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Chapter 9 Analysis and Calculation of Sinusoidal Circuit
Main content：
1 resistance and conductivity
2 Resistance Series-parallel Connection
3 Phase Graph
Key point
4.analysis of Sinusoidal Steady State Circuit
5 power of Sinusoidal Steady State Circuit
6 Complex power
7 The transmission of maximal power
nodus
Key point
8 Series connection resonance
Key point
9 .1
resistance and conductivity
U U u
Z 
I I i
 Z 
U
  u   i
I
Ohm’s
Law
The magnitude
of complex
impedance
U  IZ
complex
impedance
a
+
U
–
b
I N0
U
Z 
I
  u  i
The angle of complex impedance
conclusion：the magnitude ofＺ is the result which the rms
value of total voltage divide the rms value total current，and
the angle of Ｚis the result which total voltage’s angle minus
total current’s angle 。
impedance
1
Z  R  jL 
jC
1
 R  j (L 
)
C
 R  jX  Z  Z
I
U
admittance
U R
L
U L
C
Impedance
Reactance X＞0，Z is inductive triangle
X＜0，Z is capacitive
X=0，Z is resistant
Z
1
Y
Z
R
U C
N0
 X L  XC
R
Phasor equation:
Phasor model:
I
jL
1

U
jC
U  U R  U L  U C
if
I  I0 （reference phasor
R
U R
L
U L
U R  IR
U L  I jL 
U C
 1 



UC  I 
 j C 


C

U  I  R 

1 

j  L 

C 

The relation between
total voltage and
total current
Note：
Z  R  j X L  X C 

Z
I
is a complex number,not a phasor
R
U R
L
U L
so it doesn’t have a dot at the top 。Z is
only a operation implement in a
equation
 

U
U  IZ
C
In sinusoidal AC circuit,if the physics is expressed by
phasor and component ‘s parameter is expressed by
complex impedance ，the equations is similar to
that of DC circuit.
U C
I
Impedance triangle
Voltage triangle
Power triangle
U
S
R
U R
L
U L
C
U
U C
Q
Z

X L  XC
R
U L  U C
U R
N0
P
I
U R
L
U L
C

U
R
U C
Assume R、L、C is identified，
Whether is Circuit character
identified?
（impedance？inductance？
capacitance？）
Forbidden！
1
X L  L 、 X C 
C
ifω is different ，May come out ：
XL > XC ，or XL < XC , or XL =XC 。
9 .2
Resistance Series-parallel
Connection
（一）simple series
connection circuit
I
i
ui
Z1
U i
uo
Z
Z2
Z=Z1+Z2+…
I
U i
Z1
Z2
U o
U O 
Z2
U i  uo
Z1  Z 2
（二)Simple Parallel Connection Circuit
i
I
i1 i2
ui
Z1
Z2
I
U i
U i
1 1
1
 

Z Z1 Z 2
I2
Z1
Z2
Z
I2 
Z1 
I
Z1  Z 2
Example 9-2
If R1=10Ω， R2=1000 Ω ，
L=05H、C=10μF , US=100V,
+
ω=314 rad/s.Solve the sum of every
U10
branch’s current
I
R1
L
+

U

U S 10 -
I1
C


U

100
0
solution： S
Z C  R2
Z eq  R1  jL 
 102.11  j132.13
Z C  R2

U
I  S  0.652.3
Z eq
I2
R2


U10  182.07  20.03


U
U
I  10  0.5769.97 I  10  0.18  20.03
1
2
ZC
R2
9.3 Phase Graph of Circuit
In series current is reference
phasor，KVL
In parallel voltage is reference
phasor ，KCL I
R
U
L
C
U R
U L
U C
U L
U L  U C
U C
Voltage
triangle
U
U R
I
First draw
reference
phasor
I
example 9-4
draw the phasor
graph of example
U L
I1
U R1
+
-
I
I2
R1
U S
L
+

U10
-
U S
U10
I1
C
I2
R2
9.4 Analysis of Sinusoidal Steady State Circuit
Complemented
example
If：I1=10A、UAB =100V，
I2
 j10
C1
I
A
A
I1
solve： A 、UO
C2
B
5 j5 
UO
Two Solution：
1 Phase Operation By Virtue of Compound.
2.use phasor graph to solve
solution 1:Phase Operation By Virtue of Compound
I1
 j10
C1
A
I
A
I2
C2
5 j5 
UO
B
If： I1=10A、
UAB =100V，
solution：
the number of A、UO
Assume ：
U As reference phasor, namely：
AB
so： 
I 2  100
5  j5


U

100

0
V
AB
 10 2  45
A


I1  1090  j10 A
I  I  I  100 A  The number of A 10A
1
2
I1
 j10
C1
I
A
A
I2
C2
If： I1=10A、
UAB =100V，
B
5 j5 
UO
Solve:
the number of A、UO
I  I  I  100 A
1
2
U C1  I（  j10)   j100 V
U o  U C1  U AB  100  j100
 100 2  45 V

