A) \[\frac{1}{ln\,\,2}\sec \]
B) \[\frac{2}{ln\,\,2}\sec \]
C) \[\frac{3}{ln\,\,2}\sec \]
D) \[\frac{4}{ln\,\,2}\sec \]
Correct Answer: B
Solution :
[b] \[I={{I}_{0}}(I-{{e}^{-t/\tau }})\] where \[\tau \to \] time constant \[\therefore \,\,\,\,\frac{3}{4}{{I}_{0}}={{I}_{0}}(1-{{e}^{-t/\tau }})\Rightarrow \frac{3}{4}=l-{{e}^{-t/\tau }}\] \[\Rightarrow \,\,{{e}^{-t/\tau }}=\frac{1}{4}\] \[\,\Rightarrow \,\,\,\frac{-t}{\tau }\] In \[e=In\,\frac{1}{4}\Rightarrow \frac{-4}{\tau }=-2\] In \[2\Rightarrow \tau =\frac{2}{In\,\,\,2}\]
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