In the circuit given:

Solution:

$ \text { Draw ckt for } t\lt0 \quad[t=\overline{0}] \\ $

from fig.

$ \begin{aligned} \\ I_0 &=I(\overline{0})=2 \text { Amp } \quad \because I_0=\frac{20}{10} \\\\ \&_0 &=V_c(\overline{0})=0 \text { Volts } \\ \end{aligned} \\ $

Draw ckt. at

$t=0^{+}$

$ \text { Draw ckt for } t\gt0 \\ $

$ \begin{aligned} \\ &\text { KVL to } 100 p \\\\ &-10 i(t)-1 \cdot \frac{d i(t)}{d t}-\frac{1}{1 \times 10^{-6}} \int_0^t i(t) \cdot d t=0 \\ \end{aligned}\\ $

put $ t=0^{+} \\ $

$ -10 i\left(0^{+}\right)-1 \cdot \frac{d i\left(0^{+}\right)}{d t}-10^6 \int_0^{0^{+}} i\left(0^{+}\right) \cdot d t=0 \\ $

$ \begin{aligned} \\ -10(2)-1 \cdot \frac{d i\left(0^{+}\right)}{d t}-0=0 \\\\ \therefore \frac{d i\left(0^{+}\right)}{d t}=-20 \frac{A m p}{\mathrm{sec}} \\ \end{aligned} \\ $

differentiating eq-n (1)

$ \begin{gathered} \\ -10 \frac{d i(t)}{d t}-\frac{d^2 i(t)}{d t^2}-10^6 i(t)=0 \\\\ \text { put } t=0 \\\\ -10 \frac{d i\left(0^{+}\right)}{d t}-\frac{d^2 i\left(0^{+}\right)}{d t^2}-10^6 i\left(0^{+}\right)=0 \\\\ -10[-20]-\frac{d^2 i\left(0^{+}\right)}{d t^2}-10^6 i\left(0^{+}\right)=0 \\\\ 200-10^6(2)=\frac{d^2 i\left(0^{+}\right)}{d t^2} \\\\ \frac{d^2 i\left(0^{+}\right)}{d t}=-2 \times 10^6 \mathrm{Amp} / \mathrm{sec}^2 \\ \end{gathered} \\ $