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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Chapter 8
AC SteadyState Analysis
-So far , we have discussed the response of circuits due to DC source.
-Here , we discuss circuits with sinusoidal sources.
* Sinusoidal
- Here we study sinusoidal functions :Consider the sinusoidal function :
( ) ( )tsinXtx M=Where :
XM = amplitude
= radian ( angular ) frequency
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
XM
- X M2
2
32
t
t)X(
T
Note , X (t) repeats itself every (2 )
radius.
DEF : Period , TTime it takes the signal to repeat itself
f
1T =
7/30/2019 Elec 3202 Chap 8
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Where: f frequency in Hertz ( Hz )
since T = 2
= 2 / T = 2 f
Consider the two signals x 1 (t) & x 2 (t)
)t(sinX(t)X
)t(sinX(t)X
M22
M11
+=
+=
Subtract from both signals
t)(sinX(t)X)-t(sinX(t)X
M22
M11
= +=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
If = X1(t) & X2 (t) are in phase
If X1(t) leads X2 (t) by Or X2(t) lags X1 (t) by
t
t)(sinX(t)X M22 =)-t(sinX(t)X M11 +=
{
)t(sinXX(t) M +=Q
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )sinsincoscoscos
cos
sin
cos
sin
sin==m
mm
( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( )
( ) ( )tcosBtsinAX(t)
tcossinXcostsinXX(t)
tcossincostsinXX(t)
MM
M
+=
+=
+=
From trigonometry , we have
Where :
A = XM cos ()
B = XM sin ()Also : tan () = sin () / cos () = B / A
= tan -1 ( B / A )
B
A
X M
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( )
( )
( )
++=
+=
+=
A
BtantsinBAX(t)
tsinXX(t)
BAX
122
M
22
M
7/30/2019 Elec 3202 Chap 8
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
30)t(50cos20(t)X60)t(50sin10(t)X
2
1
+=+=
120)t(50sin20(t)X
90)30t(50sin2030)t(50cos20(t)X
2
2
+=
++=+=QFind the frequency , phase angle between X1 and X2
Subtract 120 from both sides
t)(50sin20(t)X'60)t(50sin10(t)X'
2
1
= =
Let
Note :
cos (x) = sin ( x + 90 o)
sin (x) = cos ( x 90o
)
Frequency is50 rad/sec.
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
X 1 (t) leads X 2 (t) by -60 ( X2 leads by 60 )
X 2 (t) lags X 1 (t) by -60 ( X1 lags by 60 )
* Relationship between sinusoidal and complex numbers
We know that if the complex number
z = x + j y = r e j
Where :
=
+=
x
ytan
yxr
1
22
7/30/2019 Elec 3202 Chap 8
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Consider the circuit :
V(t) = VM cos ( t)
Find i (t) ?
