Question
Download Solution PDFयदि बैंडविड्थ की सीमा 0 हर्ट्ज़ से 10 हर्ट्ज़ तक है, तो सी.आर.ओ. द्वारा पुनः प्रस्तुत की जाने वाले एक साइन तरंग के सबसे तेज वृद्धि समय (
मिली सेकेंड में) की गणना कीजिये?Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसूत्र:
BW × वृद्धि समय = 0.35
दिया गया है कि, बैंडविड्थ (B.W.) = 10 हर्ट्ज़
वृद्धि समय, \({t_r} = \frac{{0.35}}{{B.W.}} = \frac{{0.35}}{{10}} = 0.035\ s=0.035\times1000m = 35\;ms\)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.