Chapter 31 Electromagnetic Oscillations and AC Circuit Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: July 25, 2019)
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 [ it( ) Re( Ie i t ) for the current and v( t ) Re( Ve i t ) 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
1 1 U Li 2 Cv 2 2 L 2 C where q = CvC.
From the energy conservation, U remains constant with time.
dU d dv Li i Cv C 0 dt L dt L C dt
dq q Since i and v , we have L dt C C
d 2q 2q 0 (LC oscillation, simple harmonics). dt 2 with
1 LC
The total energy U is rewritten as
2 1 dq 1 U L q2 2 dt 2C
We assume the initial condition as
At t = 0,
q = Q and dq/d t = 0.
The solution of the second order differential equation is given by
q Qcos(t)
dq i Q sin(t) L dt
qêQ, iL êHQwL 1.0
0.5
Out[11]= têT 0.2 0.4 0.6 0.8 1.0 1.2 1.4
-0.5
-1.0
Fig. q /Q (red) and iL /(Q) (blue) as a function of t/T.
Energy of UE and UB
q2 Q2 U cos2 (t) E 2C 2C 1 Q2 U LQ 22 sin2 (t) sin 2 (t) B 2 2C Q2 U U U E B 2C
2 2 Fig. U E /(Q 2/ C ) (red) and U B /(Q 2/ C ) (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.
di v L L L dt v Ri R L q CvC dq i L dt
From the KVL, we have
vL vR vC 0 or the second order differential equation
d 2q R dq q 0 dt 2 L dt LC
We assume that
2 q = x, 2 = R/L, and 0 = 1/( LC )
Then we have
2 x t)(" 2 tx )(' 0 tx )( 0 with the initial conditions
x )0(' v0 and x )0( x0
The solution of this differential equation depends is classed into three types,
2 2 (1) underdamping: 0 0 2 2 (2) critical damping 0 0 2 2 (3) overdamping 0 0
(( Note )) We assume that x(t) is expressed by a solution given by exp( pt ). Then we have
2 2 2 pt x t)(" 2 tx )(' 0 tx )( ( p 2p 0 )e 0
2 2 The solution of the quadratic equation p2 p 0 0 is
2 2 2 2 p 2p (0 )
2 2 ( p ) i 0
2 2 p i 0 i1 with
2 2 1 0
The solution is given by
t tq )( tx )( e [C1 cos(1t) C2 sin(1t)]
From the initial condition, we have
C1 x0
v0 x0 C2 1
The final form is given by
v x t 0 0 tq )( e [x0 cos(1t) sin(1t)] 1
3. AC circuit theory
3.1 phasor diagram (a)
i Imax sin(t)
vR RImax sin(t) 1 I I v idt max cos(t) max sin(t ) C C C C 2 di v L LI cos(t) LI sin(t ) L dt max max 2
(b)
i Imax cos(t)
vR RImax cos(t) 1 I I v idt max sin(t) max cos(t ) C C C C 2 di v L LI sin(t) LI cos(t ) L dt max max 2
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
Vmax LImax
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
I V max max C
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
Vmax RImax
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
1 T P dtI sin( )Vt sin(t ) avg max max T 0 1 T I V dt[cos cos(2t )] max max 2T 0 1 I V cos 2 max max
I rmsVrms cos 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
1 1 I I , V V rms 2 max rms 2 max
3.3.2 inductance
Pavg 0 since =/2. No power is dissipated in an inductor.
3.3.3 Capacitance
Pavg 0 since = -/2. No power is dissipated in a capacitor.
3.3.4 Resistance
1 P RI 2 avg 2 max
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
1 T i I 2 sin2 (t)dt rms max T 0 1 T I 2 1[ cos(2t)]dt max 2T 0 I max 2
The root-mean square value of the voltage is defined as
V v max rms 2
(( 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 Vmax 120 2 = 170 V.
The power dissipated in the resistor is rewritten as
2 2 V P IR rms avg rms R
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
Vs s IZ
VR RI
VL LI 1 V I C C where I is the amplitude of the current, Vs is the amplitude of the voltage source, Zs is the total impedance defined by
1 Z R2 (L )2 s C
and the angle between Vs and I is defined as
1 1 tan (L ) R C
The average power dissipated in the system is
1 P IV cos V I cos RI 2 2 s s rms rms rms
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 = 2 f
(1) S1 and S 2 open
2 E R2 ( L)2 I s C 1 2 ( L) tan C 1 R
Es lags I1 by 1. The average power dissipated in the system is
1 P IE cos E (I ) cos R(I ) 2 2 s 1 1 s rms 1 rms 1 1 rms
where Es cos1 RI1
(2) S1 closed and S2 open
1 E R2 (L )2 I s C 2 1 (L ) tan C 2 R
Es leads I2 by 2. The average power dissipated in the system is
1 P IE cos E (I ) cos (IR ) 2 2 s 2 2 s rms 2 rms 2 2 rms
where Es cos2 RI 2
(3) S1 closed and S2 closed
E I s 3 1 L C
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?