V (t)+
-
R
Li (t)
( ) (1)dtdi(t)L(t)iRtcosV
dt
di(t)L(t)iRV(t)
M LLL+=
+=Using KVL :
Assume that the solution is a sinusoidal function
( )( ) ( )
)sin(A-A2),cos(AA1where
(2)tsinAtcosAi(t)
tcosAi(t)
21
==
+=+=
LL
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Now find A1 and A2
Plug (2) in (1)
( ) ( ) ( )
( ) ( )[ ][ ] ( ) [ ] ( )tsinALARtcosALAR
tcosAtsinA-L
tsinARtcosARtcosV
1221
21
21M
++=++
+=
ALAR0
ALARV
12
21M
=+=
Solving for A1 and A2 , we find
222
M2222
M1
LR
VLA,
LR
VRA
+=
+=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( )
2
2
2
1
1
M
M
1
2
AAA
RLtan
R
L
VR
VL
A
Atan
)sin(A-A2and)cos(AA1since
+=
=
=
=
=
==
( ) ( )( )
( ) 2222
M
2222
2
M
222
2222
2
M
22
2222
2
M
22
2
2
1
LR
V
LR
VLR
LR
VL
LR
VRAA
+=
+
+=
++
+=+
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( )
++=
+=+
=+=
R
Ltantcos
LR
Vi(t)
tcosAi(t)LR
VAAA
1222
M
222
M2
2
2
1
It is clear that it is very complicated to find the solution usingsinusoidal function
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
[ ]
[ ][ ]
VVwhereeVv(t)
eVRe
eeVRe
eVRe)t(cosVv(t)
)t(sinVj)t(cosVeV
x)sin(j(x)cose
formulasEulerUsing
)tw(cosV(t)v
thatassumedEarlier we
M
tj
tj
M
jtj
M
)t(j
MM
MM
)t(j
M
jx
M
==
==
=+=+++=
+=
+=
+
+
Dropping e jt since it
exists in all termsPhasor form
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Phasor representation :
( )
( ) )90(AtsinA
AtcosAformPhsordomainTime
o
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
V (t)+
-
R
L
o0VVisformphasorThe
t)(cosVv(t)
M
M
=
=
Assume :
tj
M
tj
M
eIi(t)
eIi(t)
)t(cosIi(t)
o==
+=
We know that
tjo
M
tj
eVtv
Vetv
0)(
)(
=
=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
KVL :
[ ] [ ] [ ]
jILIRV
ejILeIReV
eIdt
dLeIReV
dt
di(t)L(t)iRv(t)
tjtjtj
tjtjtj
+=+=
+=
+=
[ ]
+
=
+=+=
222 LR
jLRVI
jLRVIjLRIV
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( )
+
+=
+
=
R
LtanLR
LR
V
LR
jLR0VI
1222
222
M
222M
o
Example :
Convert from time domain to phasor form
( )( )
Aooo
o
o
o
3012)90(12012I
V)(-4524V
120t337sin12i(t)
45t337cos24v(t)
==
=
+==
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Convert from phasor form to time domain
zHK1fif)(-7510I
V12016V
o
o
==
=
( ) 200010002f2 ===
( )
( )o
o
75t2000cos10i(t)
120t2000cos16v(t)
=
+=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
* Phasor relationships for circuit elements
1. Resistor :
V (t)
+
-R
i (t)
Assume :
iv
tj
iM
tj
vM
)t(j
M
)t(j
M
)t(j
M
)t(j
M
&
IRV
eIReV
eIReV
i(t)Rv(t)
eVv(t)&
eIi(t)
iv
v
i
=
=
=
=
=
=
=
++
+
+
In phasor form
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
t
i (t)
v (t)
)t(j
M
)t(jM
i
v
eIi(t)
eVv(t)+
+
==
2. Inductor :
V (t)
+
-
L
i (t)Assume :
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
[ ])t(jM ieIdt
dL
dt
di(t)Lv(t)
+==
o
iv
tj
iM
tj
vM
)t(j
M
)t(j
M
90&
ILjVeILjeV
eILjeV iv
+=
= =
= ++
t
i(t)
v(t)
o90
{
The voltage leads the
Current by 90o
.
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
3. Capacitori (t)
V (t)
+
-
)t(j
M
)t(j
M
i
v
eIi(t)eVv(t) +
+
==
[ ])t(j
Mv
eVdt
dCdt
dv(t)Ci(t)
+
==
Assume :
o
iv
tjvM
tjiM
)t(j
M
)t(j
M
90
ICj
1Vor
VCjI
eVCjeI
eVCjeI vi
=
=
==
= ++
t
i(t) v(t)
o90
{
The current leadsThe voltage by 90o.
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
If the voltage across the 20 mH inductor is
v (t) = 12 cos (377 t + 20o ) find i (t) ?
