# 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 : $\left [ taken \, \, \ln 5=1.6 \right ]$Option 1)$0.324s$Option 2)$0.103s$Option 3)$0.002s$Option 4)$0.016s$

$i=i_{0}\left ( 1-e^{-\frac{t}{\tau }} \right )$

$\frac{80}{100}i_{0}=i_{0}\left ( 1-e^{-\frac{t}{\tau }} \right )$

$\Rightarrow 0.8=1-e^{-\frac{t}{\tau }}$

$\Rightarrow e^{-\frac{t}{\tau }}=0.2$

$\Rightarrow -\frac{t}{\tau }=\ln \left ( 0.2 \right )$

$\Rightarrow t=\tau \ln 5=t=\frac{L}{R _{eq} }\ln 5$

$=\frac{10\times 10^{-3}}{\left ( 0.1+0.9 \right )}\times 1.6$

$\Rightarrow t= 0.016second$

Option 1)

$0.324s$

Option 2)

$0.103s$

Option 3)

$0.002s$

Option 4)

$0.016s$

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