This document provides an overview of polyphase circuit analysis and sinusoidal steady-state analysis. It includes: - Basic information about the author Faridah Hanim Mohd Noh from the University of Tun Hussein Onn Malaysia. - An introduction to sinusoidal steady-state analysis, which is important for analyzing AC circuits and equipment ratings. Circuits are analyzed after transients have decayed. - Descriptions of sinusoidal sources and definitions of key concepts like period, frequency, angular frequency, phase angle, and RMS value as they relate to sinusoidal steady-state analysis. - Examples of applying these concepts to calculate values for circuits involving resistors, inductors, and capacitors connected

1. POLYPHASE CIRCUIT
ANALYSIS
BEF23803
FARIDAH HANIM MOHD NOH
Department of Electrical Power Engineering
Faculty of Electrical and Electronic Engineering
Universiti Tun Hussein Onn Malaysia

2. Basic Information
 Name: Faridah Hanim Binti Mohd Noh
 Room: Lecturer Building (C16-001-08)
 Contact: 07-4537388 / 0196562217, hanim@uthm.edu.my

3. Chapter 1
Sinusoidal Steady-State Analysis

4. BEF 23803
Introduction
• Many important applications in electrical
engineering involve AC (Sinusoidal) currents and
voltages.
• Electric power is distributed worldwide using AC
signals with frequency of either 50 or 60 Hz.
• The transmission and reception of electromagnetic
waves necessary in any wireless application
involves AC signals.
• Circuit response determined using steady-state
AC analysis are those found after the transients
responses have decayed to zero.

5. Introduction
• It is important for all engineers to be able to
analyze steady-state conditions, because it is the
steady-state that determines continuous ratings
of the equipment.
• Failure to correctly size equipment results in
either:
 Excessive expenditure of capital when the
plant is oversized,
 Damage when the plant is oversized
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6. Sinusoidal Sources
• A sinusoidal voltage source produces a voltage that varies
sinusoidally with time
•A sinusoidal current produce a current that varies sinusoidally with
time
•A sinusoidal varying function can be expressed in sine or cosine
function
•For this course – use cosine function

7. Sinusoidal Voltage
Sinusoidal Sources
)
cos(
)
( P 
 
 t
V
t
v
f
frequency
Angular

 2

T
f
Frequency
1

angle
Phase


The + shifts the time function to the left and - shifts
the function to the right.
 
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Φ


T
Period

8. Period of Sine Wave
• Period – the time
required for a given
sine wave to
complete one full
cycle.
• The symbol: T
• The period of a given
sine wave is the
same for each cycle.
• Unit: s (second)

9. Frequency of a Sine Wave
•Frequency is the
number of cycles that a
sine wave completes in
one seconds.
•Symbol : f
•Unit : Hertz (Hz)
•1 Hz = 1 cycle per
second
•60 Hz = 60 cycles per
second
•Higher number of cycles
per second means that
the system have higher
frequency value.

10. Angular Frequency
•ω represent the angular
velocity of the sine wave
•Unit : radian/second
s
rad
T
f /
2
2


 


11. Phase Angle
•Symbol: Φ
•It determines the value
of the sinusoidal
function at t = 0.
•Changing Φ have no
effect on VP and ω
•ωt and Φ should be in
the same unit
•ωt usually in unit radian
•Φ usually in degree

12. • The root mean square (rms) value of a periodic function is
simply the “square root of the mean value of the square
function”.
Sinusoidal Sources
 
2
)
cos(
1 0
0
2
P
T
t
t
P
rms
V
dt
t
V
T
V


 



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13. • Example 1
An electrical appliance is a pure resistor and is rated as:
120 V, 8 A, 50 Hz. Determine:
a) Peak voltage and current,
b) Time between voltage zeros,
c) Radian frequency,
d) Resistance, e
e) Power dissipated.
Sinusoidal Sources
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14. • Example 1 (Solution)
a) The stated voltage and current are rms because this is the
default and we are not told otherwise.
Then: VP = √2*Vrms = √2 x 120 = 169.7 V.
And: IP = √2*Irms = √2 x 8 = 11.3 A.
b) The time between voltage zeros is the periodic time,
T=1/f = 20 msec
c) The radian frequency is, ω = 2πf = 2 x π x 50 = 314.16
rad/s.
d) Resistance is given by Ohm’s Law as: R= 120/8 = 15 Ω
e) In ac circuits: P = Irms2R = 82 x 15 = 960 W.
Sinusoidal Sources
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15. Sinusoidal Response
RL Circuit
• The voltage source is sinusoidal, and applying
KVL to the circuit,
• When the switch is closed there will be a
transient response which eventually settles
down to some steady state value.
dt
di
L
Ri
t
V
V
V
V
P
L
R
s