 The number of UO 141V
Solution 2:use phasor
graph to solve

I1
 j10
C1
I
If： I1=10A、 UAB =100V，
A
A
solve：the number of A、UO
B
C2
I2
U o  U C1  U AB
5 j5 
UO


assume：U AB  1000 V
I1  10A 、lead 90°
I 2  100 2 2  10 2A
5 5
I2 Lag U AB 45°
UC1=I XC1=100V
uC lag i 90°
thus：
I1
I  I1  I2
45°
U C1
I2
I
U AB
U O
I=10 A、 UO =141V
Complemented example:
R1
is
If ：
ie
iL
iR 2
L
R2
C
e
is  I m sin( t  1 )
e  Em sin( t   2 )
R1、 R2、L、C
solve：current of every branch
R1
Original circuit
is
iL
iR 2
L
R2
C
e
 jX c
IL
R1
Phasor model
Is
jX L
IR 2
R2
Ie
E
solution一：node’s electric potential method
A
IL
R1
Is
jX L
 jX c
IR 2
R2
References：
Ie
E
Node’s equation:
E
 jX C

UA 
1
1
1


jX L R2  jX C
 IS 
I  I m 
S
1
2
E
E  m  2
2
jX L  jL
1
 jX C   j
C
Solve the current of every branch from
node’s electric potential：
 jX c
A
IL
R1
Is
jX L
IR 2
R2
Ie
E

U
I  A  i
L
L
jX L

U
I  A  i
R2
R2
R2
  E
U
Ie  A
 ie
 jX C
Solution 二: Thevenin’s theorem
A
IL
R1
Is
jX L
Ie
U AB
E S
Ie
E
E
B
Solve:
A
Z
R2
solve：current of
every branch
jX c
Ie
IR 2
 jX c
B
Z  jX L R2
E S  IS（jX L
IL、IR 2
R2 )
Conclusion :the solution steps of
sinusoidal AC circuit
1、Draw phasor model graph according to initial
circuit graph（stay circuit’s structure is changeless）
R  R 、 L  jX L、 C   jX C
u  U、 i  I、 e  E
2、write equations or draw phasor graph
according to phasor model
3、use complex number or phasor
graph to solve
4、transform the result to required form
9.5
Power of Sinusoidal Steady
Circuit
u
一. resistive circuit
i
1 .instantaneous equation
：
2. Rms value equation ：
3. Phase equation
u  iR
i
u
ωt
U  IR
U  I R
I
R
U
Power of Resistive Circuit
1. Instantaneous power p:instantaneous
voltage multiply instantaneous current
p  u i  Ri  u / R
2
2
minuscule
i  2 I sin ( t )
u i 2 U sin ( t )
u
ωt
Conclusion ：
p0
（consumed
energy component）
1.
pChange with time
2
2
p
3.
Is proportional to u 、i
2.
p
ωt
2. average power（ (active power)
P ：the average of the instantaneous power
over one period
1 T
1 T
P   p dt   u  i dt
T 0
T 0
1 T
2
  2UI sin t dt
capital
T 0
1 T
  UI (1  cos 2 t )dt UI
T 0
P U  I
二.inductive circuit
1、 .instantaneous equation：
2 Rms value equation
3. Phase
3.
equation
di
uL
dt
i  2 I sin ( t )
U  IL
U  jl I
u
i
90
t
IL
i
u
L
U
I
I
Power of Inductance Circuit
1. Instantaneous power p ：
p  i  u  2UI sin t cost  UI sin 2t
u
i
i
u
i
u
i
i
t
u
u
p
+
Transformati
on of energy
is reversible
p <0
p >0
p >0
Store
Energy
+
Extract
energy
p <0
t
2. average power P （real power）
Instantaneous power
p  i  u  UI sin 2t
1 T
1 T
P   0 p dt   0 U I sin ( 2t ) dt  0
T
T
Conclusion ：in a purely inductive circuit,the average
power is zone. It only exchange energy with source。
3. Reactive Power Q
Q definition ：A measure of the power associated with
purely inductive circuit;
instantaneous power。
the maximal value of
Q U I  I XL U
2
2
XL
unit： (var)
u  2 U sin ( t )
三.capacitive circuit
1、 Instantaneous equation ：
i
u
1
U 
I
C
2 Rms value
3. Phase equation
u
90
du
iC
dt
U 
1
jc
i
I
I
t
U
1.Instantaneous power p
Power of Conductance Circuit
p  i  u  U I sin 2t
uu  2 U sin ( t )
i
ωt
i
i
u
u
p
deliv
er
P<0
Extrac
t
energy
P>0
absorb
Store
Energy
i
i
u
u
deliver
absorb
2 average power P
p  i  u  U I sin 2t
P0
3. Reactive Power Q
the maximal value of instantaneous power
Q  UI
（Q is negative for capacitors）
四、 Calculation of Power of No Circuit
1. Instantaneous power
i
p  u  i  p R  p L  pC
2. average power( real power)
u
1 T
P   pdt
T 0
1 T
  ( pR  pL  pC )dt
T 0
2
 PR  U R I  I R
R
uR
L
uL
C
uC
N0
The relation among P、U、、 I ：
P  URI
U R  U cos 

U L  U C
U

U R
P  UI cos 
Total voltage Total current The angle between u and i
COS
----- power
factor
3. Reactive Power Q：
In R、L、C series connection circuit ，storing
energy component L、C do not consume energy but
they can swallow and disgorge energy ,whose scale is
expressed by reactive power Q 。
Q  QL  QC
 U L I （  U C I）
（U L  U C）
I
 IU sin 
U L  U C
U

