Ee1 chapter11 ac_circuits

IT2001PA
Engineering Essentials (1/2)

Chapter 11 – Alternating Current Circuits

Lecturer Name
lecturer_email@ite.edu.sg
Aug 17, 2012
Contact Number


Lesson Objectives
Upon completion of this topic, you should be able to:
 Explain and calculate the fundamentals of alternating
current waveform.

2
IT2001PA Engineering Essentials (1/2)


Specific Objectives
Students should be able to :
 Describe the basic construction of a simple alternator.
 Describe the characteristics of an alternating quantity as
fluctuating over a given period of time.
 Define the following terms with reference to an alternating
quantity :
 Frequency
 Average value
 Instantaneous value
 Maximum value



Specific Objectives
Students should be able to :
 State that the frequency of the generated emf is
proportional to the speed and number of poles of the
alternator.
 Explain the term Root Mean Square (RMS) value with
reference to an alternating current.
 State the following equations for a sinusoidal waveform :
RMS value = 0.707 x Maximum value
Average value = 0.637 x Maximum value



Alternating Waveform
 Alternating Current (ac)
current that is continuously reversing
direction, alternately flowing in one
direction and then in the other.
 Alternating Voltage
can be similarly described.
The designation ac is normally applied to
both current and voltage.
5


Alternating Waveform
 Vertical axis : current (I) or voltage (V)

Horizontal axis : time (t)
+V +I

t t

-V -I

6


Why not Direct Current (DC)?
 Problems of using DC as +V
power distribution system:
 suffers from rapid power loss
in the wires due to their t
resistance, which dissipates
energy as heat
 DC power stations had -V
useful ranges of about two
kilometers
 Once generated, DC power
cannot be modified

7


Power Distribution System
V=RI
 
1) 2) 3)
To prevent risk of To avoid serious For the safety of the
sparks and short heating effects, user, the voltage
circuits in the present at high needs to be low in the
generator itself, current, it needs to be home.
power needs to be transmitted at the
generated with low lowest current V low  I is high
voltages. practical.

I low  V is high  DC is not suitable
V low  I is high

8


Advantages of A.C.

--- readily available from generators
and power supply sockets in homes
and workshops.
--- can be stepped up or down by the
use of a transformer.
--- for transmission purpose.

9


Alternator

An alternator is a machine that converts
mechanical energy to electrical energy.
A simple alternator basically consists of
two parts:

a) coils or windings
b) magnetic poles

10


Generation of Alternating Current
An alternating quantity may be generated by
a) rotating a coil in a magnetic field
b) rotating a magnetic field within a stationary coil

Carbon Brushes
1, 2

Slip Rings
a, b

11


How does AC generate a sine waveform?

.
.
.
.
.

x x
x

x
x

12



13



x x

.

.

14


 AC generator consists of a magnet and a loop
of wire which rotates in the magnetic field of
the magnet.
 As the wire rotates in the magnetic field, the
changing strength of the magnetic field
through the wire produces a force which drives
the electric charges around the wire.
 The force initially generates an electric current
in one direction along the wire. Then as the
loop rotates through 180 degrees the force
reverses to give an electric current in the
opposite direction along the wire. Every time
the loop rotates through 180 degrees the
direction of the force and therefore the current
changes.
 The changing direction of the force after every
180 degrees of rotation gives the alternating
current.
 AC generator also has slip rings which make
sure that the ends of the wire are always
connected to the same side of the electric
circuit. This makes sure that the direction of the
current changes every half revolution of the
wire.
15


Generation of Alternating emf
The magnitude of the emf is given by,

e = Em sin θ

16


Sine Wave
 A common type of ac.
Also referred as
sinusoidal waveform
sinusoid
Symbol for a sine wave voltage source :

OR

17


What is an Alternating Quantity?

An alternating quantity is one which acts in
alternate directions and whose magnitude
undergoes a definite cycle of changes in
definite interval of time.

18


Polarity
 When the voltage changes polarity at its zero
crossing,the current correspondingly changes
direction.

+V Positive Alternation -
VS R
t +
-V I
+V
Negative Alternation +
I
VS R
t -

-V
19


Amplitude
 The maximum value of the sine wave.
+V
Positive Maximum (peak)

t

Negative Maximum (peak)
-V

20


What is Cycle of a Waveform?
One complete waveform is known as one cycle.
Each cycle consists of two half-cycles.
During 1st half-cycle, the quantity acts in one direction
and
during the second half-cycle, in the opposite direction.

