Question

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

𝑖= 1/2𝐿∫¹₀ 𝑡^2𝑒^−𝑡 𝑑𝑡

For 𝐿=(1 ×10−3)𝐻, Find 𝑖.

Answer 1

Let us find the energy:

$𝑖= \frac{1}{2}𝐿\int_0^1 𝑡^2𝑒^−𝑡 𝑑𝑡=|u=t^2,\ dv=e^{-t}dt,\ du=2tdt,\ v=-e^{-t}|= \frac{1}{2}𝐿(-t^2e^{-t}|_0^1+2\int_0^1te^{-t}dt)=\frac{1}{2}𝐿(-e^{-1}+2\int_0^1te^{-t}dt)=|u=t, dv=e^{-t}dt, du=dt, v=-e^{-t}|=\frac{1}{2}𝐿(-e^{-1}+2(-te^{-t}|_0^1+\int_0^1e^{-t}dt))= -e^{-t}|=\frac{1}{2}𝐿(-e^{-1}+2(-e^{-1}-e^{-t}|_0^1))= \frac{1}{2}𝐿(-e^{-1}+2(-e^{-1}-e^{-1}+1))= \frac{1}{2}𝐿(2-5e^{-1})$

For $𝐿=(1 ×10^{−3})H$ we have $i=\frac{1}{2}(2-5e^{-1})10^{-3}H.$