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$$
\begin{array}{l}{\text { Use the node-voltage method to find } V I \text { in the }} \\ {\text { circuit shown. }}\end{array}
$$

$$
a{t} \text { node } V_{1}
$$

$$
[\frac{1}{10}+\frac{1}{40}+\frac{1}{20}] V_{1}-\frac{1}{10} V_{2}-\frac{1}{20} V_{3}=0\longrightarrow (1)
$$

$$
\text { at node } V_2
$$

$$
[\frac{1}{10}+\frac{1}{30}] V_{2}-\frac{1}{10}V_1-\frac{1}{30}V_3=i\longrightarrow (2)
$$

$$
V_{2}=10 V \rightarrow(3)
$$

$$
V_{3}=-20 i \rightarrow (4)
$$

$$
V_1=24V
$$        $$
V_3=64 \mathrm{V}
$$        $$
i=-3.2 \mathrm{A}
$$

$$
\begin{array}{l}{\text { Use the node-voltage method to find } v \text { in the }} \\ {\text { circuit shown. }}\end{array}
$$

$$
\text { For node(1) }
$$

$$
\left[\frac{1}{7.5}+\frac{1}{2.5}\right] V_{1}-\frac{1}{2.5} V=4.8\longrightarrow (1)
$$

$$
\text { for noder (2) }
$$

$$
\left(\frac{1}{2.{5}}+\frac{1}{10}\right){V}-\frac{1}{2.5} V_1+[\frac{1}{2.5}+\frac{1}{1}] V_2=12 \longrightarrow (2)
$$

$$
V_{1}=i _x * 7. 5 \longrightarrow (3)
$$

$$
V_{2}-V=i_ x \longrightarrow (4)
$$

$$
V_{1}=15 \mathrm{V} \quad , \  \mathrm{V}_{2}=10 \mathrm{V} \quad , \mathrm{V}=8 \mathrm{V}
$$

$$
i_{x}=2 A
$$

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