Question
Download Solution PDFTwo identical coils A and B have 400 turns placed such that 60% of flux produced by one coil links with the other. If a current of 10A flowing in coil A produces a flux of 20 mWb in it, find the mutual inductance between coil A and B.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
The value of the mutual inductance is given by:
where, M = Mutual inductance
Calculation
Given, NB = 400
IA = 10 A
ϕA = 20 mWb
60% of the flux produced by coil A links with coil B.
ϕBA = 0.6 × ϕA
ϕBA = 0.6 × 20 mWb = 12 mWb
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.