chapter 31 electromagnetic oscillations and ac circuit
TRANSCRIPT
Chapter 31
Electromagnetic oscillations and AC circuit
Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton
(Date: July 25, 2019)
The phasor diagram is very useful in discussing the AC circuit formed of series
connection of R, L, and C. The detail of this concept will be discussed below. We note
that this method is not appropriate for the AC circuits which are formed of various kind
of combinations with connections of R, L, and C elements. In this case we need to use the
concept of impedance such as R, i L , and 1/ ( )i C in the frequency domain. We also
use the transformations [ ( ) Re( )i ti t Ie for the current and ( ) Re( )i tv t Ve for the
voltage] between the frequency domain and time domain, The quantities I and V are
complex numbers. In the frequency domain, the AC circuit can be simply solved by using the KCL and KVL laws. In general physics, in spite of convenience, this method is not
adopted.
1. LC oscillations
In a resistanceless LC circuit, the total energy U is given by
22
2
1
2
1CL CvLiU
where q = CvC.
From the energy conservation, U remains constant with time.
0dt
dvCvi
dt
dLi
dt
dU CCLL
Since dt
dqiL and
C
qvC , we have
02
2
2
qdt
qd (LC oscillation, simple harmonics).
with
LC
1
The total energy U is rewritten as
2
2
2
1
2
1q
Cdt
dqLU
We assume the initial condition as
At t = 0,
q = Q and dq/dt = 0.
The solution of the second order differential equation is given by
)cos( tQq
)sin( tQdt
dqiL
Out[11]=
0.2 0.4 0.6 0.8 1.0 1.2 1.4têT
-1.0
-0.5
0.5
1.0
qêQ, iLêHQwL
Fig. Qq / (red) and )/( QiL (blue) as a function of t/T.
Energy of UE and UB
C
QUUU
tC
QtLQU
tC
Q
C
qU
BE
B
E
2
)(sin2
)(sin2
1
)(cos22
2
22
222
222
Fig. )2//( 2 CQUE (red) and )2//( 2 CQUB (blue) as a function of t/T.
2. Damped oscillations in a RLC circuit
A circuit containing resistance, inductance, and capacitance is called an RLC circuit.
dt
dqi
Cvq
Riv
dt
diLv
L
C
LR
LL
From the KVL, we have
0 CRL vvv
or
the second order differential equation
02
2
LC
q
dt
dq
L
R
dt
qd
We assume that
q = x, 2 = R/L, and 02 = 1/(LC)
Then we have
0)()('2)("2
0 txtxtx
with the initial conditions
0)0(' vx and 0)0( xx
The solution of this differential equation depends is classed into three types,
(1) underdamping: 02
0
2
(2) critical damping 02
0
2
(3) overdamping 02
0
2
((Note))
We assume that x(t) is expressed by a solution given by exp(pt). Then we have
0)2()()('2)("2
0
22
0 ptepptxtxtx
The solution of the quadratic equation 2 2
02 0p p is
1
22
0
22
0
22
0
22
)(
)(2
iip
ip
pp
with
22
01
The solution is given by
)]sin()cos([)()( 1211 tCtCetxtq t
From the initial condition, we have
01 xC
1
002
xvC
The final form is given by
)]sin()cos([)( 1
1
0010 t
xvtxetq
t
3. AC circuit theory
3.1 phasor diagram
(a)
)sin(max tIi
)2
sin()cos(
)2
sin()cos(1
)sin(
maxmax
maxmax
max
tLItLIdt
diLv
tC
It
C
Iidt
Cv
tRIv
L
C
R
(b)
)cos(max tIi
)2
cos()sin(
)2
cos()sin(1
)cos(
maxmax
maxmax
max
tLItLIdt
diLv
tC
It
C
Iidt
Cv
tRIv
L
C
R
The phasor diagram is given by the following figure.
For more general case (more complicated AC circuits), we need to use the method of
complex number. This will be discussed in the Appendix.
3.2. Impedance of single elements in AC circuit
3.2.1 Inductance L
maxmax LIV
The relation between Vmax and Imax for the inductance is described in the following phasor
diagram.
We say that V leads to I by 90° (or I lags V by 90°).
