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$$
\begin{array}{l}{\text { Use the node-voltage method to find the steady- }} \\ {\text { state expression for } v(t) \text { in the circuit shown. The }} \\ {\text { sinusoidal sources are } i_{s}=10 \cos \omega t \mathrm{A} \text { and }} \\ {v_{s}=100 \mathrm{sin} \omega t \mathrm{V}, \text { where } \omega=50 \mathrm{krad} / \mathrm{s}}\end{array}
$$

 الخطوات

(1) المصادر $$\Rightarrow$$ phasor

$$
Z \Leftarrow C, L
$$

(2) نبدأ خطوات Node

ref

(3) هنحول إلى $$V(t)$$

$$
i_{s}=10 \cos (\omega t) \Rightarrow 10 \angle {0}
$$

$$
V_{s}=100 \sin \omega t=100 \cos (\omega t-90) \Rightarrow 100 \angle -90
$$

$$
Z=J w L=J50 * 10^{3}*100 * 10^{-6}=J5
$$

$$
Z=-J \frac{1}{w c}=\frac{-J}{50*10^{3}*9 *10^{-6}}=-2.22 J
$$

$$
\left[\frac{1}{20}+\frac{1}{5}+\frac{1}{-2.22J}+\frac{1}{5J}\right] V=\frac{100 \ \angle-90}{20}+10
$$

$$
V=31.5 \ \angle -71.6^{\circ}
$$                            (V)

$$
W=50* 10^{3}
$$

$$
V=31.5 \ \angle-71.6
$$

$$
V=31.5 \cos (\omega t-71.6)
$$

$$
W_{V}=W_{I s}=W_{V s}
$$

$$
V(t)=31.5 \cos \left[50* 10^{3}-71.6^{\circ}\right]
$$                        (V)

 

 

 

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