Download Chapter 5 Steady-State Sinusoidal Analysis

5.4 Circuit Analysis Using Phasors and
Complex 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.)
2. Replace inductances by their complex impedances ZL
= jωL. Replace capacitances by their complex
impedances ZC = 1/(jωC). Resistances have complex
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.
•Find the steady-state current for the circuit as
follows, and also find the phasor voltage across
each element and construct a phasor diagram.
Answers:
I  0.707  15o ,VR  70.7  15o
VL  106.175 ,VC  35.4  105
o
o
I  0.707  15o ,VR  70.7  15o
VL  106.175 ,VC  35.4  105
o
o
•Find the steady-state voltage uC(t) for the circuit as follows, and
also find the phasor current through each element and construct
a phasor diagram showing the currents and source voltage.
Z RC
100( j100)

 50  j50()
100  j100
Answers:
VS  10  90,VC  10  180o , I  0.1414  135o
I R  0.1  180o , I L  0.1  900
VS  10  90,VC  10  180o , I  0.1414  135o
I R  0.1  180o , I L  0.1  900
•To find u1(t) in steady state.
Answers:
V1  16.129.70
•To find i(t) in steady state; construct a phasor diagram showing
all three voltages and current; what is the phasor relationship
between us(t) and i(t)?
Answers:
I  0.0283  1350 , Vs  10  900 , VL  7.07  450
i(t )
lags
us (t ) by 450
•To find phasor voltage and the phasor current
through each element in the circuit.
-j50Ω
Answers:
j200Ω
V  277  56.3, I C  5.5533.7o
I L  1.39  146.3o , I R  2.77  56.3o
•Steady-state Response of RLC in Series
I
U
U R
R
U L
j L
U C
1
j C
Z  R  j X L  j XC
 R  j( X L  X C )
 R jX
Reactance
U
1 2
2
2
2
| Z |  R  X  R  (L 
)
I
C
1
L 
X
C
   u   i  arctan  arctan
R
R
•Steady-state Response of RLC in Series
I
U
U R
R
U L
j L
U C
U  U R  U L  U C
 IR  I j X  I j X
L
1
j C
 IZ
C

UL
U
UX


UC
U
UX

UR

I

UC
Voltage Triangle


UR
•Steady-state Response of RLC in Series
UIZ
I
U R
U
U L
U C
R
j L
1
j C
Impedances
Triangle
U X  U L  UC
Z
 IX
X

R
U R  IR
| Z | R 2  X 2
1
L 
UL  UC

C
  arctan
 arctan
R
UR
•Power and Power Factor
P U rmsI rmscos 
Q  U rms I rms sin 
S  U rms I rms
P
cos  
S
P  S cos 
Q  S sin 
Power Factor
功率因数
•Three Triangles
U  U R  (U L  U C )
2
S
2
U
U R  U cos
U X  U sin
Z  R 2  ( X L  X C )2
R  Z cos
X  Z sin
Z

