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7/30/2019 ELEC Chap3
1/42
UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Chapter (3)
Nodal and loop analysis
Consider the following circuit:
+
-
+
-
R2
R4
R5
R7
R6
R3
R1
V1
V2
I
i1
i5 i4
i3
i6
i2
a b
cd
e
fg
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Definitions:
Node :
A point where two or more circuit elements join
Ex. a,b,c,d,e,f,g
Essential Node:A node where three or more circuit element join
Ex. b,c,e,g
Path:
A trace of adjoining elements with no elements included more
than once
1. V1-R1-R5-R62. R5-R6-R4-V2 ,etc
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Branch:
A path that connects any two nodes.
Ex. R1 , V1 , R1-R5 , etc
Essential Branch:
A path that connects two essential nodes without passing throughan essential node.
Ex. V1-R1 , R5 , R2-R3 , V2-R4 ,
Loop :
A path whose last node is the same as its starting node
Ex. (1) V1-R1-R5-R3-R2
(2) V1
-R1
-R5
-R6
-R4
-V2
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Mesh :
A loop that doesnt enclose any other loops.
Ex. V1-R1-R5-R3-R2Ex. V2-R2-R3-R6-R4 ,
In chapter (2) we studied circuits containing a single loop or a
single node-pair
Such circuits can be solved easily by one algebraic equation.
Here , we will study circuit containing multiple node andmultiple loops
Hence we will introduce (2) analysis techniques :
1. Nodal analysis2. Loop analysis
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(1) Nodal Analysis :
Nodal analysis : is a technique in which KCL is used to determine
the nodes voltages at all essential nodes with respect to the
reference node.
Here , node voltage is defined as the voltage of a given node with
respect to a reference node
+
-+-
i5i3i1 i4i2Vm = 10
R1= 2 R3= 2 R5= 1
R4 = 2R2= 1
1 2
3
Vn = 5
Example:
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Essential nodes : 1,2,3
Consider (3) to be the reference node (ground)
At (1) apply KCL:
( )110V4V
02
V2V5
02
V
2
VV
2
V5
02
VV
1
V
2
V10
0R
VV
R
V
R
VV
0iii
21
21
211
1
2111
3
21
2
1
1
1m
321
KK=
=+
=+
=
=
=
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
( ) ( )
( )210V4V
05V22
V
01
5V
2
V
2
VV
21
21
2221
KK=
=
=+
0R
VV
R
V
R
VV
0iii
:KCLApply,(2)At
5
n2
4
2
3
21
543
=
=
2V
2VBAVBVA
10
10
V
V
41
14
2
1
1
2
1
=
===
=
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Note :
Number of equations = N-1
Where :N is the number of essential nodes
Example :Circuit with only independent current source
iR1
i1
R3= 5
R1= 60
R4 = 2R2= 15
1 2
3iR2
5 A = I S2iR4
V 1V 2
+
-
+
-
15 A = I S1
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Find V1,V2 and i
# of essential node = N=3Select (3) as ground (reference node)
# of KCL equations = N-1 = 2
At node (1) apply KCL
05
V
5
V
15
V
60
V
15
0R
VVRV
RV15
0iiiI
2111
3
21
2
1
1
1
1R2R11s
=+
=
=
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(1)15V
5
1V
60
17
15V51V
601412
15V5
1V
60
1
15
1
5
1
21
21
21
KK=
=++
=
++
(2)50V7V2
050V5V2V2
052
V
5
VV
05RV
RVV
0Iii
21
221
221
4
2
3
21
s2R41
KK=
==
=
=At node (2)
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
A105
1060R
VVi
V10V
V60V
50
15
V
V
72
5
1
60
17
211
2
1
2
1
===
=
=
=
Example :
Circuit with dependent current source
i21 2
3
I0
V 2
I S = 2 mA k10R3 =
V 1 k10R2 =
2 Ix
i1
Ix
k10R1 =
Find I0
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
# of essential nodes = N=3
Choose (3) to be reference node ( ground )
We have N-1 = 2 KCL equationat node(1) and( 2)
At node (1) , apply KCL
(1)A2V0.1V0.2
10kV
5kV
10kVV
10kVAm2
0R
VV
R
VAm2
0iiI
21
21211
2
21
1
1
21s
KK=
=+=
=
=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
0II2i 0x2=
At node (2) apply KCL,
0II2R
VV
0I2Ii
0x2
21
0x2
=
=
(2)0V2V1
010kV
10kV2
10kVV
R
VI
R
ViIwhere
21
2121
3
20
1
11x
KK=+
=
=
==
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Am0.4k10
4
R
VI
V4V
V8V
0
2
V2
V1
21
0.10.2
3
20
2
1
===
=
=
=
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7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(1)16V18
1V0.201388
0
18
V
18
V
48
V
8
V16
018
VV48V
8V128
1
2111
2111
KK=
=+
=
KCL at (1)
(2)7V0.20555V18
1
0
10
70V
20
V
18
VV
21
2221
KK=
=
KCL at (2)
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
V60V
V96V
7
16
V
V
0.205518
118
10.201338
2
1
2
1
=
=
=
A1
10
70Vi
A42060
20Vi
A218
6096
18
VVi
A248
96
48
Vi
A48
96128
8
V128i
2e
2d
21c
1b
1a
=
=
===
=
=
=
===
===
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
A110
70Vi
A420
60
20
Vi
A2
18
6096
18
VVi
A24896
48Vi
A48
96128
8
V128i
2e
2d
21c
1b
1a
==
===
=
=
=
===
=
=
=
Special case:
What if a branch between two essential non-reference nodecontain a voltage source ?
