dasar teknik elektro 2

8/22/2019 Dasar Teknik Elektro 2

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Dasar Teknik Elektro

EL.112 (3 sks)Enjang A.Juanda/ Lukmanul Hakim


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Silabus Mata Kuliah

Secara garis besar disajikan:1. Pengantar Teknik Elektro.

2. Dasar-dasar rangkaian listrik.3. Respon rangkaian bolak-balik pada

kondisi steady state.4. Pengantar system.5. Dasar elektronika.

6. Dasar komponen elektronika

semikonduktor.7. Pengantar analisa jaringan.8. Dasar elektronika digital &

mikroprosesor.9. Penguat OP-Amp.


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Tujuan

Setelah selesai mengikuti mata kuliah inim a h a s i s w a d i h a r a p k a n m a m p umenjelaskan dasar teknik elektro dansedapat mungkin mempraktekkan bagian-bagian yang praktisnya tentang dasart e k n i k e l e k t r o .

Evaluasi

- Kehadiran- Tugas Presentasi dan diskusi- Makalah- UTS

- UAS


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Rincian BahanI). Membahas silabus perkuliahan dan mengakomodasikanberbagai masukan dari mahasiswa untuk memberi kemungkinan

revisi terhadap pokok bahasan yang dianggap tidak penting danmemasukkan pokok bahasan yang dianggap penting. Sesuaidengan apa yang dikemukakan dalam silabus, pada pertemuanini dikemukakan pula tujuan, ruang lingkup, prosedurperkuliahan, penjelasan tentang tugas yang harus dilakukanmahasiswa, ujian yang harus diikuti termasuk jenis soal dan caramenyelesaikan/ menjawab pertanyaan, dan sumber-sumber.Terakhir, menyampaikan uraian pendahuluan tentangDasarTeknik Elektro/ Pengantar Teknik Elektro.II). Pengertian dan definisi-definisi yang terkait dengan dasarteknik elektro.

III). Rangkaian-rangkaian listrik dasar: DC dan sistem DC.IV). Rangkaian-rangkaian listrik dasar: AC dan sistem AC.V). Pengertian sistem, piranti, komponen dan kaitan satu samalain dalam teknik elektro.VI). Dasar elektronika

VII). Dasar semikonduktor dan komponen semikonduktorVIII). UTS.


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Lanjutan Rincian

IX). Pengenalan pesawat-pesawat elektronikaX). Pengantar analisa jaringan

XI). Dasar-dasar teknik dijital

XII). Komponen-komponen dijital dan dasar-dasar analisis

rangkaian dijitalXIII). Dasar-dasar rangkaian dijital

XIV). Sejarah dan dasar teknik mikroprosesor

XV). Dasar teknik mikroprosesor dan pemrogramannya

XVI). Dasar penguat Op-Amp

XVII). UAS


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Daftar Pustaka

Sumber Utama:Ralph J.Smith & Richard C.Dorf:Circui ts , Devices, and Sys tems,John Wiley & Sons,1995.

J.R.Cogdell: Foundation o fElectr ic al Engineering, PrenticeHall,1995.

David E.Johnson, JohnyR.Johnson, John L.Hilburn :

Electr ic Circu i t Analysis, PrenticeHall.


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Referensi/PengayaanReferensi:1. P.H. Smale, , Telecommunicat ion System I, Pitman Publishing

Limited, London, 1978.

2. R.Margunadi, Pengantar Umum Elektroteknik, P.T.Dian Rakyat,Jakarta, 1986.

3. Allen, Mottershead, Electronic Devices and Circuits , an introdu ct ion,Prentice-Hall of India, New Delhi, 1976.

4. Enjang A. Juanda dan Jaja Kustija, Pengantar Elektro Teknik, JPTE-FPTK-IKIP, Bandung, 1994.

5. A.P. Malvino, Electron ics Priciples, Mc.Graw-Hill Company, London,1985

6. Brian Moore and John Donaghy, Operat ional Am pli f ier Circuits,Heinemann, London, 1986.

7. Archie W.Culp,Jr (Terjemahan: Ir. Darwin Sitompul M.Eng), Pr insip-pr ins ip K onvers i Energi , Penerbit Erlangga, Jakarta, 1985.

