MCQ
For a coil having $L =2 mH$, current flows through it is $i=t^2 e^{-t}$, then the time at which emf becomes zero
- A1 s
- B3 s
- C4 s
- ✓2 s
✓
Answer
Correct option: D.
2 s
(D) 2 s
As, $i=t^2 e^{-i}$ and $|e|=L \frac{d i}{d t}$
So, emf will be zero when $\frac{ di }{ dt }=0$
$
\begin{aligned}
& \text { Now, } \frac{d i}{d t}=2 t e^{-i}-t^2 e^{-i}=0 \\
\Rightarrow & 2 t e^{-i}-t^2 e^{-i}=0 \\
\Rightarrow & t e^{-i}(t-2)=0 \\
& \text { As, } t \neq \infty \text { and } t \neq 0 \\
\therefore \quad & t=2 s
\end{aligned}
$
As, $i=t^2 e^{-i}$ and $|e|=L \frac{d i}{d t}$
So, emf will be zero when $\frac{ di }{ dt }=0$
$
\begin{aligned}
& \text { Now, } \frac{d i}{d t}=2 t e^{-i}-t^2 e^{-i}=0 \\
\Rightarrow & 2 t e^{-i}-t^2 e^{-i}=0 \\
\Rightarrow & t e^{-i}(t-2)=0 \\
& \text { As, } t \neq \infty \text { and } t \neq 0 \\
\therefore \quad & t=2 s
\end{aligned}
$
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