Download Ch. 10

Fundamentals of
Electric Circuits
Chapter 10
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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.
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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
1. Transform the circuit to the phasor or
frequency domain
2. Solve the problem using circuit techniques
3. Transform back to time domain.
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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.
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Example 1
Using nodal analysis, find v1 and v2 in the circuit of
figure below.
Answer:
v1(t) = 11.32 sin(2t + 60.01 )
V
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v2(t) = 33.02 sin(2t + 57.12 ) V
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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.
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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.
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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.
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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.
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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.
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Example 3
Calculate vo in the circuit of figure shown below
using the superposition theorem.
Vo = 4.631 sin(5t – 81.12 ) + 1.051 cos(10t – 86.24 ) V
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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:
Vs  Z s I s

Is 
Vs
Zs
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Example 4
Find Io in the circuit of figure below using the
concept of source transformation.
Io = 3.288 99.46 A
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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.
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Thevenin transform
Norton transform
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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
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