Let's solve the circuit using nodal analysis to determine $v_x$ and $i_x$.

Step 1: Label the node voltages

• The ground is the bottom node (with the 50V voltage source).
• Let's call the top node $v_1$, and the node in the middle $v_x$.

Step 2: Apply Kirchhoff's Current Law (KCL) at each node

At node $v_x$:

• The current leaving the node through the 2Ω resistor: $\frac{v_x}{2}$
• The current leaving through the dependent current source $4i_x$, where $i_x = \frac{v_x}{4}$, so $4i_x = v_x$
• The current leaving through the 5Ω resistor: $\frac{v_x}{5}$

Thus, applying KCL at node $v_x$:

$\frac{v_x}{2} + v_x + \frac{v_x}{5} = 3 \quad \text{(from the current source of 3A)}$

Simplifying the equation:

$\left( \frac{1}{2} + 1 + \frac{1}{5} \right)v_x = 3$

$\frac{10 + 20 + 4}{20}v_x = 3$

$\frac{34}{20}v_x = 3$

$v_x = \frac{3 \times 20}{34} = \frac{60}{34} = 1.76 \, \text{V}$

Step 3: Determine $i_x$

Now that we know $v_x$, we can find $i_x$:

$i_x = \frac{v_x}{4} = \frac{1.76}{4} = 0.44 \, \text{A}$

Final Answers:

• $v_x = 1.76 \, \text{V}$
• $i_x = 0.44 \, \text{A}$

Would you like further details on the steps or clarifications?

Here are 5 related questions to further your understanding:

1. How would the solution change if the 50V source was replaced by a different value?
2. What is the role of the dependent current source in this circuit?
3. Can you solve for the power dissipated in the 5Ω resistor?
4. How can you apply mesh analysis to this same circuit?
5. How do dependent sources affect the stability of the system?

Tip: When solving circuits with dependent sources, always express the controlling variable (like $i_x$) in terms of node voltages.