To solve this circuit problem using Nodal Analysis, we need to apply the following approach:

Step-by-Step Solution Outline

1. Identify Nodes:

   • We label the nodes as $A$, $B$, and $C$ as given.
   • There are sources and resistors in the network that influence the node voltages.

2. Set a Reference Node:

   • Usually, the lowest potential point (often the ground or the bottom node in the diagram) is set as the reference node.
   • Let’s assume the node where the current source $I_1$ and resistor $R_2$ meet (bottom left) is the ground (0V).

3. Define Node Voltages:

   • Let $V_A$, $V_B$, and $V_C$ be the voltages at nodes $A$, $B$, and $C$ relative to the ground.

4. Apply Kirchhoff’s Current Law (KCL) at each node except the reference node:

   • For each node, sum the currents leaving or entering the node and set the equation equal to zero.
   • Use Ohm's Law, $I = \frac{V}{R}$, to express each current in terms of the node voltages and resistances.

5. Set up Equations:

   • Formulate equations based on KCL for nodes $A$, $B$, and $C$.
   • Use the given values for resistances and sources to simplify.

6. Solve for Currents $i_1$, $i_2$, $i_3$, and $i_X$:

   • Once node voltages are determined, use them to find the currents through each branch.

Would you like me to proceed with the detailed calculations, or do you need clarification on any specific part of this process?

Here are some related questions that might further enhance your understanding of nodal analysis and circuit theory:

1. How does choosing a different reference node affect the node voltage equations?
2. How do we handle dependent sources in nodal analysis?
3. Can we apply nodal analysis in circuits with only voltage sources, and how would that change the approach?
4. What is the role of superposition in analyzing complex circuits?
5. How would mesh analysis differ from nodal analysis in solving this circuit?

Tip: When solving circuits with multiple current sources, keep track of their directions carefully, as reversing them can change the signs in your equations.