SISTEM SATU FASA
EET103 TEKNOLOGI ELEKTRIK
SISTEM SATU FASA
• Pengenalan dan ciri-ciri sistem satu
fasa
• Kebaikan dan keburukan sistem satu
fasa
• Pengiraan voltan, arus dan kuasa
PENGENALAN DAN CIRI-CIRI
SISTEM SATU FASA
• ALTERNATING CURRENT: a current or
voltage varies periodically in
magnitude and direction
PERBEZAAN AC DAN DC
Difference between DC and AC:
• DC: a direct flow of electrons through a
conductor such as a metal wire. A
battery or DC generator usually
provides a source of electrons and the
potential or voltage between the positive
(+) and negative (-) terminals.
• AC (Alternating Current): a back-and-forth
movement of electrons in a wire. When the
force of a negative (-) charge is at one end of
a wire and a positive (+) potential is at the
other end, the electrons in the wire will move
away from the (-) charge, just like in DC
electricity. But if the charges at the ends of
the wires are suddenly switched, the
electrons will reverse their direction.
AC voltage (1 cycle):
Mathematics of AC voltages
• An AC voltage v(t) can be described
mathematically as a function of time by the
following equation:
v(t )  A  sin( t )
v(t )  A  cos(t )
where
A is the amplitude in volts (also called the peak voltage),
ω is the angular frequency in radians per second
t is the time in seconds
• Since angular frequency is of more
interest to mathematicians than to
engineers, this is commonly rewritten
as:
v(t)  A  sin( 2f t )
Where f is the frequency in hertz (Hz).
AC voltage:
• Amplitude is the maximum voltage
reached by the signal. It is measured in
volts, V.
• Peak voltage (Vp) is another name for
amplitude.
• Peak-peak voltage (Vp-p) is twice the
peak voltage (amplitude). When reading
an oscilloscope trace it is usual to
measure peak-peak voltage.
RMS (effective) value:
• The root-mean-square (RMS) value of an alternating
current is the steady direct current which converts
electrical energy to other forms of energy in a given
resistance at the same rate as the alternating current
(AC).
• In power distribution work the AC voltage is nearly always
given in as a root-mean-square (rms) value, written Vrms.
For a sinusoidal voltage:
A
Vrms 
2
 0.707  A
Vrms = 0.707 × Vpeak and Vpeak = 1.414 × Vrms
Frequency
• The number of complete cycles of
alternating current or voltage completed
each second
• Frequency is always measured and
expressed in hertz (Hz).
Period
• An individual cycle of any sine wave represents
a definite amount of TIME.
• The time required to complete one cycle of a
waveform is called the PERIOD of the wave.
• The relationship between Period (T) in seconds
and frequency (f) in Hz:
1
T
f
1
f 
T
Frequency measurement:
Wavelength
• The time it takes for a sine wave to
complete one cycle is defined as the
period of the waveform. The distance
traveled by the sine wave during this
period is referred to as WAVELENGTH.
Wavelength measurement:
Alternating current values:
• PEAK AND PEAK-TO-PEAK VALUES:
Contoh (1)
Rajah berikut menunjukkan satu kitar
gelombang arus sinus. Berikan
persamaan bagi arus tersebut sebagai
fungsi masa.