2 1 2 Es R ( ) I C /1( C) tan R
Es lags I by the angle . The average power dissipated in the system is
1 P IE cos E I cos RI 2 2 s s rms rms rms where
Es cos RI
3.8. Example ( RL circuit)
2 2 Es R (L) I L tan R
Es leads I by the angle . The average power dissipated in the system is
1 P IE cos E I cos RI 2 2 s s rms rms rms where
Es cos RI
3.9 Resonance of the RLC circuit
1 E R2 (L )2 I s C or
E E I s s Z 1 s R2 (L )2 C where
1 Z R2 (L )2 (impedance) s C
The amplitude of the current has a maximum when
1 LC
We define the quality factor of the circuit by
L Q 0 R
Then the amplitude of the current can be rewritten as
E 1 I s R 1 Q2 ( 0 )2 0
We now consider the frequency dependence of I2, instead of I.
2 E 1 I 2 s R 1 Q2 ( 0 )2 0
We note that this is a scaling function of /0. In the figures we show the plot of I 2 2 2 vs , where the quality factor Q is changes as a parameter. (Es / R ) 0
Es 2êR2 1.0
0.8
0.6
0.4 Q=100 - 1000
0.2
wêw0 0.6 0.8 1.0 1.2 1.4
Es 2êR2 1.0
0.8
0.6 Q = 10 - 100
0.4
0.2
wêw0 0.6 0.8 1.0 1.2 1.4
I 2 Fig. 2 2 vs , where the quality factor Q is changes as a parameter. (Es / R ) 0
______ Suppose that 1 (<<1). 0
0 1( )1 1
Then we have
2 E 1 I 2 s R 1 Q2 1[ 1( )]2
2 E 1 I 2 s max R 1 4Q 22 1 4Q 22
When 1 2Q 1, or 2Q
I 2 I2 is equal to max . Then the width is estimated as 2
2 0 1( ) 0 1( ) 20 or
2 1 2 0 Q
Q 0 (definition of the quality factor) 2 or
2 0 (full-width at half-maximum) Q
4. Transformer
d V N B p p dt
d V N B s s dt
Then we have
V N s s (transformation of voltage) Vp N p
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
VI ss VI pp or
I V N s p p (transformation of current) I p Vs Ns
The equivalent resistance Req is the value of the load resistance as seen by the generator.
Vp Vp Is Vs N p 2 Req ( ) R I p Vs I p Is Ns
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 i 60.1 sin(2500t .0 680)
Imax = 1.60 A = 2500 rad/s
(a) sin(2500t .0 680) ,1 2500t .0 680 2 or
t = 356.30 s.
(b)
1 LC 1 L .0 0025H C 2
(c)
The total energy is conserved.
1 1 1 1 1 Li2 Cv2 Li 2 Cv 2 .0 0025 6.1 2 2.3 103 J 2 2 2 max 2 max 2
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
1 R 2 (L )2 I rms C rms
VL rms LI rms
VR rms RIrms 1 V I C rms C rms 1 (V ) (L )I LC rms C rms
1 R2 (L )2 I rms C rms I rms rms 1 R2 (L )2 C
(a) Irms = 2.58576 A
(b) VR rms RIrms = 38.7865 V
1 (c) V I = 159.202 V C rms C rms
(d) VL rms LIrms = 223.394 V
1 (e) (V ) (L )I = 64.192 V LC rms C rms
(f) rms = 75.0 V
2 (g) PR RIrms = 100.293 W
(h) PC 0
(i) PL 0
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
2 2 rms R (L) Irms
rms Irms R2 (L)2 2 R 2 P RI 2 R rms rms R rms 2 2 2 2 R (L) R (L)
(a) When L = 0,
2 1202 P rms 1000W or R = 14.4 R R R
R 2 2 1 1000W 1000W P rms rms R R2 (L)2 R (L)2 (L)2 5 1 1 R2 R2 or
(L)2 1 5 R2 or
L R 2 L 2 39.76 mH R
(b) and (c) Yes.
The inductance is replaced by a variable resistance R0.
2 2 R rms rms 1 PR 2 (R R ) R R 2 0 1( 0 ) R
R 1( 0 ) 5 R R 0 5 1 .1 2361 R or
R0 .1 2361 4.14 .17 799
(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 Imax , 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.
Em 10 V.
