Q. The current in a $LR$ circuit builds up to 3/4$^{th}$ of its steady state value in 4$s$. The time constant of this circuit is
A) $\frac{1}{\ln 2}s$
B) $\frac{2}{\ln 2}s$
C) $\frac{3}{\ln 2}s$
D) $\frac{4}{\ln 2}s$
Explanation
(b) We know that $i = i_{0}\left\lbrack 1 - e^{\frac{- Rt}{L}} \right\rbrack$ or $\frac{3}{4}i_{o} = i_{o}\left\lbrack 1 - e^{- t/\tau} \right\rbrack$
[where $\tau = \frac{L}{R} =$ time constant]
$\frac{3}{4} = 1 - e^{- t}$ or $e^{- t/\tau} = 1 - \frac{3}{4} = \frac{1}{4}$
$e^{t/\tau} = 4$ or $\frac{t}{\tau} = \ln 4$
$\Rightarrow \tau = \frac{t}{\ln 4} = \frac{4}{2\ln 2} \Rightarrow \tau = \frac{2}{\ln 2}\sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.