Question
Download Solution PDFएक विद्युत परिपथ का प्रवेश्यता को Y = (3 + j4) द्वारा दर्शाया जाता है। इस परिपथ में प्रतिरोध का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
प्रतिबाधा (Z) = \({1\over Admittance (Y)}\)
Z = R + jX
जहाँ, R = प्रतिरोध और X = प्रतिघात
Y = G + jB
जहाँ, G = चालकता & B = आग्राहिता
गणना:
दिया गया है, Y = 3 + j4
\(Z={1\over 3+j4}\)
\(Z={1\over 3+j4}\times{3-j4\over 3-j4}\)
\(Z={1\over 25}(3-j4)\)
प्रतिरोध, R = \(\frac{3}{25}\Omega\)
Last updated on May 29, 2025
-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.
-> 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.