E.E. Essential Knowledge Sereies My Online Courses: https://www.byparams.com/courses 正弦激勵, 相量與穩態分析

1. Department of Electronic Engineering
National Taipei University of Technology

2. •
• ( )
• ( )
• ( )
•
•
Department of Electronic Engineering, NTUT2/41

3. (I)
•
Department of Electronic Engineering, NTUT
( ) sinmv t V tω=
Vm
t (sec)
T
v(t)
−Vm
2
π
ω
π
ω
3
2
π
ω
2π
ω
5
2
π
ω
6
2
π
ω
0
T
ω
: Vm (volt)
: T (sec)
: f (Hz, cycle/sec)
: (rad/sec)
2 fω π=
3/41

4. (II)
• ( ) ( )sinmv t V tω φ= +
ω
: Vm (volt)
: T (sec)
: f (Hz, cycle/sec)
: (rad/sec)
2 fω π=
φ: (rad)檔案中找不到關聯識別碼 rId12 的圖像部分。
Vm
t (sec)
v(t)
−Vm
2
π
ω
π
ω
3
2
π
ω
2π
ω
5
2
π
ω
6
2
π
ω
0
T
φ
1t
Department of Electronic Engineering, NTUT4/41

5. ( )
• vR = iR
( ) ( )
• ,
•
R, L, C = +
( ) ( / )
L
di
v L
dt
=
1 t
Cv idt
C −∞
= ∫
Department of Electronic Engineering, NTUT5/41

6. 1
• RL i(t)
if (t)
,
KVL
+
−
L
Ri(t)
( ) ( )cos Vs mv t V tω=
( )
( ) cosm
di t
L Ri t V t
dt
ω+ =
( ) cos + sinfi t A t B tω ω=
( )
( ) cosm
di t
L Ri t V t
dt
ω+ =
mRA LB Vω+ = 0LA RBω− + =
( ) ( )sin cos cos sin cosmL A t B t R A t B t V tω ω ω ω ω ω ω− + + + =
, 2 2 2
mLV
B
R L
ω
ω
=
+2 2 2
mRV
A
R Lω
=
+
( )fi t
( ) ( ) ( )1
2 2 2 2 2 2 2 2 2
cos sin cos tan cos A
R
m m m
f m
RV LV V L
i t t t t I t
R L R L R L
ω ω
ω ω ω ω φ
ω ω ω
− 
= + = − = + 
+ +  +
2 2 2
m
m
V
I
R Lω
=
+
1
tan
L
R
ω
φ −
= −
( ) ( ) ( )f ni t i t i t= +( ) 1
RtL
ni t Ae−
=
( )sv t
Department of Electronic Engineering, NTUT6/41

7. 2
• RLC v(t) i(t)
KVL
+
−
+
−
5
3
R = Ω 5 HL =
( )i t
( )v t
1
F
25
C =( )sv t
( ) 17cos3sv t t=
( ) ( )
( )
( )2
2
1 sd v t dv t v tR
v t
dt L dt LC LC
+ + =
( ) ( )
( )
2
2
5 85cos3
d v t dv tR
v t t
dt L dt
+ + =
( ) ( ) ( ) ( )20cos3 +5sin3 =5 17 cos 3 2.9 5 17 cos 3 166 Vv t t t t t= − − = −
( )
( ) ( )
( ) ( ) ( ) ( )
1 1 3 17 3 17
60sin3 15cos3 cos 3 1.33 cos 3 76 A
25 25 5 5
dv t dv t
i t C t t t t
dt dt
= = = + = − = −
,
( ) 1 2cos3 sin3v t A t A t= +
( ) ( )1 2 1 24 cos3 4 sin3 85cos3A A t A A t t− + + − − =
1 24 85A A− + = 1 24 0A A− − =
1 20A = −
2 5A =
Department of Electronic Engineering, NTUT7/41