(( Solution ))
Suppose that I Imax sin( 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 RI max 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
VR RImax 20 I max
1 V( ) I 17.6839 Imax C C max
VL ( L ) I max 7.53982 Imax
The source voltage;
1 VRL2( ) 2 I 22.4255 I Z C max max
Since VZ E0 10.0 V 22.4255 I max , we have the maximum current as
E0 Imax 0.445921 [A] Vz
RI max 8.91843 [V]
LI max 3.36217 [V]
1 I 7.88562 V C max
The phase factor:
V V arctan[C L ] 26.89 . VR
Since Vs E0 sin( 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;
1 P RI 2 av 2 max
1 2 Zcos I max 2 1 ZIcos I 2 max max 1 I V cos 2 max max where cos is the power factor. RI max LI max
Vmax ZI max
1 L I C max 1 Imax C
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(t ) [V] 2 For the voltage across the capacitance 7.88562 sin(t ) [V] 2 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
Ie ti *eI ti ti )( Re[Ie ti ] 2 where (= 2 f) 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 i 0eI 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
ti titi ti )( Re[Ie ] Re[ 0eI e ] I0 cos(t )
The voltage is also described by
tv )( Re[Ve ti ]
((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
di d tv )( Re[Ve ti ] L L Re[Ie ti ] Re[iLIe ti ] dt dt or
V iLI Lei 2/ I
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
Vɺ Z iL L Iɺ
(b) Capacitance C For capacitance, one can write down a relation
dv t)( d ti )( Re[Ie ti ] C C Re[Ve ti ] Re[iCVe ti ]] dt dt
I iCV or
I 1 V ei 2/ I iC C
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
V 1 Z C I iC
(c) Resistance For resistance, one can write down a relation
tv )( Re[Ve ti ] Ri t)( RRe[Ie ti ] Re[RIe ti ]]
V RI
The relation between V and I for the resistance is described in the following phasor diagram.
The impedance of the resistance is given by
Z R R
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
P t)( titv )()( Ve ti V *e ti Ie ti *eI ti 2 2 1 (VIe 2 ti V I ** e 2 ti V * I VI * ) 4
The time average of the power P(t) over a period time T (= 2 /) is given by
1 T 1 1 1 P tP )( dt (V *I VI * ) Re[V *I] Re[ *VI ] avg T 0 4 2 2
i When I I0 and V 0eV (I0 and V0 are real numbers), then the power is rewritten as
1 P VI cos i v cos Ri 2 avg 2 00 rms rms rms 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 V iLI . Then Pavg is calculated as
1 1 1 2 P Re[V *I] Re(iLI *I) Re(i IL ) 0 avg 2 2 2
No power is dissipated in an inductor.
(b) Capacitance * * I For the capacitance, we have V . Then Pavg is calculated as iC
* 1 1 I 1 1 1 1 2 P Re[V *I] Re[( I]) Re[i *II ] Re[i I ] 0 avg 2 2 iC 2 C 2 C
No power is dissipated in a capacitor.
(c) Resistance * * For the resistance, we have V RI . Then Pavg is calculated as
2 1 1 1 2 1 2 1 V P Re[V *I] Re[RI *I] Re[ IR ] IR avg 2 2 2 2 2 R
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?
ti titi ti )( Re[Ie ] Re[ 0eI e ] I0 cos(t )
The root-mean square value of the current is defined as
1 T I 2 T i ti )]([ 2 dt 0 cos2 (t )dt rms T T 0 0 I 2 T I 0 1[ cos(2t 2)]dt 0 2T 0 2 or
I I i max rms 2 2 where Imax (=| I| = | I0|) is the maximum of the current.
Similarly, we have the root-mean value of the voltage as
V V v max rms 2 2
The power dissipated in the resistor is rewritten as
2 2 v P iR rms avg rms R
((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 = 0.707 vmax .
A4. The series RLC circuit
Circuit in the time domain
AC circuit in the frequency domain
es t)( vR t)( vL t)( vC t)(
The relation of the expressions in the time domain and the frequency domain is given by
ti es t)( Re( seE ) ti )( Re(Ie ti ) ti vL t)( Re(VLe ) ti vC t)( Re(VCe ) ti vR t)( Re(VRe )
Es, I, VL, VC, and VR are complex numbers. The phasor diagram of these will be given later.
