A) 3 s
B) 4 s
C) 1 s
D) 2 s
Correct Answer: D
Solution :
Emf across an inductor is given by \[emf=L\frac{di}{dt}\] \[=L\frac{d}{dt}({{t}^{2}}{{e}^{-t}})=0\] \[\Rightarrow \] \[\frac{d}{dt}({{t}^{2}}{{e}^{-t}})=0\] \[\Rightarrow \] \[{{e}^{-t}}\times 2t+{{t}^{2}}\times (-1)\,{{e}^{-t}}=0\] \[\Rightarrow \] \[t{{e}^{-t}}(2-t)=0\] \[\Rightarrow \] \[2-t=0\] \[\Rightarrow \] \[t=2\,s\]You need to login to perform this action.
You will be redirected in
3 sec