A) 1 s
B) 3 s
C) 4 s
D) 2 s
Correct Answer: D
Solution :
As, \[i={{t}^{2}}\,{{e}^{-\,1}}\] and \[\left| \varepsilon \right|=L\frac{di}{dt}\] So, emf will be zero when \[\frac{di}{dt}=0\] Now, \[\frac{di}{dt}=2\,t{{e}^{-\,t}}\,-{{t}^{2}}{{e}^{-\,t}}\] \[\frac{di}{dt}=0\] \[\Rightarrow \,\,\,\,\,\,\,\,2t{{e}^{-\,t}}\,-{{t}^{2}}\,{{e}^{-\,t}}\,=\,\,0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,t{{e}^{-}}^{t}\,\,(t - 2) = 0\] As, \[t\ne \infty \,\,and\,\,t\ne 0\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\operatorname{t}= 2 s\]
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