electric circuits shoubra...ch7: two-port circuits 9/6/2013 2 references mahmood nahvi, joseph a....
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Electric CircuitsElectric CircuitsElectric CircuitsElectric CircuitsByBy
Prof. Dr.Prof. Dr. Mousa AbdMousa Abd--AllahAllahDr. Mohammed EisaDr. Mohammed Eisa
ContentsCH1: Sinusoids and PhasorsCH2: Sinusoidal steady-State AnalysisCH3: AC Power AnalysisCH4: Three Phase CircuitsCH5: Transient Response of CircuitsCH6: Fourier Analysis and Circuit
ApplicationsCH7: Two-Port Circuits
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ReferencesMahmood Nahvi, Joseph A. Edminister, “Electric
Circuits”, Schaum’s Outlines, Mc Graw Hill. ISBN: 9780071633727, 2011.
James W. Nilsson, Susan A. Riedel, “Electric Circuits”, Prentice Hall, New Jersey, ISBN:9780137050512-0137050518, 2011.
Jhon Bird, “Electrical Theory and Technology”, Elsevier, Oxford, ISBN:9780080549798-0080549799, 2007.
Chapter IChapter IChapter IChapter ISinuSoidS, PhaSorS and SinuSoidS, PhaSorS and
ResonanceResonance
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sinusoids
Definitions
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Ex.1: Find the amplitude, phase, period, and frequency of the sinusoid: v(t)=12cos(50t+10o)
Solution:The amplitude is: Vm = 12 VThe phase is : = 10o
The angular frequency is = 50 rad/sThe period is: T =(2/) = (2/50) =0.1257 sThe frequency is f =(1/T) = (1/0.1257) = 7.958 Hz
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Ex.2 : Calculate the phase angle between v1 = -10 cos(t+50o) & v2 = 12 sin(t-10o)state which sinusoid is leading
Solution
Phasor diagram
)Im()sin()( )( tjmm eVtVtv
)Im()Im()( tjjtjm VeeeVtv
mj
mj
m VeVeVV )Im(
Phasors
mVV mII
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Ex. 3 : Transform these sinusoids to phasors:(a) v = - 4 sin(30t+50o)(b) i = 6 cos(50t-40o)
Solution(a) v = -4 sin(30t+50o) = 4 cos (30t+50+90o)
= 4 cos (30t + 140o)
The phasor form of v is V = 4140o
(b) i = 6 cos(50t - 40o)The phasor I = 6 -40o
Ex.4: Given i1(t)=4cos(t+30o) andi2(t)=5sin(t-20o), find their sum.
Solution: i1(t)= 4 cos(t+30o) I1 = 430o
i2(t)= 5 sin(t-20o) = 5 cos(t -20o-90o) = 5 cos(t-110o)I2 = 5 -110o
I = I1 + I2 = 430o + 5-110o = 3.464+j2-1.71-j4.698 = 1.754 –j2.698 = 3.218-56.97o
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Phasor relationship for circuit elements
(a) Resistori = Im cos(t + )
v = i R = RImcos(t + )I = Im v = R Im
(b) Inductori = Im cos(t + )v = L(di/dt) = -LIm sin(t + )v = LIm cos(t + +90o)I = Im and V = Vm+90o
The voltage lead the current by 90o → XL= j ω L
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(b) capacitor
)sin()cos( tCVtVdtdci mm
)90cos( tCVi m
mVV 90 mII
The voltage lag the current by 90o → XC= - j 1/ωC
v(t) = Vm cos(ωt+) i = C dv/dt
G conductance, B susceptance
Impedance and Admittance
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Series Circuit
Parallel Circuit
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Δ-Y-Conversion
T
rms dttvT
V0
2 )(1
T
av dttvT
V0
)(1
2m
rmsV
V
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Ex.5: consider the circuit,(a) Calculate the sinusoidal voltages v1 and v2
using phasors and voltage divider rule(b) Sketch the phasor diagram showing E, V1 and
V2.(c) Sketch the sinusoidal waveforms of e, v1 and v2.
Vjj
jV oo 69.334.78)071.70()4030()8040(
80401
Vjj
jV oo 87.829.43)071.70()4030()8040(
40302
Solution:(a) The phasor form of the voltage source is determined as
e = 100 sin(t) E = 70.710o V
and
(b) Phasor diagram
(c) v1(t) = Vm sin(t+) = 2x78.4 sin(t-33.69)= 111 sin(t-33.69)
v2(t) = Vm sin(t+) = 2x43.9 sin(t+82.87)= 62 sin(t+82.87)
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Ex. 6: Refer to the circuit;(a) Find the total impedance, ZT.(b)Determine the supply current, IT.(c) Calculate I1, I2, using current divider rule.(d)Verify Kirchhoff’s current law at node a.
Solution
201
11
11
2011.
jjZa
T kZ o
T 020
AAxxZ
VIb oo
TT 0250010250
0102005. 63
AxI
AxI
AxIc
oo
oo
oo
9050000250901020
9050000250901
020
02500250020020.
3
2
1
d. I1 + I2 + I3 = 2500 + 5000-90 + 500090 = 2500o
IT = I1 + I2 + I3 Kirchhoff’s current law is verified.
Ex. 7: Find the current I in the circuit.
Solution:
8.06.110
8168424
)42(4 jjjjjjZ an
2.310
)8(4 jjZ bn
2.36.110
)42(8 jjZ bn
Z1 = Zan+12 = 13.6 + j0.8 , Z2 = Zbn – j3 = j0.2 and Z3 = Zcn +j6+8 = 9.6+j2.8
16.1336.9
)8.26.9(2.08.06.1332
321 j
jjjj
ZZZZZZT
o
T jZVIcurrentsourceThe 2.4666.3
16.13050
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At balance V1=V2
21
21 ZZ
ZVV s
x
xs ZZ
ZVV
32
3321
2
ZZZ
ZZZ x
21
3 ZZZ
Z x
AC Bridges
Series Resonance
)1(C
LjRZ
At resonance Im(z) = 0
01
CL
CL
oo
1
sradLCo /1
HzLC
f o2
1
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22 )1(C
LR
VI
The highest power is at ωo
RVRIP o
22)(
III 707.02
RVPP
2
21 21)()(
RZ 2
RC
LR 2)1(2
The half power frequencies are at:
ω1 & ω2 can be get by setting
RC
LR 2)1(2
LCLR
LR 1
22
2
1
LCLR
LR 1
22
2
2
21 o
By solving the above equation for ω
Bandwidth “It is defined as the half power bandwidth”B = ω2 – ω1 = (R / L)
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Quality Factor Q“Measure the sharpness of resonance “
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Example: For the circuit shown:(a) Find the resonant frequency and the half power frequencies(b) Calculate the quality factor and bandwidth(c) Determine the amplitude of the current at ωo, ω1 and ω2
Solution
Parallel Resonance
At resonance Im(Y)=0
At resonance:(1)The parallel LC combination acts like open circuit.
i.e. the entire current flows through R(2) The inductor and capacitor currents can be much
more than the source current
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Example: For the circuit shown:(a) Calculate ωo, Q and B(b) Find ω1 and ω2(c) Determine the power dissipated at ωo, ω1 and ω2
Solution
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WxxRIP o3232 105.12)8000()1025.1(
WxxR
VP 3
22
1025.680002
102