A) \[\frac{{{e}^{1/2}}}{{{e}^{1/2}}-1}\]
B) \[\frac{{{e}^{2}}}{{{e}^{2}}-1}\]
C) \[1-{{e}^{-1}}\]
D) \[{{e}^{-1}}\]
Correct Answer: B
Solution :
\[i={{i}_{0}}(1-{{e}^{-Rt/L}})\] \[{{i}_{0}}=\frac{E}{R}\](Steady current) when \[t=\infty \] \[{{i}_{\infty }}=\frac{E}{R}(1-{{e}^{-\infty }})=\frac{5}{10}=1.5\] \[{{i}_{1}}=1.5(1-{{e}^{-R/L}})=1.5(1-{{e}^{-2}})\]Þ\[\frac{{{i}_{\infty }}}{{{i}_{1}}}=\frac{1}{1-{{e}^{-2}}}=\frac{{{e}^{2}}}{{{e}^{2}}-1}\]
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