8. (Complex Number)
•
θ
a
b
A a jb= +
Re
b
Im
| |A
+A a jb=
j
A A e Aθ
θ= = ∠
[ ]Re cosa A A θ= =
[ ]Im sinb A A θ= =
2 2
A a b= + 1
tan
b
a
θ −
=
1j = −
1 1 1A a jb= + 2 2 2A a jb= +
( ) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2+A A a jb a jb a a j b b= + + + = + + +
( ) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2A A a jb a jb a a j b b− = + − + = − + −
• +A a jb= A a jb∗
= −
•
( )( )
( ) ( )
2
1 2 1 1 2 2 1 2 1 2 2 1 1 2
1 2 1 2 1 2 2 1
A A a jb a jb a a ja b ja b j b b
a a b b j a b a b
= + + = + + +
= − + +
( )( ) ( )1 2 1 1 2 2 1 2 1 2A A A A A Aθ θ θ θ= ∠ ∠ = ∠ +
•
( )( )
( )( )
1 1 2 21 1 1 1 2 1 2 1 2 1 2
2 2 2 2
2 2 2 2 2 2 2 2 2 2 2
a jb a jbA a jb a a b b b a a b
j
A a jb a jb a jb a b a b
+ −+ + −
= = = +
+ + − + +
( )11
1 2
2 2
AA
A A
θ θ= ∠ −
Department of Electronic Engineering, NTUT8/41

9. (a+jb) (I)
•
(a) A = 4 + j3
,
A
(b) A = –4 + j3
A θ∠
2 2
4 3 5A = + = 1 3
= tan 36.9
4
θ −
=
5 36.9A = ∠
1 13 3
= tan 180 tan 180 36.9 143.1
4 4
θ − − 
= − = − = 
− 
( )
2 2
4 3 5A = − + =
5 143.1A = ∠
(a)
4
θ = −
tan13
4
| |A =5
Re
3
j
o
o
A
A
jA
9.365
9.364
3
tan
534
34
1
22
∠=⇒
==⇒
=+=⇒
+=
−
θ
Re
j
θ =
−
180
3
4
1
tan
−4
| |A =5
3
( )
oo
oo
A
A
jA
1.14351.143
9.361804
3
tan
534
34
1
22
∠=⇒=
−=−=⇒
=+−=⇒
+−=
−
θ
(b)
Department of Electronic Engineering, NTUT9/41

10. (a+jb) (II)
(c) A = –4 – j3
( )
o
oo
oo
A
A
jA
9.365
9.361.323
9.36360
4
3
tan
534
34
1
22
−∠=⇒
−==
−=−=⇒
=−+=⇒
−=
−
θ
j
| |A =5
Re
−3
4
θ =
−
360
3
4
1
tan
−3
Re
j
−4
| |A = 5
( ) ( )
o
oo
oo
A
A
jA
1.1435
1.1439.216
9.36180
4
3
tan
534
34
1
22
−∠=⇒
−==
+=
−
−
=⇒
=−+−=⇒
−−=
−
θ
θ =
−
180 +
3
4
1
tan
A θ∠
( ) ( )
2 2
4 3 5A = − + − =
1 13 3
= tan 180 tan
4 4
180 36.9 216.9 143.1
θ − −− 
= − 
− 
= + = = −
5 143.1A = ∠ −
(d) A = 4 – j3
( )
22
4 3 5A = + − =
1 13 3
= tan =360 tan
4 4
323.1 36.9
θ − −− 
− 
 
= = −
5 36.9A = ∠ −
(c)
(d)
Department of Electronic Engineering, NTUT10/41

11. (I)
•
•
• RL (Complex excitation)
(Real excitation)
i(t)
KVL:
cos sinj t
m m mV e V t jV tω
ω ω= +
cos Re
sin Im
j t
m m
j t
m m
V t V e
V t V e
ω
ω
ω
ω
  =  

