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ELECTRICAL CIRCUIT
ET 201
Define and explain characteristics of
sinusoidal wave, phase relationships
and phase shifting
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(CHAPTER 1.1 ~ 1.4)
SINUSOIDAL ALTERNATING
WAVEFORMS
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Understand Alternating Current
DIRECT CURRENT (DC)IS WHEN THE CURRENTFLOWS IN ONLY ONE DIRECTION. Constant flow of
electric charge
EX: BATTERY ALTERNATING CURRENT AC)THE CURRENT
FLOWS IN ONE DIRECTION THEN THE OTHER.
Electrical current whose magnitude and direction varycyclically, as opposed to direct current whose direction
remains constant.
EX: OUTLETS
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Sources of alternating current
By rotating a magnetic field within astationary coil
By rotating a coil in a magnetic field
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Generation of Alternating
Current A voltage supplied by a battery or other
DC source has a certain polarity and
remains constant.
Alternating Current (AC) varies in polarityand amplitude.
AC is an important part of electrical andelectronic systems.
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Faradays Laws of electromagneticInduction.
Induced electromotive fieldAny change in the magnetic environment of a coil of wire will cause a
voltage (emf) to be "induced" in the coil.
e.m.f, e = -N d
N = Number of turn
dt = Magnetic Flux
Lenzs law
An electromagnetic field interacting with a conductor will generate
electrical current that induces a counter magnetic field that opposes
the magnetic field generating the current.
Faradays and Lenzs Lawinvolved in generating a.c current
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Sine Wave Characteristics
The basis of an AC alternator is a loop ofwire rotated in a magnetic field.
Slip rings and brushes make continuouselectrical connections to the rotatingconductor.
The magnitude and polarity of the
generated voltage is shown on thefollowing slide.
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Sine Wave Characteristics
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Sine Wave Characteristics
The sine wave at theright consists of two,
opposite polarity,alternations.
Each alternation iscalled a half cycle.
Each half cycle has amaximum value calledthepeak value.
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Sine Wave Characteristics
Sine waves may represent voltage,current, or some other parameter.
Theperiodof a sine wave is the time fromany given point on the cycle to the same
point on the following cycle.
The period is measured in time (t), and inmost cases is measured in seconds or
fractions thereof.
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Frequency
The frequencyof a sine wave is thenumber of complete cycles that occur inone second.
Frequency is measured in hertz(Hz). Onehertz corresponds to one cycle persecond.
Frequency and period have an inverserelationship. t= 1/f, and f= 1/t.
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Peak Value
The peak value of a sine wave is themaximum voltage (or current) it reaches.
Peak voltages occur at two different pointsin the cycle.
One peak is positive, the other is negative. The positive peak occurs at 90 and the
negative peak at 270. The positive and negative have equal
amplitudes.
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Chapter 6 -13
Average Values
The average value of any measuredquantity is the sum of all of the
intermediate values.
The average value of a full sine wave iszero.
The average value of one-half cycle of asine wave is:
Vavg= 0.637Vpor Iavg= 0.637Ip
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Chapter 6 -14
rms Value
One of the most important characteristicsof a sine wave is its rms or effectivevalue.
The rms value describes the sine wave in
terms of an equivalent dc voltage. The rms value of a sine wave producesthe same heating effect in a resistance asan equal value of dc.
The abbreviation rms stands for root-mean-square, and is determined by: Vrms=0.707Vp or Irms= 0.707Ip
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Peak-to-Peak Value
Another measurement used to describe sine waves aretheir peak-to-peak values.
The peak-to-peak value is the difference between thetwo peak values.
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Form Factor
Form Factor is defined as the ratio of r.m.svalue to the average value.
Form factor = r.m.s value = 0.707 peak value
average value 0.637 peak valur = 1.11
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Peak Factor
Crest or Peak or Amplitude Factor
Peak factor is defined as the ratio of peakvoltage to r.m.s value.
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13.1 Introduction
Alternating waveforms
Alternating signal is a signal that varies with respect to time. Alternating signal can be categories into ac voltage and ac
current.
This voltage and current have positive and negative value.
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13.2 Sinusoidal AC Voltage
Characteristics and Definitions
volts or amperes
units of time
Voltage and current value is represent by vertical axis and timerepresent by horizontal axis.
In the first half, current or voltage will increase into maximum positivevalue and come back to zero.
Then in second half, current or voltage will increase into negativemaximum voltage and come back to zero.
One complete waveform is called one cycle.
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Defined Polarities and Direction
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
The voltage polarity and current direction will be for an instantin time in the positive portion of the sinusoidal waveform.
In the figure, a lowercase letter is employed for polarity and
current direction to indicate that the quantity is time dependent;that is, its magnitude will change with time.
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Defined Polarities and Direction
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
For a period of time, a voltage has one polarity, while for thenext equal period it reverses. A positive sign is applied if the
voltage is above the axis.
For a current source, the direction in the symbolcorresponds with the positive region of the waveform.
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There are several specification in sinusoidal
waveform:
1. period
2. frequency
3. instantaneous value
4. peak value
5. peak to peak value6. angular velocity
7. average value
8. effective value 22
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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Period (T) Period is defines as the amount of time is take to go through
one cycle.
Period for sinusoidal waveform is equal for each cycle.
Cycle
The portion of a waveform contained in one period of time.
Frequency (f)
Frequency is defines as number of cycles in one seconds.
