Question

solve for the variables i_x and v_0 (hint: use kirchhoffs laws)

Explanation:

Step1: Apply Kirchhoff's Current Law (KCL) at the node

The current entering the node is 45 A and the currents leaving are $i_x$ and the current through the 2 - ohm and dependent - source branch. According to KCL, $45 = i_x+\frac{v_o}{2}$. Also, from Ohm's law, $v_o = 6i_x$.

Step2: Substitute $v_o$ into the KCL equation

Substitute $v_o = 6i_x$ into $45 = i_x+\frac{v_o}{2}$. We get $45 = i_x+\frac{6i_x}{2}$.
Simplify the right - hand side: $45 = i_x + 3i_x=4i_x$.

Step3: Solve for $i_x$

Divide both sides of the equation $45 = 4i_x$ by 4. So, $i_x=\frac{45}{4}=11.25$ A.

Step4: Solve for $v_o$

Since $v_o = 6i_x$, substitute $i_x = 11.25$ A into the equation. Then $v_o=6\times11.25 = 67.5$ V.

Answer:

$i_x = 11.25$ A, $v_o = 67.5$ V