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 $

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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} $



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