11/11/2013 1 Module M2-1 Electrical Engineering TUTORIAL 8 CIRCUIT ANALYSIS IV NOVEMBER 2013 Topics Phasor Impedance 2 Impedance Nodal Analysis Mesh Analysis
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11/11/2013
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Module M2-1Electrical Engineering
T U T O R I A L 8 C I R C U I T A N A L Y S I S I V
N O V E M B E R 2 0 1 3
Topics
Phasor
Impedance
2
Impedance
Nodal Analysis
Mesh Analysis
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Sinusoids3
A sinusoid is a signal that has the form of the sine or cosine function.
)sin()( tVtv m
where Vm = the amplitude of the sinusoidω = the angular frequency in radians/sФ = the phase
Sinusoids
A periodic function is one that satisfies v(t) = v(t + nT), for all t and for all integers n.
4
g
HzT
f1
f 22
T
2sin ( ) sin sin 2 sinm m m mV t T V t V t V t
sin( ) sin cos cos sinA B A B A B
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Sinusoids
Graphical addition
5
cos sin cos( )A t B t C t 3cos 4sin 5cos( 53 1)t t t cos sin cos( )A t B t C t 3cos 4sin 5cos( 53.1)t t t
2 2 1, tanB
C A BA
Sinusoids: Example6
Given a sinusoid, , calculate its )604sin(5 ot amplitude, phase, angular frequency, period, and frequency.
Solution:
Amplitude = 5, phase = –60o, angular frequency = 4π rad/s, Period = 0.5 s, frequency = 2 Hz.
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Phasors7
A phasor is a complex number that represents the amplitude p pand phase of a sinusoid.
It can be represented in one of the following three forms:
)i( jj a Rectangular
rzjrez
)sin(cos jrjyxz a. Rectangular
b. Polar
c. Exponential
22 yxr
x
y1tan where
Phasor Properties8
• Addition & subtraction
M ltiplication
)()( 212121 yyjxxzz
• Multiplication
• Division
• Reciprocal
212121 rrzz
212
1
2
1 r
r
z
z
rz
11
• Square root
• Complex conjugate
• Euler’s identity
2 rz
jrerjyxz
sincos je j
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Phasors9
phasor domain time domain
( )j tm mV V e V ( ) cosmv t V t
Phasor will be defined from the cosine function
Phasors10
( )j tm mV V e V m m
( )j tm mI V e I
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Phasors: Example11
Transform the following sinusoids to phasors:
i = 6cos(50t – 40o) A
v = –4sin(30t + 50o) V
406
1404
A
= 4cos (30t+50o+90o) V
Transform the following phasors to sinusoids:
V 3010 V
A j12) j(5 I
= 10cos(ωt + 210o) V
22.62 13 )12
5( tan 512 122
= 13cos(ωt + 22.62o) A
Phasors
Let
12
( ) cosmv t V t
Derivative 90
( )sin cos 90
Re Re
m m
j t j tm m
dv tV t V t
dt
V e j V e
1 1 Integral
90
1 1( ) sin cos 90
1 1 Re Re
m m
j t j tm m
v t dt V t V t
V e V ej
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Phasors13
The differences between v(t) and V:
v(t) is instantaneous or time-domain representationV 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 complexg y p
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.
Phasors14
Relationship between differential, integral operation in phasor listed as follow:in phasor listed as follow:
)(tv VV
dt
dvVj
vdtj
V
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Circuit Elements15
Resistor: Inductor: Capacitor:
Circuit Elements16
Summary of voltage current relationshipSummary of voltage‐current relationship
Element Time domain Frequency domain
R
L
Riv RIV
diL LIjV
C
dtLv LIjV
dt
dvCi
Cj
IV
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Impedance and Admittance
Frequency dependence
17
LjZ
Z
Z
;
0;0
CjZ
1
0;
;0
Z
Z
Impedance: Example18
Find Z for
>> ((200 + 8*10*i)*1/(1*10*10^ 3*i))/((200 +
Find Zin for
ω = 10 rad/s
>> ((200 + 8*10*i)*1/(1*10*10^-3*i))/((200 + 8*10*i) + 1/(1*10*10^-3*i)) + 80 + 1/(0.5*10^-3*10*i)
ans =1.2950e+002 -2.9505e+002i
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Nodal Analysis19
Currents entering a node equal to currents l i th t dleaving that node
3 Steps to solve:
1. Select a node as the reference node. Assign voltages with respect to the reference node to the remaining nodes.
2. Apply KCL to each of the non-reference nodes. Use Oh ’ l h b h i f Ohm’s law to express the branch currents in terms of node voltages.
3. Solve the resulting simultaneous equations to obtain the unknown node voltages.
Nodal Analysis: Example20
Find iX in the circuit using nodal analysis.1010
tV4cos20 xI2xI
Solution sradt /40204cos20 We first convert the circuit to the frequency domain
sradt /4,0204cos20 41 jLjH
25.0 jLjH
5.21
1.0 jCj
F
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Nodal Analysis: Example21
Node 1 Node 2
10
V020xI2xI
1V 2V
5.2j
4j
2j
Apply KCL at node 145.210
20 2111
j
VV
j
VV
or 205.2)5.11( 21 VjVj (1)
22
At node 2,24
2 221
j
V
j
VVIx
Nodal Analysis: Example
24 jj
But Ix= V1/-j2.5. Substituting this gives
245.2
2 2211
j
V
j
VV
j
V
By simplifying, we get 01511 21 VV (2)
Equation (1) and (2) can be put in matrix form as
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23
Nodal Analysis: Example
205.25.11 1Vjj VV 43.1897.181
The current Ix is given by
Aj
VIx
4.10859.79052
43.1897.18
521
01511 2V 3.19891.132V
j 905.25.2
Transforming this to the time domain,
Atix )4.1084cos(59.7
24
The sum of voltages in a mesh equals to zero.
Mesh Analysis
3 Steps to solve:
1. Assign mesh currents in to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents.
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Mesh Analysis: Example25
Determine current Io in the circuit using mesh analysis.
4
A05 2j
10jV9020
8 2j
1I
2I
3I 0I
Mesh Analysis: Example26
SolutionApplying KVL to mesh 1,we obtainpp y g ,
010)2()2108( 321 IjIjIjj (1)For mesh 2,
09020)2()2()224( 312 IjIjIjj (2)
For mesh 3 I = 5 Substituting this in Eqs (1) and (2)For mesh 3, I3= 5 .Substituting this in Eqs. (1) and (2)
502)88( 21 jIjIj (3)
1020)44(2 21 jjIjIj (4)
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Mesh Analysis: Example27
Put eqs. (3) and (4) in matrix form as
30
50
442
288
2
1
j
j
I
I
jj
jj
AI 22.3512.62
The desired current isThe desired current is
AII 78.14412.620
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