ee1 chapter11 ac_circuits
TRANSCRIPT
Apr 14, 2023
Lecturer Name [email protected]
Contact Number
IT2001PAEngineering Essentials (1/2)
Chapter 11 – Alternating Current Circuits
Chapter 11 – Alternating Current Circuits
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Lesson Objectives
Upon completion of this topic, you should be able to: Explain and calculate the fundamentals of alternating
current waveform.
Chapter 11 – Alternating Current Circuits
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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
Chapter 11 – Alternating Current Circuits
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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
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Alternating Waveform
Alternating Current (ac)
current that is continuously reversing
direction, alternately flowing in one
direction and then in the other.
can be similarly described.
The designation ac is normally applied to
both current and voltage.
Alternating Voltage
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Alternating Waveform
Vertical axis : current (I) or voltage (V)
t
V+
V-
t
I+
I-
Horizontal axis : time (t)
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Why not Direct Current (DC)?
Problems of using DC as power distribution system: suffers from rapid power loss
in the wires due to their resistance, which dissipates energy as heat
DC power stations had useful ranges of about two kilometers
Once generated, DC power cannot be modified
t
V+
V-
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Power Distribution System
1)To prevent risk of sparks and short circuits in the generator itself, power needs to be generated with low voltages.
V low I is high
V=RI
2)To avoid serious heating effects, present at high current, it needs to be transmitted at the lowest current practical.
I low V is high
3)For the safety of the user, the voltage needs to be low in the home.
V low I is high
DC is not suitable
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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.
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Alternator
An alternator is a machine that converts mechanical energy to electrical energy.A simple alternator basically consists of two parts:
a) coils or windingsb) magnetic poles
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Generation of Alternating CurrentAn 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
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Generation of Alternating Current
x
xx x
x
How does AC generate a sine waveform?
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Generation of Alternating Current
How does AC generate a sine waveform?
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Generation of Alternating Current
x
x
How does AC generate a sine waveform?
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Generation of Alternating Current 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.
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Generation of Alternating emf
The magnitude of the emf is given by,
e = Em sin
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Sine Wave
A common type of ac.
Also referred assinusoidal waveformsinusoid
Symbol for a sine wave voltage source :
OR
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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.
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Polarity
When the voltage changes polarity at its zero crossing,the current correspondingly changes direction.
V+
t
V-
VS
+
-R
-
+VS R
Positive Alternation
Negative Alternation
I
I
t
V+
V-
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Amplitude
The maximum value of the sine wave.
t
V+
V-
Positive Maximum (peak)
Negative Maximum (peak)
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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 andduring the second half-cycle, in the opposite direction.
Each cycle :
consists of 2 alternations.
consists of 2 peaks.
reaches maximum amplitude 2 times.
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Cycle
A sine wave repeats itself in identical cycles.
t
V+
V- 1st cycle 2nd cycle 3rd cycle
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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.
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Period Measurement
From one zero crossing to the corresponding zero crossing in the next cycle
t
V+
V- period period period
positive zero crossing to positive zero crossing.
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Period Measurement
From one peak to the corresponding peak in the next cycle
t
V+
V- period period
positive peak to positive peak.
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Period Measurement
From any point to the corresponding point in the next cycle.
t
V+
V-
period
period
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Example 11-1
What is the period of the sine wave?
V
t(s)2 4 60
1 cycle takes 2s to complete.Therefore the period is 2s.
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Example 11-2
What is the period of the sine wave?
5 cycles takes 12s to complete
V
t(s)120
T =time taken
no. of cycles
1 cycle takes (12/5)s to complete
1 cycle takes 2.4s to completeTherefore 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.
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Frequency vs Period
Reciprocal relationship :
More cycles in 1s Lesser cycles in 1s
V
t0
1s
TV
t0
1s
T
1f
T =
- higher frequency. - lower frequency.- shorter period. - longer period.
f = 1T
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Example 11-3
Which sine wave has the higher frequency? Determine the period and the frequency of both waveforms.
V
0 1s
Waveform A
t t
V
0 1s
Waveform B
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Example 11-3
continue …V
t0 t
V
01s 1s
Waveform A Waveform B
Waveform B complete more cycles in 1s.Therefore Waveform B has the higher frequency.