3.2.2 Capacitance C
C
IV
max
max
The relation between Vmax and Imax for the capacitance is described in the following
phasor diagram.
We say that V lags I by 90° (or I leads V by 90°).
3.2.3 Resistance
For resistance, one can write down a relation
maxmax RIV
The relation between Vmax and Imax for the resistance is described in the following phasor
diagram.
3.3. Power dissipated in single elements
3.3.1 Definition
The time average of the power P(t) over a period time T (= 2/) is given by
cos
cos2
1
)]2cos([cos2
1
)sin()sin(1
maxmax
0
maxmax
0
maxmax
rmsrms
T
T
avg
VI
VI
tdtVIT
tVtdtIT
P
where cos is the power factor, Imax and Vmax are the amplitudes of current and voltage.
Note that the definition of the root-mean squares Irms and Vrms will be given later (Section
3.4), where
max
1
2rmsI I , max
1
2rmsV V
3.3.2 inductance
0avgP
since =/2. No power is dissipated in an inductor.
3.3.3 Capacitance
0avgP
since = -/2. No power is dissipated in a capacitor.
3.3.4 Resistance
2
max2
1RIPavg
Power is dissipated only in a resistor.
3.4. Root-mean square value of current and voltage
The root-mean square value of the current is defined as
2
)]2cos(1[2
1
)(sin1
max
0
2
max
0
22
max
I
dttIT
dttIT
i
T
T
rms
The root-mean square value of the voltage is defined as
2
maxVvrms
((Note))
Domestic electricity is provided at a frequency of 60 Hz in the United States, and a
residential outlet provides 120 Vac. This means the rms voltage is max 120 2V = 170 V.
The power dissipated in the resistor is rewritten as
R
VIRP rms
rmsavg
22
3.5. The series RLC circuit
Phasor diagram of the RLC circuits is given in the following way, where we use I and V
instead of Imax and Vmax (or Irms and Vrms)
where
IC
V
LIV
RIV
IZV
C
L
R
ss
1
where I is the amplitude of the current, Vs is the amplitude of the voltage source, Zs is the
total impedance defined by
22)
1(
CLRZ s
and the angle between Vs and I is defined as
)1
(1
tanC
LR
The average power dissipated in the system is
2coscos
2
1rmsrmsrmsss RIIVIVP
3.6. Example (RLC circuit)
Problem 31-46 (SP-31)
31-46
Figure shows a driven RLC circuit that contains two identical capacitors and two
switches. The emf amplitude is set at 12.0 V, and the driving frequency is set at 60.0 Hz.
With both switches open, the current leads the emf by 30.9°. With switches S1 closed and
switch S2 still open, the emf leads the current by 15.0°. With both switches closed, the
current amplitude is 447 mA. What are (a) R, (b) C, and (c) L?
((Solution))
E0 = 12 V
f = 60 Hz
= 2f
(1) S1 and S2 open
R
LC
ILC
REs
)2
(
tan
)2
(
1
1
22
Es lags I1 by 1. The average power dissipated in the system is
2
11111 )(cos)(cos2
1rmsrmsrmsss IRIEIEP
where 11cos RIEs
(2) S1 closed and S2 open
R
CL
IC
LREs
)1
(
tan
)1
(
2
2
22
Es leads I2 by 2. The average power dissipated in the system is
2
22222 )(cos)(cos2
1rmsrmsrmsss IRIEIEP
where 22cos RIE s
(3) S1 closed and S2 closed
CL
EI s
13
3.7. Example (RC circuit)
Problem 31-44 (HW-31)) Hint
An alternating emf source with a variable frequency fd is connected in series with a
50.0 resistor and 20.0 F capacitor. The emf amplitude is 12.0 V. (a) Draw a phasor
diagram for phasor VR (the potential across the resistor) and phasor VC (the potential
across the capacitor). (b) At what driving frequency fd do the two phasors have the same
length? At that driving frequency, what are (c) the phase angle in degrees, (d) the angular
speed at which the phasors rotate, and (e) the current amplitude?
