1. Fundamentals of Electric Circuits Chapter 10
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2. Overview
This chapter applies the circuit analysis introduced in the DC circuit analysis for AC circuit analysis.
Nodal and mesh analysis are discussed.
Superposition and source transformation for AC circuits are also covered.
Applications in op-amps and oscillators are reviewed.

3. Steps to Analyze an AC Circuit
There are three steps to analyzing an AC circuit.
They make use of the fact that frequency domain analysis is simpler because it can make use of the nodal and mesh techniques developed for DC
Transform the circuit to the phasor or frequency domain
Solve the problem using circuit techniques
Transform back to time domain.

4. Nodal Analysis
It is possible to use KCL to analyze a circuit in frequency domain.
The first step is to convert a time domain circuit to frequency domain by calculating the impedances of the circuit elements at the operating frequency.
Note that AC sources appear as DC sources with their values expressed as their amplitude.

5. Using nodal analysis, find v1 and v2 in the circuit of figure below.
Example 1
Using nodal analysis, find v1 and v2 in the circuit of figure below.
Answer:
v1(t) = sin(2t ) V
v2(t) = sin(2t ) V

7. Nodal Analysis II Impedances will be expressed as complex numbers.
Sources will have amplitude and phase noted.
At this point, KCL analysis can proceed as normal.
It is important to bear in mind that complex values will be calculated, but all other treatments are the same.

8. Nodal Analysis III
The final voltages and current calculated are the real component of the derived values.
The equivalency of the frequency domain treatment compared to the DC circuit analysis includes the use of supernodes.

9. Mesh Analysis
Just as in KCL, the KVL analysis also applies to phasor and frequency domain circuits.
The same rules apply: Convert to frequency domain first, then apply KVL as usual.
In KVL, supermesh analysis is also valid.

11. Superposition
Since AC circuits are linear, it is also possible to apply the principle of superposition.
This becomes particularly important is the circuit has sources operating at different frequencies.
The complication is that each source must have its own frequency domain equivalent circuit.

12. Superposition II
The reason for this is that each element has a different impedance at different frequencies.
Also, the resulting voltages and current must be converted back to time domain before being added.
This is because there is an exponential factor ejωt implicit in sinusoidal analysis.

13. Example 3
Calculate vo in the circuit of figure shown below using the superposition theorem.
Vo = sin(5t – 81.12) cos(10t – 86.24) V

16. Source Transformation
Source transformation in frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance.
Or vice versa:

17. Example 4
Find Io in the circuit of figure below using the concept of source transformation.
Io = 3.28899.46 A

19. Thevenin and Norton Equivalency
Both Thevenin and Norton’s theorems are applied to AC circuits the same way as DC.
The only difference is the fact that the calculated values will be complex.

20. Thevenin transform
Norton transform

21. Find the Thevenin equivalent at terminals a–b of the circuit below.
Example 5
Find the Thevenin equivalent at terminals a–b of the circuit below.
Zth =12.4 – j3.2 
VTH = 18.97-51.57 V