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
BEF 23803
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.
BEF 23803
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.
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
BEF 23803
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
BEF 23803
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
BEF 23803
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.
BEF 23803
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,
BEF 23803
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.
BEF 23803
)
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.
BEF 23803
)
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
BEF 23803
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
BEF 23803
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