It can derives as
23
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Hzhertz,1
Tf
f = Hz
T = seconds (s)http://modul2poli.blogspot.com/ Page 23 of 241
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The cycles within T1, T2, and T3may appear different in
the figure above, but they are all bounded by one period of time
and therefore satisfy the definition of a cycle.
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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Frequency = 1 cycle
per second
Frequency = 21/2cycles
per second
Frequency = 2 cycles
per second
1 hertz (Hz) = 1 cycle per second (cps)
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Signal with lower frequency Signal with higher frequency
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Instantaneous value
Instantaneous value is magnitude value of waveform atone specific time.
Symbol for instantaneous value of voltage is v(t) andcurrent is i(t).
26
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
V8)1.1(
V0)6.0(
V8)1.0(
v
v
v
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Peak Value
The maximum instantaneous value of a function as measuredfrom zero-volt level.
For one complete cycle, there are two peak value that ispositive peak value and negative peak value.
Symbol for peak value of voltage is Emor Vm and current is Im.
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Peak value, Vm= 8 V
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13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Peak to peak value The full voltage between positive and negative peaks of the
waveform, that is, the sum of the magnitude of the positive and
negative peaks.
Symbol for peak to peak value of voltage is Ep-por Vp-p andcurrent is Ip-pPeak to peak value, Vp-p = 16 V
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Angular velocity Angular velocity is the velocity with which the radius vector
rotates about the center.
Symbol of angular speed is and units is
radians/seconds (rad/s) Horizontal axis of waveform can be represent by time and
angular speed.
360radian2
142.3,3.572
360radian1 0
0
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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Angular velocity
Degree Radian90 (/180) x ( 90) = /2 rad
60 (/180) x ( 60) = /3 rad
30 (/180) x (30) = /6 rad
Radian Degree
/3 (180/) x ( /3) = 60
(180/) x ( ) = 180
3 /2 (180/) x (3 /2) = 270
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Plotting a sine wave versus (a) degrees and (b) radians.
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The sinusoidal wave formcan be derived from thelength of the vertical
project ionof a radius vector
rotating in a uniform circular
motion about a fixed point.
Waveform picture with respect to angular velocity
13.2 Sinusoidal AC
Voltage
Characteristics andDefinitions
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Angular velocity
Formula of angular velocity
Since () is typically provided in radians/second, theangle obtained using = tis usually in radians.
t
t
(seconds)time
)radiansordegrees(distance,degreeangular
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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34T
2 or f 2 (rad/s)
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Angular velocity
The time required to complete one cycle is equal to theperiod (T) of the sinusoidal waveform.
One cycle in radian is equal to2(360o
).
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13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Angular velocityDemonstrating the effect of on the frequency fand period T.
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Example 13.6Given = 200 rad/s, determine how long it will take the
sinusoidal waveform to pass through an angle of 90
Solut ion
t
rad2
90
ms85.7200
2/
t
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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Example 13.7Find the angle through which a sinusoidal waveform of
60 Hz will pass in a period of 5 ms.
Solut ion
rad885.11056022 3 ftt
108180
885.1
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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Average value Average value is average value for all instantaneous value in
half or one complete waveform cycle.
It can be calculate in two ways:1. Calculate the area under the graph:
Average value = area under the function in a period
period
2. Use integral method
For a symmetry waveform, area upper section equal to area
under the section, so just take half of the period only.
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
T
dttvT
valueaverage0
)(1
_
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13.2 Sinusoidal AC Voltage
Characteristics and Definitions
Average value Example: Calculate the average value of the waveform below.
Vm
Vm
rad 2
Solution:
voltvv
v
dv
dv
dttvT
valueaverage
mm
om
m
m
T
637.02
cos
sin
sin1
)(1
_
0
0
0
For a sinus waveform , average value can
be calculate by
mm
average VV
V 637.0 http://modul2poli.blogspot.com/ Page 39 of 241
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Effective value The most common method of specifying the amount of sine wave ofvoltage or current by relating it into dc voltage and current that will
produce the same heat effect.
Effective value is the equivalent dc value of a sinusoidal current or
voltage, which is 1/2 or 0.707 of its peak value. The equivalent dc value is called rms valueor effective value.
The formula of effective value for sine wave waveform is;
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
mm
mm
EEE
III
707.02
1
707.02
1
rms
rms
rmsrms
rmsrms
414.12
414.12
EEE
III
m
m
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Example 13.21The 120 V dc source delivers 3.6 W to the load. Find Emand Imof
the ac source, if the same power is to be delivered to the load.
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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Example 13.21so lu t ion
W6.3dcdc PIE mA30120
6.3
dc
dc E
PI
2dcrms
mEEE
V7.169120414.12 dc EEm
and2
dcrmsmIII
mA43.4230414.12 dc IIm
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
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Example 13.21solut ion
2dcrms
mIII 2
dcrmsmEEE
mA43.42
30414.1
2 rms
IIm
13.2 Sinusoidal AC Voltage
Characteristics and Definitions
V7.169
120414.1
2 rms
EEm
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13.5 General Format for the
Sinusoidal Voltage or Current
The basic mathematicalformat for the sinusoidal
waveform is:
where:
Am: peak value of thewaveform
: angle from the
horizontal axis
volts or amperes
Basic sine wave for current or voltage
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The general format of a sine wave can also be as:
General format for electrical quantities such as currentand voltage is:
where: and is the peak value of current
and voltage while i(t) and v(t) is the instantaneous
value of current and voltage.
sinsin mm ItIti
sinsin mm EtEte
13.5 General Format for the
Sinusoidal Voltage or Current
= t
mI
mE
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Example 13.8
Given e(t)= 5 sin, determine e(t)at = 40and = 0.8.
