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Wednesday, 20 February 2019

4:32 PM

<<ELEC2134_Circuits1_AlternatingCurrent.pdf>>

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ELEC2134: Circuits and Signals
Professor Eliathamby Ambikairajah
Alternating Current (AC)
AC current and voltage
Phase angle and Phase difference
Peak and peak-peak values
Average value
Root Mean Square (RMS) value of a sinusoid
RMS current calculation for a sinusoid
RMS calculation for general periodic functions

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AC fundamentals
In Direct Current (DC), the
electric charge flows only in
one direction.
In Alternating Current (AC),
the flow of electric charge
periodically reverses direction.
2
Prof E Ambikairajah, UNSW, Australia
Electronsflowfromnegativeto
positive
+
-
+
-
+
-
AC
AC
First this way
Then this way

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AC is more efficient and economical to transmit over
long distances
AC voltage may be increased or decreased with a
transformer
The power losses (
P
L
) in a conductor are a product of
the square of the current and the resistance of the
conductor
When transmitting a fixed power through a power
line, if the current is doubled, the power loss will be
four times greater.
The power transmitted (
P
T
) is equal to the product of
the current and the voltage
The same amount of power can be transmitted with a
lower current by increasing the voltage
Use of a higher voltage leads to significantly more
efficient transmission of power.
3
Prof E Ambikairajah, UNSW, Australia
AC transmission
Note:
DC cannot be stepped-up
or stepped-down by a
transformer
Note:
High voltage DC
transmission (HVDC) is
becoming available but tend
to be more expensive, less
efficient over short distances.

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An alternating current or voltage wave can
take several different forms but we will only
consider the
sine wave
(
sinusoidal
waveform
)
Therefore
can be re-written as
The sine wave (or sinusoid) has a pattern
that repeats. Hence it is a periodic wave.
Prof E Ambikairajah, UNSW, Australia
Characteristics of Alternating Current and voltage (AC)
v(t)
Time(t
)
-V
m
T
T/2
V
m
Onecycle
Period(T)
Positive
halfcycle
Negative
halfcycle
Positive
Peakvoltage
Instantaneous
voltage
Peak
voltage
Frequency
Time
Frequency
Angular Velocity
=

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5
Prof E Ambikairajah, UNSW, Australia
Periodic functions
Any function or waveform that satisfies
is a periodic function where
T
is the period.
Units of periodic time (
T
): seconds (s), millisecond (ms) or microsecond (
) etc.
The time taken for the waveform to repeat themselves is known as the period (T)
[Hertz: cycles per second, 1Hz = 1 cycle per second]
The electricity at your
home or university is an
alternating current (AC)
and the current is a sine
wave with a frequency
of 50 Hz.
What is the period of
this alternating current?

Embedded file printout ELEC2134_Circuits1_AlternatingCurrent.pdf Machine generated alternative text:
A sinusoidal waveform may or may
not always start at
t
= 0. It can start
earlier or later than
t
= 0.
A
delay
or
advance
in the waveform is
known as the
phase angle
(
). The
units of
are given in degrees
or
radians
.
A general expression for a sinusoidal
waveform is
6
Prof E Ambikairajah, UNSW, Australia
Phase Angle
Two sinusoids (see above
diagram) with a phase
difference
is chosen as the
reference and is shown to
begin at 0
o
.
leads
waveform
by
degrees. Therefore it
is written as
Note:
If
is not zero, the
waveforms A and B
are said to be
out of phase
. If
= 0, then the
waveforms A and B
are said to be
in phase
Note that phase angle
can take any
values between 0 to 360 degrees

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Prof E Ambikairajah, UNSW, Australia
Note:
Waveform
B
leads waveform
C
by 105
(45+60) degrees
Waveform
A
is chosen as the reference and is
shown to begin at 0
o
.
Waveform
B
leads waveform
A
by 45 degrees
Waveform
C
lags waveform
A
by 60 degrees
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Phase Difference
Note:
A sinusoid can also be expressed as a cosine function.

m
v(t)Vcos(t)
=
)
=
)
Useful Trigonometric Identities

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Prof E Ambikairajah, UNSW, Australia
Find the frequency
f
, period
T
of the voltage waveforms given below. Also find the
phase angle between
v
1
(
t
) and
v
2
(
t
)
.
Does
v
1
(
t
) lead or lag
v
2
(
t
) ?
8
Example 1.1
=
= 0.02 sec = 20 ms
Phase angle {
} between
This is similar to part (i) above. So
leads
by 60
degrees.
-

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Prof E Ambikairajah, UNSW, Australia
Find the phase angle between
and
. Does
lead or lag
?
9
Example 1.2
Therefore
leads
by 155
Therefore
lags
by 205
(155
+
50
which is the same as 360
-155
)
OR
155

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Prof E Ambikairajah, UNSW, Australia
Q1:
Consider the sinusoids
v
1
(
t
)
=
10 cos(200
t
+ 45
)and
v
2
(
t
)
=
5 sin(200
t
+ 15
)
. Determine the time by which
v
2
(
t
)
is advanced or delayed with respect to v
1
(
t
)
.
10
Exercise 1.1
Ans:
is delayed with respect to
by 10.47 ms
Q2
:
A sinusoidal current is given as
i
(
t
)
=
100 cos(5000
t -
135
)
mA
Determine the period T and the time t
0
at which the first
positive peak occurs.
Ans: T = 0.4 ms,
t
0
= 0.15 ms
Q3:
Express the voltage shown in Figure 1 in general form
given below:
v
Volts
where
Figure 1
Ans

