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ElectromagnetismElectromagnetism
Chapter 1 Electric Field
Chapter 2 Conductors
Chapter 3 Dielectrics
Chapter 4 Direct-Current Circuits
Chapter 5 Magnetic Field
Chapter 6 Electromagnetic Induction
Chapter 7 Magnetic Materials
Chapter 8 Alternating Current
Chapter 9 Electromagnetic Waves
Chapter 8 Chapter 8 Alternating Current Alternating Current
§1. Alternating Current
§2. Three Simple Circuits
§3. Complex Number and Phasor
§4. Complex Impedance
§5. Power and Power Factor
§6. Resonance
§1.§1. Alternating Current Alternating Current
Steady current : magnitude and direction not changing
Varying currentmagnitude varying , not reversing
Alternating currentSinusoidal current : i = Imcos ( t + ) , u , Three important quantities : Amplitude Im
( or rms I = Im/ )2 Angular frequency ( = 2 /T = 2f ) Initial phase ( phase t + )
Im
o t
i
Alternating CurrentAlternating Current
Features of sinusoidal quantities : derivative and integral are still sinusoidal any periodic quantities can expand as a sum of sinusoidal functions with different frequency
Denotation : instantaneous : little case i , u rms : capital I , U amplitude : subscript m Im, Um
( rms : root-mean-square )
§2.§2. Three Simple Circuits Three Simple Circuits
1. Introduction
2. Pure Resistance
3. Pure Capacitance
4. Pure Inductance
1. Introduction1. Introduction DC R act on current
L short circuit ( ideal , no resistance ) C open circuit ( ideal , no current ) AC R 、 L 、 C all act on current
L self-induced emf
C charge/discharge Relationship between i and u
i = Imcos ( t + i )
u = Umcos ( t + u )
)cos(2 itI )cos(2 utU
To study :(1) U / I = ? Ratio of rms
(2) u - i = ? Difference of phase
2. Pure Resistance2. Pure Resistance
u(t) = i(t) R
or
R
iu
)cos(2 itIR )cos(2 utU
U = I R u = i
or U / I = R u - i = 0
0 t
iu
3. Pure Capacitance3. Pure Capacitance
Left plate q = Cu
iu)cos(2 itI
)]sin([2 utUC
I = UC i = u + / 2
or U / I = 1/C u - i = - / 2
Capacitive reactance : XC = 1/C
0 t
u
Ct
qi
d
d
t
uC
d
d
)2
cos(2 utCU
i
Pure Capacitance : current leads voltage by / 2
4. Pure Inductance4. Pure Inductance
i u
0 t
ui
L acts as an emf u(t) = - S(t)
)]sin([2 itIL
)2
cos(2 itIL
U = LI u = i + / 2
or U / I = L u - i = / 2
Inductive reactance : XL = L
t
iL
d
dS
)cos(2 utU
L
自t
iLu
d
d
Pure Inductance : current lags voltage by / 2
§3.§3. Complex Number and Phasor Complex Number and Phasor
1. Complex Numbers
● Expressions
● Calculations
2. Complex Number Method
3. Phasors
4. Complex Form of Relations between u and i
● Pure Resistance
● Pure Capacitance
● Pure Inductance
5. Examples
Expressions of Complex NumbersExpressions of Complex Numbers
Algebraic : = a + jb a = Re()
b = Im() )1( j
a
b
bar
1
22
tan
)
0 +1
+j(a,b)
a
br
Phasor : r = | | modulus
a = r cos
b = r sin
Trigonometric : = r cos + j rsin Exponential : = r e j
( Euler formula : e j = cos + j sin )
Multiplication : Division :
Calculations of Complex NumbersCalculations of Complex Numbers
