Electro Magnetic Induction And Alternating Currents Question 189
Question: The current in an L-R circuit builds up to $ {{(3/4)}^{th}} $ of its steady state value in 4 seconds. The time constant of this circuit is
Options:
A) $ \frac{1}{ln2}\sec $
B) $ \frac{2}{ln2}\sec $
C) $ \frac{3}{ln2}\sec $
D) $ \frac{4}{ln2}\sec $
Show Answer
Answer:
Correct Answer: B
Solution:
-
$ I=I_{0}(I-{{e}^{-t/\tau }}) $
where $ \tau \to $ time constant
$ \therefore \frac{3}{4}I_{0}=I_{0}(1-{{e}^{-t/\tau }})\Rightarrow \frac{3}{4}=l-{{e}^{-t/\tau }} $
$ \Rightarrow {{e}^{-t/\tau }}=\frac{1}{4} $ $ ,\Rightarrow ,\frac{-t}{\tau } $
In $ e=In,\frac{1}{4}\Rightarrow \frac{-4}{\tau }=-2 $
In $ 2\Rightarrow \tau =\frac{2}{In,2} $