 =  
1
j t
mv V e ω
= ( ) [ ]1cos Reg mv t V t vω= =
+
−
L
Ri(t)cosmV tω
1
1+ j t
m
di
L Ri V e
dt
ω
=
1
j t
i Ae ω
=
( ) j t j t
mj L R Ae V eω ω
ω + =
-1
tan
2 2 2
L
j
m m R
V V
A e
R j L R L
ω
ω ω
−
= =
+ +
1
tan
1 2 2 2
L
j t
RmV
i e
R L
ω
ω
ω
− 
− 
 
=
+
i1 i(t)
⇒⇒⇒⇒
Department of Electronic Engineering, NTUT11/41

12. (II)
6 RL
i(t) i1 ( vg(t) v1 )
i1 v1 if (t)= Re[i1] vg = Re[v1]
1
tan
1 2 2 2
L
j t
RmV
i e
R L
ω
ω
ω
− 
− 
 
=
+
( ) [ ]
1
tan
1
1 2 2 2 2 2 2
Re Re cos tan
L
j t
Rm mV V L
i t i e t
RR L R L
ω
ω ω
ω
ω ω
− 
−  − 
    
= = = −   
 + +  
[ ]1
1 1Re Re
di
L Ri v
dt
  
+ =  
  
[ ]( ) [ ]( )1 1Re Re cosm
d
L i R i V t
dt
ω+ = ( ) ( ) [ ]1Refi t i t i= =
Department of Electronic Engineering, NTUT12/41

13. • 11 :
• ?
KVL: 1
1+ j t
m
di
L Ri V e
dt
ω
= 1
j t
i Ae ω
=
( ) j t j t
mj L R Ae V eω ω
ω + =
-1
tan
2 2 2
L
j
m m R
V V
A e
R j L R L
ω
ω ω
−
= =
+ +
1
tan
1 2 2 2
L
j t
RmV
i e
R L
ω
ω
ω
− 
− 
 
=
+
A
” ”
” ”
j t
e ω
j t
e ω
jω
j t
e ω
1 jω
Department of Electronic Engineering, NTUT13/41

14. (I)
• (Euler’s formula)
(Phasor)
• rms
• ”−”
( ) ( )cos Re Rej j t j t
m mv t V t V e e Veθ ω ω
ω θ    = + = =   
j
m mV V e Vθ
θ= = ∠
2
j m
rms rms
V
V V e Vθ
θ θ= = ∠ = ∠
j
m mV e Vθ
θ= = ∠V j
m mV V e Vθ
θ= = ∠
j t
e ω
j t
e ω
1:
Department of Electronic Engineering, NTUT14/41

15. (II)
I
R
+
=
−
V RI
i
R
+
=
−
v Ri
I
+
−
i
L
+
−
di
v L
dt
= V j LIω= j Lω
+
−
V
1
I j CV V I
j C
ω
ω
= ⇒ =
1
j CωC
+
−
v
dv
i C
dt
=
R R
L j Lω
C
1
j Cω
( )
( )
2:
Department of Electronic Engineering, NTUT15/41

16. (III)
+
−
L
Ri(t)cosmV tω +
−
RI0mV V= ∠
j Lω
j LI R Vω + =1
2 2 2
0
tan
L
m mV VV L
I
R j L R j L RR
ω
ω ω ω
−∠
= = = ∠ −
+ + +
( )
1 1
tan tan
1
2 2 2 2 2 2 2 2 2
Re Re cos tan
L L
j j t
j tR Rm m mV V V L
i t e e e t
RR L R L R L
ω ω
ω
ω ω
ω
ω ω ω
− −   
− −    −   
     
= = = −     
 + + +      
2:
1:
, !
3:!
!
Department of Electronic Engineering, NTUT16/41

17. R L C
I
R
+
=
−
V RI
(a)
i
R
+
=
−
v Ri
t
v i,
v
i
(b)
i
L
+
−
I
+
−
(a) (b)
di
v L
dt
= V j LIω= j Lω
v
i
t
v i,
v
i
t
v i,
Department of Electronic Engineering, NTUT
C
+
−
v
+
−
I
1
j Cω
dv
i C
dt
=
(a) (b)
1
V I
j Cω
=
17/41