Addition/Subtraction :1 2 = ( a1 a2 )+ j ( b1 b2 )
( parallelogram rule )
111 jba 1e1jr 222 jba 2e2
jr
21 ee 2121 jj rr )(
2121e jrr
)(
2
1
2
1 21e j
r
r
Information : rms, initial phase Steps of calculation : i, u calculating result of of i, u
real complex take real part 4 theorems :( Next page )
2. Complex Number Method2. Complex Number Method
Instantaneous : i = Imcos ( t + ) = Re[ Ime j ( t + ) ]
where Ime j ( t + ) jtj Iee 2 Ie tj 2
jIeI
UI , UI ,
Complex rms Definition :
Four TheoremsFour Theorems
Complex rms of (ki) ( k any real constant )Ik
21 II Ij jI /
]2Re[ Ie tj ]2Re[ )2/( jtj eIe
Ij
]/2Re[ )2/( jtj IeejI /
Complex rms of( i1 i2 )
Complex rms of di/dt Complex rms of idtPro. : i = Imcos ( t + )
di/dt = Imcos ( t + + /2 )
complex rms of di/dt = I e j e j/2
idt = (1/)Imcos ( t + - /2 )
complex rms of idt = I e j e -j/2 /
3. Phasors3. Phasors
complex rms phasorjIeI
0 +1
+j
I
21 II Ij
jI /
complex rms of di/dt
complex rms of idt
length = I ( rms ) angle = ( phase )
parallelogram rule
times of length , rotate counterclockwise /2
1/ times of length , rotate clockwise /2
4. Complex Form of 4. Complex Form of uu , , i i RelationsRelations
● Pure Resistance
● Pure Capacitance
● Pure Inductance
U
Pure ResistancePure Resistance
Instantaneous : u = i R
Complex rms :or
U = I R
u = i
RIU
RIeUe iu jj
I0
R
iu
Pure CapacitancePure Capacitance
Instantaneous : or
U = I /C u = i - / 2
U
ICj
U 1
iu jj IeCj
Ue
1
I0
t
uCi
d
d ti
Cu d
1
Complex rms :
or
iu
C
)2
(1
ijIe
C
Acturely, 1/jC includes all information
about relationship between u and i
( ratio of rms and difference of phase )Complex capacitive reactance : - j XC = - j /C
Pure InductancePure Inductance
i u
L
自
t
iLu
d
d
U = LI
u = i + / 2
Instantaneous :
Complex rms :
or
IjLU
iu jj LIejUe )
2(
ijLIe U
I0
Complex inductive reactance : j XL = jL
Example 1 Example 1 (( p.330p.330/[Ex./[Ex.11]])) (1)(1)
Series RL circuit, relation between u and i.
Sol. : u = u1 + u2
i
u
u2
u1 R
L
21 UUU ILjRI ILjR )( jzeLjR exponential :
where
222 LRz
R
L 1tan
0 +1
+j
R
L
zIzeU j 222/ LRzIU
iu
)( iu jj zIeUeIf i = Imcost is known, can get u = zImcos(t+ )
ILjU 2
IRU 1
Example 1 Example 1 (( p.330p.330/[Ex./[Ex.11]])) (2)(2)
Phasor :first draw
I0
U
2U
1U
then I
same phase with I
leads by / 2I
and U2 /U1 = L/R
then 21 UUU
get 22
21 UUU
iu
ILR 222
1
21tanU
UR
L1tan
= 3 110 2
Example 2Example 2 (( p.332p.332/[Ex./[Ex.22]]))Fluorescent lamp ( daylight lamp ) : tube R , ballast L , in series , emf 220 V , tube U1=110 V. Find U2 of ballast.
Sol. :u
u2
u121 UUU
U22 = U 2 - U1
2
22
21 UUU
R
L~
31102 U )(V190
= 220 2 - 110 2
Example 3Example 3 (( p.332p.332/[Ex./[Ex.33]]))RC in parallel. Find relation between i1and i2 .
Sol. : phasor
in parallel , draw first
i
ui2i1
RUI /1
UCjI 2
and I2 /I1 = CR
i2 leads i1 by / 2
R C
same phase with U
leads by / 2U
U0
I
2I
1I
U
Example 4Example 4 (( p.332p.332/[Ex./[Ex.44]]))Continue Ex.3, find phase difference between i and i1 .