Each cycle :
consists of 2 alternations.
consists of 2 peaks.
reaches maximum amplitude 2 times.
21


Cycle
 A sine wave repeats itself in identical cycles.

+V

t

-V 1st cycle 2nd cycle 3rd cycle

22


Period

 The time required for a given sine wave to
complete one full cycle.

symbol - T
unit - second (s)
Time taken is the same for each cycle;
thus fixed value for a given sine wave.

23


Period Measurement
 From one zero crossing to the corresponding zero
crossing in the next cycle
positive zero crossing to positive zero crossing.
+V

t

-V period period period

24


Period Measurement
 From one peak to the corresponding peak in the
next cycle
positive peak to positive peak.
+V

t

-V period period

25


Period Measurement
 From any point to the corresponding point in the
next cycle.

+V
period

t

-V period

26


Example 11-1
 What is the period of the sine wave?

V

0
t(s)
2 4 6

1 cycle takes 2s to complete.
Therefore the period is 2s.

27


Example 11-2
 What is the period of the sine wave?
V

0
t(s)
12

5 cycles takes 12s to complete
1 cycle takes (12/5)s to complete time taken
1 cycle takes 2.4s to complete T =
no. of cycles
Therefore the period is 2.4s.
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Frequency

 The number of cycles a sine wave can
complete in 1 second.

symbol : f
unit - Hertz (Hz)
Examples :
160Hz - 160 complete cycles in 1s.
50Hz - 50 complete cycles in 1s.
29


Frequency vs Period
 Reciprocal relationship :
1 1
f = T =
T f
More cycles in 1s Lesser cycles in 1s
- higher frequency. - lower frequency.
- shorter period. - longer period.
V T V T

0
t t
0
1s 1s
30


Example 11-3
 Which sine wave has the higher frequency?
Determine the period and the frequency of
both waveforms.

V V

0
t 0
t
1s 1s
Waveform A Waveform B

31


Example 11-3
 continue …
V V

0
t 0
t
1s 1s
Waveform A Waveform B

Waveform B complete more cycles in 1s.
Therefore Waveform B has the higher frequency.

32


Example 11-3
 continue … 3 cycles in 1s
-- frequency is 3Hz
V
period (T) = 1/f = 1/3Hz
t = 0.333s
0 1s
Waveform A 3 cycles take 1s
OR
-- 1 cycles takes (1s/3) = 0.333s
-- period is 0.333s
frequency (f) = 1/T = 1/0.333s
= 3Hz
33


Example 11-3
 continue … 5 cycles in 1s
-- frequency is 5Hz
V period (T) = 1/f = 1/5Hz
= 0.2s
t
0 1s OR
Waveform B 5 cycles take 1s
-- 1 cycles takes (1s/5) = 0.2s
-- period is 0.2s
frequency (f) = 1/T = 1/0.2s
= 5Hz
34


Example 11-4
 The period of a sine wave is 10ms. What is
the frequency?

period (T) = 10ms
frequency (f) = 1/T
= 1/10ms
= 100Hz

35


Example 11-5
 The frequency of a sine wave is 60Hz. What
is the period?

frequency (f) = 60Hz
period (T) = 1/f
= 1/60Hz
= 16.7ms

36


Voltage Sources
If an ac voltage is applied to a
circuit, an ac current flows.
The voltage and current will have
the same frequency.
+V +I

I
t t

-V -I
37


Sine Wave Angles
 The horizontal axis can be replaced by
angular measurement (degrees).

+ V ½ cycle
one cycle

angle
0o

90 o 180 o 270 o 360 o
-V peak zero peak zero
38


Sine Wave Values
 Ways to express and measure the value of a
sine wave :

1) instantaneous value.
2) peak value.
3) peak-to-peak value.
4) root-mean-square value.
5) average value.

39


Instantaneous Value

 The voltage or current value of a
waveform at a given instant in time.