U R
4Apparent Power S：
In circuit the rms value of total voltage multiply the rms value of total
current 。
S  UI
unit：VA、KVA
notice： S＝U I is used to measure to the maximal
power provided by generator（rated voltage×rated
current）
5. Power triangle：
Real power P
Reactive power Q
Apparent Power S
P  UI cos 
Q  UI sin 
S  UI
S

Q
P
（help
remember）
The relation of power in circuit
if i leading u ，（capacitive circuit）
i
u
t
UI sin 
+
_
_
p
+
UI cos 
I
Impedance triangle
Voltage triangle
Power triangle
S
U
R
U R
L
U L
C
U
Q
Z

X L  XC
R
U L  U C
U R
P
U C
Judge Right Or Wrong
in R-L-C series circuit
U  U R  U L  U C  IR  I  X L  X C  ？
Neglect the phase of U
I
R
U
L
C
U  U R  U L  U C
U R
U L
U L
U L  U C
U C
U

U
U R
I
Judge Right Or Wrong



U  IZ ？
U、I
Reflect sinusoid voltage or current，
But complex impedance is a operational sign。
Z can not add “•”
Judge Right Or Wrong
in Ｒ－Ｌ－Ｃ sinusoidal AC circuit
U
I

Z
？
u
i
Z
？
U
I  ？
Z

U
I 

Z
？

U

I 
Z
？
9. 6 Compound Power
U  U u ,
I
I  I i
j
load
_
S
Q

power is describer as complex number
P  UI cos( u   i )
 UI Re[e
+
U
一. Compound Power:
+1
 Re[Ue
j u

U
P
S  U I *  UI( u   i )  UI
 UIcos  jUIsin
 P  jQ
Ie
j( u  i )
- j i
]
]  Re [U I* ]
*

I
S  UI*
def
Q  UI sinφ
Reactive Q
unit: var
Real power is conservation; reactive power is
conservation; complex power is conservation
Complex power is conservation ,but apparent power is
not conservation
S  UI*  (U 1  U 2 ) I *
 U I * U I*  S  S
I
+
U
_
1
+ U _
1
+
U 2
_
S  UI
S1  U1 I
U  U1  U 2
 S  S1  S 2
Complex power is expressed :
2
1
2
S2  U 2 I
b
S   Sk
k 1
*
*
2




S U I  Z I I  I Z
*
*
* *
2 *






S  U I  U (UY )  UU Y  U Y
Increasing of Alternative Current Power and Power Factor
Total current
P  UI cos 
Total voltage
Power factor
COS
U L
The angle between u and i

The meaning of Increasing of
Alternative Current Power and
Power Factor ：
（1）make the best of the
capacity of electric of
source
U
U L  U C

I

UR
U C
S=UI If S=1000KVA，and
COS =0.4，P=400KW
COS =0.9，P=900KW
（2）reduce the energy loss and voltage
Put out questions：
Usually many loads is inductive
and its Equivalent circuit and the phasor relation are as
follows:
i
u
R
uR
L
uL
U
U L

U R
I
Consume the real power：
P = PR = UICOS 
When U、P is certain，

 Hope increase COS 
COS 
I
The relation between power factor cosΦ
and circuit parameter
i
u
Z
X L  XC

load
R
X L  XC
  tg
R
1
Note：
cos  is defined by the character of loads。
It relates to circuit parameter, frequency
but do not relate to circuit voltage and current
Example：
40W incandescent lamp
P  UI cos 
COS   1
P
40
I

 0.182 A
U 220
40W daylight lamp
COS  0.5
P
40
I

 0.364 A
U cos  220  0.5
Require the capacity
of generating and
supplying-electric
facility is large
power station usually ask the user COS  0.85
or the users will be punished。
The power factor of usual circuit
Purely resistive circuit
Purely inductive circuit
or purely capacitive circuits
R-L-C series circuit
electromotor none load
full load
daylight lamp
（R-L-C series circuit）
COS   1
(  0)
COS  0 (   90)
0  COS  1
(90    90)
COS  0.2 ~ 0.3
COS  0.7 ~ 0.9
COS  0.5 ~ 0.6
Principals of Increasing Power Factor
Make sure initial loads’ state is constant ,
Namely :load voltage and real power is constant
the measurement of increasing power
factor :
i
Parallel capacitive
R
u
L
uR
uL
C
Calculation of Parallel- Connection Capacity
if initial circuit’s power factor is cos L，
require to compensate cos L to cos ,
solve the parallel connection C ？
（if U、P is certain）
IC
i
R
u
L
uR
uL
C
U
I
IRL

L
Analysis Basis： in all course P、U are constant。
From the phasor diagram：
IC
I C  I RL sin  L  I sin 
U
 P  UIRL cos L
P  UI cos 
IC  U
XC
 U C