- Jurnal1. IEEE, Telecommunication Transactions.- InternetDosen dapat dihubungi melalui:Alamat rumah dan telpon: Jl. Suryalaya IX No.31 Bandung 40265-

T.7310350Alamat e-mail: [email protected]


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Appersepsi


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ARUS SEARAH

(DC)

(Arus dan Tegangan Listrik)


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Arus ListrikQ = 1.6 X 10 19 coulomb

i = dq/dt

1 A = 1 coulomb/det

Tipe Besar Arus:

Stasiun Pembangkit : 1000 A

Starter Mobil : 100 A

Lampu Dop : 1 A

Radio Mini : 10 mA

Jam Tangan : 1 mikroA


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DC Circuits: Review Current: The rate of flow of electric charge past a

point in a circuit Measured in amperes (A)

1 A = 1 C/s = 6.25 1018 electrons per second

Current direction taken as directionpositive charges flow

Analogous to volume flow rate (volume/unit time) of waterin a pipe

Voltage: Electrical potential energy per unit charge Measured in volts (V): 1 V = 1 J/C

Groundis the 0 V reference point

Analogous to water pressure

Resistance: Restriction to charge flow Measured in ohms (W)

Analogous to obstacles that restrict water flow


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A Simple DC Circuit

Resistors have a constant resistance over a broad

range of voltages and currents

Then withR = constant (Ohms law)

Power = rate energy is delivered to the resistor =

rate energy is dissipated by the

resistor

IRV

V V

R

VRIIVP

22


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Hukum Ohm

Bahwa

I V (hub. linier)

Ditulis: V = I R

Atau : R = V/I


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Bahan Non-Linier

i

v0


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Power (Daya):energi yang diberikan pada elektron tiap satuan waktu

P = v dq/dt

= v I

1 watt = 1 volt x 1 A

Contoh Daya :

Generator : 300 MW

Radiator : 1000 W

Lampu Senter : 6 W

Jam Tangan : 10 mikroW


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Ideal Voltage and Current Sources An ideal voltage source is a source of voltage with

zero internal resistance (a perfect battery) Supply the same voltage regardless of the amount of

current drawn from it

An ideal current source supplies a constant current

regardless of what load it is connected to Has infinite internal resistance

Transistors can be represented by ideal current sources

(Introductory Electronics, Simpson, 2nd Ed.)


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Ideal Voltage and Current Sources Load resistanceR

Lconnected to terminals of a real

current source: Larger current is throughthe smaller resistance

Current sources can always be converted to voltage

sources

Terminals AB actelectrically exactly

like terminals AB


(Introductory Electronics, Simpson, 2nd

Ed.)


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Contoh:Berapa arus yang mengalir pada resistor???

a. VA = 6 V

VB = 2 V

VAB = 4 V

I = 4 A

b.


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Komponen Rangkaian DCa) baterai, b) resistor dan c) kabel penghubung

Resistor standar:

Toleransi 10%

10,12,15,18,22,27,33,39,

47,56,68,dan 82

Short circuit (hubung

singkat: V =0 (R = 0)

Open Circuit (hubung

terbuka): I = 0 (R = ~)


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Hukum Kirchhoff

I. Total arus pada

suatu titik

cabang = 0

I = 0

II. Total penurunan

tegangan pada

rangkaiantertutup = 0

V = 0


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Resistor Seri dan Paralel

Seri: masing-masing

dilewati arus

yang sama

RT = R1 +R2 +R3

Paralel: masing-masing

mendapattegangan yang

sama

1/RT = 1/R1 + 1/R2 + 1/R3


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Penyederhanaan Rangkaian


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Bagaimana resistansi

tergantung pada dimensi

R = l/A

(tergantung

dimensi)

Seri?

Paralel?


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Pembagi Potensial

(Potential Divider)


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Voltage Divider Voltage divider: Circuit that produces a predictable

fraction of the input voltage as the output voltage Schematic:

Current (same everywhere) is:

Output voltage (Vout) is then given by:

R1

R2

21

in

RRVI

in21

2

2out

VRR

RIRV

(Student Manual for The Art of

Electronics, Hayes and Horowitz,

2nd Ed.)


