ET201 Electrical Circuits

7/23/2019 ET201 Electrical Circuits

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ELECTRICAL CIRCUIT

ET 201

Define and explain characteristics of

sinusoidal wave, phase relationships

and phase shifting

http://modul2poli.blogspot.com/ Page 1 of 241


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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

http://en.wikipedia.org/wiki/Electric_chargehttp://en.wikipedia.org/wiki/Electric_charge


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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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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 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.

Frequency-to-period and period-to-frequency conversions are common in

electronic calculations.http://modul2poli.blogspot.com/ Page 11 of 241


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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.


13 2 Si id l AC V lt


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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



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



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






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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



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.






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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)



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



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.



Peak value, Vm= 8 V




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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


13 2 Sinusoidal AC Voltage


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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






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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






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Plotting a sine wave versus (a) degrees and (b) radians.


13 2 Si id l AC


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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






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34T

2 or f 2 (rad/s)



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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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






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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






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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.



T

dttvT

valueaverage0

)(1

_




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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;



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.






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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






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Example 13.21solut ion

2dcrms

mIII 2

dcrmsmEEE

mA43.42

30414.1

2 rms

IIm



V7.169

120414.1

2 rms

EEm


13 5 General Format for the


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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



= 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






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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.






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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



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



(a) But is in radian, so must be calculate in radian:

(b) Given, , sot

t

rad73.242.156rad412.058.23377

2

1

t


13 6 Phase Relationship


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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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tAa m sin

The unshifted sinusoidal waveform is

represented by the expression:

t




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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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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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t

If the waveform passes through the horizontal axis

with a positive-going slope after 0

:

tAa m sin




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t

2

cos90cossin

cos2

sin90sin

ttt

ttt




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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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+ cos

+ sin

- cos

- sin

cos (-90o)

sin (+90o)

90sin270sincos

180sinsin90cossin

90sincos

Start at + sin position;

Note:

sin (- ) = - sin

cos(- ) = cos




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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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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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70sin5

30sin10(a)

ti

tv

Example 13.2so lu t ion

ileadsvby 40or

vlagsiby 40




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Example 13.2so lu t ion (contd)

ileadsvby 80or

vlagsiby 80

20sin10

60sin15(b)

tv

ti




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Example 13.2solut ion (contd)

ileadsvby 110or

vlagsiby 110

10sin3

10cos2(c)

tv

ti




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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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2

(CHAPTER 1.7)

OSCILLOSCOPE TOMEASURE WAVEFORMS


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
http://en.wikipedia.org/wiki/Waveformhttp://en.wikipedia.org/wiki/Waveformhttp://en.wikipedia.org/wiki/Electronic_test_instrumenthttp://en.wikipedia.org/wiki/Electronic_test_instrumenthttp://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/Voltagehttp://en.wikipedia.org/wiki/Electronic_test_instrumenthttp://en.wikipedia.org/wiki/Electronic_test_instrumenthttp://en.wikipedia.org/wiki/Waveformhttp://en.wikipedia.org/wiki/Waveform


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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.


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


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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4

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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5

g

Example 14.13(a)

Sketch the complex number C= 3 +j4in the

complex plane

Solut ion




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6

g

Example 14.13(b)

Sketch the complex number C= 0j6in the

complex plane

Solut ion




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7

g

Example 14.13(c)

Sketch the complex number C= -10

j20in

the complex plane

Solut ion


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


2.2 Polar Form


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Example 14.14(a)

305C

Counterclockwise (CCW)


2.2 Polar Form


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Example 14.14(b)

1207 C

Clockwise (CW)


2.2 Polar Form 180 qq ZZC


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Example 14.14(c)

602.4 C

180602.4

2402.4

180 qq ZZC


14.9 Conversion Between Forms


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1. Rectangular to Polar




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2. Polar to Rectangular




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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



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Example 14.16

Convert C= 1045to rectangular form

07.745cos10 X

Solut ion

07.745sin10 Y

07.707.7 jC




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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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Example 14.18

Convert C= 10 230to rectangular form

43.6230cos10

Solut ion

66.7230sin10 Y

66.743.6 jC


2.4 Mathematical Operations with


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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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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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Complex Conjugate

In rectangular form, the

conjugate of:

C= 2 +j3

is 2j3

Complex Numbers




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Complex Conjugate

In polar form, the

conjugate of:

C= 2 30o

is 2 30o

Complex Numbers




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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



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



C l N b


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Example 14.19(a)

13;4221 jj CC

143221 jCC

55 j

Find C1+ C2.