I
IRL
L
P
P
UC 
sin  L 
sin 
U cos  L
U cos 
P
P
UC 
sin  L 
sin 
U cos  L
U cos 
P
C
(
tg


tg

)
L
2
U
i
R
u
L
uR
uL
C
Node （1）after paralleling
connection capacity ,initial loads’ state
is constant , namely ，I1 and cos1
are constant；
（2）total current I reduce，because
IX reduce but IR is constant
Which is better for compensating power
factor to inductive or capacitive？
IC


U Lack of
I
IRL
inductive（
compensate
I'C
Excessive
compensate
C larger
I

U
IC smaller）
IRL
Capacitive （
I'Clarger）
Conclusion :when  is the same ，if compensate to capacitive ,it
requires the capacity of capacitance is large ,and it is not
economical .so circuits usual work in lack of compensation state
By paralleling capacitive to compensate，the real power of
circuit is whether change？
IC
U
I I2
R
 jX C jX L
P  UI cos 
I1L
I < IRL
 <L
U
I
IRL

L
And cos  、 I
Through computing we know total power is constant。
qualitative illustration：the resistances in circuits do
not change ，so consumed powers do not change。
what else ways to increase power factor except paralleling
capacitance ？
Before Compensate
I
U
After compensate Whether can
Series
I C
capacitance？
R
L
U
U RL
L
U RL
U RL
U
U RL  U

R
I
U C
I
 0
If external voltage is constant ，load can not get rated work
voltage and current。
Example.
If：f=50Hz, U=380V, P=20kW, cos1=0.6(lagging)。Require
increase power factor is 0.9 ,solve parallel capacitance C。
IC
I
+
U
_
solution：
P=20kW
cos1=0.6
C
+
U
_
By cos φ1  0.6 conclude : φ1  53.13o
By cos φ2  0.9 conclude : φ2  25.84o
P
C
(tgφ 1  tgφ 2 )
2
U
20  10 3



(
tg
53
.
13

tg
25
.
84
)
2
314  380
 375 F
R
I L
C
L
1 2
U
I
I L
IC
Supplement: Power Measurement
i
+
u
-
*
Average torque :
*
W
Power instrument
*
i2
Voltage line
Z
*
i1
R
Current line
U
M K
I cos   K 'UI cos   K ' P
R
measure：the measure of P = U’s measure I’s
measurecos
When measure，P、U、I must not exceed each measure。
I
9-7 Maximum Power Transmission
Zi
Zi= Ri + jXi， ZL= RL + jXL


US
I 
, I 
Zi  ZL
-
( Ri  RL ) 2  ( X i  X L ) 2
2
R
U
L S
realpower P  RLI 2 
( Ri  RL ) 2  ( X i  X L ) 2
P
0
X L
P
0
RL
ZL
US
US

+
RL and XLcan
change at will
The Conditions of making ZL get
maximal power：
Best Match
RL= Ri
Pmax 
ZL= Zi*，即
XL =-Xi
U S2
4Ri
supplement example
5
R
1
+
uS(t)
5
–
us (t )  2 sin( 2t  45o ) V
If :
2.5H
C
Hope R to get the
maximal power ,solve C？
Solution 1：Use thevenin’s theorem equivalent circuit：
U S  1  45 o V
2.5
Z
i
+
–
Zi 
1

US
2
j5

C
+
1
–
C
Uoc
2.5  j5
 2  j1Ω
2.5  j5
1/(j2C) = -j1
1
 1, C  0.5 F
2C
Supplement :Work Principal of Lamp
Double metal
electrode
Fixed electrode
filament
Ballast L
link
Light tube
+
-
Impose Line voltage on two electrode
Generate splendent light and
release electricity
Double metal electrode is heated and get through
circuit（L、filament、started electrode）
Current
Electrode eject a
Temperature rise 800 C
is large
large of electron
--1000 C
Start up implement
Release electricity end
get through
Double metal electrode
get cold and cut off
di
ul  l dt
Ballast generate high
voltage pulse
Electrode shoot a large of electron，
because of voltage pulse filament
breakdown and release electricity
Mercury atom impact with
high energy electron and
generate ultraviolet ray
Ultraviolet ray irradiate
fluorescent powder and
generate visible light
Sometimes use
9-8
Resonance of Series Connection Circuit
Sometimes avoid
In L、C circuit ，if the port voltage and current
of circuit are in phase ,it is called resonance 。
Define
resonance
一、 Resonance of Series
Connection Circuit

I
R
+
j L

U
_
1、the conditions of resonance：
Z  R  j(ωL  1 )  R  j( X L  X C )
ωC
 R  jX
1
jω C When
ω0 
1
LC
ω 0 L  1 ，Circuit arise resonance
 0C
resonant angular frequency
Fixed Frequency
the condition of RLC series circuit arising resonance:
（1）. L C constant ,change w 。
w0 is
decided by circuit parameter ;a RLC series connection
circuit only have a correspondingw0 ; when frequency equal the
resonance frequency ,circuit arise resonance。
（2）. The voltage frequency is constant ,change ，L or C
( usually change C )。
Usually radios choose broadcasting station ,in essence
choose different frequent single through changing c to make
circuit arise resonance.
２、the feature of Resonance of RLC Series
Connection Circuit：


UL

(1). U and I Are in the same phase

UR
(2). Input port Z is purely resistance 。
In circuit the value of |Z| is minimal。