)
cos( 

BEF 23803
)
cos(
)
cos(
2
2
2
)
/
(
2
2
2














 
t
L
R
V
e
L
R
V
i
m
t
L
R
m
Transient
component
Steady-state
component

16. Sinusoidal Response
RL Circuit
• Characteristic of steady state solution:
 Steady state solution is a sinusoidal function
 The frequency of response signal is identical
to the source signal. This is true in a linear
circuit
 For maximum of the steady state response
in differs from maximum amplitude of the
source. The maximum amplitude of the
response is
The amplitude of the source = Vp.
 The phase angle difference of the response
is difference with the source.
BEF 23803
R
L

 1
tan

2
2
2
L
R
V
I p
p




17. Phasors Analysis
Passive Circuit Elements
• Dealing with AC circuits, Ohm’s Law must be re-stated in
terms of phasors.
• Since, the voltage and current are complex numbers, the
quantity is call impedance, Z which consist of real part due
to the resistance and imaginary part is associated with
inductance/capacitance.
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I
V
Z 

18. Phasors Analysis
Impedance of Resistor
• If the waveform switched on at peak values ( and the
peak is where I is Irms )
• Therefore,
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0


I
2
t
I
R
t
v
t
I
t
i
R
R


cos
2
)
(
cos
2
)
(


Converting into rms phasors


0
0




RI
V
I
I
R
R

19. Phasors Analysis
Impedance of an Inductor
• For ideal inductor the voltage and current are varies with
time.
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)
90
cos(
2
)
(
)
90
cos(
2
sin
2
,
cos
2
)
(









t
I
L
t
v
t
I
t
I
dt
di
Then
t
I
t
i
L
L
L







Converting into rms phasors


90
0




LI
V
I
I
L
L

• Therefore,

20. Phasors Analysis
Impedance of a Capacitor
• For ideal capacitor the voltage and current are varies with
time.
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)
90
cos(
2
)
(
)
90
cos(
2
sin
2
,
cos
2
)
(









t
V
C
t
i
And
t
V
t
V
dt
dv
Then
t
V
t
v
C
c
C







Converting into rms phasors


90
0




CV
I
V
V
C
C

• Therefore,

21. Example 2
• An inductance of 35 mH has a winding resistance of 7.2 ohm
that appears in series with the inductance. It is connected
across a 240 V, 50 Hz supply. Determine:
a) Impedance
b) Current drawn
Phasors Analysis
BEF 23803

22. Example 2 (Solution)
a)
b)
Phasors Analysis
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

















79
.
56
15
.
13
11
2
.
7
11
035
.
0
16
.
314
/
16
.
314
50
2
2
Z
or
j
jX
R
Z
and
X
s
rad
f
L
L



)
(
79
.
56
25
.
18
79
.
56
15
.
13
0
240
value
rms
A
Z
V
I 








23. Example 3
• A 20 uF capacitor is connected in parallel with a 100 ohm
resistor and the combination is placed across the supply of
240 V, 50 Hz. Determine:
a) Impedance
b) Current drawn
Phasors Analysis
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24. Example 3 (Solution)
a)
b)
Phasors Analysis
BEF 23803


























13
.
32
68
.
84
45
72
.
71
,
10
28
.
6
01
.
0
15
.
159
1
100
1
1
1
1
,
15
.
159
10
20
16
.
314
1
/
16
.
314
50
2
2
3
6
Z
or
j
Z
Therefore
j
j
jX
R
Z
Hence
X
s
rad
f
C
C



)
(
13
.
32
8
.
2
13
.
32
68
.
84
0
240
value
rms
A
Z
V
I 