Symbol :
voltage - v
current - i

40


Instantaneous Value
 Different at different points of time.
+V +I
8.5
5.5
5
t3 3 t3
t t
t1 t2 t1 t2
-6 -4.5

-V -I
41


Instantaneous Value
 Different at different points of angle.
+V +I
Vp Ip
v i
degree degree
θ θ

-V v = Vmax sin θ -I i = Imax sin θ
v = Vmax sin 2πft i = Imax sin 2πft
42


Example 11-6
 A current sine wave has a maximum value of
200A. How much is the current at the instant
of 30 degrees of the cycle?

i = Imax sin θ
i = 200 sin 30 o
= 200 x 0.5
= 100A
43


Peak Value

 The voltage or current value of a waveform at
its maximum positive or negative points.

Symbol :
voltage - Vp or Vmax
current - Ip or Imax

44


Peak Value
 The peaks are equal for a sine wave and is
characterized by a single peak value.

+V
Positive Maximum (peak)

t

Negative Maximum (peak)
-V
45


Peak-to-Peak Value
 The voltage or current value of a waveform
measured from its minimum to its maximum
points.

Symbol :
voltage - Vpp
current - Ipp

46


Peak-to-Peak Value

+V Vpp = 2Vmax

Vpp = 2Vmax
Vp
t 1
Vpp
Vp 2 Vpp = Vmax
1
Vmax = Vpp
-V 2
Vmax = 0.5Vpp
47


Root Mean Square Value
 Equal to the dc voltage that produces the same
amount of heat in a resistance as does the AC (sine
wave) voltage.

+ +
VS R Vdc R
- I radiated - I same
heat amount of
Vmax = 10 V radiated heat.
Vrms = 7.07V = Vdc
48



 Also known as the effective value.

Symbol :
voltage - Vrms
current - Irms
The voltage and current values
given are usually as rms unless
otherwise stated.
49



+V

Vrms Vrms = 0.707Vmax
Vp
t 1
Vrms = Vmax
√2
√2 Vrms = Vmax
-V Vmax = √2 Vrms
Vmax = 1.414 Vrms
50


Average Value

The average of a sine wave over half-
cycle.

The average value taken over a
complete cycle is always zero.

Symbol :
voltage - Vavg
current - Iavg
51


Average Value
+V

Vavg = 0.637 Vmax
Vavg 1
Vp
t 0.637 V = V
avg max
1
Vmax = 0.637 Vavg
-V
Vmax = 1.57 Vavg

52


Form & Peak Factor for Sine Wave
Form factor = Rms value
Avg value
0.707 X Max value
= 0.637 X Max value
= 1.11
Max value
Peak factor =
Rms value
Max value
= 0.707 X Max value
= 1.414
53


Example 11-7
 Determine Vmax when :
Vpp = 3V ; Vrms = 5V ; Vavg = 4V.

Vmax = 0.5 Vpp Vmax = 1.414 Vrms
= 0.5 (3) = 1.414 (5)
= 1.5V = 7.07V

Vmax = 1.57 Vavg
= 1.57 (4)
= 6.28V
54


Example 11-8
 Determine Vrms when
Vp = 20V ; Vpp = 10V ; Vavg = 30V.

Vrms= 0.707 Vmax Vmax= 0.5 Vpp Vmax = 1.57 Vavg
= 0.707 (20) = 0.5 (10) = 1.57 (30)
=14.14V = 5V = 47.1V
Vrms = 0.707 Vmax Vrms= 0.707 Vmax
= 0.707 (5) = 0.707 (47.1)
= 3.535V = 33.33V

55


Example 11-9

 Determine Vavg when Vmax = 37V ; Vpp = 28V ;
Vrms = 46V.

Vavg= 0.637 Vmax Vmax= 0.5 Vpp Vmax=1.414 Vrms
= 0.637 (37) = 0.5 (28) = 1.414 (46)
= 23.569V = 14V = 65. V
Vavg= 0.637 Vp Vavg= 0.637 Vmax
= 0.637 (14) = 0.637 (65.)
= 8.918V = 41.5V

56


Sine Wave Values

Vpp = 2Vmax Vmax = 0.5Vpp

Vrms = 0.707Vmax Vmax = 1.414 Vrms

Vavg = 0.637 Vmax Vmax = 1.57 Vavg
Note:
1 1
0.707 = 1.414 = 1.57
0.637
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Summary

58


Next Lesson

59

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