Q. A coil of self inductance $10\, mH$ and resistance $0.1\, \Omega $ is connected through a switch to a battery of internal resistance $0.9\, \Omega $. After the switch is closed, the time taken for the current to attain $80\%$ of the saturation value is : (Take $ln5 = 1.6$)
Solution:
$i = i_{0} \left(1- e^{-t/\tau}\right)$
$ \frac{80}{100} i_{0} =i_{0} \left(1-e^{-t/\tau}\right) $
$ 0.8 = 1 - e^{-t/\tau}$
$ e^{-t/\tau} = 0.2 = \frac{1}{5} $
$ - \frac{t}{\tau} = ln\left(\frac{1}{5}\right) $
$ - \frac{t}{\tau} = -In \left(5\right) $
$ t= \tau. ln\left(5\right) $
$ = \frac{L}{R_{eq}} .ln \left(5\right) = \frac{10\times10^{-3}}{\left(0.1+0.9\right)} \times1.6$
$ t = 1.6 \times10^{-2} $
$t = 0.016\, s$