18. (I)
Z (Impedance)
( ) ( )1cosmv t V tω φ= +
( ) ( )2cosmi t I tω φ= + 2mI I φ= ∠
1mV V φ= ∠
( ) ( )1 2= m
m
VV
Z Z R jX
I I
θ φ φ= = ∠ ∠ − = + Ω
(a)
+
−
+
−
V
I
(b) ( )
( )v t
( )i t
[ ]ImX Z=
[ ]ReR Z=
(Reactance)
(Resistance)
Z ω
ω
( )Z jω
( )R R ω=
( )X X ω=
θ
R
X
Z R jX= +
Re
Im
2 2
Z R X= +
1
tan
X
R
θ −
=
cosR Z θ=
sinX Z θ=
Department of Electronic Engineering, NTUT18/41

19. (II)
•
• (Admittance)
• Z Y Y
Y = G + jB G = Re[Y] B = Im[Y]
(Susceptance) Y G B
R
V
Z R
I
= =
LX Lω= L L
V
Z jX j L
I
ω= = =
1
CX
Cω
= −
1
C C
V
Z jX j
I Cω
= = = −
1
Y
Z
=
( )( ) 2 2 2 2
1 1
=
R jX R X
Y j G jB
Z R jX R jX R jX R X R X
−
= = = − = +
+ + − + +
Department of Electronic Engineering, NTUT19/41

20. • (a) RL (b)
KVL:
+
−
L
Ri( )cos VmV tω
(a)
+
−
R
j Lω
I0mV ∠
(b)
0L mZ I RI V+ = ∠ ( ) 0mj L R I Vω + = ∠
( ) 1
2 2 2
cos tanmV L
i t t
RR L
ω
ω
ω
− 
= − 
 +
1
2 2 2
0
tanm mV V L
I
R j L RR L
ω
ω ω
−∠  
= = ∠ − 
+  +
Z j L Rω= +
0mVV
I
Z R j Lω
∠
= =
+
Department of Electronic Engineering, NTUT20/41

21. 3
• (a) v(t) i(t)
1. (a) (b)
2. (c)
3. (c)
10 0V = ∠ 2 rad secω =
( )Z 2L j L jω= = Ω ( )
1
1CZ j j
Cω
= − = − Ω
1 1.5 2Z j= +
( )( )
2
1 1
0.5 0.5
1 1
j
Z j
j
−
= = −
− +
( ) ( )0.283sin 2 81.9 Vv t t= −
1 2
10 0 10 0
4 36.9
2.5 36.9
TI
Z Z
∠ ∠
= = = ∠−
+ ∠
( )
1 4 36.9
2.83 8.1 A
1 1 2 45
TI I
j
∠ −
= = = ∠
− ∠−
( ) ( ) ( )2.83sin 2 8.1 Ai t t= +
+
−
+
−( )10sin2 Vt
1.5 Ω 1 H
0.5 F 1 Ω
( )Ti t ( )i t
( )v t
(a)
+
−
+
−
10 0∠ 1 Ω
2j Ω
1j− Ω
1.5 Ω
TI
I
V
(b)
+
−
+
−
TI
10 0∠ V
1Z
2Z
(c)
,
,
,
( ) ( )
2
1 2
0.5 0.5
10 0 10 0
1.5 2 0.5 0.5
Z j
V
Z Z j j
−
= × ∠ = × ∠
+ + + −
0.283 81.9 V= ∠−
Department of Electronic Engineering, NTUT21/41

22. 4
• (a) i(t)
1. (a) (b)
a KCL:
+
−
a
1
2
1
8F
+
−
v1
i t( )
4Ω( )10cos4 At v1
(a)
+
−
a
1
2
−j2 Ω
+
−
V1
I
4Ω( )10 0 A∠ V1
(b)
( )
1 1
2
1
4
8
C
j
Z j j
Cω
= − = − = − Ω
 
 
 
