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• Notes

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

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

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

$\text { at node } V_2$

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

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

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

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

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

$\text { For node(1) }$

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

$\text { for noder (2) }$

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

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

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

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

$i_{x}=2 A$