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Example 11-3
continue …
V
t0 1s
Waveform A
3 cycles in 1s -- frequency is 3Hz
3 cycles take 1s-- 1 cycles takes (1s/3) = 0.333s-- period is 0.333s
OR
period (T) = 1/f = 1/3Hz = 0.333s
frequency (f) = 1/T = 1/0.333s = 3Hz
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Example 11-3
continue … 5 cycles in 1s -- frequency is 5Hz
t
V
0 1s
Waveform B 5 cycles take 1s-- 1 cycles takes (1s/5) = 0.2s-- period is 0.2s
period (T) = 1/f = 1/5Hz = 0.2s
OR
frequency (f) = 1/T = 1/0.2s = 5Hz
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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
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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
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Voltage Sources
If an ac voltage is applied to a circuit, an ac current flows.The voltage and current will have the same frequency.
t
V+
V-
t
I+
I-
I
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Sine Wave Angles
The horizontal axis can be replaced by angular measurement (degrees).
angle
V+
V-
0 o
½ cycleone cycle
90 o
peak270 o
peak360o
zero180o
zero
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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.
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Instantaneous Value
The voltage or current value of a waveform at a given instant in time.
Symbol :voltage - vcurrent - i
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Instantaneous Value
Different at different points of time.
t
V+
V-
t
I+
I-
5.53
8.5
5
-6
t1 t2
t3
-4.5
t1 t2
t3
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Instantaneous Value Different at different points of angle.
degree
V+
V- v = Vmax sin
degree
I+
I- i = Imax sin
Vp
v
Ip
i
v = Vmax sin 2ft i = Imax sin 2ft
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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
i = 200 sin 30 o
= 200 x 0.5 = 100A
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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
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Peak Value
The peaks are equal for a sine wave and is characterized by a single peak value.
t
V+
V-
Positive Maximum (peak)
Negative Maximum (peak)
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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
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Peak-to-Peak Value
Vpp = 2Vmax
t
V+
V-
Vp
Vp
Vpp
Vpp = 2Vmax
12 Vpp = Vmax 1
2 Vmax = Vpp
Vmax = 0.5Vpp
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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.
I
+
-Vdc R
sameamount of
radiated heat.
I
+
-VS R
radiatedheat
Vmax = 10 V
Vrms = 7.07V = Vdc
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Root Mean Square Value
Also known as the effective value.
Symbol :voltage - Vrms
current - Irms
The voltage and current values given are usually as rms unless otherwise stated.
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Root Mean Square Value
Vrms = Vmax
12
t
V+
V-
Vp
VrmsVrms = 0.707Vmax
2 Vrms = Vmax
Vmax = 2 Vrms
Vmax = 1.414 Vrms
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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
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Average Value
Vavg = 0.637 Vmax
t
V+
V-
Vp
Vavg
10.637 Vmax = Vavg
Vmax = 1.57 Vavg
10.637 Vavg = Vmax
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Form & Peak Factor for Sine Wave
Form factor =
= 1.11
Avg value 0.707 X Max value0.637 X Max value=
Peak factor =
= 1.414
Rms value Max value
Max value0.707 X Max value=
Rms value
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Example 11-7
Determine Vmax when :
Vpp = 3V ; Vrms = 5V ; Vavg = 4V.
Vmax = 0.5 Vpp
= 0.5 (3) = 1.5V
Vmax = 1.414 Vrms
= 1.414 (5) = 7.07V
Vmax = 1.57 Vavg
= 1.57 (4) = 6.28V
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Example 11-8
Determine Vrms when
Vp = 20V ; Vpp = 10V ; Vavg = 30V.
Vrms= 0.707 Vmax
= 0.707 (20) =14.14V
Vmax= 0.5 Vpp
= 0.5 (10) = 5V
Vmax = 1.57 Vavg
= 1.57 (30) = 47.1V
Vrms = 0.707 Vmax
= 0.707 (5) = 3.535V
Vrms= 0.707 Vmax
= 0.707 (47.1) = 33.33V
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Example 11-9
Determine Vavg when Vmax = 37V ; Vpp = 28V ;
Vrms = 46V.
Vavg= 0.637 Vmax
= 0.637 (37) = 23.569V
Vmax= 0.5 Vpp
= 0.5 (28) = 14V
Vmax=1.414 Vrms
= 1.414 (46) = 65. V
Vavg= 0.637 Vmax
= 0.637 (65.) = 41.5V
Vavg= 0.637 Vp
= 0.637 (14) = 8.918V
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Sine Wave Values
Vmax = 0.5Vpp
Vmax = 1.414 Vrms
Vmax = 1.57 Vavg
Vpp = 2Vmax
Vrms = 0.707Vmax
Vavg = 0.637 Vmax
10.637 = 1.57
10.707 = 1.414
Note:
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Summary
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