4rad secω =10 0I = ∠
1 1
1
2 10 0
2
V V
I
j
−
+ = ∠
−
( )1 1 10 0I j+ = ∠ ( )
10 0
7.07 45 A
1 1
I
j
∠
= = ∠ −
+
( ) ( )( )7.07cos 4 45 Ai t t= −
, ,
Department of Electronic Engineering, NTUT22/41

23. (Phasor Diagram)
• (Vector)
•
Vg= VR + VL + VC
(b) VR , VL , VC
V|VL| |VC|
+
−
Vg
R
+ −VR + −VL
+
−
VC
1
j Cω
j Lω
VL VC+
VR
(a) RLC
I
(a) RLC
0I I= ∠
I I I
(b)
VR
VL
VC
(a)
VL
VC (c) VR + VL + VC = Vg
Department of Electronic Engineering, NTUT23/41

24. Department of Electronic Engineering, NTUT24/41

25. 5 −
• (a) i(t)
1. (a) (b)
2.
V V + 3000I
KVL:
4.
3. KCL:
(a)
+−
+
−
1
k
2
Ω
2 kΩ
1
F
5
µ
( )3000 Vi t
( ) ( )3000v t i t+
2 kΩ
1
F
5
µ
( ) ( )4cos5000 Vsv t t=
( )sv t
I V
+
−
+−
1
k
2
Ω
3000V I+
3000I
4 0sV = ∠ ( )
2
1 2 k
5
j− Ω ( )2 1 kj− Ω
(b)
( ) ( )( ) ( )( )
4 3000
0
1 2 2 1 103103 1 2 103
2 5
V V V I
jj
− +
+ + =
−−
( )
4
1
103
2
V
I
−
=
3
24 10 53.1 A=24 53.1 mAI − °
= × ∠ ∠
( ) ( )24cos 5000 53.1 mAi t t °
= +
Department of Electronic Engineering, NTUT25/41
( )i t ( )v t

26. 6 −
• (a) i1(t) i2(t)
1. ω = 2 rad/s : (b)
(b)
+
− 1I 2I
3 Ω 2j Ω 1 Ω
1j Ω
2j− Ω
1 Ω18 0sV = ∠
2.
( ) 1 24 2 18 0j I I °
+ − = ∠
( )1 22 1 0I j I− + − =
2 2 0 AI °
= ∠
1 4.47 26.6 AI °
= ∠ −
3.
( )2 2sin2 Ai t t=
( ) ( )1 4.47sin 2 26.6 Ai t t °
= −
Department of Electronic Engineering, NTUT26/41
+
−
3 Ω
(a)
1 H 1 Ω
1 Ω
( ) ( )18sin 2 Vsv t t=
( )sv t ( )1i t
1
H
2
1
F
4
( )2i t

27. 7 −
• (a)
1. (b) I i(t)
I = I1 + I2
2. I1 I2 (c) (d)
3. (c)
+
−
+
−
5 Ω
(a)
4 Ω1 H( )i t
1
F
20
( )1sv t
( ) ( )1 20sin 2 Vsv t t= ( ) ( )2 10sin 2 Vsv t t=
( )2sv t
+
−
+
−
5 Ω 4 Ω2j Ω
20 0∠
10j− Ω
10 0∠
I
(b)
+
−
5 Ω 4 Ω2j Ω
20 0∠
10j− Ω
1I
(c)
+
−
5 Ω 4 Ω2j Ω
10j− Ω
10 0∠
2I
(d)
( ) ( )
1
10 4 2 10 4 2
5 5 10
10 4 2 4 8
j j j j
Z
j j j
− + − +
= + = + = Ω
− + + −
1
1
20 0 20 0
2 0 A
10
I
Z
° °
°∠ ∠
= = = ∠
Department of Electronic Engineering, NTUT27/41

