Question
Download Solution PDFThe total inductance of two coupled coils in the ‘series aiding’ and ‘series opposing’ connections are 4 H and 2 H, respectively. The value of mutual inductance will be _________.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
Series adding:
\(L_{eq}=L_1+L_2+2M\)
Series opposing:
\(L_{eq}=L_1+L_2-2M\)
Calculation:
Given, \(4=L_1+L_2+2M\) ......(i)
\(2=L_1+L_2-2M\) ........(ii)
Adding equations (i) and (ii), we get:
\(3=L_1+L_2\)
Putting value in equation (i):
\(4=3+2M\)
M = 0.5 H
Additional Information 1.) Parallel adding:
\(L_{eq}={L_1L_2-M^2\over L_1+L_2-2M}\)
2.) Parallel opposing:
\(L_{eq}={L_1L_2-M^2\over L_1+L_2+2M}\)
Last updated on Jun 16, 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.