Sol. : I2 /I1 = CR
i2 leads i1 by /2
R = 138 k = 1.38 10 5 C = 1000 pF = 10 -9 F
= 2f = 2 2000
CR 1.73
/3 ( i leads i1 )
1
21tanI
I
U0
I
2I
1I
CR1tan
§4.§4. Complex Impedance Complex Impedance
1. Three Ideal Elements
2. Two-Terminal Net without emf
3. Exponential Formula and Algebraic Formula
●Exponential Formula
●Algebraic Formula
●Impedance Triangle
1. Three Ideal Elements1. Three Ideal Elements
Resistor u = Ri
Capacitor
Inductor
IRU
tiC
u d1
IC
jU 1
t
ILu
d
d ILjU
RZ
CjZ1
LjZ
introduce Complex Impedance Z so that IZU Z determined by R 、 L 、 C and , not U 、 I Z represents relation between i and u
( U/I and u - i )
2. Two-Terminal Net without emf2. Two-Terminal Net without emf
orIZU IUZ /
u
i2i1
i
i
uILjRU )(
LjRIUZ /
Ex.2 : RC in parallel i = i1+ i2
Ex.1 : RL in series
21 III
RUI /1
Cj
UI
/2
UCj
)1
( CjR
UI
1)1
( CjR
Z CRj
R
1
222
2
1 RC
CRjR
jxr
3. Exponential and Algebraic Formulae3. Exponential and Algebraic Formulae
Exponential Formula : Z = ze j
z impedance —— modulus of Z
phase constant —— angle of Z
I
U
)( iujeI
U
Algebraic Formula : Z = r + j x
r effective resistance > 0, not necessarily = R Ex. x effective reactance > 0 for inductive net
< 0 for capacitive net
= 0 for resistive net
IUz /
iu
Z represents relations for i and u ( U/I and u - I )
Complex Form of LawsComplex Form of Laws
1. Ohm’s Law
DC : U = IR U = - IR
AC : ZIU ZIU
21
111
ZZZ
0)( I )()( ZI
2. Kirchhoff’s Rules
DC : ( I ) = 0 ( ) = ( IR )
AC :
series connection : Z = Z1 + Z2 + ···
parallel connection :
ExampleExample (( p.336p.336/[Ex.]/[Ex.] ))Condition for balancing an AC bridge.
Sol. : uAC = uAD
3311 ZIZI
21 II
and
4
3
2
1
Z
Z
Z
Z
ADAC UU
4422 ZIZI
43 II
or)1/( 44
3
2
1
CRjR
R
R
LjR
4
3
2
1
R
R
R
R 32RCRL
Maxwell Bridge , for
measuring L
A B
C
D
LR2
R3
R4
R1
CG
~
i1
i4
i2
i3
§5.§5. Power and Power Factor Power and Power Factor
1. Instantaneous Power, Average Power
and Power Factor
2. Significance of Raising Power Factor
3. Method to Raise Power Factor
Average power
Pure Resistance Pure Inductance Pure Capacitance Two-Terminal Net without emf
1. Power and Power Factor1. Power and Power Factor
DC : P = IU keep constant
AC : p(t) = i(t)u(t) instantaneous power
( for AC with f = 50 Hz , average is important )
T
ttpT
P0
d)(1
Pure ResistancePure Resistance
Resistance : i = Imsin t u = iR p = iu = i 2R
T
tRtIT
P0
22m dsin
1
T
ttRIT 0
22m dsin
1
T
tt
RIT 0
2m d
2
2cos11
RI 2m2
1 RI 2 IU
Resistor : non-energy-storing , energy heat
m2
1rms II :
0 tT
iuP
I
p
0 T/4 and T/2 3T/4 : p > 0, absorb energy and store it in M field
Pure InductancePure Inductance
Inductance : voltage leads current by /2
i = Imsin t u = Umsin( t + /2 ) = Umcos t
tUIiup 2sin2
1mm
T
tpT
P0
d1
T
ttUIT 0mm d2sin
2
11 0
2m2
1LIW
p
0 tT
T/4 T/2 and 3T/4 T : p < 0, release energy, field disappear ( i : Im 0 )
External energy M Field energy
Never dissipated at all !