R
C
IC
REs
)/1(tan
)1
(22
Es lags I by the angle . The average power dissipated in the system is
2coscos
2
1rmsrmsrmsss RIIEIEP
where
RIEs cos
3.8. Example (RL circuit)
R
L
ILREs
tan
)( 22
Es leads I by the angle . The average power dissipated in the system is
2coscos
2
1rmsrmsrmsss RIIEIEP
where
RIEs cos
3.9 Resonance of the RLC circuit
IC
LREs
22)
1(
or
22 )1
(C
LR
E
Z
EI s
s
s
where
22)
1(
CLRZs
(impedance)
The amplitude of the current has a maximum when
LC
1
We define the quality factor of the circuit by
R
LQ 0
Then the amplitude of the current can be rewritten as
20
0
2)(1
1
QR
EI s
We now consider the frequency dependence of I2, instead of I.
20
0
2
2
2
)(1
1
Q
R
EI s
We note that this is a scaling function of /0. In the figures we show the plot of
)/( 22
2
REs
Ivs
0
, where the quality factor Q is changes as a parameter.
Q=100 - 1000
0.6 0.8 1.0 1.2 1.4wêw0
0.2
0.4
0.6
0.8
1.0Es2êR2
Q = 10 - 100
0.6 0.8 1.0 1.2 1.4wêw0
0.2
0.4
0.6
0.8
1.0Es
2êR2
Fig. )/( 22
2
REs
Ivs
0
, where the quality factor Q is changes as a parameter.
_______________________________________________________________________
Suppose that
10
(<<1).
1)1( 10
Then we have
22
2
max
22
2
22
2
2
4141
1
)]1(1[1
1
Q
I
QR
E
QR
EI
s
s
When
12 Q , or Q2
1
I2 is equal to 2
2
maxI. Then the width is estimated as
000 2)1()1(2
or
Q
12
2
0
2
0Q (definition of the quality factor)
or
Q
02
(full-width at half-maximum)
4. Transformer
dt
dNV
dt
dNV
Bss
Bpp
Then we have
p
s
p
s
N
N
V
V (transformation of voltage)
If Ns>Np, the transformer is called a step-up transformer. If Np>Ns, the transformer is
called a step-down transformer
From the energy conservation, we have
ppss VIVI
or
s
p
s
p
p
s
N
N
V
V
I
I (transformation of current)
The equivalent resistance Req is the value of the load resistance as seen by the generator.
RN
N
I
V
I
I
V
V
I
VR
s
p
s
s
p
s
s
p
p
p
eq
2)(
5. Typical examples
5.1 Problem 31-17 (SP-31)
In an oscillating LC circuit with C = 64.0 F, the current is given by i = 1.60 sin(2500
t + 0.680), where t is in seconds, i in amperes, and the phase constant in radians. (a) How
soon after t = 0 will the current reach its maximum value? What are (b) the inductance L
and (c) the total energy?
((Solution))
C = 64.0 F )680.02500sin(60.1 ti
Imax = 1.60 A
= 2500 rad/s
(a) ,1)680.02500sin( t 2
680.02500
t
or
t = 356.30 s.
(b)
HC
L
LC
0025.01
1
2
(c)
The total energy is conserved.
JCvLiCvLi 322
max
2
max
22 102.36.10025.02
1
2
1
2
1
2
1
2
1
5.2 Problem 31-61 (SP-31)
In Fig. R = 15.0 , C = 4.70 F, and L = 25.0 mH. The generator provides an emf
with rms voltage 75.0 V and frequency 550 Hz. (a) What is the rms current? What is the
rms voltage across (b) R, (c) C, (d) L, (e) C, and L together, and (f) R, C, and L together?
At what average rate is energy dissipated by (g) R, (h) C, and (i) L?
