Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An inductor of is connected to a battery through a resistor of <img alt="10 \mathrm{k} \Omega" src="/latex-im

An inductor of 10 \mathrm{mH} is connected to a 20 \mathrm{~V} battery through a resistor of 10 \mathrm{k} \Omega and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 \mu \mathrm{s} is \frac{\mathrm{x}}{100} \mathrm{~mA}. Then \mathrm{x} is equal to _______. \left(\right. Take\, \, \mathrm{e}^{-1}=0.37 )
 

Answers (1)

best_answer

L= 10\times 10^{-3}H
R= 10k\Omega = 10^{4}\Omega

After a long time \left ( t\rightarrow \infty \right ),Current\, in \, the \, circuit = I_{0}= \frac{V}{R}= 2\times 10^{-3}A
(\because Inductor will behave as short  ckt or straight conducting wire)
As the switch is opened, there will be a decay of current
I_{t}= I_{0}\, e^{\frac{-t}{t}}
t= \frac{L}{R}= \frac{10^{-2}}{10^{4}}= 10^{-6}
at t= 10^{-6}s
I_{t}= I_{0}\, e^{-1}
    = 2\times 10^{-3}\times 0\cdot 37
   = 0\cdot 74\times 10^{-3}A
I_{t}= \frac{74}{100}mA
\therefore X= 74

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main session 2 registration 2025 begins at jeemain.nta.nic.in; documents required, eligibility criteria
anam-khan

JEE Main 2025: BTech mechanical engineering cut-offs for women at NITs remain high despite low enrolment
anam-khan

JEE Main Results 2025: January 28 shift 1 rated ‘toughest’; what is a good score for session 1 exam

Latest Question