Question

In an inductor of self-inductance L = 2 mH, current changes with time according to relation i = $t^{2}e^{-t}$. At what time emf is zero?(a) 4s(b) 3s(c) 2s(d) 1s

Answer 1

Hey mate..

here's the answer....

In an inductor of self-inductance L = 2 mH, current changes with time according to relation i = $t^{2}e^{-t}$. At what time emf is zero?(a) 4s(b) 3s(c) 2s(d) 1s

$\huge{Option \:B}$

Hope this helps❤

Answer 2

Answer:

The time is 2 sec.

(c) is correct option.

Explanation:

Given that,

Self inductance L=2 mH

Emf = 0

The relation of current

$i = t^2e^{-t}$

We need to calculate the time

Using formula of emf

$\epsilon=-L\dfrac{di}{dt}$

Put the value into the formula

$0=-L\dfrac{d}{dt}(t^2e^{-t})$

$0=-2(2te^{-t}-t^2e^{-t})$

$-4te^{-t}+2t^2e^{-t}=0$

$-4+2t=0$

$t=2\ sec$

Hence, The time is 2 sec.