A) \[0.8\,ln2\]
B) \[\ln \,2\]
C) \[0.125\,\ln \,2\]
D) \[0.25\,\ln \,2\]
E) \[12\,\ln \,2\]
Correct Answer: C
Solution :
The instantaneous value of current is given by \[i={{i}_{0}}(1-{{e}^{-R/{{L}^{t}}}})\] Given, \[\frac{i}{{{i}_{0}}}=\frac{1}{2},R=0.4\,\Omega ,\] \[L=50\text{ }mH=50\times {{10}^{-3}}H\]. \[\therefore \] \[\frac{1}{2}=1-{{e}^{-\frac{0.4t}{50\times {{10}^{-3}}}}}\] \[\Rightarrow \] \[{{e}^{\frac{0.4t}{50\times {{10}^{-3}}}}}=\frac{1}{2}\] Taking log \[\frac{0.4t}{50\times {{10}^{-3}}}=-\log 2\] \[\Rightarrow \,\,\,\,\,\,t=0.125\,\ln \,2\]
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