Need Help?

Subscribe to Circuit

Subscribe
  • Notes
  • Comments & Questions

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

 

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

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) 

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

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

 

No comments yet

Join the conversation

Join Notatee Today!