A coil of self inductance $10\mathrm{ }\mathrm{m}\mathrm{H}$ and resistance of $0.1\mathrm{ }\mathrm{\Omega }$ is connected through a switch to a battery of internal resistance $0.9\mathrm{ }\mathrm{\Omega }$ . After the switch is closed, the time taken for the current to attain $80\mathrm{\%}$ of the saturation value is: $\left[\ln 5=1.6\right]$

The instantaneous current in an LR circuit is given by the equation :
(instantaneous current) $\mathrm{I}={\mathrm{I}}_s\left(1-e^{-\frac{Rt}{L}}\right)$

Where $\mathrm{I}, {\mathrm{I}}_{\mathrm{s}}, \mathrm{R}, \mathrm{t}, \mathrm{L}$ are the instantaneous current, saturation current, resistance, time and inductance respectively.

Given that the current becomes so percent of saturation current.
$\Rightarrow \left(0.8\right){\mathrm{ }\mathrm{I}}_s={\mathrm{I}}_s\left(1-e^{-\frac{Rt}{L}}\right)$

$R=0.1\text{Ω}+0.9\text{Ω}=1\text{Ω}$
$\Rightarrow 0.8=\left(1-e^{-\frac{1\times t}{10\times {10}^{-3}}}\right)$
$\Rightarrow 0.8=1-e^{-100t}\mathrm{ }$
Therefore, the time is:
$\mathrm{t}=\frac{l\mathrm{n}\left(\frac{1}{1-0.8}\right)}{100}=\frac{\mathit{ln}\left(5\right)}{100}=0.016\mathrm{ }\mathrm{s}$