0 votes
174 views
in Physics by (98.9k points)
edited

A coil of inductance 300 mH and resistance 2 is connected to a source of voltage 2 V. The current reaches half of its steady state value in

(a) 0.15 s

(b) 0.3 s

(c) 0.05 s

(d) 0.1 s

1 Answer

0 votes
by (98.9k points)
selected by
 
Best answer

Correct answer is option d) 0.1 s

Explaination:

During growth of charge in an inductance, I = I0 (1 – e –Rt/L)

or \(\frac{I_0}{2} = I_0(1 – e^{ –\frac{Rt}{L}})\)

\( e^{ –\frac{Rt}{L}}=\frac{1}{2}=2^{-1}\)

or \( \frac{Rt}{L} =ln2\) 

⇒t = (\(\frac{L}{R}\))In 2

t =[\(\frac{(300 \times 10^{-3})}{2}\)] x (0.693)

or t = 0.1 sec

Related questions

0 votes
1 answer 200 views

Doubtly is an online community for engineering students, offering:

  • Free viva questions PDFs
  • Previous year question papers (PYQs)
  • Academic doubt solutions
  • Expert-guided solutions

Get the pro version for free by logging in!

5.7k questions

5.1k answers

108 comments

497 users

...