Solut ion

Complex Numbers



C l N b


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Example 14.19(b)

36;63 21 jj CC

366321

jCC

93 j

FindC

1+ C

2

Solut ion

Complex Numbers



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


C l N b


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Example 14.20(a)

41;6421 jj CC

461421

jCC

23 j

FindC

1- C

2

Solut ion

Complex Numbers


2.4 Mathematical Operations withC l N b


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Example 14.20(b)

52;3321 jj CC

532321

jCC

25 j

FindC

1- C

2

Solut ion

Complex Numbers




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Example 14.21(a)

455453452

Complex Numbers

NOTE



unless they differ only by multiples of 180http://modul2poli.blogspot.com/ Page 98 of 241



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Complex Numbers

06180402

NOTE



unless they differ only by multiples of 180

Example 14.21(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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Example 14.22(a)

105;3221 jj CC

Find C1C2.Solut ion

1053221 jj CC

3520 j

Complex Numbers




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Example 14.22(b)

64;3221 jj CC

Find C1C2.

Solut ion

643221 jj CC

1802626

Complex Numbers


2.4 Mathematical Operations withComplex Numbers


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Example 14.23(a)

3010;20521 CC

Find C1C2.

Solut ion

302010521

CC

5050

Complex Numbers




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Example 14.23(b)

1207;40221

CC

Find C1C2.

Solut ion

120407221

CC

8014

Complex 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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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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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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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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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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)()( 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.


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


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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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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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


15.3 Series AC Circuit


117/241

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


15.3 Series Configuration


118/241

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


14.5 Power FactorF f tP


119/241

9

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


14.5 Power Factor


120/241

10

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




121/241

11

R-L

1. Phasor Notation

te sin4.141 0V100 E

Series R-Lcircuit Apply phasor notation




122/241

12

R-L

2. ZT

Impedance diagram:

43

)904()03(

21

j

T

ZZZ

13.535 T

Z




123/241

13

R-L

3. I

13.535

0V100

T

Z

E

I

13.53A20 I


R L



124/241

14

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


R L



125/241

15

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


R L



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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


R L



127/241

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


R L



128/241

18

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


R C



129/241

19

R-C

1. Phasor Notation

A13.53sin07.7 ti A13.535 I

Series R-Ccircuit Apply phasor notation


R-C



130/241

20

R-C

2. ZT

Impedance diagram:

86

)908()06(

21

j

T

ZZZ

13.5310

TZ


R C



131/241

21

R-C

3. E

)13.5310)(13.535(

T

IZE

V050

E


R C



132/241

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


R C



133/241

23

R-C


Or;

0 CR VVEV

CR VVE


R C



134/241

24

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


R C



135/241

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


R C



136/241

26

R-C

Power:The total power delivered to the circuit is

Or;

W150

13.53cos)5)(50(

cos

T

EIP

W1506522

RIP


R C



137/241

27

R-C

Power factor:

Or;

13.53cos

cos

Tp

F

leading6.0p

F

T

TP

Z

RF cos

leading6.0

10

6

PF


R L C



138/241

28

R-L-C

1. Phasor Notation

TIME DOMAIN

PHASOR DOMAIN


R L C Impedance diagram:



139/241

29

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



140/241

30

R-L-C

3. I

13.535

050

TZ

E

I

A13.5310

I


R-L-C



141/241

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


R-L-C



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32

R-L-C


Or;

0 CLR VVVEV

CLR VVVE


R-L-C



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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


R-L-C



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34

R L C

Time domain:


R-L-C



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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


The basic format for the VDR in AC circuits is

15.4 Voltage Divider Rule


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36

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


Example 15.11(a)



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37

Example 15.11(a)

Calculate I, VR, VLand VCin phasor form.