Solut ion
For = 40, V21.340sin5 te
For = 0.8
144180
8.0
V94.2144sin5 te
13.5 General Format for the
Sinusoidal Voltage or Current
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Example 13.9
(a) Determine the angle at which the magnitude of the
sinusoidal function v(t)= 10 sin 377tis 4 V.
(b) Determine the time
at which the magnitude
is attained.
13.5 General Format for the
Sinusoidal Voltage or Current
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Example 13.9 - so lu t ion
Vsin tVtv m rad/s377V;10 mV
Hence, V377sin10 ttv When v(t)= 4 V,
t377sin104
4.0104sin377sin t
58.234.0sin 11
13.5 General Format for the
Sinusoidal Voltage or Current
42.15658.231802
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Example 13.9solut ion (contd)
ms24.7
377
73.2
ms1.09377
412.0
2
1
t
t
13.5 General Format for the
Sinusoidal Voltage or Current
(a) But is in radian, so must be calculate in radian:
(b) Given, , sot
t
rad73.242.156rad412.058.23377
2
1
t
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13.6 Phase Relationship
Phase angle
Phase angle is a shifted angle waveform from referenceorigin.
Phase angle is been represent by symbol or
Units is degreeor radian
Two waveform is called in phase if its have a samephase degree or different phase is zero
Two waveform is called out of phase if its have adifferent phase.
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13.6 Phase Relationship
tAa m sin
The unshifted sinusoidal waveform is
represented by the expression:
t
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13.6 Phase Relationship
where is the angle (in degrees or radians) that
the waveform has been shifted.
Sinusoidal waveform which is shiftedto the
right or left of 0is represented by theexpression:
tAa m sin
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13.6 Phase Relationship
If the wave form passes through the horizontal axis
with apositive-going(increasing with the time)slope before 0:
tAa m sin
t
tAa m sin
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13.6 Phase Relationship
t
If the waveform passes through the horizontal axis
with a positive-going slope after 0
:
tAa m sin
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13.6 Phase Relationship
t
2
cos90cossin
cos2
sin90sin
ttt
ttt
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13.6 Phase Relationship
The terms leadingand laggingare used to
indicate the relationship between two sinusoidalwaveforms of the same frequency for angular
velocity )plotted on the same set of axes.
The cosine curve is said to leadthe sine curveby 90.
The sine curve is said to lagthe cosine curve
by 90. 90is referred to as the phase angle between
the two waveforms.
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13.6 Phase Relationship
+ cos
+ sin
- cos
- sin
cos (-90o)
sin (+90o)
90sin270sincos
180sinsin90cossin
90sincos
Start at + sin position;
Note:
sin (- ) = - sin
cos(- ) = cos
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13.6 Phase Relationship
If a sinusoidal expression should appear as
the negative sign is associated with the sine
portion of the expression, not the peak value Em,i.e.
And, since;
tEe m sin
tEetEe mm sinsin
180sinsin tt
180sinsin tEtEmm
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13.6 Phase Relationship
Determine the phase relationship between the following waveforms
70sin530sin10(a)
titv
Example 13.2
20sin10
60sin15(b)
tv
ti
10sin310cos2(c)
tvti
10sin230sin(d)
tvti
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13.6 Phase Relationship
70sin5
30sin10(a)
ti
tv
Example 13.2so lu t ion
ileadsvby 40or
vlagsiby 40
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13.6 Phase Relationship
Example 13.2so lu t ion (contd)
ileadsvby 80or
vlagsiby 80
20sin10
60sin15(b)
tv
ti
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13.6 Phase Relationship
Example 13.2solut ion (contd)
ileadsvby 110or
vlagsiby 110
10sin3
10cos2(c)
tv
ti
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13.6 Phase Relationship
Example 13.2so lu t ion (contd)
10sin2
30sin(d)
tv
ti
ORvleadsiby 160
Or
ilagsvby 160
ileadsvby 200Or
vlagsiby 200
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1
ELECTRICAL CIRCUIT
ET 201
- Identify common oscilloscope controls- Use an oscilloscope to measure the
amplitude of a waveform
- Use an oscilloscope to measure theperiod and frequency of a waveform
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(CHAPTER 1.7)
OSCILLOSCOPE TOMEASURE WAVEFORMS
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Oscilloscope
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OscilloscopeOscilloscopes are
commonly used to
observe the
exact wave
shapeof an
electrical signal.
Type of electronic
test instrumentthat
allows observation
of constantly
varying
signal voltageshttp://modul2poli.blogspot.com/ Page 66 of 241
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Focus control
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Focus control
This control adjusts CRT focus to obtain
the sharpest, most-detailed trace. I Intensity control
This adjusts trace brightness. Slow traces
on CRT 'scopes need less, and fast ones,especially if they don't repeat very often.
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ET 201 ELECTRICAL
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ET 201 ~ ELECTRICAL
CIRCUITSCOMPLEX NUMBER SYSTEM
Define and explain complex numberRectangular form
Polar form
Mathematical operations
(CHAPTER 2)
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COMPLEX
NUMBERS
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2. Complex Numbers
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p
A complex number
represents a point ina two-dimensionalplane located withreference to two
distinct axes.
This point can alsodetermine a radius
vector drawn from theorigin to the point.