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Voltage Divider Easier way to calculate Vout: Notice the voltage drops

are proportional to the resistances For example, ifR1 =R2then Vout = Vin / 2

Another example: IfR1 = 4 W andR2 = 6 W,

then Vout = (0.6)Vin

Now attach a load resistorRLacrossthe output:

You can modelR1 andRL as one resistor (parallel

combination), then calculate Vout for this new voltage divider

R1

R2

R1

R2 RL

R1

= R2 RL


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Tugas :

Tentukan besarnya v3 !


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Pembagi tegangan terbebani


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Penyederhanaan Rangkaian


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Voltage Division

v1 isR1

v2 isR2

and

vs v1 v2 is(R1R2)

is

vs

R1 R2

v1 vsR1

R1 R2v2 vs

R2

R1 R2

Applying KVL to the loop,

Combining these yields the basic voltage division formula:

and


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v1 10 V8 kW

8 kW 2 kW 8.00 V

Using the derived equations

with the indicated values,

v2 10 V2 kW

8 kW 2 kW 2.00 V

Design Note: Voltage division only applies when both

resistors are carrying the same current.

Voltage Division (cont.)


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Teorema TheveninJika suatu kumpulan rangkaian sumber potensial dan resistor

dihubungkan dengan dua terminal keluaran, maka rangkaian

tersebut dapat digantikan dengan sebuah rangkaian seri dari

sebuah sumber potensial rangkaian terbuka dan sebuah resistor


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Thevenin and Norton Equivalent

Circuits


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Find the Thevenin Equivalent

VoltageProblem: Find the Thevenin

equivalent voltage at the output.

Solution:

Known Information and Given

Data: Circuit topology andvalues in figure.

Unknowns: Theveninequivalent voltage vTH.

Approach: Voltage source vTH

is defined as the output voltagewith no load.

Assumptions: None.

Analysis:Next slide

Th i Th


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Thevenins Theorem Thevenins Theorem: Any combination of voltage

sources and resistors with 2 terminals is electrically

equivalent to an ideal voltage source in series with a

single resistor

Terminals AB electrically equivalent to terminals AB

Thevenin equivalentV

Th andR

Th given by:)circuitopen(Th VV

(output voltage with no load attached) )circuitshort(

)circuitopen(Th

I

VR

I(short circuit) = current when the output is

shorted directly to ground

(Introductory Electronics,

Simpson, 2nd Ed.)

RTh

VTh

Th i Th


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Thevenins Theorem Thevenins theorem applied to a voltage divider:

Thevenin equivalent circuit:

Note thatRTh =R1R2

Imagine mentally shorting out the voltage source

ThenR1 is in parallel withR2

Only works for constant (independent) voltage sources (batteries)

R1

R2

in

21

22outTh VRR

RIRVV

1

in)circuitshort( R

VI

21

21ThTh

)circuitshort()circuitshort(

)circuitopen(

RR

RR

I

V

I

VR

RTh

VTh


(a load resistanceRL

can then be attached

between terminals A

and B, in series withRTh)


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Contoh:Dengan menggunakan teorema Thevenin, hitung besarnya arus I2

pada rangkaian di bawah ini


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Menurut

Thevenin

Vo/c = 5,455 V

RP = 5,455 k Ohm

I2 = 0,397 mA

E l P bl #1 9


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Example Problem #1.9

Solution (details given in class):

(a) 15 V

(b) 10 V

(c) VTh = 15 V,RTh = 5k

(d) 10 V

(e) PL = 0.01 W,PR2 = 0.01 W,PR1 = 0.04 W

(The Art of Electronics, Horowitz and Hill, 2nd Ed.)

E l P bl #1 7


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Example Problem #1.7

Solution (details given in class):

1V source: 0.667 V

10k10k voltage divider: 0.4 V


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THVENIN EQUIVALENT

CIRCUITS


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The Thvenin voltage is equal to the open-circuit

phasor voltage of the original circuit.

ocVV t

We can find the Thvenin impedance by

zeroing the independent sources and

determining the impedance looking into thecircuit terminals.


