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$\begin{array}{l}{\text { Use the mesh-current method to find the power dissi- }} \\ {\text { pated in the } 20 \Omega \text { resistor in the circuit }}\end{array}$

$\left(\sum R_{1}\right) I_{1}-R_{12} I_{2}-R_{13} I_{3} \pm V_{B}=0$

Mesh (1) $(20+3+2) I_1-3I_2-20I_3-135=0 \rightarrow (1)$

Mesh (2) $(3+4+5) I_2-3I_1-4I_3=0 \rightarrow (2)$

Mesh (3) $(1+20+4) I 3-20 I_{1}-4I_2+10i=0 \rightarrow (3)$

$i=I_{2}-I_{1} \rightarrow(4)$

Solve (1), (2), (3), (4)

$I_{1}=64.8 A_{1}, I_{2}=39 A, I_{3}=68.4 A \quad i=-25.8 A$

$P_{20 \Omega}= I^{2} R=(I_3-I_1)^2*20=(68.4-64.8)^2*20$

$P_{20 \Omega}=259.2 w$

$\left(\sum \frac{1}{R}\right)_{1} V_{1}-\frac{1}{R_{12}} V_{2}-\frac{1}{R_{13}} V_{3}=I_{in}$

Node (1) $\left(\frac{1}{3} + \frac{1}{5}+\frac{1}{2}\right) V_{1}-\frac{V_{2}}{3}-\frac{V_3}{5}=\frac{135}{2} \rightarrow (1)$

Node (2) $\left(\frac{1}{4}+\frac{1}{3}+\frac{1}{20}\right)V_2-\frac{V_3}{4}-\frac{V_1}{3}=0 \rightarrow (2)$

Node (3) $\left(\frac{1}{4}+\frac{1}{1}+\frac{1}{5}\right) V_3-\frac{V_2}{4}-\frac{V_1}{5}=10i \rightarrow (3)$

$i=\frac{فرق \ الجهد}{المقاومة}=\frac{V_{2}-V_1}{3} \rightarrow (4)$

Solve (1), (2), (3), (4)

$V_{1}=5.4 \mathrm{v}, V_{2}=-72 \mathrm{v}, V_{3}=189.6\mathrm{v}$

$P_{20 \Omega}=\frac{V^2}{R}=\frac{V_2^2}{20}=\frac{(72)^2}{20}=259.2w$