The horizontal axis iscalled the realaxis,while the vertical axisis called the
imaginary ( j )axis.http://modul2poli.blogspot.com/ Page 71 of 241
2.1 Rectangular Form
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g
The format for the
rectangular formis
The letter Cwas
chosen from the
word complex
The bold face (C)
notation is for anynumber with
magnitude and
direction.
The italicnotation isfor magnitudeonly.
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2.1 Rectangular Form
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g
Example 14.13(a)
Sketch the complex number C= 3 +j4in the
complex plane
Solut ion
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2.1 Rectangular Form
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g
Example 14.13(b)
Sketch the complex number C= 0j6in the
complex plane
Solut ion
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2.1 Rectangular Form
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7
g
Example 14.13(c)
Sketch the complex number C= -10
j20in
the complex plane
Solut ion
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2.2 Polar Form
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The format for thepolar formis
Where:
Z : magnitude only
q: angle measuredcounterclockwise(CCW)from the
positive real axis.
Angles measured in
the clockwisedirectionfrom the positive realaxis must have anegative sign
associated with them.http://modul2poli.blogspot.com/ Page 76 of 241
2.2 Polar Form
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180 qq ZZC
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2.2 Polar Form
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Example 14.14(a)
305C
Counterclockwise (CCW)
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2.2 Polar Form
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11
Example 14.14(b)
1207 C
Clockwise (CW)
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2.2 Polar Form 180 qq ZZC
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12
Example 14.14(c)
602.4 C
180602.4
2402.4
180 qq ZZC
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14.9 Conversion Between Forms
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1. Rectangular to Polar
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14.9 Conversion Between Forms
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14
2. Polar to Rectangular
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2.3 Conversion Between Forms
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15
Example 14.15
Convert C= 4 +j4to polar form
543 23 Z
Solut ion
13.533
4
tan
1
q
13.535Chttp://modul2poli.blogspot.com/ Page 83 of 241
2.3 Conversion Between Forms
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16
Example 14.16
Convert C= 1045to rectangular form
07.745cos10 X
Solut ion
07.745sin10 Y
07.707.7 jC
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2.3 Conversion Between Forms
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17
Example 14.17
Convert C= - 6 +j3to polar form
71.636 22 Z
Solut ion
6
3
tan180
1
q
43.153
43.15371.6 C
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2.3 Conversion Between Forms
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18
Example 14.18
Convert C= 10 230to rectangular form
43.6230cos10
Solut ion
66.7230sin10 Y
66.743.6 jC
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2.4 Mathematical Operations with
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19
Complex Numbers
Complex numbers lend themselves readily tothe basic mathematical operations of addition,
subtraction, multiplication, and division.
A few basic rules and definitions must be
understood before considering these
operations:
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2.4 Mathematical Operations with
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Complex Conjugate
The conjugateor complex conjugateof
a complex number can be found by simplychanging the signof the imaginary part in
the rectangular formor by using the
negative of the angle of the polar form
Complex Numbers
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2.4 Mathematical Operations with
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21
Complex Conjugate
In rectangular form, the
conjugate of:
C= 2 +j3
is 2j3
Complex Numbers
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2.4 Mathematical Operations with
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22
Complex Conjugate
In polar form, the
conjugate of:
C= 2 30o
is 2 30o
Complex Numbers
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2.4 Mathematical Operations with
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23
Reciprocal
The reciprocalof a complex number is 1
divided by the complex number.
In rectangular form, the reciprocal of:
In polar form, the reciprocal of:
jYXC isjYX
1
qZC isqZ
1
Complex Numbers
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2.4 Mathematical Operations with
C l N b
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Addition
To add two or more complex numbers, simply
add the real and imaginary parts separately.
Complex Numbers
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2.4 Mathematical Operations with
C l N b
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25
Example 14.19(a)
13;4221 jj CC
143221 jCC
55 j
Find C1+ C2.
Solut ion
Complex Numbers
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2.4 Mathematical Operations with
C l N b
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26
Example 14.19(b)
36;63 21 jj CC
366321
jCC
93 j
FindC
1+ C
2
Solut ion
Complex Numbers
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2.4 Mathematical Operations with
C l N b
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Subtraction
In subtraction, the real and imaginary parts are
again considered separately .
Complex Numbers
NOTE
Addition or subtraction cannot be performed in polar form
unless the complex numbers have the same angle or
unless they differ only by multiples of 180http://modul2poli.blogspot.com/ Page 95 of 241
2.4 Mathematical Operations with
C l N b
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28
Example 14.20(a)
41;6421 jj CC
461421
jCC
23 j
FindC
1- C
2
Solut ion
Complex Numbers
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2.4 Mathematical Operations withC l N b
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29
Example 14.20(b)
52;3321 jj CC
532321
jCC
25 j
FindC
1- C
2
Solut ion
Complex Numbers
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2.4 Mathematical Operations withC l N b
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30
Example 14.21(a)
455453452
Complex Numbers
NOTE
Addition or subtraction cannot be performed in polar form
unless the complex numbers have the same angle or
unless they differ only by multiples of 180http://modul2poli.blogspot.com/ Page 98 of 241
2.4 Mathematical Operations withC l N b
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31
Complex Numbers
06180402
NOTE
Addition or subtraction cannot be performed in polar form
unless the complex numbers have the same angle or
unless they differ only by multiples of 180
Example 14.21(b)
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2.4 Mathematical Operations withC l N b
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Multiplication To multiply two complex numbers in rectangular
form, multiply the real and imaginary parts of one
in turn by the real and imaginary parts of the
other.
