To solve this problem, let's analyze the circuit step by step. We need to:

1. Find all node voltages.
2. Find the voltage across the 0.5 $I_x$ dependent current source.

Given Information:

• We have a circuit with a voltage source of 30 V, resistors of 5 Ω, 10 Ω, and 20 Ω.
• There are two dependent sources:
  • A dependent current source providing 0.5 $I_x$.
  • A dependent voltage source with 6$V_z$.

Solution Outline:

1. Define Node Voltages:

   • Let $V_a$ be the voltage at the node directly after the 30 V source and before the 5 Ω resistor.
   • Let $V_b$ be the voltage at the node between the two 10 Ω resistors.
   • Let $V_c$ be the voltage at the node after the 6$V_z$ source and before the 20 Ω resistor.

2. Current Definitions:

   • Define $I_x$ as the current through the 20 Ω resistor.
   • Define $V_x$ as the voltage across the leftmost 10 Ω resistor.
   • Define $V_z$ as the voltage across the central 10 Ω resistor.

3. Write Node Equations Using Kirchhoff’s Current Law (KCL):

   For node $V_b$:

   • $\frac{V_b - 30}{10} + \frac{V_b - 0}{10} + \frac{V_b - V_c}{20} = 0$
   • Simplify to solve for $V_b$ in terms of other known variables.

   For node $V_c$:

   • $\frac{V_c - V_b}{20} + \frac{V_c}{10} = I_x$
   • $V_c = 6V_z = 6 \times \frac{V_b - 0}{10}$
   • Simplify and solve for $V_c$.

4. Dependent Source Calculations:

   • Once $V_b$ and $V_c$ are calculated, substitute them into the expression for the dependent sources.
   • For the voltage across 0.5 $I_x$, use Ohm’s Law on the resistors in that branch.

5. Final Calculations and Answers:

Let me calculate the values to find the node voltages and the required voltage across the 0.5 $I_x$ dependent current source.It appears that the initial approach didn't yield results due to a setup issue with the equations. I’ll refine the equations for accuracy in calculating the node voltages and voltage across the dependent source. Let me correct and simplify this step.The solution for the node voltages is as follows:

• $V_b = \frac{150}{11} \approx 13.64 \, \text{V}$
• $V_c = \frac{90}{11} \approx 8.18 \, \text{V}$
• $V_z = \frac{15}{11} \approx 1.36 \, \text{V}$

Voltage across the 0.5 $I_x$ dependent current source

The dependent current source is 0.5 $I_x$, where $I_x$ is the current through the 20 Ω resistor, calculated as:

$I_x = \frac{V_c}{20} = \frac{90/11}{20} = \frac{90}{220} \approx 0.41 \, \text{A}$

Then, the voltage across the dependent current source, 0.5 $I_x$, would be:

$V_{\text{source}} = 0.5 \times I_x \times 20 \approx 0.5 \times 0.41 \times 20 \approx 4.09 \, \text{V}$

Final Answers:

1. Node voltages:

   • $V_b \approx 13.64 \, \text{V}$
   • $V_c \approx 8.18 \, \text{V}$

2. Voltage across the 0.5 $I_x$ dependent current source:

   • $V_{\text{source}} \approx 4.09 \, \text{V}$

Would you like further explanation on any step?

Additional Questions to Explore:

1. How does one apply Kirchhoff's Voltage Law (KVL) to verify these node voltages?
2. What would change in the solution if the dependent voltage source was altered?
3. How do dependent sources affect the analysis in circuits compared to independent sources?
4. Can you solve for the power delivered by each source in this circuit?
5. What is the significance of grounding in solving for node voltages?

Tip:

In circuits with dependent sources, clearly define each current and voltage based on known values and dependencies before setting up equations for simplicity and accuracy.