iu
Pure CapacitancePure Capacitance
u
i
0 T/4 and T/2 3T/4 : p > 0, absorb energy and store it in E field
Capacitance : current leads voltage by /2
u = Umsin t i = Imcos t
tUIiup 2sin2
1mm
T
tpT
P0
d1
0
T/4 T/2 and 3T/4 T : p < 0, release energy, field disappear ( u : Um 0)
p
2m2
1CUW
0 tT
External energy E Field energy
Never dissipated at all !
Two-Terminal Net without emfTwo-Terminal Net without emf
u = Umsin t i = Imsin( t - )
)]2cos([cos2
1mm tUIiup
T
tpT
P0
d1
Resistor : = 0 P = IU Inductor : = /2 P = 0 Capacitor : =- /2 P = 0
0cos2
10mm T
dtUIT
cos2
1mmUI cosIU
( Trigonometric : cos(- )- cos(+ ) = 2sin sin )
u
0 tT
i
p
cos —— Power factor
Lost on cable (1) voltage U’ = IR (2) power P’ = I 2R
Reduce lost : R , thick wire , cost more I , not decrease consumer’s power P = IUcos
2. Significance of Raising Power Factor2. Significance of Raising Power Factor
S = IU visual power P = IUcos work power Q = IUsin workless power
~ ZR
RI
—— increase power factor cos
Ex. : inductive load, i lags u by U
I
PI
QIWorkless current : I Q= I sin
work current : I P= I cos
3. Method to Raise Power Factor3. Method to Raise Power Factor
Workless current : I Q= I sin
work current : I P= I cos
then P = IUcos = I PU
I Q= I sin useless to P ,
but a part of total current I ,
and a part of energy lost on cable P’ = I 2R
increase cos to reduce I Q
inductive net
u
iCi
i’
U
I
'I
CI
’
cos’ > cos
P = IUcos = I’Ucos’
add capacitace
§6.§6. Resonance Resonance
Resonance : series RLC circuit
I
U
CU
LU
CL UU
RUI
U
CU
LU
CL UU
RUI
U
CU
RU
LU
UL > UC
u leads i
Inductive
UL = UC
u and i in phase
resistive
UL < UC
u lags i
capacitive
CL
1
C
L
1
CL
1
Resonance
R L C
uuR uL uC
Resonance in Series Circuit Resonance in Series Circuit (( 11 )) Current : Complex impedance : )
1(
CLjRZ
22 )1
(C
LRz
Impedance :
Current :z
UI
22 )/1( CLR
U
Resonance :C
L
1
R
UI 0 maximum current :
LC
10 Resonance frequency of RLC circuit :
when = 0 , I = I 0 maximum , resonance
Resonance in Series Circuit Resonance in Series Circuit (( 22 )) Voltages : UL = IL UC = I/C
Resonance : UL0 = I00L UC0= I0/0C
Let
then UL0 = QU UC0 = QU if R << 0L , Q very large ~ 10 2 ( good , bad ) UL0 = UC0 = QU > U Quality factor :
LR
U0
CR
U
0
1
R
LQ 0
CR0
1
C
L
RQ
1
Resonance in Series Circuit Resonance in Series Circuit (( 33 )) Resonance curve : Relation for I ~
keep R 、 L 、 C 、 U constant
LC
10
Selectivity : to select the wanted program
—— modulate for a radio adjust C change
when 0 matches 1of a signal
for example ( 0 = 1 )
then I1 >> I i ( i 1 )
0
II0
0
~ ~ ~