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Physics Question on Alternating current

For a coil having L = 2 mH, current flow through it is I=t2e−lI= t^2e^{-l} then, the time at which emf become zero

A

2 sec

B

1 sec

C

4 sec

D

3 sec

Answer

2 sec

Explanation

Solution

I=t2e−tI=t^2e^{-t}
∣∈∣=LdIdt here emf is zero when dIdt=0|\in|=L\frac{dI}{dt} \, here \, emf \, is \, zero \, when \, \frac{dI}{dt}=0
dIdt=2te−t−t2e−t=0;2te−t=t2e−t\frac{dI}{dt}=2te^{-t}-t^2e^{-t}=0;2te^{-t}=t^2e^{-t}
i.e. te−t(t−2)=0⇒t≠8    t≠0,t=2te^{-t}(t-2)=0 \Rightarrow t\ne8 \, \, \, \, t \ne 0,t=2 sec.