A) 0.103 s
B) 0.016 s
C) 0.002 s
D) 0.324 s
Correct Answer: B
Solution :
\[i={{i}_{0}}(1-{{e}^{-t/\tau }})\] \[\frac{80}{100}{{i}_{0}}={{i}_{0}}(1-{{e}^{-t/\tau }})\] \[0.8=1-{{e}^{-t/\tau }}\] \[{{e}^{-t/\tau }}=0.2=\frac{1}{5}\] \[-\frac{t}{\tau }=\ln \left( \frac{1}{5} \right)\] \[-\frac{t}{\tau }=-\ln \left( 5 \right)\] \[t=\tau .\ln \left( 5 \right)\] \[=\frac{L}{{{\operatorname{R}}_{eq}}}.\ln \left( 5 \right)\] \[=\frac{10\times {{10}^{-3}}}{(0.1+0.9)}\times 1.6\] \[t=1.6\times {{10}^{-2}}\] \[t=0.016s\]You need to login to perform this action.
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