AC Steady-State Analysis(Using the Nodal Analysis)

Problem-Solving Strategy for AC Steady-State Analysis

STEP 1. For relatively simple circuits (e.g., those with a single source), use

- Ohm’s law for ac analysis—that is, $\mathbf V = \mathbf I\mathbf Z$

- The rules for combining $\mathbf Z_s$ and $\mathbf Y_p$

- KCL and KVL

- Current and voltage division

STEP 2. For more complicated circuits with multiple sources, use

- Nodal analysis

- Loop analysis

- Superposition

- Source transformation

- Thévenin’s and Norton’s theorems

PROBLEM:

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed. 

If there are any errors in solution, please let me know in the comments

SOLUTION:

The KCL equation for middle of the node is

$\frac{\mathbf V_o-4}{1}+\frac{5\mathbf V_o-4}{1}+\frac{\mathbf V_o}{-j}=0$

Thus, $V_o=\frac{8}{6+j}=1.297-0.216j=1.315\angle-9.462\,^{\circ}V$

Related post: Nodal Analysis Technique

                  AC Steady-State Analysis(Using the Loop Analysis)

                  AC Steady-State Analysis(Using the Superposition)

                  AC Steady-State Analysis(Using the Source Transformation)

                  AC Steady-State Analysis(Using the Thévenin's Theorem)

                  AC Steady-State Analysis(Using the Norton's Theorem)