Transcript Chapter 5 - Electrical and Computer Engineering

Chapter 5
Steady-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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3. Compute power for steady-state ac
circuits.
4. Find Thévenin and Norton
equivalent circuits. Lightly.
5. Determine load impedances for
maximum power transfer.
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SINUSOIDAL CURRENTS
AND VOLTAGES
Vt = Vm cos(ωt +θ)
Vm is the peak value
ω is the angular frequency in radians
per second
θ is the phase angle
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1
f 
T
Frequency
Angular frequency
2

T
  2f
sin z   cosz  90
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

Root-Mean-Square Values
Vrms
1

T
Pavg
T
 v t dt
2
0
2
rms
V

R
T
I rms
1
2



i
t
dt

T 0
Pavg  I
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2
rms
R
RMS Value of a Sinusoid
Vrms
Vm

2
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
Time function : v1 t   V1 cosωt  θ1 
Phasor : V1  V1θ1
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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 
v1 t   20 cost  45

v2 t   10 cost  60

V1  20  45

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V2  10  30
Chapter 5
Steady-State Sinusoidal Analysis


Vs  V1  V2
 20  45  10  30
 14.14  j14.14  8.660  j5
 23.06  j19.14

 29.97  39.7


v s t   29.97 cost  39.7
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

Sinusoids can be visualized as the realaxis 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 V2 lags
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 is
θ = (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
VL  jL  I L
Z L  jL  L90
VL  Z L I L
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
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VC  Z C I C
1
1
1

ZC   j


  90
C jC C
VR  RI R
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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 complex
impedances ZL = jωL. Replace
capacitances by their complex impedances
ZC = 1/(jωC). Resistances have impedances
equal to their resistances.
3. Analyze the circuit using any of the
techniques studied earlier in Chapter 2,
performing the calculations with complex
arithmetic.
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“SPOOKY” And “SPECTRE”
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AC-130 GUNSHIP
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AC Power
Chapter 5
Steady-State Sinusoidal Analysis
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AC Power Calculations
P  Vrms I rms cos 
PF  cos 
   v  i
Q  Vrms I rms sin  
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apparent power  Vrms I rms
P  Q  Vrms I rms 
2
PI
QI
2
rms
2
rms
R
X
2
2
P
Q
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V
2
Rrms
R
V
2
Xrms
X
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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.
Vt  Voc
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.
Voc Vt
Z t

I sc I sc
I n  I sc
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Maximum 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 Connection
Three-phase sources and loads can be
connected either in a wye configuration or in a
delta configuration.
The key to understanding the various threephase
configurations is a careful examination of the
wye–wye circuit.
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Pavg  p t   3VYrms I Lrms cos 
VY I L
Q3
sin    3VYrms I Lrms sin  
2
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Z   3ZY
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