28. 7 − ( )
3. (d)
:
4. I
( )
2
10 5
4 2 4 2 4 2 8
5 10
j
Z j j j
j
−
= + + = + + − =
− +
−
5 Ω 4 Ω2j Ω
10j− Ω
10 0∠
2I
(d)
3I
( )3
10 0 5
0 A
8 4
I
∠
= = ∠
( )2 3
10 2 1 5
153.4 A
5 10 2 2
j j
I I
j
− − +
= − = = ∠
−
( )1 2
2 1
2 0 1 0.5 1.12 26.6 A
2
j
I I I j
− +
= + = ∠ + = + = ∠
( ) ( ) ( )1.12sin 2 26.6 Ai t t= +
Department of Electronic Engineering, NTUT28/41

29. •
Zth
( ) Zth
: V Z Ioc th sc=
a
b
a
b
Zth
Eth Voc= +
−
a
b
a
b
IN Isc= Zth
Department of Electronic Engineering, NTUT29/41

30. 8 −
• (a) a-b
1. Voc
a
b
+
−
4 Ω 10j Ω
6 Ω
I
4j− Ω
(a)
( )6 0 V∠
a
b
+
−
4 Ω 10j Ω
4j− Ω
(b)
( )6 0 V∠
+
−
ocV
a
b
4 Ω 10j Ω
4j− Ω
(c)
thZ
4
6 0 3 3 4.24 45 V
4 4
oc
j
V j
j
° °−
= × ∠ = − = ∠ −
−
2. Zth
( )4 4
10 2 8 8.246 75.96
4 4
th
j
Z j j
j
°−
= + = + Ω = ∠ Ω
−
4.24 45 4.24 45
0.375 90 A
2 8 6 11.3 45
I
j
° °
°
°
∠ − ∠ −
= = = ∠ −
+ + ∠
3.
a
b
+
−
2 8j+ Ω
6 Ω
I
4.24 45 V°
∠−
Department of Electronic Engineering, NTUT30/41

31. 9 −
• (a) a-b I
1. Voc Zth
2. 6
a
b
+
−
4 Ω 10j Ω
6 Ω
I
4j− Ω
(a)
( )6 0 V∠
4.24 45
0.515 120.96 A
8.246 75.96
oc
th
V
I
Z
°
°
°
∠ −
= = = ∠ −
∠
( )2 8
0.515 120.96 0.375 90 A
2 8 6
j
I
j
° °+
= ∠ − = ∠ −
+ +
b
a
2 8j+ Ω 6 Ω
I
0.515 120.96 A°
∠ −
Department of Electronic Engineering, NTUT31/41

32. (AC Steady-State Power) (I)
• ( ) ( ) ( ) ( ) ( )1 1cos cosm mp t v t i t V t I tω φ ω φ= ⋅ = + × +
( ) ( ) ( )1 20 0
cos cos
T T
m mt t
W p t dt V t I t dtω φ ω φ
= =
= = + × +∫ ∫
( ) ( )1 2 1 2
0
1 1 1
sin 2 cos cos
2 2 2
T
m m m m
t
V I t t V I Tω φ φ φ φ θ
ω =
 
= − + + + − =  
( )1 2
1 1
cos cos cos cos cos
2 2 2 2
m m
m m m m rms rms
V IW
P V I V I V I S
T
φ φ θ θ θ θ= = − = = = =
( ) ( )1 2 1 20
1
cos 2 cos
2
T
m mt
V I t dtω φ φ φ φ
=
= + + + −  ∫
Peak RMS ( )
cosθ : Power Factor (PF)
( )1 2θ φ φ= −
( )
S:
+
−
v(t)
i(t)
Department of Electronic Engineering, NTUT32/41

33. (II)
V I ( )
(a)
( )
( )
1 1
2 2
V
A
rms
rms
V V V
I I I
φ φ
φ φ
= ∠ = ∠

= ∠ = ∠
1
cos cos
2
m mP VI V Iθ θ= =
cos
R
Z
θ = ( ) 2
cos
R
P VI Z I I I R
Z
θ
 
= = =  
 
I
+
−
V
Z
+
−
v(t)
i(t)
j
θ
(a)
Z R jX Z θ= + = ∠
Z
[ ]Re Z R=
[ ]Im Z X=θ Z
Department of Electronic Engineering, NTUT33/41