Example 15.11(a) - Solut ion



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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


Example 15.11(a) Solut ion (contd)



149/241

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





150/241

40


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





151/241

41


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


15.3 Series ConfigurationExample 15.11(b)


152/241

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


Example 15.11(c)



153/241

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


Example 15.11(d)



154/241

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


Example 15.11(e)



155/241

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)



156/241

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


15.6 Summaries of Series AC Circuits

F i AC i it ith ti l t


157/241

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.



(continued )


158/241

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.



(continued )


159/241

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.



(continued )


160/241

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.



161/241


162/241

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


163/241

RESONANCE

Understandresonance in series

and parallel

circuits

Resonanphenomeno

functio

Effect of chthe frequ

Graph impedanc

frequen

Resonanfrequency e

Determine Q


164/241


165/241

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


166/241

Current will be maximum & offering min

impedance.


167/241

Current will be minimum & offering max

impedance.


168/241

Resonance circuit serves as stable freque

source.

Resonance circuit serves as filter.

The

circu


169/241

A series RLC circuits reactance changes you change the voltage sources frequen

Its total impedance also changes.


170/241

At low frequencies, Xc > XL

and the circuprimarily capacitive.

At high frequencies, XL > Xc and the circ

primarily inductive.


171/241

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.


172/241

A series RLC circuit contains both inductive reacand capacitive reactance (Xc).

Since XL and Xc have opposite phase angles, the


173/241

The smaller reactance dominates, since

reactance results in a larger branch curr


174/241


175/241


176/241

(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:


177/241

PARALLEL CIRCUIT:

(a) Quality factor: the ratio of the circulating branch currents t

current .

(b) Frequency bandwidth, B = f2 f1:


178/241

Bandwidth, f is measured between the 70.7% amof series resonant circuit.

The difference between the two half-powe

Lower cut-off frequency ( ):


179/241

Lower cut-off frequency, (L):

Upper cut-off frequency, (H):


180/241

BW = fc/Q

Where:

fc = resonan

Q = quality f


181/241


182/241

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:


183/241

- The ratio of the power loss in a dielectric material topower transmitted through the dielectric.


184/241

CHARACTERISTIC SERIES CIRCUIT PARALLEL C

Resonant frequency,

fr

Quality factor,Q

Bandwidth, BW

Half power

frequency,fL & fH&


185/241

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.


186/241

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


187/241

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


188/241

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


189/241

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


190/241

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


191/241

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


INTRODUCTION


192/241

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


ADVANTAGES OF 3SYSTEMvs 1SYSTEM


193/241

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


GENERATION OF 3SUPPLY

There are two ways to generate 3supply.

Moving magnetic field while coil is permanent


194/241

Moving magnetic field while coil is permanent.

Moving coil while magnetic field is permanent.




195/241

Generation for R phase (same as single phase generation)




196/241

Generation for Y phase (coil rotate 120 so that it will equal with Rphase, thus Y lags R by 120)




197/241

Generation for B phase (coil rotate 240 so that it will equal with Rphase, thus B lags R by 240)




198/241

Phasor diagram for 3 phase system


CONNECTION IN 3system

Physically 3 system


199/241

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



a) STAR/ WYE Connection


200/241

Physical connection Conventional representation

diagramhttp://modul2poli.blogspot.com/ Page 200 of 241


b) DELTA Connection


201/241

Physical connectionConventional representation

diagram


REVIEW


202/241

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.


CALCULATION


203/241

STAR CONNECTION DELTA CONNECTION

Relationship between VLand Vph Relationship between ILand Iph


FORMULA USE IN THREE PHASE

CALCULATION


204/241



205/241

EXERCISE



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ET201 Electrical Circuits

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Transcript of ET201 Electrical Circuits