In rectangular form:
In polar form:
Complex Numbers
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2.4 Mathematical Operations withC l N b
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33
Example 14.22(a)
105;3221 jj CC
Find C1C2.Solut ion
1053221 jj CC
3520 j
Complex Numbers
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2.4 Mathematical Operations withC l N b
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34
Example 14.22(b)
64;3221 jj CC
Find C1C2.
Solut ion
643221 jj CC
1802626
Complex Numbers
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2.4 Mathematical Operations withComplex Numbers
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35
Example 14.23(a)
3010;20521 CC
Find C1C2.
Solut ion
302010521
CC
5050
Complex Numbers
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14.10 Mathematical Operations withComplex Numbers
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36
Example 14.23(b)
1207;40221
CC
Find C1C2.
Solut ion
120407221
CC
8014
Complex Numbers
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14.10 Mathematical Operations withComplex Numbers
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Division To divide two complex numbers in rectangular
form, multiply the numerator and denominator
by the conjugate of the denominator and the
resulting real and imaginary parts collected.
In rectangular form:
In polar form:
Complex Numbers
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2.4 Mathematical Operations withComplex Numbers
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38
Example 14.24(a)
54;4121 jj CC Find
Solut ion
2
1
C
C
5454
5441
54
54
54
41
2
1
jj
jj
j
j
j
j
C
C
27.059.02516
1124j
j
Complex Numbers
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2.4 Mathematical Operations withComplex Numbers
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39
Example 14.24(b)
16;8421 jj CC Find
Solut ion
2
1
C
C
1616
1684
16
16
16
84
2
1
jj
jj
j
j
j
j
C
C
41.143.0136
5216j
j
Complex Numbers
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2.4 Mathematical Operations withComplex Numbers
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40
Example 14.25(a)
72;101521 CC Find
Solut ion
2
1
C
C
7102
15
72
1015
2
1
C
C
33.7
Complex Numbers
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2.4 Mathematical Operations withComplex Numbers
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41
Example 14.25(b)
5016;120821 CC Find
Solut ion
2
1
C
C
5012016
8
5016
1208
2
1
C
C
1705.0
Complex Numbers
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2.4 Mathematical Operations withComplex Numbers
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42
)()( 212121 yyjxxzz
)()( 212121 yyjxxzz
212121 rrzz
21
2
1
2
1 r
r
z
z
rz
11
jrerjyxz
sincos je j
Addition
Subtraction
Multiplication
Division
Reciprocal
Complex conjugate
Eulers identity
Complex Numbers
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1
ELECTRICAL TECHNOLOGYET 201
Define series impedances and analyzeseries AC circuits using circuit
techniques.
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14.3 Response of Basic R, L and C Elementsto a Sinusoidal Voltage or Current (review)
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g ( )
FIG. 15.46 Reviewing the frequency response of the basic elements.http://modul2poli.blogspot.com/ Page 112 of 241
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3
(CHAPTER 15)
SERIES
AC CIRCUITS
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15.3 Series Impedances
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The overall properties of series AC circuits are
the same as those for DC circuits. For instance, the total impedance of a system is
the sum of the individual impedances:
[]
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15.3 Series Impedances
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5
Example 15.7
Draw the impedance diagramand find the total impedance.
84
900
21
j
jXR
XR
L
L
T
ZZZ
34.6394.8
TZ
Solut ion
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15.3 Series Impedances
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6
26
12106
90900
321
j
jj
jXjXR
XXR
CL
CL
T
ZZZZ
Example 15.8
Draw the impedance diagramand find the total impedance.
43.1832.6
TZ
Solut ion
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15.3 Series AC Circuit
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7
In a series AC configuration having two
impedances, the current Iis the samethrough
each element (as it was for the series DC circuit)
The current is determined by Ohms Law:
21 ZZZ T
????,21
VV
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15.3 Series Configuration
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8
Kirchhoffs Voltage Law can be applied in thesame manner as it is employed for a DC circuit.
The power to the circuit can be determined by:
Where
E, I : effective values (Erms, Irms)
T : phase angle between Eand I
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14.5 Power FactorF f tP
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For a purely resistive load;
Hence;
For purely inductive or purely capacitive load;
Hence;
TpF cosfactorPower
0T
1cos TPF
Trmsrms IEP cos
rmsrmsTrmsrms IEIEP cos
90T
0cos Trmsrms
IEP
0cos TP
F
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14.5 Power Factor
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Power factor can be lagging orleading.
Defined by the current through the load.
Lagging power factor:
Current lags voltage
Inductive circuit
Leading power factor:
Current leads voltage Capacitive circuit
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15.3 Series Configuration
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11
R-L
1. Phasor Notation
te sin4.141 0V100 E
Series R-Lcircuit Apply phasor notation
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15.3 Series Configuration
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12
R-L
2. ZT
Impedance diagram:
43
)904()03(
21
j
T
ZZZ
13.535 T
Z
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15.3 Series Configuration
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13
R-L
3. I
13.535
0V100
T
Z
E
I
13.53A20 I
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R L
15.3 Series Configuration
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R-L
4. VRand VL
Ohms Law:
)03)(13.53A20(
RR ZIV
V13.5360
R
V
)904)(13.53A20(
LL ZIV
V87.3680
L
V
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R L
15.3 Series Configuration
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R-L
Kirchhoffs voltage law:
Or;
In rectangular form,
0 LR VVEV
LR VVE
V;483613.53V60 jR
V
V486487.36V80 jL
V
LR VVE
0V100
0100)4864()4836(
jjj
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R L
15.3 Series Configuration
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16
R-L
Phasor diagram:
I is in phase with the VRand lags the VLby 90o.