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Prof E Ambikairajah, UNSW, Australia
From a sinusoidal waveform we can measure the following values:
Peak value
Peak-peak value
Average value
Root Mean Square (RMS) value
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Measurements of AC quantities

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Prof E Ambikairajah, UNSW, Australia
Thepeakvalueofasinewaveis
knownasthemaximumvoltage(or
current)ofthewaveform.
Peakvoltage(orcurrent)occursat
twodifferentpositionsinacycle
(onepositivepeakandone
negativepeak).
Thepositiveandnegativepeaks
haveequalmagnitudes.
Thepositivepeakoccursat90
and
negativepeakoccursat270
.
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Peak Value
Peak
value
t
90
0
270
0
Peak-Peak Value
The peak-to-peak (pk-pk) value of
an AC waveform is the total height
between opposite peaks.

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Prof E Ambikairajah, UNSW, Australia
13
Average Value
Average value
or
of a sine
wave is the average of all its
instantaneous values over a full
cycle.
=
=
Sincethe positive and negative
alternations are identical, the average is
zero.
Thus, the average value of a sine wave,
taken over an entire cycle is zero.
t
V
m
Period=T
V(t)
v(
=
As expected the net area of the positive and negative half cycles is zero.

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Prof E Ambikairajah, UNSW, Australia
14
Example 1.3
What is the average value for each of the following waveforms?
Period (T)
Note: Average value
=
Period (T)
=
=

Embedded file printout ELEC2134_Circuits1_AlternatingCurrent.pdf Machine generated alternative text:
V
m
t
Period(T)
v(t)
V
ave
Prof E Ambikairajah, UNSW, Australia
15
Example 1.4
Find the average value for the following
half-wave rectified sine wave.
Find the average value for the
following full-wave rectified sine wave.
=
t
Period(T)
v(t)
V
ave

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Prof E Ambikairajah, UNSW, Australia
The root mean square value or
rms
value is actually a measure of
the heating effect of a sine wave.
It can be defined as the equivalent value of DC that will produce
the same amount of heat (power) in a load (
R
L
) as a given sine
wave.
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R
oot
M
ean
S
quare Value (RMS)
R
L
Vs
v
s
R
L
volts
t
120V
t
Volts
170V
DC Voltage
DC circuit
AC voltage
AC circuit
A sine wave with a peak value of 170V ( = 120 V rms) is applied across a resistor
(
R
L
), a certain amount of heat is produced.
If a DC voltage with a steady-state value of 120 V is applied to the resistor (
R
L
), the
same amount of heat is produced.
Note:
When rms
value of the
sinusoidal source
voltage is equal to
the DC source
voltage, then the
AC source will
supply the same
average power as
the DC source.

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Prof E Ambikairajah, UNSW, Australia
The
of a periodic current waveform
i
(
t
) is defined as the
current which produces heat (power) in a given resistor R at the
same average rate as a direct current
.
Consider a periodic current waveform
RMS value or Effective Value of a periodic Waveform
i
(
t
)
instantaneous current
p
(
t
)
instantaneous power
R
+
-
i(t)
Mean
Square
Root

Machine generated alternative text:
Prof E Ambikairajah, UNSW, Australia
RMS current calculation for a sinusoid
=
=
We have
T
=
Note:
We usually quote most AC
voltages as RMS rather than peak
value. If
is 240 V, then
Note:
Australia: 240
;
UK: 220
; USA: 110
;

Machine generated alternative text:
Prof E Ambikairajah, UNSW, Australia
So the
rms
value of a
sine
wave of voltage or current is :
=
and
=
RMS value of a
sinusoid
Then the RMS voltage (
V
rms
) of a
sinusoidal waveform
is determined by
multiplying the peak voltage value by
0.707
(equivalent to dividing it by the
square root of two)
The RMS voltage or current, which is also known as the effective value, of a
sinusoid depends on the magnitude of the waveform and is independent of
the waveform frequency and its phase angle.
v(

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Prof E Ambikairajah, UNSW, Australia
20
Example 1.5
Derive an expression for the rmsvalue of a sawtooth voltage waveform shown
below.
t
0
v
m
T
T/2
v(t)
-v
m
-T
Note:
Consider a waveform consisting of a sum of sinusoids of different frequencies: The rmsvalue is
equal to the square root of the sum of the squares of the rmsvalues of each sinusoid. Assume that
the ac current
i
(
t
) is given by
i
+
Therefore

Machine generated alternative text:
Prof E Ambikairajah, UNSW, Australia
20ohms
+
-
v(
v(
20ohms
25V
Exercise 1.6
Q1:
Derive an expression for the rmsvalue of a triangular
voltage waveform given below:
Q2
:
A sine wave has a peak value of 45 V. Calculate its rms
and peak-to-peak values.
Q3:
The sources in the AC and DC circuits shown below
supply equal values of average power. Determine the values
of average power and of the amplitude of the sinusoidal
voltage.
Ans: 31.8 V and 90 V pk-pk
Ans: P
ave
= 31.25 W, V
m
= 35.355 V
0
T
t
T/2
-T/2
-v
m
v
m
v(t)

Machine generated alternative text:
[1] Alexander, C. K., & Sadiku
Circuits
th
edition, McGraw Hill.
th
edition, Wiley & sons.
School of Electrical engineering and Telecommunications, UNSW,
Australia.
[4] Hambley
College Publishing.
References

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