Question

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

• A. 2 sec
• B. 1 sec
• C. 4 sec
• D. 3 sec

Answer 1

The Correct Option is A

To determine the time at which the electromotive force (emf) across the coil becomes zero, we need to understand the relationship between inductance, current, and emf in an inductor.

For an inductor, the emf ($E$) is given by:

E = -L \dfrac{dI}{dt}

where:

• $L$ is the inductance (2 mH in this case).
• $I = t^2 e^{-t}$ is the current flowing through the coil.
• \dfrac{dI}{dt} is the derivative of the current with respect to time.

To find the time when the emf becomes zero, we set E = 0:

0 = -L \dfrac{dI}{dt}

Thus, \dfrac{dI}{dt} = 0.

Let's find \dfrac{dI}{dt} for the given current I = t^2 e^{-t} by using the product rule of differentiation, since it is a product of t^2 and e^{-t}:

\dfrac{dI}{dt} = \dfrac{d}{dt}(t^2 e^{-t})

Using the product rule:

\dfrac{dI}{dt} = \dfrac{d}{dt}(t^2) \cdot e^{-t} + t^2 \cdot \dfrac{d}{dt}(e^{-t})

\dfrac{dI}{dt} = 2t \cdot e^{-t} + t^2 \cdot (-e^{-t})

\dfrac{dI}{dt} = (2t e^{-t} - t^2 e^{-t})

Factor out e^{-t}:

\dfrac{dI}{dt} = e^{-t}(2t - t^2)

Set \dfrac{dI}{dt} = 0:

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

Since e^{-t} is never zero, we solve:

2t - t^2 = 0

Factorizing gives:

t(2 - t) = 0

This gives solutions t = 0 and t = 2 seconds.

Since we are looking for the nontrivial time when the emf becomes zero, the correct solution is t = 2 seconds.

Therefore, the time at which the emf becomes zero is 2 seconds.