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$$
\text { The switch in the circuit shown in Fig. has been closed for a long time before it is }
$$

$$
\text { opened at } t=0 . \text { Find the following: }
$$

$$
\begin{array}{l}{\text { a) } i_{\mathrm{L}}(t) \text { for } t \geq 0} \\ {\text { b) } i_{\mathrm{o}}(t) \text { for } t \geq 0} \\ {\text { c) } v_{\mathrm{o}}(t) \text { for } t \geq 0}\end{array}
$$

$$
\text { For } t<0
$$
قبل فتح المفتاح

DC

$$
\frac{d i}{d t}=0
$$                                    $$
V=0 \Rightarrow S \cdot C
$$

(1) $$
I_{0}=20 \mathrm{A}
$$

$$
\text { For }t>0
$$
بعد فتح المفتاح

(2)

(3) $$
\tau=\frac{L}{R}=\frac{2}{10}=0.2 sec
$$

(4) $$
i(t)=I_{0} e^{\frac{-t}{\tau}}=20 e^{-\frac{t}{0.2}}=20 e^{-5 t} \quad (A)
$$

$$
i_{L}(t)=20 e^{-5 t}
$$

$$
i_{1} (t)=\frac{10}{40+10} i_{L}(t)
$$

$$
i_{1} (t)=\frac{10}{50} * 20 e^{-5 t}
$$

$$
i_{1}(t)=4 e^{-5 t}
$$

$$
i_{0}(t)=-i_{1}(t)=-4 e^{-5 t}
$$

$$
V_{0}(t)=i_{0}(t) R
$$

$$
V_{0}(t)=-4 e^{-5 t} * 40
$$

$$
V_{0}(t)=-160 e^{-5 t} \quad (V)
$$

 

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