I lags Eby 53.13o.
13.53A20 I
V13.5360
R
V
V87.3680
L
V
0V100 E
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R L
15.3 Series Configuration
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17
R-L
Power: The total power delivered to the circuit is
Where
E, I : effective values;
T : phase angle between E and I
Or;
W1200
13.53cos)20)(100(
cos
TT EIP
W120032022
RIPT
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R L
15.3 Series Configuration
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R-L
Power factor:
13.53cos
cos
Tp
F
lagging6.0p
F
TZ
R
IE
R
E
IR
EI
RI
EI
P
EIP
2
cos
cos
T
TP
Z
RF cos
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R C
15.3 Series Configuration
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R-C
1. Phasor Notation
A13.53sin07.7 ti A13.535 I
Series R-Ccircuit Apply phasor notation
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R-C
15.3 Series Configuration
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20
R-C
2. ZT
Impedance diagram:
86
)908()06(
21
j
T
ZZZ
13.5310
TZ
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R C
15.3 Series Configuration
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21
R-C
3. E
)13.5310)(13.535(
T
IZE
V050
E
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R C
15.3 Series Configuration
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22
R-C
4. VRand VC
Ohms Law:
)06)(13.535(
RR
ZIV
V13.5330
R
V
)908)(13.535(
CC ZIV
V87.3640
C
V
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R C
15.3 Series Configuration
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23
R-C
Kirchhoffs voltage law:
Or;
0 CR VVEV
CR VVE
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R C
15.3 Series Configuration
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R-C
Phasor diagram:
I is in phase with the VRand leads the VCby 90o.
I leads Eby 53.13o.
A13.535
I
V050
E
V13.5330
R
V
V87.3640
CV
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R C
15.3 Series Configuration
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25
R-C
Time domain: V050
E
V13.5330
R
V
V87.3640
C
V
Vsin7.70 te
V13.53sin42.42 tvR
V87.36sin56.56 tvC
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R C
15.3 Series Configuration
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R-C
Power:The total power delivered to the circuit is
Or;
W150
13.53cos)5)(50(
cos
T
EIP
W1506522
RIP
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R C
15.3 Series Configuration
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27
R-C
Power factor:
Or;
13.53cos
cos
Tp
F
leading6.0p
F
T
TP
Z
RF cos
leading6.0
10
6
PF
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R L C
15.3 Series Configuration
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R-L-C
1. Phasor Notation
TIME DOMAIN
PHASOR DOMAIN
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R L C Impedance diagram:
15.3 Series Configuration
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R-L-C Impedance diagram:
2. ZT
43
373
90900
321
j
jj
XXR CL
T
ZZZZ
13.535T
Zhttp://modul2poli.blogspot.com/ Page 139 of 241
R-L C
15.3 Series Configuration
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30
R-L-C
3. I
13.535
050
TZ
E
I
A13.5310
I
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R-L-C
15.3 Series Configuration
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31
R-L-C
4. VR, VLand VC
Ohms Law:
V13.5330
R
V
V13.14330
C
V
)907)(13.5310(
LL
IZV
)03)(13.5310(
RR
IZV
)903)(13.5310(
CC
IZV
V87.3670
L
V
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R-L-C
15.3 Series Configuration
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R-L-C
Kirchhoffs voltage law:
Or;
0 CLR VVVEV
CLR VVVE
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R-L-C
15.3 Series Configuration
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33
R-L-C
Phasor diagram:
I is in phase with the VR, lags the VLby 90o, leads the VC by 90
o
I lags Eby 53.13o.
A13.5310
I
V13.5330
R
V
V13.14330
CV
V87.3670
L
V
V050 E
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R-L-C
15.3 Series Configuration
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34
R L C
Time domain:
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R-L-C
15.3 Series Configuration
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35
R L C
Power:The total power delivered to the circuit is
Or;
Power factor:
W30013.53cos)10)(50(cos
TT EIP
W30031022
RIPT
13.53coscos Tp
F
lagging6.0p
F
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The basic format for the VDR in AC circuits is
15.4 Voltage Divider Rule
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The basic format for the VDR in AC circuits is
exactly the same as that for the DC circuits.
Where
Vx: voltage across one or more elements in a series that
have total impedance ZxE: total voltage appearing across the series circuit.
ZT: total impedance of the series circuit.
E
Z
Z
V
T
x
x
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Example 15.11(a)
15.3 Series Configuration
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Example 15.11(a)
Calculate I, VR, VLand VCin phasor form.
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Example 15.11(a) - Solut ion
15.3 Series Configuration
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38
Example 15.11(a) Solut ion
Combined the Rs, Ls and Cs.
T TCT
10 0.1 H 100 mF
202sin377tv i
H1.005.005.0
21
LLLT
1046
21 RRR
T
21
111
CCCT
F100200200
200200
21
21m
CC
CCC
T
e
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Example 15.11(a) Solut ion (contd)
15.3 Series Configuration
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39
Example 15.11(a) Solut ion (cont d)
Find the reactances.
