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THEVENIN EQUIVALENT CIRCUIT
Thevenin equivalent circuit consist of anindependent voltage source, VTh in serieswith a resistor RTh.
Thevenin equivalent circuit
ThV
ThR
a
b
Thevenin voltage, VTh = Open circuit voltage inthe original circuit.
Thevenin resistance, RTh is the ratio of open-circuit voltage to the short-circuit current.
sc
ThTh
Th
Thsc
i
VR
R
Vi
Example
V25
5
a
b
20
4
A3
1V
abV
Step 1
Node-voltage equation for open-circuit
ThVVV
VV
32
03205
25
1
11
Step 2
Short-circuit condition at terminal a-b.
V25
5 a
b
20
4
A3
2V scI
Node-voltage equation for short-circuit
VV
VVV
16
04
3205
25
2
222
84
32
sc
ThTh
I
VR
AI sc 44
16
Short-circuit current
Thevenin resistance
Thevenin equivalent circuit
V32
8
a
b
NORTON EQUIVALENT CIRCUIT
A Norton equivalent circuit consists of anindependent current source in parallel with theNorton equivalent resistance.
Can be derive from a Thevenin equivalentcircuit simply by making a sourcetransformation.
Norton current, IN = the short-circuit current atthe terminal of interest.
Norton resistance, RN = Thevenin resistance,RTh
Example
V25
5
a
b
20
4
A3
Step 1Source transformation.
A5 5
a
b
20
4
A3
Step 2Parallel sources and parallel resistors combined.
a
b
4
A8 4
Step 3Source transformation, series resistorscombined, producing the Thevenin equivalentcircuit.
V32
8
a
b
Step 4Source transformation, producing the Nortonequivalent circuit.
a
b
A4 8
Example
Step 1
Deactivated all sources except voltage source
V0 is calculated using voltage divider
Vk
kV 54
1020
Step 2
Deactivated all sources except current source
V0 is calculated by using current divider
VkmV
mAmk
ki
2)2)(1(
1)2(4
2
0
0
V0 =2+5=7V.
Step 3
Sum all the resulting voltages:
Question For Thevenin
Open-circuit voltage, Voc
Node-voltage equation for Voc
VV
V
VV
VV
oc
oc
ococ
ococ
10
202
0424
0222
24
Thevenin resistance, RTh
5422THR
Thevenin equivalent circuit
VV 88.6)10(16
110
Question For Norton
Open-circuit current, Isc
AIsc 64
123
Norton resistance, RN
RN = 4Ω
Norton equivalent circuit
VV 18)3(612460
PLT 121
CHAPTER 2
AC Circuit Analysis
• Sinusoids’ features
• Phasors
• Phasor relationships for circuit elements
• Impedance and admittance
• Kirchhoff’s laws in the frequency domain
• Impedance combinations
SINUSOIDAL STEADY-STATE ANALYSIS SINUSOIDAL AND PHASOR
• Basic Approach
• Nodal Analysis
• Mesh Analysis
• Thevenin Equivalent Circuits
• A sinusoid is a signal that has the form ofthe sine or cosine function.
• A general expression for the sinusoid,
whereVm = the amplitude of the sinusoidω = the angular frequency in radians/sФ = the phase
Sinusoids
)sin()( tVtv m
A periodic function is one that satisfiesv(t) = v(t + nT), for all t and for all integers n.
2T Hz
Tf
1
f 2
• Only two sinusoidal values with the samefrequency can be compared by theiramplitude and phase difference.
• If phase difference is zero, they are inphase; if phase difference is not zero, theyare out of phase.
Example 1
Given a sinusoid, , calculate itsamplitude, phase, angular frequency, period,and frequency.
SolutionAmplitude = 5, phase = –60o, angular frequency = 4 rad/s, Period = 0.5 s, frequency = 2 Hz.
)604sin(5 ot
Example 2
Find the phase angle between and ,
does i1 lead or lag i2?
)25377sin(41
oti )40377cos(52
oti
Solution
Since sin(ωt+90o) = cos ωt
therefore, i1 leads i2 155o.
)50377sin(5)9040377sin(52
ooo tti
)205377sin(4)25180377sin(4)25377sin(41
oooo ttti
PHASORS
• A phasor is a complex number that represents the amplitude and phase of a sinusoid.
• It can be represented in one of the following three forms:
rzjrez
)sin(cos jrjyxza.Rectangular
b.Polar
c.Exponential
22 yxr
x
y1tanwhere
Example 3
• Evaluate the following complex numbers:
a.
b.
