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Transcript of Ee602 Ac Circuit
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CIRCUITANALYSIS
INTRODUCTION TOAC CIRCUITS
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CHAPTER GUIDELINE
Introduction to alternating current
Characteristics of Sinusoidal Function
Phasor representation of sinusoids
Impedance and reactance
Parallel and series combinations in the frequencydomain
Ohms law and Kirchoffs law in the frequency domain
Delta-Wye Transformation
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COURSE OUTCOMES (CO)
Understand the concept of ac Identify the characteristics of ac signal & sinusoidalfunctions
Understand the concept of phasor and its application
in ac circuits Understand the concept of impedance, admittance,reactance and susceptance
Understand the Ohms law and Kirchoffs law in accircuit and relate them with dc circuit
Construct an AC circuit to investigate the relationbetween sinusoidal signals and phasor
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PHASOR REPRESENTATION OF SINUSOIDS
Sinusoids are easily expressed in terms of phasors
A phasor is a complex number that represents the amplitude and phaseof a sinusoid.
It can be represented in one of the following three forms:
rzjrez
)sin(cos jrjyxz a. Rectangular
b. Polar
c. Exponential 22 yx
zofmagnituder
xy
zofphasethe
1tan
where
1j
cosrx
sinry
jj
1
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PHASOR REPRESENTATION OF SINUSOIDS
Mathematic operation of complex number:1. Addition
2. Subtraction
3. Multiplication
4. Division
5. Reciprocal
6. Square root
7. Complex conjugate
8. Eulers identity
)()( 212121 yyjxxzz
)()(212121 yyjxxzz
212121 rrzz
21
2
1
2
1 r
r
z
z
rz
11
2 rz
jrerjyxz
sincos je j
(Rectangular form)
(Rectangular form)
(Polar form)
(Polar form)
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EXAMPLE
Evaluate the following complex numbers:
a.
b.
Solution:
a. 15.5 + j13.67
b. 8.293 + j2.2
]605j4)1j2)([(5o
oo
3010j43
403j510
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PHASOR REPRESENTATION OF SINUSOIDS
Transform a sinusoid to and from the time domain to thephasor domain:
(time domain) (phasordomain)
)cos()(
tVtv m
mVV
Amplitude and phase difference are two principal concerns in thestudy of voltage and current sinusoids.
Phasor will be defined from the cosine function in all our proceeding
study. If a voltage or current expression is in the form of a sine, it will
be changed to a cosine by subtracting from the phase.
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PHASOR REPRESENTATION OF
SINUSOIDS
Time domain
representation
Phasor domain
representation
)cos( tVm
)sin( tVm
)cos( tIm
)sin( tIm
mV
o
mV 90
mI
omI 90
Sinusoid-phasor transformation
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EXAMPLE
Transform the following sinusoids to phasors:i = 6cos(50t 40o) A
v =4sin(30t + 50o) V
Solution:
a. I A
b. Since sin(A) = cos(A+90o);
v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V
Transform to phasor => V V
406
1404
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EXERCISE:
Find the sinusoids corresponding to these phasors:a.
b. V3010 V
Aj12)j(5 I
Solution:
a) v(t) = 10cos(t + 210o) V
b) Since
i(t) = 13cos(t + 22.62o) A
22.6213)
12
5(tan512j512 122I
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PHASOR REPRESENTATION OF SINUSOIDS
The differences between v(t) and V:
v(t) is instantaneous or time-domainrepresentationV is the frequency or phasor-domain representation.
v(t) is time dependent, V is not. v(t) is always real with no complex term, V is
generally complex.
Note: Phasor analysis applies only when frequency is constant; whenit is applied to two or more sinusoid signals only if they have
the same frequency.
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Relationship between differential, integral operation
in phasor listed as follow:
)(tv VV
dt
dvVj
vdt jV
PHASOR REPRESENTATION OF SINUSOIDS
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So now we know how to represent a voltage or current in the phasor or
frequency domain
But how can we apply this to circuits involving the passive elements R,
L and C?
transform the voltage-current relationship from the time domain to the
frequency domain for each element
PHASOR REPRESENTATION OF SINUSOIDS
Time
domain
Frequency
domain Timedomain
Frequency
domain
Time
domain
Frequency
domain
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Summary of voltage-current relationship
Element Time domain Frequency domain
R
L
C
Riv IRV
dt
diLv IV Lj
dt
dvCi
Cj
Cj
IV
VI
PHASOR REPRESENTATION OF SINUSOIDS
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EXAMPLE
If voltage v(t) = 12cos(60t +45o) is applied to a 0.1 Hinductor, calculate the current, i(t), through the inductor.