1. Transform the circuit into phasor domain.
7.37
)1.0(377TL
LX
53.26)10100(377
11
6
T
C
CX
V377sin220 te V020
E
i I
T L C
10 37.7 26.53
200V
VI
E
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Example 15.11(a) Solut ion (contd)
15.3 Series Configuration
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40
Example 15.11(a) Solut ion (cont d)
2. Determine the total impedance.
3. Calculate I.
17.1110
53.267.3710
j
jj
jXjXRCLTT
Z
16.4815
TZ
T L C
10 37.7 26.53
200VV
I
16.4815
020
TZ
EI A16.481.33
I
E
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Example 15.11(a) Solut ion (contd)
15.3 Series Configuration
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41
Example 15.11(a) Solut ion (cont d)
4. Calculate VR, VLand VC
V16.483.13
R
V
)010)(16.4833.1(
RR
IZV
T L C
10 37.7 26.53
200VV
I
V84.4114.50
L
V
)907.37)(16.4833.1(
LL
IZV
V16.13828.35
C
V
)9053.26)(16.4833.1(
CC
IZV
E
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15.3 Series ConfigurationExample 15.11(b)
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42
Example 15.11(b)
Calculate the total power factor.
Solut ion
Angle between Eand Iis
16.48coscos Tp
F
lagging667.0p
F
A16.481.33
IV020
E
16.48
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Example 15.11(c)
15.3 Series Configuration
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43
a p e 5 (c)
Calculate the average power delivered to the circuit.
Solut ion
16.48cos)33.1)(20(cos TT EIP
W74.17T
P
A16.481.33
IV020
E
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Example 15.11(d)
15.3 Series Configuration
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44
p ( )
Draw the phasor diagram.
Solut ion
A16.481.33
I
V16.483.13
R
V
V84.4114.50
L
V
V16.13828.35
C
V
V020
E
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Example 15.11(e)
15.3 Series Configuration
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45
p ( )Obtain the phasor sum of VR, VLand VCand show
that it equals the input voltage E.
Solut ion
V933.9894.8V16.483.13 jR
V
V446.33355.37V84.4114.50 jL
V
V534.23284.26V16.13828.35 jC
V
534.23446.33933.9284.26355.37894.8 jjj
CLR
VVVE
V020020021.0965.19
jjEhttp://modul2poli.blogspot.com/ Page 155 of 241
Example 15.11(f)
15.3 Series Configuration
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46
p ( )Find VRand VCusing voltage divider rule.
Solut ion
T L C
10 37.7 26.53
200VV
IE
16.4815
TZ
)020(
16.4815
010
E
Z
ZV
T
R
RV16.483.13
R
V
)020(16.4815
9053.26
E
Z
Z
V
T
C
CV16.13837.35
C
V
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15.6 Summaries of Series AC Circuits
F i AC i it ith ti l t
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47
For a series AC circuits with reactive elements:
The total impedance will be frequency dependent.
The impedance of any one element can begreater than the total impedance of the network.
The inductive and capacitive reactances arealways in direct opposition on an impedancediagram.
Depending on the frequency applied, the samecircuit can be either predominantly inductive orpredominantly capacitive.
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15.6 Summaries of Series AC Circuits
(continued )
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48
(continued)
At lower frequencies, the capacitive elementswill usually have the most impact on the total
impedance.
At high frequencies, the inductive elements willusually have the most impact on the total
impedance.
The magnitude of the voltage across any oneelement can be greater than the applied voltage.
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15.6 Summaries of Series AC Circuits
(continued )
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49
(continued)
The magnitude of the voltage across an elementas compared to the other elements of the circuitis directly related to the magnitude of itsimpedance; that is, the larger the impedance of
an element , the larger the magnitude of thevoltage across the element.
The voltages across an inductor or capacitor arealways in direct opposition on a phasor diagram.
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15.6 Summaries of Series AC Circuits
(continued )
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50
(continued)
The current is always in phase with the voltageacross the resistive elements, lags the voltageacross all the inductive elements by 90, andleads the voltage across the capacitive elements
by 90.
The larger the resistive element of a circuit
compared to the net reactive impedance, thecloser the power factor is to unity.
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1. Explain AC circuit concept and their
analysis using AC circuit law.
2. Apply the knowledge of AC circuit in so
problem related to AC electrical circuit
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RESONANCE
Understandresonance in series
and parallel
circuits
Resonanphenomeno
functio
Effect of chthe frequ
Graph impedanc
frequen
Resonanfrequency e
Determine Q
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Resonance is a condition in RLC circuit in which the cap
inductive reactance are equal in magnitude, thereby res
purely resistive impedance.
Z = R + j( ) ; note: = 0
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Current will be maximum & offering min
impedance.
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Current will be minimum & offering max
impedance.
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Resonance circuit serves as stable freque
source.
Resonance circuit serves as filter.
The
circu
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A series RLC circuits reactance changes you change the voltage sources frequen
Its total impedance also changes.
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At low frequencies, Xc > XL
and the circuprimarily capacitive.
At high frequencies, XL > Xc and the circ
primarily inductive.
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Reactance change as you change the vol
sources frequency.
At low frequencies, XL < Xc and the circu
primarily inductive.
At high frequencies, Xc< XL and the circu
primarily capacitive.
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A series RLC circuit contains both inductive reacand capacitive reactance (Xc).