Es VR VL VC s IZ
VR RI
VL iLI 1 V I C iC or
E I s Zs E R V s R Z s siE L VL Zs
Es /1( iC) VC Zs where Zs is the total impedance defined by
1 1 Z R iL R i(L ) s iC C 1 R2 (L )2 (cos isin) C 1 R2 (L )2 ei C i Zs e where is an angle and ei is the phase factor,
1 L tan C R and
1 Z R2 (L )2 s C 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.
i HAL 1.0
0.5
t Hsec L -0.01 0.01 0.02 0.03
-0.5
-1.0
v 200 vS vC vR
vL 100
t Hsec L -0.01 0.01 0.02 0.03
-100
-200
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
zz 02 z0 z1 z2 z0
We can solve for z0 to get
z z 2 z 1 1 zz 0 2 4 21
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 z1 iL and z2 /(1 iC ) . We define the complex current In and voltage Vn.
In the above Fig. we get the following recursion relation,
Vn Vn Vn1 1Iz n z1 z0 or
V z z z n1 1 1 0 1 Vn z0 z0 or
n Vn
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
z z 2 z 1 1 zz 0 2 4 21
iL L L22 2 C 4
Then we have
L L22 iL z z 0 1 C 4 2 22 z0 L L iL C 4 2 ix 1 x2 ix 1 x2
2 where x and 0 0 LC
»a» 1.0
0.8
0.6 Out[23]= 0.4
0.2
wêw0 1 2 3 4 5
»a»
0.1
0.001 Out[21]=
10 -5
wêw0 1 10 100 1000
Fig. as a function of x . 0
Re @aD, Im @aD 1.0
0.5 Re HaL
0.0 wêw0 1 2 3 4 5
-0.5 Im HaL
-1.0
Fig. Re[ ] and Im[ ] as a function of x . 0
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 ( Z L RL iX L ), what impedance will result in the maximum average power P being absorbed by the load?
V I s Z Z 0 L Z VsL V Z LI Z0 Z L
The power P (absorbed by the lad) is given by
1 1 P Re[VI * ] Re[V *I] 2 2 where
2 V * Z V Z V *VI s sL sL * (Z0 ZL ) Z0 ZL Z0 Z L and
Z R iX L L L Z0 ZL (R0 RL ) (Xi 0 X L )
Then P is given by
2 Vs RL P 2 2 [(2 R0 RL ) (X 0 X L ])
First we assume that RL is constant. Only XL is a unknown parameter. P has a maximum when
X L X 0
Then P obtained as
2 Vs RL P 2 [(2 R0 RL ) has a maximum at RL = R0. In conclusion, in order to get maximum power to ZL, we select
* Z L R0 iX 0 Z0
______B. Complex number
B1. Imaginary unit The imaginary unit is defined by i 1 : i2 1.
5 5i
20 20i 2 5i
These numbers are called a pure imaginary number .
B2. Properties of complex number For real numbers a, b, c, and d,
a bi c di → a = c and b = d.
a bi 0 → a = 0 and b = 0.
(a bi) (c di) (a c) (b )id (a bi) (c di) (a c) (b )id (a bi)(c di) ac bd (ad bc)i (a bi)(a bi) a2 (bi)2 a2 ( )1 b2 a2 b2 a bi (a bi)(c di) (ac bd) (bc ad)i c di (c di)(c di) c2 d 2
B3. Simplifying powers of i The expression of in , 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+i y
A complex number in x+i y 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
z x iy x2 y2 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+2 i, b = -2+ i a+b = 1+3 i, a - b=5+ i,
y
a+b
a
a-b b
O x
B5. Complex conjugate
The complex number z and z* are complex conjugate.
z a ib
z* a ib where a and b are real.
zz* (a ib)(a ib) a2 b2 .
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:
y tan x
x = r cos θ, y = r sin θ
Multiplying the last expression throughout by i gives us:
yi ir sin
So we can write the polar form of a complex number as:
z x iy r(cos isin ) where r is the absolute value (or modulus) of the complex number, and θ is the argument of the complex number.
Imaginary
R
Q
q +f P f
q Real O
Points P, Q and R on the unit circle in the complex plane, are denoted by complex numbers z1, z2, and z1z2.
z1 cos isin
z2 cos isin
zz 21 (cos isin )(cos isin) (cos cos sin sin) i(sin cos cos sin) cos( ) isin( )
The complex conjugate of z1 is
z * cos isin 1 cos( ) isin( )
((Example))
We make a plot of zn (n = 0, 1, 2, 3, …), where
z r(cos isin).
We choose r = 1.05 and = 5º.
B7. Euler’s formula
Euler's formula states that, for any real number ,
ei cos isin 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 z x iy r(cos isin ) rei
n zn rei er inn r n[cos(n ) isin(n )]. where
r x2 y2 y tan x
(( Note )) (a) The expression of z-n
n zn rei r nein
r n[cos(n ) isin(n )] r n[cos(n ) isin(n )]
(b) The expression of z* (complex conjugate of z)
z* r(cos isin ) r[cos( ) isin()] rei
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.