Sarthaks Test
0 votes
2.5k views
in Physics by (63.1k points)

A coil has a time constant of 1 second and an inductance of 8 H. If the coil is connected to a 100 V d.c. source, determine : 

(i) the rate of rise of current at the instant of switching 

(ii) the steady value of the current and 

(iii) the time taken by the current to reach 60% of the steady value of the current.

1 Answer

+1 vote
by (64.6k points)
selected by
 
Best answer

λ = L/R ; R = L/λ = 8/1 = 8 ohm

(i) Initial di/dt = V/L = 100/8 = 12.5 A/s 

(ii) IM = V/R = 100/8 = 12.5A

(iii) Here, i = 60% of 12.5 = 7.5 A

Now, i = Im (1 −e−t/λ)

∴ 7.5 = 12.5 (1 −e−t/1);

t = 0.915 second

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...