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 series combination of the $12V$ voltage source and $-j\Omega$ resistor is converted to a $12jA$ current source in parallel with the $-j\Omega$ resistor.
$\mathbf I=\frac{12}{-j}=12jA$
Now we can combine $-j\Omega$ and $1\Omega$
$1||-j=\frac{-j}{1-j}=(0.5-0.5j)\Omega$
Also, combining $12jA$ and $-2A$,
$(12j-2)A$
The parallel combination of the $(12j-2)A$ current source and $(0.5-0.5j)\Omega$ resistor is converted to a $(5+7j)V$ voltage source in series with the $(0.5-0.5j)\Omega$ resistor.
$\mathbf V=(12j-2)\times(0.5-0.5j)=(5+7j)V$
Combining $6V$ and $(5+7j)V$,
$(11+7j)V$
Therefore, $\mathbf V_o$ is
$\mathbf V_o=(11+7j)\times\frac{2}{2+0.5-0.5j}=7.385+7.077j=10.228\angle 43.781^{\circ}V$
Related posts: Source Transformation
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