((Solution))
R = 15
C = 4.70 F
L = 25.0 mH
= 75.0 V
f = 550 Hz
Phasor diagram
rmsrmsLC
rmsrmsC
rmsrmsR
rmsrmsL
rmsrms
IC
LV
IC
V
RIV
LIV
IC
LR
)1
()(
1
)1
(22
22
22
)1
(
)1
(
CLR
I
IC
LR
rmsrms
rmsrms
(a) Irms = 2.58576 A
(b) rmsrmsR RIV = 38.7865 V
(c) rmsrmsC IC
V1
= 159.202 V
(d) rmsrmsL LIV = 223.394 V
(e) rmsrmsLC IC
LV )1
()(
= 64.192 V
(f) rms = 75.0 V
(g) 2
rmsR RIP = 100.293 W
(h) 0CP
(i) 0LP
5.3 Problem 31-60 (SP-31) A typical light dinner used to dim the stage lights in a theater consists of a variable
inductor L (whose inductance is adjustable between 0 and Lmax) connected in series with a lightbulb B as shown in Fig. The electrical supply is 120 V (rms) at 60.0 Hz; the light-
bulb is rated at 120 V, 1000 W. (a) What Lmax is required if the rate of energy dissipation in the lightbulb is to be varied by a factor of 5 from its upper limit of 1000 W? Assume
that the resistance of the lighbulb is indepenent of its temperature. (b) Could one use a variable resistor (adjustable between zero and Rmax) instead of an inductor? (c) If so, what
Rmax is required? (d) Why isn’t this done?
\ ((Solution))
L = 0 – Lmax 120 V (rms) at 60 Hz
Light bulb 1000 W at 120 V
rms = 120 V
f = 60 Hz
22
22
22
2
22
22
)()(
)(
)(
LR
R
LRRRIP
LRI
ILR
rmsrmsrmsR
rmsrms
rmsrms
(a) When L = 0,
RW
RP rms
R
22120
1000
or R = 14.4
5
1000
)(1
1000
)(1
1
)(2
2
2
2
2
22
2W
R
L
W
R
LRLR
RP rmsrms
R
or
5)(
12
2
R
L
or
2R
L mH
RL 39.762
(b) and (c) Yes.
The inductance is replaced by a variable resistance R0.
20
2
2
0
2
)1(
1
)(
R
RRRR
RP rmsrms
R
2361.115
5)1(
0
0
R
R
R
R
or
799.174.142361.10R
(d) This is not done because we lose the energy in the variable resistance.
5.4 Phasor diagram of RLC circuit
G. Gladding, M. Selen, and T. Steltzer, SmartPhysics; Electricity and
Magnetism (II) (W.H. Freeman, 2011)
Consider the driven LCR circuit. The source voltage has a maximum voltage of 10 V
and oscillates at a frequency of 60 Hz, which converts to an angular frequency of 377
rad/s. The values for the circuit components are 20 , 20 mH, and 150 F, respectively.
Our goal is to determine maxI , the maximum current, and , the phase angle between
the current and the source voltage. Once we know these two quantities at any time. We
can determine both of these quantities by simply constructing the phasor diagram.
Fig. RLC circuit. f 60 Hz. L = 20 mH. C = 150 F. 2 f 376.991 rad/s.
10mE V.
((Solution))
Suppose that max sin( )I I t . The voltages across the capacitance, inductance, and
resistance are drawn in the x-y plane, as well as the current.
Fig. Phasor diagram for voltages of the circuit. We assume that maxRI is directed along
the x axis. The voltage source is actually expressed by m sin( )t . So it is
necessary to rotate the phasor diagram by appropriate angle at the last stage.
Note that
max max20 RV RI I
max
1( )CV I
C 17.6839 Imax
max( )LV L I 7.53982 Imax
The source voltage;
2 2
max max
1( ) 22.4255 ZV R L I I
C
Since 0 max10.0 V 22.4255 ZV E I , we have the maximum current as
0max 0.445921
z
EI
V [A]
max 8.91843RI [V]
max 3.36217LI [V]
max
17.88562I
C V
The phase factor:
arctan[ ] 26.89C L
R
V V
V
.
Since 0 sin( )sV E t , we need to rotate the phasor diagram (in the x-y plane) around the
origin by the angle in counter clockwise. So we get the final phasor diagram
The average power dissipated in the circuit;
2
max
2
max
max max
max max
1
2
1cos
2
1cos
2
1cos
2
avP RI
Z I
ZI I
I V
where cos is the power factor.
RImaxLImax
1
CImax
1
CL Imax
Vmax ZImax
Fig. Phasor diagram of the RLC circuit. 26.89 .
For the voltage source 10 sin( )t [V]
For the voltage across the resistance: 8.91834 sin( )t [V]
For the voltage across the inductance 3.36217 sin( )2
t
[V]
For the voltage across the capacitance 7.88562 sin( )2
t
[V]
For the current 0.445921 sin( )t [A]
________________________________________________________________________
Appendix A AC circuit theory based on the complex plane
Here we discuss the AC circuit theory based on the complex number. This theory is mathematically correct.