I

UC
|Z|
(3). current I have the maximal
value I0=U/R (U is certain)。
R
0
O


(4). LC series connection total
voltage is zone
U U
 0,

L
I
+

C

Electric source voltage is
imposed on resistance
UR=U0
U
_
R
+ U
R
_ +

UL
_
+
UC_
j L
1
jω C
In Series resonance circuit ，inductive
voltage and capacitive voltage is same in
value ,but the direction is reverse .So series
connection resonance is called Voltage
Resonance
When 0L=1/(0C )>>R ，
UL= UC >>U 。

UL

UR

I

UC
The phasor
diagram in
resonance circuit
(5). power
P=RI02=U2/R，resistance power is up to the
maximal value。
Land C exchange energy ，loads and electric
source do not exchange energy。
二、Characteristic Impedance and Quality Factor
1. characteristic impedance 
When arise resonance ,inductive
reactance equal capacitive
reactance
   0 L  1  L units：
 0C
C
 do not relative to response frequency ,only is defined by circuit parameter。
2. quality factor Q

ω0 L
1
1
Q



R
R
ω0 RC
R
L
C
None units
It is one of targets which illustration performance of
circuit ,and it is decided by circuit parameter。
The meanings of quality factor Q
：

UL
(a) The relation of voltage ：
Q
ω 0 L ω 0 LI 0
R

RI 0
U L0 U C 0


U
U

UR

UC
In resonance circuit Q is equal to the result that inductive voltage
UL0(or capacitive voltageUC0) divide electric source voltage 。It
show the multiple that voltage is magnified.
UL0andUC0 are the external voltage Q multiples，if
w0L=1/(w0C )>>R ，Q is high，L and C arise high
voltage ,sometimes it can be useful ,but sometimes it should
be avoided。

I
Example ： a radio C=150pF，L=250mH，
R=20

L  1290 Ω
C
Q

R
 65
If a single voltage is 10mV , inductance voltage is
650mV ,what we need.
In power system ,because electric source voltage
is high ,and when arise resonance the overwhelm
voltage
breakdown
insulater
and
destroy
equipments .what should be avoid.
Thought：
How do the common radios choose
broadcasting station？
What’s that we choose broadcasting station
in essence ？
• Why some radios’ noise is large ,while some is small？
• Why we can hear several broadcast of broadcasting
stations？
• What is the interruption that power system affect
communication ？
三. Selectivity and general resonance curve
(a) selectivity
The current is large to resonance single ,while the
current is small to other single .the capacity of choosing
different single is called selectivity.
I( )
|Z|
R
O
0

O
0

Supplement :example.
R
The circuit parameter of a radios：
+
u1
_
+
u2
_
+
u3
_
L=250H,
R=20, C=150ph,
L
U1=U2= U3 =10V,  0=5.5106 rad/s,
C f0=820 kHz.
f (kHz)
L
1
ωC
X
I=U/|Z| (A)
beijing
820
CC
640
1290
1000
1290
0
I0=0.5
–1660
– 660
I1=0.0152
Economy of beijing
1026
1611
1034
577
I2=0.0173
I=U/|Z| (A)
I0=0.5
I1=0.0152
I2=0.0173
I(f )
I2
I1
 3.46% (Very small)
 3.04%
I0
I0
∴receive the broadcast of beijing820kHz。
0
640 820 1200 f (kHz)
Choose the single of  0 from several frequent single ，
namely selectivity。
The good or bad of selectivity is relative to resonance curve，
the more pointed of resonance ,the better of selectivity 。
If LC id constant ，R is large，the curve is smooth，
the selectivity is bad。
The affection which Q to selectivity ：the change of R
affects selectivity in essence Q affect selectivity。
(b) general resonance curve
In order to compare with different resonance loop easily,
we usually make current resonance curve’s abscissa and
ordinate divide w0 and I(w0) respectively.
I (ω)
ω  ω  η , I (ω) 

ω0
I (ω0 )
I (ω) U / | Z |
R



I (ω0 )
U/R
1 2
2
R  (ω L 
)
ωC
1


ω0 L ω
ω0 2
1
1 (


 )
R ω0 ω0 RC ω
I (η )

I0
1
1
1  Q (η  )2
2
η
I (η )
I0
1
ωL
1 2
1 (

)
R ω RC
1
ω0 2
ω
1  (Q 
Q )
ω0
ω
General
Resonance
Curve
I (η )
I0
Q=0.5
0.707
Q=1
Q=10
0
1 1
2


o
The larger Q，resonance curve is more pointed 。If leave the
resonance point a little ,the curve will descend abruptly ，circuit has
strong capability of restraining to non-resonance frequency current ,
so its selectivity is good。
So ,Q is a important target which illuminate the character
of resonance circuit 。
at the place of I / I 0  1 /
2  0.707
draw a level which intersects with each
resonace curve at two points whose
abscissa is η1and η respective ly .
I (η )
I0
ω1
η1 
,
ω0
ω2
η2 
,
ω0
ω2  ω1.
0.707
Q=0.5
Q=1
Q=10
0
1
1
2

ω2  ω1
Called BW (Band Width)
* The frequent character of UL(ω ) and UC(ω)
U
U L (ω)  ω LI  ω L 

|Z|
ω LU
R 2  (ω L  1 ) 2
ωC
QU

1
2
1 )2

Q
(
1

η2
η2
I
U C (ω) 