i (mA)
170
20
0
t (ms)
10
-170
Penyelesaian:
• Persamaan umum bagi gelombang sinus
ialah
i(t) = Im sint
• Dari rajah gelombang diketahui;
Im = 170 mA dan
T = 20 ms = 0.02 s
• Maka, frekuensi dapat dikira seperti berikut:
f = 1/T = 1/0.02 = 50 Hz
• Seterusnya persamaan arus menjadi,
i(t) = Imsint
= 170 sin(2ft)
= 170 sin(100t) mA
Contoh (2)
Satu voltan ulang-alik bentuk sin
mempunyai frekuensi 2500 Hz dan nilai
puncak 15 V. Lukiskan bentuk satu kitar
gelombang voltan tersebut.
Penyelesaian
• Dari soalan, diketahui nilai-nilai berikut:
Vm = 15 V dan T = 0.4 ms
• Rajah umum gelombang sinus adalah
seperti berikut:
v (V)
15
0.4
0
t (ms)
0.2
-15
Contoh (3)
Satu voltan ulang-alik sinus diberikan
oleh persamaan:
v(t) = 156kos(800t)V
Lukiskan rajah satu kitar gelombang
voltan tersebut.
Penyelesaian:
Diberi, v(t) = Vm kos(t)
= 156 kos(800t) volt
Maka, nilai puncak voltan, Vm = 156
 = 2f = 800
f = 400
Tempoh bagi gelombang ini ialah;
T = 1/f = 1/400 = 2.5 ms
Gambarajah gelombang voltan:
v (V)
156
1.25
0
t (ms)
2.5
0.625
-156
1.875
INSTANTANEOUS VALUE
• The INSTANTANEOUS value of an
alternating voltage or current is the
value of voltage or current at one
particular instant
• There are actually an infinite number of
instantaneous values between zero and
the peak value.
AVERAGE VALUE
• The AVERAGE value of an alternating
current or voltage is the average of
ALL the INSTANTANEOUS values
during ONE alternation.
Vavg  0.637  V p
I avg  0.637  I p
Effective
and
average
value
Phase angle (sudut fasa):
• Apabila sesuatu gelombang sinus tidak
melepasi nilai kosong pada t=0, maka
persamaan gelombang tersebut mempunyai
sudut fasa yang perlu dipertimbangkan.
• Sudut fasa menunjukkan ANJAKAN sesuatu
gelombang dari sifar.
• Gelombang sinus boleh bermula dari apa-apa
nilai seperti berikut. Ia tidak semestinya
bermula dari sifar (atau dari nilai puncak bagi
fungsi kos).
Persamaan gelombang berikut:
y = Ymsin(x + a)
ao
adalah
sudut fasa
(phase
angle)
a
Ym
180
0
90
-Ym
270
x ()
360
Sine Waves In Phase (sefasa)
• When two sine waves are precisely in
step with one another, they are said to
be IN PHASE. To be in phase, the two
sine waves must go through their
maximum and minimum points at the
same time and in the same direction.
Voltage
and
current
are in
phase
Sine Waves Out Of Phase
(tidak sefasa)
• Voltage wave E1 which is considered to start at
0° (time one). As voltage wave E1 reaches its
positive peak, voltage wave E2 starts its rise
(time two). Since these voltage waves do not
go through their maximum and minimum points
at the same instant of time, a PHASE
DIFFERENCE exists between the two waves.
The two waves are said to be OUT OF PHASE.
Phase
differenc
e is 90o
GELOMBANG TIDAK SEFASA
• Dua gelombang sinus yang tidak sefasa
boleh diwakilkan dengan persamaan:
v(t) = Vm kost ; i(t) = Im kos(t + )
• Arus i(t) MENDAHULU (leading) voltan v(t)
dengan sudut .
• Dalam sebutan masa, arus mendahului voltan
dengan tempoh (T/360) saat.
• Boleh juga disebut voltan MENGEKOR
(lagging) arus dengan sudut 
Dua Gelombang Tidak Sefasa:
Θ is
phase
difference
v, i
Vm
v