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The Thvenin impedance equals the open-circuit

voltage divided by the short-circuit current.

scsc

oc

I

V

I

V ttZ

sc

II n

CURRENT DEVIDER


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CURRENT DEVIDER

(Pembagi Arus)

io = iI (G2/G1+G2)

C t


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Current

Division (cont.)

i1 5 ma

3 kW

2 kW 3 kW 3.00 mA

Using the derived equations

with the indicated values,

Design Note: Current division only applies when the same

voltage appears across both resistors.

i2 5 ma2 kW

2 kW 3 kW 2.00 mA


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Current Division

is i1 i2

and

i1 isR2

R1 R2

Combining and solving for vs,

Combining these yields the basic current division formula:

where

i2 vs

R2

i1

vs

R1

and

vs i

s1

1

R1

1

R2

is

R1R2

R1 R2 i

sR1 || R2

i2 i

s

R1

R1 R

2

Nortons Theorem


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Norton s Theorem Nortons Theorem: Any combination of voltage

sources and resistors with 2 terminals is electrically

equivalent to an ideal current source in parallel witha single resistor

Terminals AB electrically equivalent to terminals AB

Norton equivalentINand

RNgiven by:

)circuitshort(

)circuitopen(Th

I

VRRN

(same as Thevenin equivalent resistance)

(Introductory Electronics,

Simpson, 2nd

Ed.)

IN R

N

N

NR

VI

)circuitopen(

Nortons Theorem


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Nortons theorem applied to a voltage divider:

Norton equivalent circuit:

The Norton equivalent circuit is just as good as the

Thevenin equivalent circuit, and vice versa

Norton s Theorem

R1

R2

1

in

R

V

IN

21

21)circuitopen(

RR

RR

I

VR

N

N

RNI

N


(a load resistanceRL

can then be attached

between terminals A

and B, inparallelwithRN

)


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TRANSMISI LISTRIK


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TRANSMISI LISTRIK

K t k i T f t


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Konstruksi Transformator

Persamaan gelombang sinus:

)(sin tay


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Persamaan gelombang sinus: )(sin tay

ty sin4

ty sin2

2 gelombang dengan amplitudo berbeda tetapi berfase awal sama


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2 gelombang dengan amplitudo sama tetapi berfaseawal berbeda

ty sin4

)4/(sin4 ty


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)2/(sin41 ty

)2/(sin42 ty

2 gelombang dengan amplitudo sama tetapi berfaseawal berbeda

Bagaimana jika y1 + y2 ?


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Superposisi dua gelombang

Diagram Phasor:


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Rangkaian resistor murni

i dan v sefase

tIi

tVv

sin

sin


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Rangkaian Kapasitor Murni

i mendahului v sebesar 90o

CXc

tXc

Vi

tVv

/1

cos

sin


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Rangkaian Induktor Murni

v mendahului i sebesar 90o

LXL

tXLIvtIi

cossin


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Rangkaian RC untuk


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Rangkaian RC untuk

Tapis Lolos Rendah

vi = vR + vC

Sudut fase diambil dari input ke output


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Perbandingan keluaran dan masukan:

2

2

)/(1/1/

)/(

)(/1

)(1/1/

oio

o

o

io

VV

tg

RXCsaatRCuntuk

CRVV

o

o

o

io

i i i

i i

iuntukVVBagaimana

)

)

)/

Grafik vo/vi terhadap frekuensi


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o i p

(di kertas semilog: linier-logaritmik

Frekuensi 3 dB


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Frekuensi 3 dB

Besarnya Penguatan (Gain) biasa

dinyatakan dengan dB, dengan definisi:

Untuk : io VVdB /log20 10

dBfrekuensidisebutatau

dBGaindiperoleh

RXsaatVV

o

Cioo

3

3

2/1/

Rangkaian RC untuk Tapis


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Rangkaian RC untuk Tapis

Lolos Tinggi

vi = vR + vC

Sudut fase diambil dari input ke output


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Perbandingan keluaran dan masukan:

2

2

)/(1/1/

)/(

)(/1

)/1(1/1/

oio

o

o

io

VV

tg

RXCsaatRCuntuk

CRVV

o

o

o

io

i i i

i i

iuntukVVBagaimana

)

)

)/

Grafik vo/vi terhadap frekuensi


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G a o/ i te adap e ue s


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Steady-State Sinusoidal Analysis

Sinusoidal Currents and Voltages

PhasorsComplex Impedances

Circuit Analysis with Phasors&Complex Impedances

Power in AC Circuits

Thevenin and Norton Equivalent CircuitsBalanced Three-Phase Circuits


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SINUSOIDAL CURRENTS


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SINUSOIDAL CURRENTSAND VOLTAGES

Vmis the peak value

is the angular frequency in radiansper second

is the phase angle

Tis the period

1


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T

2

f 2

90cossin zz

Frequency

Tf

1

Angular frequency


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Root-Mean-Square Values

dttvT

V

T

2

0

rms

1

R

VP

2

rms

avg

dttiT

I

T

2

0

rms

1

RIP

2

rmsavg


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RMS Value of a Sinusoid

2rms

mVV

The rms value for a sinusoid is the peakvalue divided by the square root of two.

