A) 4s
B) 3s
C) 2s
D) 1s
Correct Answer: C
Solution :
(c) 2s \[L=2\,mH=2\times {{10}^{-3}}H\] \[l={{t}^{2}}{{e}^{-t}}\] \[\Rightarrow \frac{dl}{dt}={{t}^{2}}{{e}^{-t}}(-1)+{{e}^{-t}}(2t)=t{{e}^{-t}}(-t+2)\] emf \[\varepsilon =L\frac{dl}{dt}=2\times {{10}^{-3}}\times t{{e}^{-t}}(-t+2)\] Now, emf = 0, when \[(-t+2)=0\] or \[t=2s\].You need to login to perform this action.
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