i
Im
t
0
T
-Im
-Vm
Gelombang i
mencapai nilai
puncak di t1
Gelombang V
mencapai nilai
puncak di t2
CONTOH (4)
Lukiskan satu kitar gelombang arus sinus
yang diberikan oleh persamaan
i(t) = 70sin(8000t + 0.943 rad) mA.
Tandakan nilai-nilai kritikal.
PENYELESAIAN
Bentuk am gelombang arus ialah:
i(t) = Im sin(t + )
= 70 sin(8000t + 0.943 rad)
Dari persamaan diatas, nilai-nilai
kritikal ialah:
Im = 70;
 = 2f = 8000;
f = 4000 Hz = 4 kHz;
Maka,
T = 1/f = 1/4000 = 0.25 ms;
Dan,
 = 0.943 rad = 54
180
1rad 


Rajah gelombang sinus bagi
arus:
i (mA)
54
70
57
0
0.125
t (ms)
0.25
-70
SISTEM SATU FASA
• Pengenalan dan ciri-ciri sistem satu
fasa
• Kebaikan dan keburukan sistem satu
fasa
• Pengiraan voltan, arus dan kuasa
KEBAIKAN DAN KEBURUKAN
SISTEM SATU FASA
• Direct current has several disadvantages
compared to alternating current. Direct
current must be generated at the voltage
level required by the load. Alternating
current, however, can be generated at a
high level and stepped down at the
consumer end (through the use of a
transformer) to whatever voltage level is
required by the load.
• The major advantage that AC electricity has
over DC is that AC voltages can be
transformed to higher or lower voltages. This
means that the high voltages used to send
electricity over great distances from the
power station could be reduced to a safer
voltage for use in the house.
• This is done by the use of a transformer. This
device uses properties of AC electromagnets
to change the voltages.
• It is easy to convert AC to DC but expensive
to convert DC to AC.
SISTEM SATU FASA
• Pengenalan dan ciri-ciri sistem satu
fasa
• Kebaikan dan keburukan sistem satu
fasa
• Pengiraan voltan, arus dan kuasa
COMPLEX NUMBER
1. POLAR FORM
Z= Z 
Z
:magnitude of Z
 : angle of Z
2. RECTANGULAR FORM
Z = R + jX
R: real value of Z
X: imaginary value of Z
 j : operator valued √-1
3. EXPONENTIAL FORM
Z = rej
 r : magnitude
 : angle
RECTANGULAR
FORM TO POLAR FORM
Z  R  jX L @ R  jX c
Z 
R  X : magnitud
2
   tan
2
1
X

 : sudut
 R 
Z  Z 
POLAR FORM TO
RECTANGULAR FORM
Z  Z 
R Z cosθ : real value
X j Z sin  : imaginary value
Z  R  jX L @ R  jX c
COMPLEX NUMBER
ALGEBRA OPERATION
• To ADD two number:
–Transform to Rectangular
• To MULTIPLY two number:
–Transform to Polar
• To SUBTRACT two number:
–Transform to Polar
SINUSOID-PHASOR TRANSFORMATION
Time domain
representation
Frequency domain
representation
Vm cos(t   )
Vm 
Vm sin( t   )
Vm   90
I m cos(t   )
I m 
I m sin( t   )
I m   90


Ohm’s Law in AC
circuit
RESISTOR IN AC
IN FREQUENCY DOMAIN:
VR  R I
 R I m i
 Vm i
V
V and I waveform for resistor :
v, i
IN PHASE
v
i
t
NO PHASE DIFFERENCE BETWEEN
VOLTAGE AND CURRENT
INDUCTOR
• IN TIME DOMAIN:
di
vL  L
dt
• IN FREQUENCY DOMAIN:
VL  jLI me
ji
 jLI
GELOMBANG LITAR L
v, i
v
i
t
90º
ARUS MENGEKOR VOLTAN (ELI)
BEZA FASA SEBANYAK 90º = (1/4)T = 1/4f
KAPASITOR
• IN TIME DOMAIN:
1
v C   i dt
C
• IN FREQUENCY DOMAIN:
1
ji
VC 
Ime
jC
1

I
jC
GELOMBANG LITAR C
v, i
v
i
t
Arus mendulu voltan sebanyak 90° (ICE)
GELOMBANG v-i BAGI
R,L,C v v, i
V,I
SEFASA
i
t
I MENGEKOR V
SBYK 90º
v
v, i
v, i
v
i
t
90º
I MENDAHULU
V SBYK 90º
i
t
Penentuan
Mendulu/Mengekor
ARUS MENGEKOR VOLTAN
Dengan menganggap gelombang Arus sbg RUJUKAN:
Jika gelombang voltan mpy sudut fasa positif
Terletak disebelah kiri gelombang arus
Gelombang voltan mendulu gelombang arus.
ARUS MENDULU VOLTAN
Jika gelombang voltan mpy sudut fasa negatif
Terletak disebelah kanan gelombang arus
gelombang voltan mengekor gelombang arus.
AC CIRCUIT
ELEMENT
REACTANCE AND
IMPEDANCE
REACTANCE
All elements in AC circuit
(resistor, inductor and
capacitor) should have same
unit before you do the analysis.
• Inductor value (in henry) and capasitor
value (in farad) must be transform into
ohms ().
• Inductance and capacitance value in ohms
are known as reactance which is inductive
reactance for inductor, and capacitive
reactance for capacitor.
• Symbol for inductive reactance is XL
and for capacitive reactance is Xc..
• Formula to get XL dan Xc:
X L  jL  j2fL
HAFAL!
1
1
1
XC 
 j
 j
j C
C
2fC
CAPACITOR
• Series capacitors:
Ceq 
1
1
1
1

