Q. When a coil carrying a steady current is short circuited, the current in it, decreases $\eta$ time in time $t_{0}$. The time constant of the circuit is
A) $\frac{t_{0}}{In\ \eta}\ $
B) $\frac{t_{0}}{\eta - 1}$
C) $t_{0}\ $In $\eta$
D) $\frac{t_{0}}{\eta}\ $
Explanation
(a) The current in a coil is given by
$i = i_{0}e^{- t/\tau}$
Now, $i\ = \frac{i_{0}}{\eta}\ \ \ in\ t = t_{0}\ $
$\therefore\ \ \ \ \ \ \ \frac{i_{0}}{\eta}\ = i_{0}e^{- t_{0}/\tau}$
$e^{- t_{0}/\tau}\ = \ \eta^{- 1}\ $
Taking log of both sides,
$- \frac{t_{0}}{\tau}\log_{e}{e = \ - 1\log_{e}\eta}$
$\frac{t_{0}}{\tau}\ \ = \log_{e}\eta$
$\tau\ \ \ = t_{0}/\log_{e}{\eta = t_{0}/In\eta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.