A) −1
B) \[\frac{1-e}{e}\]
C) \[\frac{e}{1-e}\]
D) 1
Correct Answer: A
Solution :
\[i={{i}_{0}}{{e}^{\frac{-tR}{L}}}\] \[\therefore \]\[{{V}_{R}}=iR={{i}_{0}}{{e}^{\frac{-tR}{L}}}\] \[{{V}_{R}}=R{{i}_{0}}{{e}^{-1}}\]at\[t=\frac{L}{R}\] \[{{V}_{L}}=-L\left( \frac{di}{dt} \right)\] \[=L{{i}_{0}}\left( \frac{-R}{L} \right){{e}^{\frac{-tR}{L}}}\] \[=-R{{i}_{0}}{{e}^{-1}}\]at\[t=\frac{L}{R}\] \[\therefore \]\[\frac{{{V}_{R}}}{{{V}_{L}}}=-1\]You need to login to perform this action.
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