previous Year Papers to increase their chances of selection for the Technician post.