question_answer

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 \[\ln 5=1.6\]) [JEE Main 10-4-2019 Afternoon]

A) 0.103 s

B)              0.016 s

C) 0.002 s                        

D) 0.324 s

Correct Answer: B

Solution :

\[i={{i}_{0}}(1-{{e}^{-t/\tau }})\] \[\frac{80}{100}{{i}_{0}}={{i}_{0}}(1-{{e}^{-t/\tau }})\] \[0.8=1-{{e}^{-t/\tau }}\] \[{{e}^{-t/\tau }}=0.2=\frac{1}{5}\] \[-\frac{t}{\tau }=\ln \left( \frac{1}{5} \right)\] \[-\frac{t}{\tau }=-\ln \left( 5 \right)\] \[t=\tau .\ln \left( 5 \right)\] \[=\frac{L}{{{\operatorname{R}}_{eq}}}.\ln \left( 5 \right)\] \[=\frac{10\times {{10}^{-3}}}{(0.1+0.9)}\times 1.6\] \[t=1.6\times {{10}^{-2}}\] \[t=0.016s\]