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$$
\begin{array}{l}{\text { The switch in the circuit shown in Fig. has been in positiona for a long time. At } t=0 \text { , the switch moves from }} \\ {\text { position a to position } b \text { . The switch is a make-before-break type; that is, connection at position } b \text { is established }} \\ {\text { before the connection at position a is broken, }} \\ {\text { so there is no intersion at position a is broken, }} \\ {\text { a) Find the spression for } i(t) \text { for } t \geq 0 .} \\ {\text { b) What is the initial voltagacross the inductor }} \\ {\text { just after the switch has been moved to position b? }}\end{array}
$$

خطوات الحل

(1) نحدد $$I_0$$ قبل تغير المفتاح

(2) نحسب $$\tau$$

(3) نعوض في علاقات $$I, V$$

$$
t<0
$$
 $$
\text { switch at position } a
$$

$$
D C \quad \frac{d i}{d t}=0 \quad V_{L}=0 \quad {\Rightarrow S \cdot C}
$$

$$
I_{0}=-8 A
$$

$$
t<0
$$
    $$
\text { switch at position b }
$$

$$
R=2 \Omega
$$, $$
V_{s}=24 \mathrm{V}
$$, $$
\tau=\frac{L}{R}=\frac{200 * 10^{-3}}{2}=0.1 \mathrm{sec}
$$

$$
i_{l}(t)=\frac{V_{s}}{R}+\left(I_{0}-\frac{V_{s}}{R}\right) e^{\frac{-t}{\tau}}
$$

$$
i_l(t)=\frac{24}{2}+[-8-\frac{24}{2}]e^{\frac{-t}{0.1}}=12=20e^{-10t}
$$

$$
t \geq 0
$$

$$
V_{l}(t)=L \frac{di(t)}{d t}
$$

$$
V_{L}(t)=200*10^{-3} *[-20*-10e^{-10t}]
$$

$$
V_{l}(t)=40 e^{-10t} \quad (volt)
$$            $$
t \geq 0
$$

$$
\left(t=0^{t}\right)
$$            $$
V (0^{t})=40 e^{-10*0}=40 \mathrm{volt}
$$

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