AC Steady-State Analysis(Using the Superposition)

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

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

$\mathbf V_o'=6V$

$\frac{\mathbf V_1}{-j}-4+\frac{\mathbf V_1}{2}=0$

$\mathbf V_1=\frac{4}{j+0.5}=(1.6-3.2j)V$

So, $\mathbf V_o''=\frac{\mathbf V_1}{2}=(0.8-1.6j)V$

Then, $\mathbf V_o=\mathbf V_o'+\mathbf V_o''=6+0.8-1.6j=6.9857\angle-13.2405\,^{\circ}V$