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Chapter 5Steady-State Sinusoidal Analysis
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Chapter 5Steady-State Sinusoidal Analysis
1. Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal.
2. Solve steady-state ac circuits using phasors and complex impedances.
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4. Find Thévenin and Norton equivalent circuits.
5. Determine load impedances for maximum power transfer.
6. Solve balanced three-phase circuits.
3. Compute power for steady-state ac circuits.
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SINUSOIDAL CURRENTS AND VOLTAGES
Vm is the peak value
ω is the angular frequency in radians per second
θ is the phase angle
T is the period
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Tπω 2
=
fπω 2=
( ) ( )o90cossin −= zz
Frequency T
f 1=
Angular frequency
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Root-Mean-Square Values
( )dttvT
VT
2
0rms
1∫=
RVP
2rms
avg =
( )dttiT
IT
2
0rms
1∫=
RIP 2rmsavg =
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RMS Value of a Sinusoid
2rmsmVV =
The rms value for a sinusoid is the peak value divided by the square root of two. This is not true for other periodic waveforms such as square waves or triangular waves.
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Phasor Definition
( ) ( )111 cos :function Time θtωVtv +=
111 :Phasor θV∠=V
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Adding Sinusoids Using Phasors
Step 1: Determine the phasor for each term.
Step 2: Add the phasors using complex arithmetic.Step 3: Convert the sum to polar form.
Step 4: Write the result as a time function.
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Using Phasors to Add Sinusoids
( ) ( )o45cos201 −= ttv ω
( ) ( )o60cos102 += ttv ω
o45201 −∠=V
o30102 −∠=V
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o
oo
7.3997.2914.1906.23
5660.814.1414.1430104520
21s
−∠=
−=−+−=
−∠+−∠=
+=
jjj
VVV
( ) ( )o7.39cos97.29 −= ttvs ω
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Sinusoids can be visualized as the real-axis projection of vectors rotating in the complex plane. The phasor for a sinusoid is a snapshot of the corresponding rotating vector at t = 0.
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Phase Relationships
To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise. Then when standing at a fixed point, if V1 arrives first followed by V2 after a rotation of θ , we say that V1 leads V2 by θ . Alternatively, we could say that V2lags V1 by θ . (Usually, we take θ as the smaller angle between the two phasors.)
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To determine phase relationships between sinusoids from their plots versus time, find the shortest time interval tp between positive peaks of the two waveforms. Then, the phase angle isθ = (tp/T ) × 360°. If the peak of v1(t) occurs first, we say that v1(t) leads v2(t) or that v2(t)lags v1(t).
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COMPLEX IMPEDANCES
LL Lj IV ×= ω
o90∠== LLjZL ωω
LLL Z IV =
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CCC Z IV =
o90111−∠==−=
CCjCjZC ωωω
RR RIV =
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Kirchhoff’s Laws in Phasor Form
We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.
The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.
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Circuit Analysis Using Phasors and Impedances
1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)
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2. Replace inductances by their compleximpedances ZL = jωL. Replacecapacitances by their complex impedances ZC = 1/(jωC). Resistances have impedances equal to their resistances.
3. Analyze the circuit using any of the techniquesstudied earlier in Chapter 2, performing the calculations with complex arithmetic.
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AC Power Calculations( )θcosrmsrmsIVP =
( )θcosPF =
iv θθθ −=
( )θsinrmsrmsIVQ =
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rmsrmspower apparent IV=
( )2rmsrms22 IVQP =+
RIP 2rms=
XIQ 2rms=
RVP R
2rms=
XVQ X
2rms=
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THÉVENIN EQUIVALENT CIRCUITS
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The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.
ocVV =t
We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.
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The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.
scsc
oc
IV
IV t
tZ ==
scII =n
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Maximum Average Power Transfer
If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.
If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.
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BALANCED THREE-PHASE CIRCUITS
Much of the power used by business and industry is supplied by three-phase distribution systems. Plant engineers need to be familiar with three-phase power.
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Phase Sequence
Three-phase sources can have either a positive or negative phase sequence.The direction of rotation of certain three-phase motors can be reversed by changing the phase sequence.
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Wye–Wye ConnectionThree-phase sources and loads can be connected either in a wye configuration or in a delta configuration.
The key to understanding the various three-phaseconfigurations is a careful examination of the wye–wye circuit.
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( ) ( )θcos3 rmsrmsavg LY IVtpP ==
( ) ( )θθ sin3sin2
3 rmsrms LYLY IVIVQ ==
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YZZ 3=Δ
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