chapter 5 steady-state sinusoidal analysis

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chapter 5 steady-state sinusoidal analysis
CHAPTER 5
STEADY-STATE SINUSOIDAL ANALYSIS
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Allan R. Hambley
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Figure 5.1 A sinusoidal voltage waveform given by v(t) = Vm cos (ωt + θ). Note: Assuming that θ is in degrees,
we have tmax= –θ × T. For the waveform shown, θ is −45˚.
360
€
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Figure 5.2
Voltage and power versus time for Example 5.1.
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Figure 5.3
Triangular voltage waveform of Example 5.2.
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Figure 5.4
Answer for Exercise 5.1(c).
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Figure 5.5
plane.
A sinusoid can be represented as the real part of a vector rotating counterclockwise in the complex
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Figure 5.6
60˚).
Because the vectors rotate counterclockwise, V1 leads V2 by 60˚ (or, equivalently, V2 lags V1 by
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Figure 5.7
The peaks of v1(t) occur 60˚ before the peaks of v2(t). In other words, v1(t) leads v2(t) by 60˚.
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Figure 5.8
Current lags voltage by 90˚ in a pure inductance.
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Figure 5.9
Current leads voltage by 90˚ in a pure capacitance.
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Figure 5.10
For a pure resistance, current and voltage are in phase.
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Figure 5.11 Answers for Exercises 5.6, 5.7, and 5.8. The scale has been expanded for the currents compared
with the voltages so the current phasors can be easily seen.
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Figure 5.12
Circuit for Example 5.4.
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Figure 5.13
Phasor diagram for Example 5.4.
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Figure 5.14
Circuit for Example 5.5.
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Figure 5.15
Phasor diagram for Example 5.5.
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Figure 5.16
Circuit for Example 5.6.
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Figure 5.17
Circuit and phasor diagram for Exercise 5.9.
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Figure 5.18
Circuit for Exercise 5.10.
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Figure 5.19
Circuit for Exercise 5.11.
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Figure 5.20
A voltage source delivering power to a load impedance
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Figure 5.21
Current, voltage, and power versus time for a purely resistive load.
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Figure 5.22
Current, voltage, and power versus time for pure energy-storage elements.
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Figure 5.23
Power triangles for inductive and capacitive loads.
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Figure 5.24
The load impedance in the complex plane.
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Figure 5.25
The complex power delivered to this circuit element is
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S=
1
VI*
2
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Figure 5.26
Circuit and currents for Example 5.7.
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Figure 5.27
Circuit for Example 5.8.
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Figure 5.28
Power triangles for loads A and B of Example 5.8.
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Figure 5.29
Phasor diagram for Example 5.8.
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Figure 5.30 The Thévenin equivalent for an ac circuit consists of a phasor voltage source Vt in series with a
complex impedance Zt .
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Figure 5.31 The Norton equivalent circuit consists of a phasor current source In in parallel with the complex
impedance Zt .
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Figure 5.32
Circuit of Example 5.10.
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Figure 5.33
Thévenin and Norton equivalents for the circuit of Figure 5.32(a).
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Figure 5.34
The Thévenin equivalent of a two-terminal circuit delivering power to a load impedance.
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Figure 5.35
Thévenin equivalent circuit and loads of Example 5.11.
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Figure 5.36
Circuit of Exercises 5.14 and 5.15.
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Figure 5.37
A balanced three-phase voltage source.
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Figure 5.38
A three-phase wye–wye connection with neutral.
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Figure 5.39 Six wires are needed to connect three single-phase sources to three loads. In a three-phase
system, the same power transfer can be accomplished with three wires.
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Figure 5.40 Phasor diagram showing the relationship between the line-to-line voltage Vab and the line-toneutral voltages Van and Vbn.
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Figure 5.41
Phasor diagram showing line-to-line voltages and line-to-neutral voltages.
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Figure 5.42
Circuit and phasor diagram for Example 5.12.
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Figure 5.43
Delta-connected three-phase source.
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Figure 5.44
Loads can be either wye connected or delta connected.
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Figure 5.45
A delta-connected source delivering power to a delta-connected load.
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Figure 5.46
Circuit of Example 5.13.
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Figure 5.47
Circuit for Example 5.14.
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Figure 5.48
Circuit for Exercise 5.18.
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Figure P5.12
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