 .. 
C1 C2
CN
X CT  X C1  X C2  ..  X C N
• Parallel capacitors:
Ceq  C1  C2  ..  C N
X CT 
1
1
1
1

 .. 
X C1 X C2
X CN
INDUCTOR
• Series inductors:
Leq  L1  L2  ..  LN
X LT  X L1  X L2  ..  X LN
• Parallel inductors: L 
eq
X LT 
1
1 1
1
  .. 
L1 L2
LN
1
1
1
1

 .. 
X L1 X L2
X LN
IMPEDANCE
Impedance is a element
connecting the resistance,
inductive reactance and
capacitive reactance in
time domain.
• Impedance is represent by Z symbol:
V V 
Z    
I  I
(ohms)
•  is angle between voltage and current
• Impedance is complex number
because of magnitude and angle
• Resistance impedance, ZR:
VR VR 
ZR 

I
I
VR

0
I
R
• Inductive impedance, ZL

VL
VL 0
ZL 


I
I  90
VL


90
I
 jX L
• Capacitive impedance, Zc
Vc Vc 0
ZC 

I
I90
Vc


  90
I
 1
1

  j
jX c
 Xc



KIRCHHOFF LAW IN AC ANALYSIS
• Transform circuit into frequency domain
using these equation.
• Use KVL and KCL…
I  Ime
ji
 I m i
A
VR  RI  RI mi V
VL  X L I  LI m(i  90)

V
1

VC  X C I 
I m (i  90)
C
V
R-L CIRCUIT IN FREQUENCY
DOMAIN
R
+ VR -
Vs
+IR
j
V

VL
=
R
L
V
-m


+ L
VL
-
CONTOH (1) : LITAR RL
Given Vs = 40 cos (2000t - 60º). Find,
a) Phasor voltage across R;
b) Phasor voltage across L;
c) Current, i(t) expression
R=1000
+ VR -
Vs
+
VL
-
L=0.25H
Step 1: Circuit in frequency domain
R=1000
Vs=40-60°
+ VR -
+
VL
-
j L=j1570.8
Step 2: Find circuit impedance
Z  R  jL
 1000  j1570.8
 1570.8 
 1000  1570.8  tan 

 1000 

 1862.157.5

2
2
1
Step 3: Calculate current
V
I 
Z

40  60

1862.157.5
40



  60  57.5
1862.1
 0.0215  117.5 A
 21.5  117.5
mA
Phasor current is;
I  21.5  117.5 mA
atau

I  21.5e
 j117.5
mA
(a) Phasor voltage across R,
VR  RI
 1000  21.5  117.5 mA

 21.5  117.5 V

(b) Phasor voltage across L,
VL   jL I
 ( j2000  0.25)  (21.5  117.5 )mA

 (1570.890 )  (21.5  117.5 )mA


 33.8  27.5 V

(c) Current, i(t) expression:
»» Transform I in time domain
I  21.5  117.5 mA

i  21.5cos(2000t  117.5 )mA

CONTOH (2) : LITAR RC
Given VC = 60 cos(1000t - 33º). Find,
a) Phasor current, I
b) Phasor voltage across R
c) Voltage source, vs(t).
R  2200
 VR 
Vs

VC

Step 1: Circuit in frequency
domain
R  2200
 VR 
Vs

VC
1
  j1447
jC

VC  60  33
Step 2: Circuit impedance (Z)
1
Z
Rj
 2200  j1447
C
magnitud dan sudut Z:
Z  2200  (1447)  2633
2
2
  1447 

  tan 
  33.3
 2200 
1
Step 3: Find I, VR,VS
(a)
Phasor current;
VC
I 
j 1
 C 
(b)
60  33



0
.
0415

57
A

1447  90
Phasor voltage across R;
VR  IR
 2200  0.0415 57  A
 91.357 V

(c)
Source phasor voltage;
V  ZI
 2633  33.3   0.041557 A


 109.323.7 V

Transform V into time domain;
v  109.3 cos(10000t  23.7 )V

IMPEDANCE
TRIANGLE
IMPEDANCE TRIANGLE
• We also can get impedance by using
impedance triangle:
Galangan,Z 