A1. Impedance of single elements in AC circuit
We represent all voltages and current (sinusoidally change with time) by complex numbers, using the exponential notation. The time-varying current is written by
2]Re[)(
* tititi eIIe
Ieti
where (= 2f) is the angular frequency, I represents a complex number that is
independent of t, and I* is the complex conjugate. The complex number I is described by ieI0 where I0 is the amplitude and is the phase. For convenience, we use the phasor
diagram in the complex plane, where for I is plotted as
)cos(]Re[]Re[)( 00 tIeeIIeti tititi
The voltage is also described by
]Re[)( tiVetv
((Note))
The absolute values V and I are the amplitudes of the real voltage and current.
(a) Inductance L
For inductance one can write down the relation
]Re[]Re[]Re[)( tititi LIeiIedt
dL
dt
diLVetv
or
ILeLIiV i 2/
The relation between V and I for the inductance is described in the following phasor diagram.
We say that V leads to I by 90° (or I lags V by 90°). The impedance of the inductance is given by
LiI
VZL
ɺ
ɺ
(b) Capacitance C
For capacitance, one can write down a relation
]]Re[]Re[)(
]Re[)( tititi CVeiVedt
dC
dt
tdvCIeti
CViI
or
IeCCi
IV i 2/1
The relation between V and I for the capacitance is described in the following phasor diagram.
We say that V lags I by 90° (or I lags V by 90°). The impedance of the capacitance is
given by
CiI
VZC
1
(c) Resistance
For resistance, one can write down a relation
]]Re[]Re[)(]Re[)( tititi RIeIeRtRiVetv
RIV
The relation between V and I for the resistance is described in the following phasor diagram.
The impedance of the resistance is given by
RZR
In summary
((Note)) The analysis of the AC theory without the concept of the complex number will be given
in the Appendix B.
A2. Power dissipated in single elements
The power is given by
)(4
1
22
)()()(
**2**2
**
VIIVeIVVIe
eIIeeVVe
titvtP
titi
titititi
The time average of the power P(t) over a period time T (= 2/) is given by
]Re[2
1]Re[
2
1)(
4
1)(
1 ****
0
VIIVVIIVdttPT
P
T
avg
When 0II and ieVV 0 (I0 and V0 are real numbers), then the power is rewritten as
2
00 coscos2
1rmsrmsrmsavg RiviVIP
where cos is the power factor, and the definition of the root-mean squares irms and vrms will given later.
(a) inductance
For the capacitance, we have ** LIiV . Then Pavg is calculated as
0)Re(2
1)Re(
2
1]Re[
2
1 2** ILiILIiIVPavg
No power is dissipated in an inductor.
(b) Capacitance
For the capacitance, we have Ci
IV
** . Then Pavg is calculated as
0]1
Re[2
1]
1Re[
2
1])Re[(
2
1]Re[
2
1 2**
* IC
iIIC
iICi
IIVPavg
No power is dissipated in a capacitor.
(c) Resistance
For the resistance, we have ** RIV . Then Pavg is calculated as
R
VIRIRIRIIVPavg
2
22**
2
1
2
1]Re[
2
1]Re[
2
1]Re[
2
1
Power is dissipated in a resistor.
A3. Root-mean square value of current and voltage
What is the root-mean square of the current and voltage?
)cos(]Re[]Re[)( 00 tIeeIIeti tititi
The root-mean square value of the current is defined as
2)]22cos(1[
2
)(cos)]([1
0
0
2
0
0
2
2
0
0
2
Idtt
T
I
dttT
Idtti
Ti
T
TT
rms
or
22
maxIIirms
where Imax (=|I| = |I0|) is the maximum of the current.
Similarly, we have the root-mean value of the voltage as
22
maxVVvrms
The power dissipated in the resistor is rewritten as
R
viRP rmsrmsavg
22
((Note))
The rms value of any quantity that varies as sin(t) or cos(t) is always 1/ 2 times the
maximum value, so
vrms = 0.707vmax.
