Given figure: Find$ \text { Obtain i (t) for } \mathrm{t}\gt0 $

Solution:

$ \text { ckt for } t\lt0 $

from fig

$ I_0=0 \\ $

Amp

$ V_0=0 \\ $

volts

ckt. for

$ t \geqslant 0 $ in S domain form

$ \begin{aligned} \\\\ &K V L \quad \text { to } 100 p \\\\\\ &\frac{2 \cdot e^{-2 S}}{S^2}-5 I(S)-S I(S)-\frac{6}{S} I(S)=0 \\\\ &I(S)\left[5+S+\frac{6}{S}\right]=\frac{2 \cdot e^{-2 S}}{S^2} \\ \end{aligned} \\ $

$ \begin{aligned} \\ &I(S)\left[\frac{s^2+5 S+6}{\frac{S}{-2 S}}\right]=\frac{2 \cdot e^{-2}}{s^2}\\\\ &I(s)=\frac{2 \cdot e^{-2 s}}{(s)\left(s^2+5 s+6\right)}=e^{-2 s}\left[\frac{2}{(s)\left(s^2+5 s+6\right)}\right]\\\\ &\begin{aligned} \\ &=e^{-2 s}\left[I_1(s)\right] ....(1)\\\\ &\text { Where } I_1(s)=\frac{2}{(s)\left(s^2+5 s+6\right)}=\frac{2}{(s)(s+2)(s+3)} \\ \end{aligned}\\\\ &=\frac{A}{S}+\frac{B}{S+2}+\frac{C}{S+3}....(2)\\\\ &A=\left.\frac{2}{(s+2)(s+3)}\right|_{s=0}=\frac{2}{(2)(3)}=\frac{1}{3}\\\\ &B=\left.\frac{2}{(s)(s+3)}\right|_{s=-2}=\frac{2}{(-2)(-2+3)}=-1\\\\ &c=\left.\frac{2}{(s)(s+2)}\right|_{s=-3}=\frac{2}{(-5)(-3+2)}=\frac{2}{3}\\\\ &\text { from (2) } \quad I_1(s)=\frac{1 / 3}{s}+\frac{-1}{s+2}+\frac{2 / 3}{s+3}-\text { (3) }....(3) \\ \end{aligned} \\ $

put (3) in (1)

$ \begin{aligned} \\ I(s) &=e^{-2 s}\left[\frac{1 / 3}{s}-\frac{1}{s+2}+\frac{2 / 3}{s+3}\right] \\\\ &=\frac{1}{3} \cdot \frac{e^{-2 s}}{s}-1 \cdot \frac{e^{-2 s}}{s+2}+\frac{2}{3} \frac{e^{-2 s}}{s+3} \\ \end{aligned} \\ $

Taking. I.L.T.

$ \begin{aligned} \\ &I(s)=\frac{1}{3} u(t-2)-e^{-2(t-2)} \cdot u(t-2)+\frac{2}{3} \cdot e^{-3(t-2)} \cdot u(t-2) \\\\ &I(s)=\left[\frac{1}{3}-e^{-2(t-2)}+\frac{2}{3} \cdot e^{-3(t-2)}\right] u(t-2) \\ \end{aligned} \\ $