Question

When inductor is short circuited then current drops to $1/n$ times the maximum value. The time constant of the circuit is

• A. $\tau=\frac{t}{ln\,n}$
• B. $\tau=\frac{ln\,n}{t}$
• C. $\tau=t\,ln\,n$
• D. $\tau=ln\,n$

Answer 1

Answer: (A)

During decay, the value of current at any time $t$ in $R-L$ circuit is given by
$I=I_{0^{e^{-t/ \tau}}}$
where $ \tau$ is the time constant of $R-L$ circuit and $I_{0}$ is the maximum current
$\therefore \, \frac{I_{0}}{n}=I_{0^{e^{-t \tau}}} \ldots\left(i\right)$
Taking natural logarithms on both sides of eqn $\left(i\right)$, we get
ln $n=\frac{t}{\tau}$ or $ \tau=\frac{t}{ln\,n}$