Solution:

i) Inductor:- consider inductor of L Henry with Initial current Io. The voltage across the inductor is given by

$ V_L(t)=L \cdot \frac{d i(t)}{d t} \\ $

Taking L.T. on both sides

$ \begin{aligned} V_L(S) &=L[S I(S)-I(0)] \\ &=L S \cdot I(S)-L I(0) \\ \therefore V_L(S) &=L S \cdot I(S)+\left[-L I_0\right] \end{aligned} \\ $

This can be represented in equivalent circuit form as follows

ii) Capacitor:-

consider capacitor of C ferry with initial voltage Vo.

The voltage across capacitor is given

$ V_c(t)=\frac{1}{c} \int_{-\infty}^t i(t) \cdot d t \\ $

Taking L.T.on both sides

$ \begin{aligned} \\ V_c(s) &=\frac{1}{C}\left[\frac{I(S)}{S}+\frac{Q(\bar{O})}{S}\right] \\\\ &=\frac{I(S)}{C S}+\frac{Q(\bar{O})}{C S} \\\\ V_c(s) &=\frac{1}{C S} \cdot I(S)+\frac{V_0}{S} \quad(\because Q=C \cdot V\\\\ \left.\therefore V_0=\frac{Q_0}{C}\right) \\ \end{aligned} \\ $

This can be represented, in equivalent circuit form as follows