ωC

U
2
1
ω C R  (ω L 
)
ωC
QU
2
η 2  Q 2 (η 2  1) 2
UC(Cm)
QU
U
U( )
UL( )
UC( )
UL( )：
0
Cm 1Lm

 =0， UL( )=0； 0<<0， UL( ) become large ； =0，
UL( )= QU；  > 0，current begin to become small，but the speech
is slow. XL continue to become large ，UL still have the trend of
becoming large ，but when UL( ) reach the maximal value ,then it
begin to reduce 。  ，XL， UL()=U。
Similarly we can discuss UC( )。
Base on math analysis ，when  = Cm，UC() reach maximal
value；when  = Lm，UL() reach maximal value 。
UC( Cm)=UL( Lm)。
( the condition Q  1 / 2 )
ωcm  ω0
1
1
 ω0
2
2Q
ωLm  ω0
2Q 2
 ω0
2
2Q  1
U C (ωcm )  U L (ωLm ) 
 Lm• Cm = 0。
QU
1
1
2
4Q
 QU
The higher Q，the more Lm and
Cm is close to 0。
9. 9
Parallel- connection Resonance Circuit
一、similar G、C、L parallel connection circuit
+


IS
U
C
G
L
_
R L C series connection
Z  R  j(ωL  1 )
ωC
ω0 
1
LC
G C L parallel connection
Y  G  j(ωC  1 )
ωL
ω0 
1
LC
R L C series connection
G C L parallel connection
|Z|
|Y|
R
G
O
I( )

0
O
U( )
IS/G
U/R

0
O

0
O


IC
UL



UR  U I



IG  IS U

UC

0

IL
R L C series connection
Voltage
resonance
U L  U C  0
G C L parallel
connection
IL  IC  0
Current resonance
IL( 0) =IC( 0) =QIS
UL( 0)=UC ( 0)=QU
ω0 L
Q
 1  1 L
R
ω0 RC R C
ω0 C
Q
 1  1
G
ω0 GL G
C
L
deduce:
U
I L (ω0 ) I C (ω0 ) ω0 L
Q


 1
U
ω0GL
IS
IS
R
ω0C 1 C


G L
G
二 、Inductance Loop Paralleled Connected Capacity
In fact the preceding current resonance do not realize，because
inductance loop exist resistance ,circuit become series and parallel
connection circuit ,resonance phenomenon is complexity.
Y  jω C 
R
C
L

1
R  jω L
R
ωL

j
(
ω
C

)
2
2
2
2
R  (ω L)
R  (ω L)
 G  jB
Arise resonance B=0，
Solve
ω0 
ω0 L
ω0 C  2
0
2
R  (ω0 L)
1  ( R )2
LC
L
Is decided by circuit
parameter。
Only under a certain condition ,circuit can arise
resonance ,if the parameter is not appropriate it can not
arise resonance 。
When circuit parameter is certain ,whether circuit can
arise resonance by changing source frequency ? It is
decided by the following conditions ：
1
R 2
L
when
( ) , R
, Arise resonance
LC
L
C
L
when R 
, Cann' t arise resonance , so ω0 is a imagiary number .
C
When circuit arise resonance ,the circuit is equal to a resistance：
R  (ω0 L)
Z (ω0 )  R0 
 L
R
RC
2
2
Supplement: Resonance of Series-Parallel Connection Circuit
Example:
( a) circuit can arise series connection resonance (X=0)，
also can arise parallel connection resonance (X=)。Solve the
series-parallel connection resonance frequency by solving input
impedance。
L3
L(a)
1
C
To (a) circuit ， L1 、 C2 parallel circuit ， in low frequency it is
inductive 。 With frequency becoming large ， at a 1arise parallel
connection resonance。 >1，the parallel part is capacitive ，at a 2
can arise series connection resonance with L3。
Quantitative analysis ：
jωL1 (
1
)

 L1 
jωC 2

Z (ω )  jωL3 
 jωL3  2
1
ω L1C 2  1 

jωL1 
jωC 2
ω 3 L1 L3C 2  ω ( L1  L3 )
j
ω 2 L1C 2  1
When Z( )=0，namely numerator =0：
ω L1 L3 C 2  ω2 ( L1  L3 )  0
3
2
ω2  0
ω2 
(resign)
L1  L3
L1L3C2
solution：
(series connection resonance)
L3
When Y( )=0，namely denominator=0，：
ω12 L1C 2  1  0
ω1 
1
L1C2
L1
(parallel connection resonance)
So，  1< 2。
C2
To (b) circuit we can do similar
qualitative analysis 。
L1 、 C2 parallel connection ， at low
frequency it is inductive 。
At 1it can arise series connection
resonance with 。When  >1，with the
frequency becoming large ，parallel part
become capacitive from inductive，
At 2 arise
resonance。
parallel
connection
C3
L1
C2
(b)
jωL 1
1
jωC 2
Z (ω1 ) 

 1
jωC 3 jωL  1
jωC
1
jωC 2
1  ω 2 L1 (C 2  C 3 )
 j
ωC 3 (1  ω 2 L1 C 2 )
1
3
jωL 1

