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.)
(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
(Introductory Electronics, Simpson, 2nd Ed.)
(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
(Introductory Electronics, Simpson, 2nd Ed.)
(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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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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