( ) ( )
( ) ( )
( )o
o
o
o
o
70t377cos1.59i(t)
)07(1.5990m20377
2012I
Im20377j2012
ILjV
=
=
=
=
=
L
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
If the voltage across the capacitor is
v (t) = 100 cos (314 t + 15o ) find i (t) ??
C = 100 F
( ) ( ) ( )
( ) ( )( )( )o
o6-4
6-
105t314cos3.14i(t)
1053.14)9015(1010314I
1510*100100314j
VCjI
+==+=
=
=
oo
o
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
In summary
1. Resistor : RR IRV =
LL ILjV =
CC ICj
1V =
2. Inductor :
3. Capacitor
* Impedance and admittance :
Definition : impedance : ( Z )The ratio of the voltage over the current
)(I
V
I
V
I
VZ
V
I
Z
1
y&I
V
Z
iV
m
m
im
Vm =
==
===
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
ivz
m
m &
I
VZwhere ==
In complex form :
( )
( )
=
+=
=
=
+=
R
Xtan
XRZ
sinZX
cosZRwhere
XjRZ
1
z
22
z
z
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
In summary :
==
+==
+==
R
Xtan
XRZI
VZ
XjRZzZZ
formcomplexformphasor
1
zivz
22
m
m
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Impedance of R,L,C
C
1j
Cj
10Z)90(
C
1ZC
Lj0Z90LZL
0jRZ0RZR
form)(complexZform)(phasorZElement
C
o
C
L
o
L
R
o
R
=+==
+==
+==
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
* Parallel and series connections of ( Z )
Series : Z 1 Z 2 Z 3 Z n
Z eq
Z 1 Z 2 Z n
Parallel
Z eq
n21eq Z
1
Z
1
Z
1
Z
1
+++= L
n21eq ZZZZ +++= L
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Definition : admittance ( y )
It is the reciprocal of Z
BjGy
xRxj
xRRy
xR
xjR
xjR
1
Z
1y
V
I
Z
1y
2222
22
+=
++
+=
+=
+==
== [S] siemens
Conductance Susceptance
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
* In terms of R ,L ,and C
o
CC
o
L
LL
R
RR
90CCjyCj
1Z-
)90(L
1
Lj
1
Z
1yLjZ-
GR
1
Z
1yRZ-
===
====
====
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
* Parallel and series connections of ( y )
Series : y 1 y 2 y 3 y n
y S
y 1 y 2 y n
Parallel
y P
n21S y
11
y
1
y
1+++= L
y
n21P yyyy +++= L
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Find the equivalent impedance
sec/rad2
F2C,F1C
H2L,2R
H1L,1R
21
22
11
=
==
==
==
L 2
L 1 C 1
C 2
R1
R2
Z eq
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( ) ( )
4
1j
C
1jZ,
2
1j
C
1jZ
4jLjZ,2j12jLjZ
2RZ,1RZ
2
C2
1
C1
2L21L1
2R21R1
====
=====
====
( )( ) ( )
22.33.2430
37j30
30
845j3Z
2
3j3
154j
2
3j3
415j
1
22
1j2j1
41j4j
41j4j
ZZZZZ//ZZ
o
eq
R2C1L1R1C2L2eq
=+=
+=
++=++=
+++
=
++++=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
v (t)
40 m H
20+
- F50
i (t)
Example :
Find the current i (t) in the circuit
( )( )( )
sec/rad377
V)-30(120V
30t377cos120v(t)
9060t377cos120v(t)
60t377sin120v(t)
o
o
oo
o
==
=
+=
+=
20RZ
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
V
Z L
R+
-CZ
I
( ) ( )
( )( )
53.05jZ
503771j
C1jZ
15.08jZ,m40377jLjZ
20RZ
C
C
LL
R
=
==
===
==
Z eq+
-V
( )( ) ( )
( )oo
o
o
o
CLReq
eq
39.23t377cos3.87i(t)
A)39.23(3.879.2330.96
)30(120I
9.2330.9653.05j//15.08j20
Z//ZZZ
Z