Perintang, R
Reaktan, X
• Dari segitiga galangan;
Z  R X
2
2
• Ini bermakna, dengan mengetahui
Galangan (Z), kita boleh juga
mengetahui nilai perintang (R) dan
Reaktan (X).
HUBUNGAN Z, R DAN X
• Hubungan Z dan R;
R
kosθ 
Z
• Hubungan Z dan X;
X
sin θ 
Z
AC Power
Calculation
POWER IN AC CIRCUIT
i( t )

v( t )

Source
Load Z
INSTANTENAOUS POWER
(KUASA SEKETIKA)
• Kuasa seketika yg diserap oleh setiap
peranti elektrik adalah hasil darab
Voltan seketika yg merintanginya dan
Arus seketika yang melaluinya.
p( t )  v( t )  i ( t )
RMS
VALUE
• Semua nilai Voltan dan Arus dalam
pengiraan kuasa adalah menggunakan
nilai RMS (root mean square). Iaitu:
Vrms
Vm

2
I rms
Im

2
• Dimana, Vm dan Im adalah Nilai Puncak
(peak) bagi voltan dan arus.
1. AVERAGE POWER (KUASA PURATA)
• Kuasa purata atau kuasa sebenar
ditakrifkan sebagai:
Vm I m
Kuasa Purata  P 

cos 
2 2
Vm I m
P
cos  ( Watt )
2
• Unit bagi kuasa purata ialah Watt.
• Kuasa purata pada Perintang boleh juga
diwakili oleh persamaan berikut:
Vm I m
Vm I m
P
cos  
 Vrms I rms
2
2
• Kita tahu V=IR, maka kuasa purata turut
diwakili oleh:
2
P  Vrms I rms
Vrms
2

 I rms R
R rms
• Kuasa purata ialah kuasa berguna yg.
disebabkan oleh elemen Perintang.
• Dengan itu, kuasa purata bagi beban
reaktif (L atau C) adalah Kosong.
•  ialah Sudut antara Voltan dan Arus,
ATAU dikenali juga sebagai Sudut
Galangan,Z.
Nilai  diperolehi seperti berikut:
V  
Z
 Z - 
I
Z  Z
CONTOH (1)
Voltan sinus mempunyai amplitud
maksimum 625 Volt dikenakan pada
terminal yg mempunyai Perintang 50.
Dapatkan kuasa purata yang dihantar
kepada perintang tersebut.
PENYELESAIAN
• Nilai rms: Vrms
625

 441.94 V
2
• Kuasa purata bagi perintang diperolehi:

V
441.94 
P

 3906.25 W
R
50
2
2
2. REACTIVE POWER (KUASA REAKTIF)
• Kuasa Reaktif ditakrifkan sebagai:
Vm I m
Kuasa reaktif  Q 

sin 
2 2
Vm I m

sin 
2
Q  Vrms I rms sin  (Var )
• Kuasa reaktif juga merupakan kuasa yg
disimpan oleh elemen reaktif iaitu L atau C.
• Maka, kuasa reaktif boleh juga dicari
dengan:
2
L
V
Q L  I X L  IVL 
2
XL
Dan ,
Q C  I X C  IVC 
2
2
C
V
XC
CONTOH (2)
Diberi v= 100 kos (t + 15º) dan i= 4 sin (t - 15º),
Pada terminal rangkaian, dapatkan:
(a)Kuasa purata
(b)Kuasa reaktif
i
+
v
-
Rangkaian
PENYELESAIAN
• Tukarkan persamaan arus, i dalam bentuk
kos:
i= 4 kos (t - 105º)
• Dari persamaan Kuasa Purata:
Vm I m
P
cosθ (Watt)
2
• Maka,
1
P  1004 cos15   105 W
2
 100 W
• Dari pers. Kuasa reaktif:
Vm I m
Q
sin θ (Var)
2
• Maka,
1
P  1004sin 15   105 W
2
 173.21 Var
• Nilai kuasa purata, P= -100 W:
Bermaksud kuasa purata telah dihantar
balik dari beban kepada terminal bekalan.
• Nilai kuasa reaktif, Q=173.21 Var:
Bermaksud kuasa reaktif telah diserap
oleh beban.
AC POWER FORMULA
• AC power calculation can be done by using
these formula:
Kuasa Purata  P  Vrms I rms cos  ( Watt )
Kuasa Re aktif  Q  Vrms I rms sin  (Var )