Since XL and Xc have opposite phase angles, the
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The smaller reactance dominates, since
reactance results in a larger branch curr
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(a) Q factor:
- Q is the ratio of power stored to power dissipated
reactance and resistance.- Q is the ratio of its resonant frequency to its band
SERIES CIRCUIT:
IF;
PARALLEL CIRCUIT:
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PARALLEL CIRCUIT:
(a) Quality factor: the ratio of the circulating branch currents t
current .
(b) Frequency bandwidth, B = f2 f1:
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Bandwidth, f is measured between the 70.7% amof series resonant circuit.
The difference between the two half-powe
Lower cut-off frequency ( ):
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Lower cut-off frequency, (L):
Upper cut-off frequency, (H):
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BW = fc/Q
Where:
fc = resonan
Q = quality f
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In Figure above, the 100% current point is 50 mA. The 70.7% level is 0
mA)=35.4 mA. The upper and lower band edges read from the curve are 291 Hz for f
for fh. The bandwidth is 64 Hz, and the half power points are 32 Hz o
resonant frequency
BW = f = fh-fl = 355-291 = 64
fl = fc - f/2 = 323-32 = 291
(c) The dissipation factor, D:
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- The ratio of the power loss in a dielectric material topower transmitted through the dielectric.
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CHARACTERISTIC SERIES CIRCUIT PARALLEL C
Resonant frequency,
fr
Quality factor,Q
Bandwidth, BW
Half power
frequency,fL & fH&
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A series resonance network consisting of a resistor of 30, a of 2uF and an inductor of 20mH is connected across a sinuso
voltage which has a constant output of 9 volts at all frequen
Calculate, the resonant frequency, the current at resonance
voltage across the inductor and capacitor at resonance, the factor and the bandwidth of the circuit. Also sketch the corr
current waveform for all frequencies.
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Resonant Frequency,r
Circuit Current at Resonance, Im
Inductive Reactance at Resonance, XL
Voltages across the inductor and the
capacitor, VL, VC
Bandwidth, BW
The upper and low
points, H and L
Current Waveform
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A series circuit consists of a resistance of 4, an inductanc
a variable capacitance connected across a 100V, 50Hz supp
the capacitance require to give series resonance and the v
generated across both the inductor and the capacitor.
Solution:
Resonant Frequency, r
Voltages across the inductor and the capacitor, VL, VC
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A parallel resonance network consisting of a resistor of 60, a 120uF and an inductor of 200mH is connected across a sinusoid
voltage which has a constant output of 100 volts at all frequen
Calculate, the resonant frequency, the quality factor and the b
the circuit, the circuit current at resonance and current magni
The upper and lowepoints and
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Resonant Frequency, r
Inductive Reactance at Resonance,XL
Quality factor, Q
Bandwidth, BW
points, H and L
Circuit Current at RAt resonance the dyimpedance of the ciR
Current Magnificatio
We can check this vcalculating the curr
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For resonance to occur in any circuit it must have at
one inductor and one capacitor.
Resonance is the result of oscillations in a circuit as energy is passed from the inductor to the capacitor.
Resonance occurs when XL = XC and the imaginary pa
transfer function is zero.
At resonance the impedance of the circuit is equal to
resistance value as Z = R.
LOGO
CHAPTER 5 : THREE PHASE
SYSTEM
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Objectives:
Know basic principles of 3system
List advantages and application of 3
system compared to 1systemKnow 3e.m.f generation
Identify star & delta connection
Determine VPH, IPH, VL,IL & power in
3system
SYSTEM
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INTRODUCTION
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3system is a combination of three 1system
In 3system balanced system, power
comes from 3AC generator
3generators have 3 coils fixed at 120 to
each other rotating in magnetic field.
3system are use for transmission and
distribution of electricity and also in industry
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ADVANTAGES OF 3SYSTEMvs 1SYSTEM
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Weight less than 1circuit of same powerrating
Have wide range of voltages
Smaller in size &higher power factor thus
more efficient
Steady torque output and ability to self start
Inherent benefits for high power transmission
Produce constant amount of power in theload
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GENERATION OF 3SUPPLY
There are two ways to generate 3supply.
Moving magnetic field while coil is permanent
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Moving magnetic field while coil is permanent.
Moving coil while magnetic field is permanent.
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GENERATION OF 3SUPPLY
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Generation for R phase (same as single phase generation)
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GENERATION OF 3SUPPLY
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Generation for Y phase (coil rotate 120 so that it will equal with Rphase, thus Y lags R by 120)
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GENERATION OF 3SUPPLY
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Generation for B phase (coil rotate 240 so that it will equal with Rphase, thus B lags R by 240)
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GENERATION OF 3SUPPLY
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Phasor diagram for 3 phase system
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CONNECTION IN 3system
Physically 3 system
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Physically 3system
consist ofthree different coils.
Each phase coils have 2
terminal and required 2
conductor as connection
So 6 conductors will be used
as
in 3connection
This kind of connection is
difficult and will cost more.
To overcome this problem,
3
supply usually connected in
DELTAor STARconnection
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CONNECTION IN 3system
a) STAR/ WYE Connection
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Physical connection Conventional representation
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CONNECTION IN 3system
b) DELTA Connection
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Physical connectionConventional representation
diagram
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REVIEW
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Two ways to generate AC rotating coil,permanent magnet field or vice versa.
Each voltage separed by 120
IN equal to zero when load is balanced.
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CALCULATION
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STAR CONNECTION DELTA CONNECTION
Relationship between VLand Vph Relationship between ILand Iph
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FORMULA USE IN THREE PHASE
CALCULATION
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EXERCISE
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