A) 0.63\[{{I}_{0}}\]
B) 0.50\[{{I}_{0}}\]
C) 0.37 \[{{I}_{0}}\]
D) \[{{I}_{0}}\]
Correct Answer: A
Solution :
Current at any instant of time \[t\] after closing an \[L-R\] circuit is given by\[I={{I}_{0}}\left[ 1-{{e}^{\frac{-R}{L}t}} \right]\] Time constant\[t=\frac{L}{R}\] \[\therefore \] \[I={{I}_{0}}\left[ 1-{{e}^{\frac{-R}{L}\times \frac{L}{R}}} \right]={{I}_{0}}(1-{{e}^{-1}})\] \[={{I}_{0}}\left( 1-\frac{1}{e} \right)\] \[={{I}_{0}}\left( 1-\frac{1}{2.718} \right)=0.63{{I}_{0}}=63%\,\,of\,\,{{I}_{0}}\]
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