Atti
domaintimetothisConverting
VLj
o
o
o
o
)4560cos(2)(
,
452906
4512
VI
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EXERCISE
If voltage v = 10 cos (100t+30o) is applied to a 50F
capacitor, calculate the current through the capacitor.
Answer: 50 cos (100t + 120o) mA
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IMPEDANCE AND REACTANCE
The impedance Z of a circuit is the ratio of the phasor
voltage V to the phasor current I, measured in ohms .
where R = resistance
X = reactance
The admittance Y is the reciprocal of impedance,
measured in siemens (S).
jXRIVZ
V
I
ZY
1
Positive X is for LNegative X is for C
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RY
1
LjY
1
CjY
Impedances and admittances of passive elements
Element Impedance Admittance
R
L
C
RZ
LjZ
CjZ
1
IMPEDANCE AND REACTANCE
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OHMSS LAW AND KIRCHOFFS LAW IN THE FREQUENCY
DOMAIN
N series connected shown in figure impedances
Applying KVL:
V = V1 + V2+ + VN = I(Z1+Z2++ZN)
similar to the series connection of resistances
If N = 2 as shown in figure below,
1 2
VI
Z Z
1 1 2 2, V Z I V Z I
1 2
1
1 2 1 2
2
Z ZV V, V = V
Z Z Z Z
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OHMSS LAW AND KIRCHOFFS LAW IN THE FREQUENCY
DOMAIN
N Parallel connected impedances
Voltage across each impedances is same
Applying KCL:
1 2
1 2
1 2
1 2
.....
1 1 1......
1 1 1 1......
....
N
N
eq N
eq N
I I I I
V Z Z Z
I
Z V Z Z Z
Y Y Y Y
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OHMSS LAW AND KIRCHOFFS LAW IN THE FREQUENCY
DOMAIN
N Parallel connected impedances
If N = 2 as shown in figure below,
1 2
1 2 1 2 1 2
1 1 2 2
2 1
1
1 2 1 2
1 1
1/ 1/
eq
eq
eq
2
Z Z1Z =
Y Y Y Z Z Z Z
V IZ I Z I Z
Z ZI I, I = I
Z Z Z Z
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DELTA-WYE TRANSFORMATION
The delta-to-wye and wye-to-delta transformations that we applied to
resistive circuits are also valid for impedances
1 2 2 3 3 1
1
1 2 2 3 3 1
2
1 2 2 3 3 13
3
a
b
Z Z Z Z Z ZZ
Z
Z Z Z Z Z ZZ
Z
Z Z Z Z Z ZZ
Z
Y conversion: - Y conversion:
1
2
3
( )
( )
( )
b c
a b c
c a
a b c
a b
a b c
Z ZZ
Z Z Z
Z ZZ
Z Z Z
Z ZZ
Z Z Z
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EXAMPLE:Find the input impedance of the circuit shownbelow. Assume that the circuit operates at =50 rad/s.
1 3
2 3
3
2 2
1 110
50 2 10
1 13 3 (3 2)
50 10 10
8 8 ( 50 0.2) (8 10)
(3 2)(8 10)1 ( 2 || 3) 10
11 8(44 14)(11 8)
10 10 3.22 1.0711 8
3.22 11.07in
Z jj C j x x
Z jj C j x x
Z j L j x j
j jZin Z Z Z j
jj j
j j j
Z j
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EXERCISE:
Determine vo(t) in the circuit shown below.
Tips: to do the analysis in frequency domain, we must first transform the time-
domain circuit to the phasor domain equivalent
Answer: vo(t) = 17.15 coz (4t +15.96o)V
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SOLUTION
3
1
2
2
1 2
20cos(4 15 ) 20 15
1 1104 10 10
5 4 5 20
60
25 2025 || 20 100
25 20
100(20 15 )
60 100(0.8575 30.96 )(20 15 ) 17.15 15.96
( ) 17.15cos(4 15.
o o
x
o
o s
o o o
o
v t Vs V
mFj C j x x
H j L j x j
Z
j xjZ j j j
j j
Z jV V
Z Z jV
v t t
96 )o V
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Calculate vo in the circuit shown below.
EXERCISE
Answer: vo(t) = 7.01 cos (10t 60o)V