एक विलंबित पूर्ण-तरंग दिष्टकारी ज्यावक्रीय धारा का औसत मान इसके अधिकतम मान एक तिहाई के बराबर होता है विलंब कोण ज्ञात कीजिए।

This question was previously asked in
SSC JE EE Previous Paper 8 (Held on: 28 Oct 2020 Morning)
View all SSC JE EE Papers >
  1. cos-10.047
  2. cos-10.678
  3. cos-10.866
  4. cos-10.386

Answer (Detailed Solution Below)

Option 1 : cos-10.047
Free
Electrical Machine for All AE/JE EE Exams Mock Test
7.7 K Users
20 Questions 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

अवधारणा:

माना Vm परिवर्तक के AC इनपुट वोल्टेज का अधिकतम मान है और V0 परिवर्तक की औसत आउटपुट वोल्टेज है और α विलंब कोण है।

एकल-फेज अर्ध परिवर्तक या विलंबित पूर्ण-तरंग दिष्टकारी के लिए,

\({V_0} = \frac{{{V_m}}}{π }\left( {1 + cosα } \right) \)

गणना:

दिया गया है कि,

\({V_0} = \frac{{{V_m}}}{3} \)

इसलिए, समीकरण बनती है,

\(\frac{{{V_m}}}{3} = \frac{{{V_m}}}{π }\left( {1 + cosα } \right)\)

or,

\(\frac{π }{3} = \left( {1 + cosα } \right) \)

cos α = 0.047
α = cos-1(0.047)

Latest SSC JE EE Updates

Last updated on Jul 1, 2025

-> SSC JE Electrical 2025 Notification is released on June 30 for the post of Junior Engineer Electrical, Civil & Mechanical.

-> There are a total 1340 No of vacancies have been announced. Categtory wise vacancy distribution will be announced later.

-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.

-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31. 

-> Candidates with a degree/diploma in engineering are eligible for this post.

-> The selection process includes Paper I and Paper II online exams, followed by document verification.

-> Prepare for the exam using SSC JE EE Previous Year Papers.

More Single Phase Full Wave Converters Questions

More Phase Controlled Rectifiers Questions

Get Free Access Now
Hot Links: teen patti wala game teen patti live teen patti master download teen patti bodhi teen patti master online