A4. The series RLC circuit
Circuit in the time domain
AC circuit in the frequency domain
)()()()( tvtvtvte CLRs
The relation of the expressions in the time domain and the frequency domain is given by
)Re()(
)Re()(
)Re()(
)Re()(
)Re()(
ti
RR
ti
CC
ti
LL
ti
ti
ss
eVtv
eVtv
eVtv
Ieti
eEte
Es, I, VL, VC, and VR are complex numbers. The phasor diagram of these will be given
later.
ICi
V
LIiV
RIV
IZVVVE
C
L
R
sCLRs
1
or
s
sC
s
sL
s
sR
s
s
Z
CiEV
Z
LiEV
Z
REV
Z
EI
)/1(
where Zs is the total impedance defined by
i
s
i
s
eZ
eC
LR
iC
LR
CLiR
CiLiRZ
22
22
)1
(
)sin(cos)1
(
)1
(1
where is an angle and ie is the phase factor,
R
CL
1
tan
and
22)
1(
CLRZ s
or
A5. Phasor diagram of the RLC circuits
((Example))
A series RLC circuit has R = 150 , L = 0.25 H, and C = 16.0 F. It is driven by 60 Hz
voltage of peak value 170 V. Determine the time dependence of the voltages across each
element.
-0.01 0.01 0.02 0.03t HsecL
-1.0
-0.5
0.5
1.0
i HAL
vL
vR
vCvS
-0.01 0.01 0.02 0.03t HsecL
-200
-100
100
200v
A6. Infinite ladder network (from the book of Feynman)
What would happen if the following network we keep on adding a pair of the
impedances (z1 and z2) forever, as we indicate by the dashed lines. Can we solve such an
infinite network?
First, we notice that such an infinite network is unchanged even if we add one more
section at the front end. If we add one more section to an infinite network it is still the
same infinite network. Suppose we call the impedance between the two terminals a and b
of the infinite network z0. Then the impedance of all the stuff to the right of the two
terminals c and d is also z0. Taking into account of the similarity of the infinite network,
we get an equation to determine the value of z0 as
02
0210
zz
zzzz
We can solve for z0 to get
21
2
110
42zz
zzz
This impedance z0 is called the characteristic impedance of the infinite network.
Now we consider a specific example where an AC source is connected between the
terminals a and b, and Liz 1 and )/(12 Ciz . We define the complex current In and
voltage Vn.
In the above Fig. we get the following recursion relation,
0
111z
VzIzVV n
nnn
or
0
10
0
11 1z
zz
z
z
V
V
n
n
or
n
nV
We can call this ratio the propagation factor for one section of the ladder; we will call it .
It is the same for all sections. Note that
42
42
22
21
2
110
L
C
LLi
zzzz
z
Then we have
2
2
22
22
0
10
1
1
24
24
xix
xix
LiL
C
L
LiL
C
L
z
zz
where 0
x and LC
20
Out[23]=
1 2 3 4 5wêw0
0.2
0.4
0.6
0.8
1.0
»a»
Out[21]=
1 10 100 1000wêw0
10-5
0.001
0.1
»a»
Fig. as a function of 0
x .
ReHaL
ImHaL
1 2 3 4 5wêw0
-1.0
-0.5
0.0
0.5
1.0Re@aD, Im@aD
Fig. Re[] and Im[] as a function of 0
x .
In summary, the network propagates energy for <0 and blocks it for >0. We say
that the network passes low frequencies and reject or filters out the high frequencies. Any
network designed to have its characteristics vary in a prescribed way wit frequency is
called a filter. In this case, we have a low pass filter.
A7. Impedance matching
Given a circuit having a load impedance ( LLL iXRZ ), what impedance will result in
the maximum average power P being absorbed by the load?
L
sLL
L
s
ZZ
VZIZV
ZZ
VI
0
0
The power P (absorbed by the lad) is given by
]Re[2
1]Re[
2
1 ** IVVIP
where
L
sL
L
sL
L
s
ZZ
VZ
ZZ
VZ
ZZ
VVI
0
2
0
*
0
*
*
)(
and
)()( 000 LLL
LLL
XXiRRZZ
iXRZ
Then P is given by
])()[(2 2
0
2
0
2
LL
Ls
XXRR
RVP
First we assume that RL is constant. Only XL is a unknown parameter. P has a maximum
when
0XX L
Then P obtained as
2
0
2
)[(2 L
Ls
RR
RVP
has a maximum at RL = R0.