This is not true for other periodic

waveforms such as square waves ortriangular waves.


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Phasor Definition

111 cos:functionTime tVtv

111:Phasor VV

Adding Sinusoids Using


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dd g S uso ds Us gPhasors

Step 1: Determine the phasor for each term.

Step 2: Add the phasors using complex

arithmetic.Step 3: Convert the sum to polar form.

Step 4: Write the result as a time function.

Using Phasors to Add


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Us g aso s to ddSinusoids

45cos201 ttv

60cos102 ttv

45201 V

30102 V

VVV


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7.3997.29

14.1906.23

5660.814.1414.1430104520

21s

j

jj

VVV

7.39cos97.29 ttvs


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Sinusoids can be visualized as the real-

axis projection of vectors rotating in the

complex plane. The phasor for a sinusoid

is a snapshot of the correspondingrotating vector at t= 0.

Ph R l ti hi


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Phase Relationships

To determine phase relationships from aphasor diagram, consider the phasors to

rotate counterclockwise. Then when standing

at afixed point, ifV1 arrives first followed by V2after a rotation of, we say that V1 leads V2by . Alternatively, we could say that V2 lags

V1 by . (Usually, we take as the smallerangle between the two phasors.)


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To determine phase relationships betweensinusoids from their plots versus time, find

the shortest time interval tpbetween positive

peaks of the two waveforms. Then, the

phase angle is

= (tp/T) 360. If the peak ofv1(t)occurs

first, we say that v1(t)leads v2(t)or that v2(t)

lags v1(t).


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COMPLEX IMPEDANCES

LL Lj IV

90 LLjZL

LLL Z IV


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93/135

CCC Z IV

90111

CCjC

jZC

RR RIV


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Kirchhoffs Laws in Phasor


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Form

We can apply KVL directly to phasors.

The sum of the phasor voltages equals

zero for any closed path.

The sum of the phasor currents entering a

node must equal the sum of the phasor

currents leaving.

Circuit Analysis Using


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Circuit Analysis Using

Phasors and Impedances

1. Replace the time descriptions of thevoltage and current sources with the

corresponding phasors. (All of the

sources must have the same frequency.)

2. Replace inductances by their complex


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p y p

impedances ZL=jL. Replace

capacitances by their complex impedancesZC= 1/(jC). Resistances have impedances

equal to their resistances.

3.Analyze the circuit using any of the techniquesstudied earlier in Chapter 2, performing thecalculations with complex arithmetic.


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AC Power Calculations

cosrmsrmsIVP

cosPF

iv

sinrmsrmsIVQ

rmsrmspowerapparent IV


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rmsrmsppp

2

rmsrms

22

IVQP

RIP

2

rms

XIQ2

rms

R

V

PR

2

rms

X

V

QX

2

rms


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Maximum Average PowerT f


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Transfer

If the load can take on any complex value,

maximum power transfer is attained for a load

impedance equal to the complex conjugate ofthe Thvenin impedance.

If the load is required to be a pure

resistance, maximum power transfer isattained for a load resistance equal to the

magnitude of the Thvenin impedance.


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BALANCED THREE-PHASE


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BALANCED THREE PHASECIRCUITS

Much of the power used by business and

industry is supplied by three-phasedistribution systems. Plant engineers need to

be familiar with three-phase power.


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Ph S


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Phase Sequence

Three-phase sources can have either

a positive or negative phase

sequence.The direction of rotation of certain

three-phase motors can be reversed

by changing the phase sequence.


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W W C ti


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WyeWye ConnectionThree-phase sources and loads can beconnected either in a wye configuration or in a

delta configuration.

The key to understanding the various three-

phase

configurations is a careful examination of thewyewye circuit.


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cos3 rmsrmsavg LY IVtpP

sin3sin2

3 rmsrms LYLY IV

IVQ


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ZZ 3


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YZZ 3


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dasar teknik elektro 2

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