VI
=
=
=
=+=+=
=
4j
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Find the equivalent impedance :
4
Z eq
2
2
2j
6j
4j-
( ) ( )[ ]
( ) ( )[ ]( ) ( )
( )17.653.551.076j3.38Z
2
4j6
2j42j2
22j4//2j2
24j6j4//2j2Z
o
eq
eq
=+=
++
++=
+++=
+++=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Phasor diagram :
A diagram that shows the magnitude and phase of variousvoltage and currents in the circuit
For example :
o
1
o2
o
1
1307I
605V
4510V
==
=
V 1 V 1I 1
60o
45o
130o10
57
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Draw the phasor diagram of all currents and voltage
I 1 I 2
j42 =Z= 21Z
o
454I =
+
-
V
A18.433.574j2
4j454
ZZ
ZII
454I
oo
21
21
o
=
=
+
=
=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( ) V18.437.1418.433.572ZIV
A108.431.78
4j2
2454
ZZ
ZII
oo
11
oo
21
12
===
=
=
+
=
| I | = 4|V | = 7.14
| I2
| = 1.79
45o108.4o
| I1 | = 3.57
18.43o
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Circuit Analysis Technique
Note :All analysis techniques discussed before can be used in AC steady-
state analysis
Example :Find Io using nodal analysis ?
I 1I 2
1j
V2+-
A901 o
I o
V11j
1j
o01
1
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Here , we have a supernode :
Applying KCL at supernode
( ) ( ) (1)901V450.707V450.7071j1
1
1j
1V
1j1
V901
1j1
V
1j
V
1j1
V901
III901
o
2
o
1
o
21o
221o
2o1
o
LL=+
++
=
++=
++=
)2(01VV
01VV
o
21
o12
LL=
+=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
A)18.44(1.58
1j
VI
V71.561.58V
V108.41.58V
o2o
o2
o
1
==
=
=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
use mesh analysis to find Vo ?
I 2
+
-
2
+
-
2j
2j
2
A902
o
o024V o
I 1
Supermesh
KVL around supermesh( )
( ) (1)024I2j2I2
0I2I2jI2024
o
21
221
o
LL=+
=++
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
V36.0210.88I2V
A36.025.44I
A15.254.56I
(2)902II
902II
o
20
o2
o
1
o21
o
12
===
=
=+
=
LL
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
find Vo ?
I 2
+
-
2 1j
V012
o
A06 o
V o+
-
V 1 V 2
2j 1
2
I 3I 1
03
V
2j
V
1j
VV
0III
2221
321
=
=
V012V o1 =
KCL at node V2
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( )
V33.76.65V3
1
21
1VV
V33.719.969V
j2
13
19012
2j1
31
9012V
90122j
21
3
1V
9012j
012
3
1
j2
1
j
1V
03
V
2j
V
1j
V012
o
22o
o
2
oo
2
o
2
oo
2
222
o
==
+=
=
+=
=
=
++
=
=
++
=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Use Thevenin Theorm to find Vo ?
+
-
V012 o
1 2 1
1j1j
Vo
+
-
First find VOC
+
-
V012 o
1 2 1
j 1j Voc
+
-
I1
I3
I2
Vx
KCL around Vx
0III
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
0III 321 =
V33.693.32j2
j1VV
V)-29.7(7.44V3.69j6.460.8j1.4
12
V
5
11j
5
21V012
14
j2j1V012
j2
1j1V012
0j2
V
j
V
1
V012
o
xoc
o
xx
x
o
x
o
x
o
xxx
o
=
+=
==+=
+
+=
+
++=
+
++=
=+
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1 Vo
+
-
+
-
ZTH
Voc
V12.541.3V
33.71.661
1
33.693.32
Z11VV
o
o
o
o
TH
OCo
=
+=
+=
Find ZTH:
33.71.661}1||]2)||1{[( oTH =++= jJZ