Need Help?

Subscribe to Circuit

Subscribe
  • Notes
  • Comments & Questions

$$
\begin{array}{l}{\text { The switch in the circuit shown in Fig. has been in position } x \text { for a long time. At } t=0 \text { , }} \\ {\text { The switch moves instantaneously to position } y \text { . Find the following: }}\end{array}
$$

$$
\begin{array}{l}{\text { a) } v_{\mathrm{C}}(t) \text { for } t \geq 0} \\ {\text { b) } v_{\mathrm{O}}(t) \text { for } t \geq 0} \\ {\text { c) } i_{\mathrm{O}}(t) \text { for } r \geq 0^{+}}\end{array}
$$

$$
\text { at position } x
$$

DC

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

KVL

$$
-100+v_{0}=0
$$

$$
V_0=100V
$$

$$
\text { at position y }
$$

$$
\frac{60 * 240}{60+240}=48
$$

$$
32+48=80 k \Omega
$$

$$
\tau =R C=80*0.5 \mu F=0.04 sec
$$

$$
V_0 (t)=V_{0} e^{\frac{-t}{2}}=100e^{\frac{-t}{0.0}}
$$

$$
V_{c}(t)=100 e^{-25 t} \quad (V)
$$

$$
V_{0}=V_ c * \frac{48}{32+48}=10 e^{-25 t} * \frac{48}{32+48}
$$

$$
V_0(t)=60 e^{-25t}
$$

$$
i_{0}=\frac{V_{0}}{R}
$$

$$
i_{0}=\frac{60 e^{-25 t}}{60}
$$

$$
i_{0} (t)=e^{-25 t} \quad(A)
$$

No comments yet

Join the conversation

Join Notatee Today!