In conclusion, in order to get maximum power to ZL, we select
*
000 ZiXRZL
_______________________________________________________________________
B. Complex number
B1. Imaginary unit
The imaginary unit is defined by 1i : 12 i .
ii
i
522020
55
These numbers are called a pure imaginary number.
B2. Properties of complex number
For real numbers a, b, c, and d,
dicbia → a = c and b = d.
0 bia → a = 0 and b = 0.
22
222222
)()(
))((
))((
)1()())((
)())((
)()()()(
)()()()(
dc
iadbcbdac
dicdic
dicbia
dic
bia
bababiabiabia
ibcadbdacdicbia
idbcadicbia
idbcadicbia
B3. Simplifying powers of i
The expression of ni , where n is a positive whole number, can be reduced to either ±1
or ±i.
i0 = 1, i1 = i i2 = -1, i3 = i2i=-i
i4 = 1, i5 = i4 i =i i6 = i4 i2=i2 = -1 i7 = i4i3=-i
In general, we have
i4n = 1, i4n+1 = i i4n+2 = -1, i4n+1 = -i
where n is an integer.
B4. Graphing complex number: complex plane
(a) Definition
z = x+iy
A complex number in x+iy form can be graphed in the complex plane by measuring x
units along the horizontal (real axis) and y units along the vertical (imaginary axis).
We define the absolute value of the complex number as
22
yxiyxz
which is the distance between the ponits of z (= x+i y) and the origin.
(b) Expression of the complex number in the real-imaginary plane
We make a plot of the complex numbers in the complex plane.
a = 3+2i, b = -2+i
a+b = 1+3i, a - b=5+i,
a
b
a+b
a-b
x
y
O
B5. Complex conjugate
The complex number z and z* are complex conjugate.
ibaz
ibaz
*
where a and b are real.
22* ))(( baibaibazz .
B6. Angles and polar coordinates of complex number
We find the real (horizontal) and imaginary (vertical) components in terms of r (the
length of the vector) and θ (the angle made with the real axis):
From Pythagoras, we have: r2 = x2 + y2 and basic trigonometry gives us:
x
ytan
x = r cosθ, y = r sinθ
Multiplying the last expression throughout by i gives us:
siniryi
So we can write the polar form of a complex number as:
)sin(cos iriyxz
where r is the absolute value (or modulus) of the complex number, and θ is the
argument of the complex number.
q
f
q +f
O
P
Q
R
Real
Imaginary
Points P, Q and R on the unit circle in the complex plane, are denoted by complex
numbers z1, z2, and z1z2.
)sin()cos(
)sincoscos(sin)sinsincos(cos
)sin)(cossin(cos
sincos
sincos
21
2
1
i
i
iizz
iz
iz
The complex conjugate of z1 is
)sin()cos(
sincos*
1
i
iz
((Example))
We make a plot of zn (n = 0, 1, 2, 3, …), where
).sin(cos irz
We choose r = 1.05 and = 5º.
B7. Euler’s formula
Euler's formula states that, for any real number ,
sincos iei
where e is the base of the natural logarithm, i is the imaginary unit. Richard Feynman
called Euler's formula "our jewel" and "the most remarkable formula in mathematics".
B8. de Moivre’s theorem
When ireiriyxz )sin(cos
)]sin()[cos( ninrerrez ninnnin .
where
x
y
yxr
tan
22
((Note))
(a) The expression of z-n
)]sin()[cos(
)]sin()[cos(
ninr
ninr
er
rez
n
n
inn
nin
(b) The expression of z* (complex conjugate of z)
ire
ir
irz
)]sin()[cos(
)sin(cos*
APPENDIX
Single-phase rms voltage and frequency over the world
Europe (Great Britain, France, Germany, Belgium, Netherland, Denmark, Sweden) 230 V 50 Hz
Australia 230 V 50 Hz U.S.A. 120 V 60 Hz
Equador 120 V 60 Hz China 220 V 50 Hz
Malawi 220 V 50 Hz
Japan: 100 V 50 (East)/60Hz (West); regions of west and east separated by Fossa Magna.