1  ω 2 L1 C 2
C3
Respectively numerator =0、 denominator=0：
L1
ω1 
1
Series resonance
L1 (C 2  C 3 )
ω2 
1
L1C 2
ω1  ω2
Parallel resonance
C2
Frequency Characteristic of Resistance
Z ( )=jX( )
X( )
(a)
O
1
2

2

X( )
(b)
O
1
Supplement: Application of LC Series-Parallel Connection Circuit
Compose kinds of passive filter circuit。
example：
source u1(t)，including two frequency 1、2
(1<2)：
u1(t) =u11(1)+u12(2)
Require response u2(t) only include the voltage with 1。
How to come true?
+
u1(t)
_
u2(t)
by the following passive filter circuit to come true:
C3
+
u1(t)
_
ω2 
ω1 
L1
C2
1
L1 C 2
1
L1 (C 2  C 3 )
R
+
u2(t)
_
Parallel resonance，open circuit
Series resonance，short circuit
1 single is added to loads directly 。
in the circuit 2 >1 ，filter high frequency，
get low frequency。
Other Forms of Filter Circuit ：
L1
L3 C3
C1
L2
C2
band-pass filter
L3
L1
C1
L2
C3
C2
band elimination filter
Conclusion: Exercises of Sine Steady Circuit
Requirements：
1.the basis concepts of sinusoid ：three key factor
of sinusoid、phase difference、waving and so.
2. complex impedance、complex admittance
3. quantitative computer：phasor method
4. qualitative analysis ：phasor diagram
5. power computer：real power、reactive power、
apparent power、power factor 、complex power and so
一、
i
+
u
–
Circuit shown in the figure，if：
Review
u( t )  10 sin( 400π t  60 ) V
Z  Z φ
1
i (t )  
cos(400π t  150  ) A
2
200Hz, T=_______.
0.005s
400 rad/s, f=_____
(a) voltage source’s  =_________

7.07V
(b) voltage’s rms value U=_______current’s
rms value I=________.
0.5A
 60
(c) voltage and current’s phase differenceu–i =_________.

capacitive
14.14

60
(d) this load is ______, |Z|=_________,  =_________.
Compare to phase ,must change sinusoid into standard sinusoid:
1
1


i(t ) 
cos( 400πt  150  180 ) 
cos( 400πt  30 )
2
2
1
1



sin( 400πt  30  90 ) 
sin( 400πt  120 ) A
2
2
 =  u–i=60º–120º= –60º
二、judge the result right or wrong ,if it is wrong ,please
correct 。
1.

I

U
(1) I 
R  jω L
j L

+ UL –
+

+

U
( 2) I  2
R  (ω L )2
(3) u  uR  uL
2
(6) P  I 2 R
2.
i
+
u
–
2
R
( 4) U  U  U
U
(5) P  R
R
(7)| Z|  R 2  (L) 2
R UR
–
U
–
2
L
2
If u( t )  311 sin( t  45 )V , Ζ  2560 Ω
so i  u  311 sin( t  45)  12.44 sin(t  45  60)A
Z
2560
311

45 o

U
2
Z
o
I 

8
.
8


15
A
o
Z
2560
 
i  8.8 2 sin(t  15 o ) A


phsor=sinusoid
三、
is6
Circuit shown as follows ,write loop
current equation and node’s
voltage equation in phasor。
i1
+
us1
–
C2
L4
C3
is3
i3
i2
Loop method：
R5




I 1  I S6
I 2  I S3


1
1
( j
 j L4  R5 ) I 3  (  j
 j L4 ) I 1
 C2
 C2


 ( j L4  R5 ) I 2  U S1
is6
Nodes methods


U2

Un1
+
C2
us1
–



U3
L4
C3
is3

U 1  U S1


 jC 2 U 1 ( jC 2 


1
1 

)U 2 
U 3  I S3
jL4
jL4

1 
1
1 

U2  (

) U 3  I S6
jL4
jL4 R5
R5
四、
40Ω


24Ω j18Ω
I

+
+
US

U1
IR
–

 j30Ω
IC
+
–
A

U2
–
 j50Ω
+

U3
–
If：A is 1.5A(rms value)。
solve：(1)US=? (2)solve circuit absorb P and Q
solve：If IR  1.50 A

U

2

I


2

90
 j2A


C

So U 2  40 1.50  600 V
 j30
I  I  I  1.5  j 2  2.553.1 A
R
C
U 1  (24  j18) I  (24  j18)  2.553.1  7590  j75V
U 3  (  j50) I  (  j50)  2.553.1  125  36.9   100  j75V
I  2.553.1 A

40Ω


I
+
24Ω j18Ω
+

U1
IR
–
A

 j30Ω
IC
US
+
–

U2
–
 j50Ω
+

U3
–
U S  U 1  U 2  U 3  j75  60  100  j75  1600  V
PA  US I cos φ  160  2.5  0.6  240 W
Pabsord  24I 2  40I R2  24  (2.5) 2  40  (1.5) 2  240 W
Qabsord  U S I sin φ  160  2.5  (0.8)  320 Var
Qabsord  18I 2  30 I C2  50 I 2
 18  (2.5) 2  30  22  50  (2.5) 2  320 Var
Electric source release：
S  160  2.5  53.1  240  j320VA
五、 
I
L
+
+
US 1
–
US 2
If ： U S1  110  30  V, U S2  11030  V,
L  1.5H, f  50Hz .
Solve ：the P and Q is released by two
–
source respectively。




