Chapter 6 – Steady State AC Circuit Analysis
Chapter 6
Steady State AC Circuit Analysis
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AC Mesh Analysis
AC Nodal Analysis
Norton and Thevenin’s Equivalent Circuit
Power Equations
Steady-state Analysis of AC circuit
1. Nodal Analysis
The analysis method is the same as that for dc circuit. Consider the following
example.
E.g.1 Find ix in the circuit of Fig. 10.1 using nodal analysis.
Fig. 10.1
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Chapter 6 – Steady State AC Circuit Analysis
Solution
We first convert the circuit to the frequency domain:
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Chapter 6 – Steady State AC Circuit Analysis
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Chapter 6 – Steady State AC Circuit Analysis
2. Mesh Analysis
The analysis method is the same as that for dc circuit analysis.
E.g. 2 Determine Io in the circuit of Fig. 10.7 using mesh analysis.
Solution
The desired current is Io = -I2 = 6.12∠144.78o .
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Chapter 6 – Steady State AC Circuit Analysis
3. Superposition Theorem
Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it
applies to dc circuits. The theorem becomes important if the circuit has sources operating at
different frequencies. In this case, since the impedances depend on frequency, we must have a
different frequency-domain circuit for each frequency. The total response must be obtained by
adding the individual responses in the time domain.
E.g. 3 Use the superposition theorem to find Io in the circuit in Fig. 10.7.
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Chapter 6 – Steady State AC Circuit Analysis
4. Source Transformation
Source transformation in the frequency domain involves transforming a
voltage source in series with an impedance to a current source in parallel with an
impedance, or vice versa.
E.g. 4 Calculate Vx in the circuit of Fig. 10.17 using source transformation.
Solution
Fig. 10.17
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Chapter 6 – Steady State AC Circuit Analysis
5. Thevenin and Norton Equivalent circuits
Thevenin’s and Norton’s theorems are applied to ac circuits in the same way
as they are to dc circuits. The only additional effort arises from the need to
manipulate complex numbers. The following figures depict the Thevenin’s and
Norton’s equivalent circuits.
(a) Thevenin equivalent
(b) Norton equivalent
Keep in mind that the two equivalent circuits are related as
v
v
VTH = Z N IN , ZTH = Z N
v
v
Just as in source transformation. VTH is the open-circuit voltage while IN is
the short-circuit current.
If the circuit has sources operating at different frequencies, the Thevenin’s or
Norton equivalent circuit must be determined at each frequency. This leads to
entirely different equivalent circuits, one for each frequency, NOT one
equivalent circuit with equivalent sources and equivalent impedances.
E.g. 5 Obtain the Thevenin’s equivalent at terminals a-b of the following circuit.
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Chapter 6 – Steady State AC Circuit Analysis
Solution
The Thevenin’s impedance is
v v
ZTH = Vs Is = 2(6 − j ) 3 = 4 − j 0.6667 Ω
Fig. 10.26
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Chapter 6 – Steady State AC Circuit Analysis
E.g. 6 Obtain the Thevenin’s equivalent at terminals a-b of the following circuit.
Solution
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Chapter 6 – Steady State AC Circuit Analysis
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