Question

1. The energy 𝑖, of an inductor with inductance 𝐿 is given by

𝑖= 12𝐿∫𝑡2𝑒−𝑡10𝑑𝑡

For 𝐿=(1 ×10−3)𝐻, Find 𝑖.

Answer 1

Ans:-

$\Rightarrow 𝑖= 12𝐿\intop_0 ^1𝑡^2𝑒^{−𝑡}𝑑t$ where $L=1\times 10^{-3}H$

Firstly $\intop t^2e^{-t}= t^2(-e^{-t})-\ \intop-e^{-t}(2t)dt$

On Simplification

$\Rightarrow -t^2e^{-t}+2[-te^{-t}+\ \intop-e^{-t}dt]\\ \Rightarrow -t^2e^{-t}\ -2te^{-t}-2e^{-t}$

So put the value of integration in the first expression So that we can find the value of $i$

$\Rightarrow i=12L[− t^2e^{−t}−2te^{−t}−2e^{−t}]_0 ^1$

$\Rightarrow i=12L(-5e^{-1}+2)$

$\Rightarrow i=1.9272L$

$\Rightarrow i =1.92\times 10^{-3} A$