Electro Magnetic Induction And Alternating Currents Question 349
Question: The current in a LR circuit builds up to $ \frac{3}{4} $ th of its steady state value in $ 4s $ . The time constant of this circuit is [Roorkee 2000]
Options:
A) $ \frac{1}{\ln 2}s $
B) $ \frac{2}{\ln 2}s $
C) $ \frac{3}{\ln 2}s $
D) $ \frac{4}{\ln 2}s $
Show Answer
Answer:
Correct Answer: B
Solution:
We know that $ i=i_{o}[ 1-{{e}^{\frac{-Rt}{L}}} ] $
or $ \frac{3}{4}i_{o}=i_{o}[ 1-{{e}^{-t/\tau }} ] $ (where $ \tau =\frac{L}{R}= $
time constant) $ \frac{3}{4}=1-{{e}^{-t/\tau }} $ or $ {{e}^{-t/\tau }}=1-\frac{3}{4}=\frac{1}{4} $
$ {{e}^{t/\tau }}=4 $
or $ \frac{t}{\tau }= $ ln 4
$ \Rightarrow \tau =\frac{t}{\text{ln},4}=\frac{4}{2,\text{ln},2}\Rightarrow ,\tau =\frac{2}{\text{ln},2}sec. $