Solution
a. –15.5 + j13.67
b. 8.293 + j2.2
]605j4)1j2)([(5 o
oo
3010j43
403j510
)()( 212121 yyjxxzz )()( 212121 yyjxxzz
212121 rrzz
21
2
1
2
1 r
r
z
z
rz
11
2 rz
jrerjyxz
sincos je j
Mathematic operation of complex number
• Addition
• Subtraction
• Multiplication
• Division
• Reciprocal
• Square root
• Complex conjugate
• Euler’s identity
• Transform a sinusoid to and from the time domain to the phasor domain:
(time domain) (phasor domain)
)cos()( tVtv m mVV
• Amplitude and phase difference are two principal concerns in the study 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.
Example 4
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
Example 5
Transform the sinusoids corresponding to phasors:
a.
b.
V 3010 V
A j12) j(5 I
Solution
a) v(t) = 10cos(t + 210o) V
b) Since
i(t) = 13cos(t + 22.62o) A
22.62 13 )12
5( tan 512 j512 122
I
The differences between v(t) and V:
• v(t) is instantaneous or time-domainrepresentation. V is the frequency orphasor-domain representation.
• v(t) is time dependent, V is not.
• v(t) is always real with no complexterm, V is generally complex.
Note: Phasor analysis applies only when
frequency is constant; when it is
applied to two or more sinusoid
signals only if they have the same
frequency.
Relationship between differential, integral
operation in phasor listed as follow:
)(tv
dt
dvVj
vdtj
V
VV
Example 6
Use phasor approach, determine thecurrent i(t) in a circuit described bythe integro-differential equation.
Answer: i(t) = 4.642 cos(2t + 143.2o) A
)752cos(50384 tdt
diidti
Phasors Relationship for circuit elements
Resistor Inductor Capacitor
Summary of voltage-current relationship
Element Time domain Frequency
domain
R
L
C
Riv RIV
dt
diLv LIjV
dt
dvCi Cj
IV
Example 7
If voltage v(t) = 6 cos(100t – 30o) is applied to
a 50 μF capacitor, calculate the current, i(t),
through the capacitor.
Answer: i(t) = 30 cos(100t + 60o) mA
IMPEDANCE AND ADMITTANCE
• The impedance, Z of a circuit is the ratio ofthe phasor voltage V to the phasor currentI, measured in ohms Ω.
• where R = Re, Z is the resistance andX = Im, Z is the reactance. Positive X isfor L and negative X is for C.
• The admittance, Y is the reciprocal ofimpedance, measured in siemens (S).
jXRI
VZ
V
I
ZY
1
RY
1
LjY
1
CjY
Impedances and admittances of passive elements
Element Impedance Admittance
R
L
C
RZ
LjZ
CjZ
1
LjZ
CjZ
1
Z
Z
;
0;0
0;
;0
Z
Z
Example 8
Refer to Figure below, determine v(t) and i(t).
Answers: v(t) = 2.236sin(10t + 63.43o) V
i(t) = 1.118sin(10t - 26.57o) A
)10sin(5 tvs
Kirchhoff’s Law in the Frequency Domain
• Both KVL and KCL are hold in the phasor
domain or more commonly called frequency
domain.
• Moreover, the variables to be handled are
phasors, which are complex numbers.
• All the mathematical operations involved
are now in complex domain.
IMPEDANCE COMBINATIONS
• The following principles usedfor DC circuit analysis all applyto AC circuit.
• For example:– voltage division
– current division
– circuit reduction
– impedance equivalence
Example 9
Determine the input impedance of the circuit in
figure below at ω =10 rad/s.
Answer: Zin = 32.38 – j73.76
• Basic Approach
• Nodal Analysis
• Mesh Analysis
• Thevenin Equivalent Circuits
BASIC APPROACH
1. Transform the circuit to the phasor or
frequency domain.
2. Solve the problem using circuit techniques
(nodal analysis, mesh analysis, etc.).
3. Transform the resulting phasor to the time
domain.
Time to Freq Solve
variables in Freq
Freq to Time
Steps to Analyze AC Circuits:
NODAL ANALYSIS
Exercise 1
Calculate V1 and V2 in the circuit shown in
figure below .
Answer
V1 = 19.3669.67 V V2 = 3.376165.7 V
MESH ANALYSIS
Exercise 2
Find Io in the following figure using mesh
analysis.
Answer : Io = 1.19465.44 A
THEVENIN EQUIVALENT CIRCUITS
Thevenin transform
Answer : Zth =12.4 – j3.2
VTH = 18.97-51.57 V
Find the Thevenin equivalent at terminalsa–b of the circuit below.
Assignment 1.3