U

U
110


30

110

30
 j110
solve： 
S1
S2
I


 0.234 A
jω L1
j314  1.5
j471
Voltage sourceUs1
P1release  U S1I cos( 30  180 )  110  0.234  (0.866)  22.3 W
Q1releae  U S1 I sin( 30  180 )  110  0.234  0.5  12.9 Var
Voltage Us2
P2 release  U S2 I cos(30  180 )  110  0.234  (0.866)  22.3 W
Q2release  U S 2 I sin( 30  180 )  110  0.234  (0.5)  12.9 Var
六、

+ I
1
A1
A2

U
–

A3

R2 I 3
I2
Solve ：assumeZ3 =|Z3| 3
Z3
If ： U=220V ， f=50HZ ， A1 is 4A,
A2is 2A，A3is 3A，Z3is inductive
load。solve：R2 、Z3。
so R2 
U 220

 110 Ω
I2
2
Solve Z3. Solution 一：draw phase graph ,voltage is reference phasor
Base on cos theorem ：
I2
U
42= 32+ 22–232cos
3
2
2
2
2

4

3

2
1
o
3 
cos
θ




,
θ

104
.
5
I3
4
242
4

I1
φ 3  180 o  θ  180 o  144.5 o  75.5 o
U 220
| Z 3 |

 73.3 Ω, Z 3  73.375.5o Ω  18.4  j71Ω
I3
3


Solve Z3 solution二：.
U 220
| Z 3 |

 73.3 Ω
I3
3 


If U  2200 V, So I2  20 A, I3  3  φ3 A,



I1  I2  I 3 即
I1  4φ A
4φ  20   3  φ 3
4cos +j4sin =2+3cos 3–j3sin 3
Thus：
4cos =2+3cos 3
(1)
4sin =–3sin 3
(2)
from (1)2+(2)2 ：16 =(2+3cos 3)2+(–3sin 3)2
=4+12cos 3+9(cos 3)2+9(sin 3)2
= 4+12cos 3+9
3 1
cosφ 3 
 , φ 3  75.5o
12 4
Z 3 | Z 3 |  3  73.375.5 o Ω  18.4  j71Ω

七、
V1

I1
US
*
* W
+
+
R
jX 1
U 1
V
+
–
A2
jX 2
–

I2
A3
jX 3

I3
U 2
–
In Analysis of Sinusoidal Steady State Circuit shown
as follows ，if V is220V，V1 is 100 2 V，A2 is 30A，A3 is
20A ，W is 1000W(P)。solve R、X1、X2 、X3。
phase method，assume：U 2  U 2 0  V
so： I2   j30A, I3  j20A, I1  I2  I3   j10A
P  I 1 R, R  P / I 1  1000 / 10 2  10 Ω
2
2

V1
I1
*
*
US
+
+
–
+
W
V
R
U 1
jX 1
–
A1
jX 2

I2
A2
jX 3
U2 can use phase graph
U 2

I3
–
U 1 100 2
If： Z1  R  jX1  Z1 φ1 So Z 1 

 10 2 Ω
I1
10
2
X1
2
2
2
X 1  Z 1  R  (10 2 )  10  10Ω φ 1  arctg
 45 
R
 I1   j10  10  90  A  U 1  100 2  45  V
U S  U 1  U 2  100  j100  U 2  100  U 2  j100
U S2  (100  U 2 ) 2  100 2 , U 2  220 2  100 2  100  96V
X 2 U 2 / I 2  96 / 30  3.2 Ω, X 3  U 2 / I 3  96 / 20  4.8 Ω

V1
I1
*
*
US
+
W
+
+
V
–
R
U 1
jX 1
–
A1
jX 2

I2
A2
jX 3
U 2

I3
–
U S  U 1  U 2
U 1  U R  U X 1
I  I  I
1
2
3
U 2
I1
I3
U S
I2
135°
U R
U 1
U X 1
U S2  U 22  U 12  2U 1U 2 cos 135
2
U  2  100 2  ( 
)U 2  (100 2 ) 2  220 2  0
2
U 22  200U 2  28400  0 , U 2  96V
2
2
（abandon negative value）
• Assignments：9- 8. 10. 15. 24. 26
（A）.28. 32 、38
Problems in homework：notice writing
Judge Right Or Wrong
in R-L-C series circuit，assume 
I  I0
U  U U U
2
R
2
L
U
 I R  X L  X C 
2
2
2
C
？
？
U 
 IR  j  X L  X C  ？
Judge Right Or Wrong
in R-L-C series circuit，assume I  I0
U  U U U
2
R
2
L
U
 I R  X L  X C 
2
2
2
C
？
？
U 
 IR  j  X L  X C  ？
Judge right
or wrong
  I0
I
in R-L-C series circuit，assume
X L  XC
1 U L  U C

 tg
  tg
？
U
R
1
？
U L UC
1 L  C


tg

 tg
？
？
R
UR
1