34. 10
• (a)
(a)
+
−
( )i t 10 Ω
0.5 H( )sv t
( ) ( )40sin 2 Vsv t t=
+
−
10 Ω
10j Ω( )
40
0 V
2
sV = ∠
(b)
I
40
0 V
2
V °
= ∠
( )( )10 20 0.5 10 10Z R j L j jω= + = + = +
10 2 45 , 45θ° °
= ∠ Ω =
( )
40 2 0
2 45 A
10 2 45
I
°
°
°
∠
= = ∠ −
∠
( )
40
cos 2 cos45 40W
2
rms rmsP V I θ ° 
= = = 
 
Department of Electronic Engineering, NTUT34/41

35. (Complex Power)
• S = P + jQ S VA P W
Q VAR
(a)
Q
S
θ
P
j
Q
S
θ
Pj
(b)
cosP S θ=( ):
sinQ S θ=( ):
:( )cosθ
:sinθ
( ) ( )rms rms
rms
V VV
I I
Z Z Z
φ
φ θ φ θ
θ
∠
= = = ∠ − = ∠ −
∠
0θ >
0θ <
( ) ( )rms rmsI I Iφ θ θ φ∗
= ∠ − − = ∠ −
rms rmsS S V I VIθ θ ∗
= ∠ = ∠ =
( )rms rms rms rmsVI V I V I Sφ θ φ θ θ∗
= ∠ ⋅ ∠ − = ∠ = ∠
ReP VI∗
 =   ImQ VI∗
 =  S VI∗
=
Department of Electronic Engineering, NTUT35/41

36. R L C
• R θ = 0
=
• ( )
θ +90 – 90
( )
( )
cos0 1PF = =
2
2 rms
rms rms rms
V
P S V I I R
R
= = = =
( )cos 90 0PF = ± =
j Lω
1
j
Cω
−
Department of Electronic Engineering, NTUT36/41

37. R
Z
j
θ
−XC
(a)
90 0θ− < <CZ R jX= − 0 90θ< <LZ R jX= +
I
V
θ
(b)
RC RL
j
XL
Z
R
θ
(c)
I
V
θ
(d)
cosθ θ RC RL
( )
I V
I V
Department of Electronic Engineering, NTUT37/41

38. 11
•
(1).
(2).
(3).
(4).
(5).
(6).
(7).
10 15Z °
= ∠ Ω 50 0V °
= ∠
50 0
5 15 A
10 15
V
I
Z
°
°
°
∠
= = = ∠ −
∠
5 15 AI∗ °
= ∠
( )( )50 0 5 15 250 15S VI∗ ° ° °
= = ∠ ∠ = ∠
( )250 VAS S= =
15θ °
=
cos15 0.966PF °
= =
( )cos 250cos15 241.5 WP S θ °
= = =
( )sin 250sin15 64.7 VARQ S θ °
= = =
Department of Electronic Engineering, NTUT38/41

39. • ZL Zth θL = –θth
ZL (
)
•
Ia
b
+
−
( Eth )
2
2
2 2
max
2 4
th th
L L L th th
th th
E E
P I R I R R
R R
 
= = = = 
 
( )L thZ Z=
L thθ θ= −
L thZ Z∗
=
thE
th th thZ Z θ= ∠
L L LZ Z θ= ∠
Department of Electronic Engineering, NTUT39/41

40. 12
•
W
( )( )500 0 2000 90
/ / 485.07 14.04 470.58 117.68
500 2000
th LZ R X j
j
° °
°
∠ ∠
= = = ∠ = +
+
470.58 117.68L L CZ R jX j= − = − Ω
2 2
max
120
7.65W
4 4 470.58
th
th
E
P
R
= = =
×
a
b
RL= ?
X = ?
XL= 2kΩ
R = 05. kΩ
RMS
+
−120 0thE = ∠
Department of Electronic Engineering, NTUT40/41

41. •
•
Department of Electronic Engineering, NTUT
j t
e ω
j t
e ω
jω
j t
e ω
jω
41/41