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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