X L  XC U L  UC
R
P
UR
S  P Q
P  S cos
Q  S sin
2
2
Q
•Example 5.6: Computer the power and reactive
power taken from the source and each element in the
circuit.
PS  0.5(W ), QS  0.5(Var ),
Answers:
QL  1(Var ), QC  0.5(Var ), QR  0
PL  0, PC  0, PR  0.5(W )
PS  PR , QS  QL  QC
•Example 5.7:Find the power, reactive power and power
factor for the source, find the phasor current i.
P  PA  PB  10(kW )
Q  QA  QB  3.56(kVar),
Answers:
Q
  arctan( )  19.590 , cos   0.94
P
I  1549.590 ( A)
5.6 THÉVENIN EQUIVALENT CIRCUITS
5.6 THÉVENIN EQUIVALENT CIRCUITS
•The Thévenin voltage is equal to the opencircuit phasor voltage of the original circuit.
Vt  Voc
•We can find the Thévenin equivalent impedance
by zeroing the independent sources and
determining the complex impedance looking into
the circuit terminals.
5.6 THÉVENIN EQUIVALENT CIRCUITS
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
•Example 5.9: Find Thevenin equivalent circuit for the circuit.
Answers:
Zt  50  j 50(),Vt  100  900 (V ), I n  1.414  450 ( A)
•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.
•Example 5.10: Determine the maximum power delivered to a
load (a) the load can have any complex value; (b) the load must
be a pure resistance.
Answers:
(a) whenZload  50  j50(), P  25(W )
(b) whenZload  50 2  70.71(), P  20.71(W )
5.7 BALANCED THREE-PHASE CIRCUITS
•Much of the power used by business and
industry is supplied by three-phase distribution
systems.
BALANCED THREE-PHASE CIRCUITS：
Three equal-amplitude ac voltages have phases that are
1200 apart.
Chapter 17 tells us how three-phase voltages are generated.
Wye (Y)connected
phaseA,Van  VY 00 ,
Line a、b、c
火线、相线
Neutral n
零线、中线
Phase voltage 相电压VY
phaseB,Vbn  VY   1200 ,
1200
phaseC ,Vcn  VY 1200
Positive phase sequence：
a→b→c
•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.
•Wye–Wye Connection
•Three-phase sources and loads can be
connected either in a wye (Y) configuration
or in a delta (Δ) configuration.
•The key to understanding the various
three-phase configurations is a careful
examination of the wye–wye (Y-Y) circuit.
sources
Line currents
线电流
Balanced loads:
All three load impedances are equal.
Balanced loads
对称负载
Neutral currents
中线电流
Four-wire connection
•Under balanced three-phase sources and loads,
INn  IaA  IbB  IcC  0 Therefore, we can omit the neutral wire.
•Wye–Wye Connection
In a balanced three-phase system, neutral current is zero.
we can eliminate the neutral wire. Then, compared with single phase
circuit, only three wires are needed to connect the sources to the
loads , it is less expensive.
•Wye–Wye Connection
Compared with single-phase circuit, the total
instantaneous power in a balanced three-phase system
is constant rather than pulsating.
Pavg  p(t )  3VYrms I Lrms cos 
•Reactive power
VY I L
Q3
sin    3VYrms I Lrms sin  
2
(线电压)
A(相电压)
B(相电压)
• A balanced positive-sequence wye-connected 60 Hz three-phase source has
phase voltage UY=1000V. Each phase of the load consists of a 0.1-H
inductance in series with a 50-Ω resistance.
• Find the line currents, the line voltages, the power and the reactive power
delivered to the load. Draw a phasor diagram showing line voltages, phase
voltages and the line currents. Assuming that the phase angle of Uan is zero.
Z  R  jL  50  j37.7  62.6237 0
  37 0
Van
I aA 
 15.97  370
Z
 I bB  15.97  157 0 , I cC  15.97830
Z  R  jL  50  j37.7  62.6237
  37 0
0
Van
 15.97  370
Z
 I bB  15.97  157 0 , I cC  15.97830
I aA 
Vab  Van  3300  1732300 Vbc  1732  900 ,Vca  17321500
Pavg  3VYrms I Lrms cos   19.13kW
VY I L
Q3
sin   14.42 kVar
2
•Delta (Δ)-connected Sources
According to KVL,
Vab  Vbc  Vca  0
Thus, the current circulating in the delta is zero.
•Wye (Y) and Delta (Δ)-connected Loads
whenZ  3ZY Two balanced loads are equal.
•Wye (Y) and Delta (Δ)-connected Loads
Z   3ZY
•Delta - Delta (Δ- Δ) connection
VL
I 
Z
I L  3I 
Line current
Assuming line voltage
Vab  VL 300
Then, we get phase current
I AB
Vab
VAB


 I  300  
Z   Z  
IaA  IAB  ICA  IAB  3  300
I  I  3  300 ,
bB
BC
IcC  ICA  3  300
A delta-connected source supplies power to a deltaconnected load through wires having impedances of
Zline=0.3+j0.4Ω, the load impedance are ZΔ=30+j6 Ω, the
balanced source A voltage is Vab  1000300 (V )
Find the line current, the line volatge at the load, the
current in each phase of the load, the power delivered
to the load, and dissipated in the line.
•Homework 5






P5.24
P5.34
P5.43
P5.53
P5.68
P5.72