A coil of self inductance $10 \mathrm{mH}$ and resistance $0.1 /$ is connected through a switch to a battery of internal resistance $0.9$. After the switch is closed, the time taken for the current to attain $80 \%$ of the saturation value is
$[$ take $\ln 5=1.6]$
Correct Option: , 4
(4) $I=I_{0}\left(1-e^{-\frac{R t}{L}}\right)$ Here $\mathrm{R}=\mathrm{R}_{\mathrm{L}}+\mathrm{r}=1 \Omega$
$0.8 I_{0}=I_{0}\left(1-e^{\frac{t}{.01}}\right)$
$\Rightarrow 0.8=1-e^{-100 t}$
$\Rightarrow e^{-100 t}=0.2=\left(\frac{1}{5}\right)$
$\Rightarrow 100 \mathrm{t}=\ln 5 \Rightarrow \mathrm{t}=\frac{1}{100} \ln 5=0.016 \mathrm{~s}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.