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The network shown in fig. has attained a steady state with the switch closed for $t<0$. At $t=0$, the switch is opened. obtain $i(t)$ for $t>0$

The network shown in fig:

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Solution:

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

enter image description here

Using $ C \cdot D \cdot R \\ $.

$ \begin{aligned} \\ I_0 =\frac{(3.6)(2.857)}{2.857+4} \\\\ I_0=1.5 \text { Amp } \\ \end{aligned} \\ $

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

enter image description here

This is simple R-L ckt. Hence current through inductor is given by

$ \begin{aligned} \\ i(t) &=I_0 \cdot e^{-\frac{R}{L} \cdot t} \\\\ &=1 \cdot 5 \cdot e^{-\frac{8}{2} t} \\\\ i(t) &=1 \cdot 5 \cdot e^{-4 t} \text { Amp. } \\ \end{aligned} \\ $

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