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$\begin{array}{l}{\text { For the circuit shown, } v\left(0^{+}\right)=12 \mathrm{V}, \text { and } i_{l}\left(0^{+}\right)=30 \mathrm{mA}} \\ {\text { a) Find the initial current in each branch of the circuit. }} \\ {\text { b) Find the initial value of } d v / d t} \\ {\text { c) Find the expression for } v(t)}\end{array}$

$i_{l}(0)=30 \mathrm{mA}$

$i_{R}(0)=\frac{V_{0}(0)}{R}=\frac{12}{200}=60 \mathrm{mA}$

KCL

$I_{c}+I_{R}+I_{c}=0$

$I_{c}(0)=I_{R}(0)-I_{c}(0)$

$I_{C}(0)=-30-60=-90 \mathrm{mA}$

$i_{c}= C \frac{d v(0)}{d t}$

$-90 \times 10^{-3}=0.2 \times 10^{-6}* \frac{dv(0)}{d t}$

$\frac{d v(0)}{d t}= -l150 \ kv/sec$

$S^{2}+\frac{S}{R _C}+\frac{1}{L_ C}=0$

$S^{2}+\frac{S}{200 \times 0.2 \times 10^{-6}}+\frac{1}{50 \times 10^{-3} \times 0.2 \times 10^{-6}}=0$

$S^{2}+25000S+10^{8}=0$

$S_{1}=-5000 \quad , S_{2}=-20000$

$\text {*real*} -V e * S_{1} \neq S_{2} \Rightarrow \text { overdomped }$

$V(t)=A_{1} e^{s_1 t}+A_2e^{s_2 t}$

$V(t)=A_1 e^{-5000 t}+A_{2} e^{-20000t}$

$V(0)=12 \mathrm{V} \quad \frac{d V(0)}{d t}=-450*10^3 \mathrm{V} / \mathrm{sec}$

$t=0 \quad V(0)=12 V \Rightarrow V(t)=A_1 e^{-5000 t}+A_{2} e^{-20000 t}$

$12=A_{1}+A_{2} \rightarrow (1)$

$\frac{d v(t)}{d t}=-5000 A_1 \mathrm{e}^{-5000t}+-20000 \mathrm{A_2 e}^{-20000t}$

$\frac{d v(0)}{d t}=-450 \times 10^{3} \mathrm{v} / \mathrm{sec}$                        $t=0 \Rightarrow$

$-450 \times 10^{-3}=-5000 A_{1}-20000 \mathrm{A_2}\rightarrow(2)$

$A_1=-14 V$                $A_{2}=26 \mathrm{V}$

$V(t)=-14e^{-5000 t}+26 e^{-20000 t}$