Find the value of equivalent inductance across terminal AB in the following circuit.

F2 ENG Savita 17-1-24 D1

This question was previously asked in
SSC JE Electrical 10 Oct 2023 Shift 2 Official Paper-I
View all SSC JE EE Papers >
  1. 36 H
  2. 26 H
  3. 23 H
  4. 50 H

Answer (Detailed Solution Below)

Option 3 : 23 H
Free
Electrical Machine for All AE/JE EE Exams Mock Test
7.8 K Users
20 Questions 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

Concept

When two inductors are connected in series, the equivalent inductance is given by:

\(L_{eq}=L_1+L_2\)

When two inductors are connected in parallel, the equivalent inductance is given by:

\(L_{eq}={L_1L_2\over L_1+L_2}\)

Calculation

3H and 1H are connected are series.

L = 3 + 1 = 4H

4H and 12H are connected in parallel.

\(L={4\times 12\over 4+12}=3H\)

5H, 3H and 15H are connected in series.

Leq = 23 H

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.

Get Free Access Now
Hot Links: teen patti master list master teen patti teen patti royal teen patti real cash 2024 teen patti wala game