Need Help?

Subscribe to Circuit

Subscribe
  • Notes
  • Comments & Questions

$$
\begin{array}{l}{\text { Use the mesh-current method to find the }} \\ {\text { power dissipated in the } 1 \Omega \text { resistor in the cir- }} \\ {\text { cuit shown. }}\end{array}
$$

$$
\text { super mesh }
$$

$$
2=I_{a}-I_{b} \longrightarrow (1)
$$

$$
2I_a-2I_c+4I_b-2I_c-10=0\longrightarrow (2)
$$

$$
\text { mesh (c) }
$$

$$
(1+2+2) I_{c}-2I_a-2I_b-6=0 \longrightarrow (3)
$$

$$
I_a=7A \quad , I_b=5A \quad, I_c=6A
$$
 $$
\text { direction all correct }
$$

$$
P_{1 \Omega}=I^2 R=I_{c}^{2}*1=(6)^2*1=36w
$$
 $$
\text { dissipated }
$$

No comments yet

Join the conversation

Join Notatee Today!