This case is called super node" case.
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Find I0
# of essential nodes = N = 3 Super node: is the voltage source and the two connecting nodes
# of equations = N-1-1 = 3-1-1=1
Reference node Super node
But we need (2) equations to find the two unknowns V1 and V2There is an equation that describe the super node.
1 2
3i2
V 2V 1
i1
i3
+ -
6 Vk6
Io
k6 k6k21k21
k6
supernode
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Apply KCL at the super node :
(1)06k
V
6k
V
012k
V
12k
V
k12
V
k12
V0Iiii
21
2211
0321
KK
=+
=+++=+++
Am0.25
12k
3
12k
V2I
V3V
V3V
6
0
V2
V1
116k
1
6k
1)2(V6V
0
2
1
12
=
==
==
=
=+ KK
The super node is described by:
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example:
Circuits with dependent voltage sources
Find I0?
N = 3 N-1 = 2 equations
Io1 2
3
i2
V 2V 1 5
i1
i3
+-20 V
i4
22
02 01
8 Io
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(1)10V5
1V
4
3
10V5
1V
5
1
20
1
2
1
21
21
KK=
=
++
5
V
5
V
20
V
2
V10
5
VV
20
V
2
V20
Iii
2111
2111
021
+=
+=
+=
KCL at node (1):
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
2
V
10
V
5
VV5
5
VV4
2
V
10
V
5
V
5
V
2
I8V
10
V
5
VV
iiI
2221
212221
02221
430
+=
+=
+=
+=
(2)0V
5
8V
0V2
1
10
11V
2
V
10
VVV
21
21
2221
KK=
=
++
+=
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
A1.25
6
5
VVI
V10V
V16V
0
10
V2
V1
581
5
1
4
3
210
2
1
==
=
=
=
=
Loop Analysis (Mesh)Mesh analysis : It is a technique in which KVL is used to
determine the current in all meshes
# of equations needed = # of meshes = be-(ne-1)Where :
be : # of essential branches
ne: # of essential nodes
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
(1)40i0i8i10
0)ii(82i40
321
211
KK=+=++KVL left loop :
6
+
-
+
- 20 V
42
8
40 V
6
+
-
Voi1i3
i2
Find V0?We have (3) meshes
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
A0.8i,A2i,A5.6i
20
0
40
i
i
i
1060
6208
0810
321
3
2
1
===
=
V28.8V
(3.6)82)(5.68)ii(8V
0
210
=
===
0i6i20i8
0)i(i6i6)i(i8
321
32212
=+
=++
20i10i6i0
04i)i(i620
321
323
=+
=++
KVL middle loop:
KVL right loop :
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Special Case :
What happens if a current source is located between two meshes
Super mesh
i2 2
i1 i3+
-50 V
3
10
+
-
100 V
5 A
46
You dont know the voltage across the current source !!
Remove the whole branch that includes the current source
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
i2 2
i1i3
+
- 50 V
3
10
+
-
100 V
46
Apply KVL around the super mesh
(1)50i6i5i9
0i6i450)i(i2)i(i3100
321
132321
KK=+
=+++++
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(2)0i2i15i30)i(i3)i(i2i10
321
12322
KK=+=++
(3)5ii0i
A5ii
321
13
KK=++
=
A6.75i
A1.25i
A1.75i
3
2
1
==
=
KVL around the upper loop:
We also know
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example:
Nodal Analysis :
001
+
-
128 V
501
256 V
+
-
005502
200 400
300
i
i50
Use nodal analysis and loop analysis to find power in the 300 ()
resistor ?
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7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(1)1.7067V0.00333V0.01V0.0250 321 KK=
( )
(2)0.256V0.0118V0.004V0.0033
500
128
300
1
500
1
400
1
250
1V
250
V
300
V
0300
V1V
500
128V
400
V
150
VV
0iiii
321
321
33332
654
KK=+
=
++
=
=KCL at node (3):
You can notice that
(3)0V0.1667V1V0.166
V61V
61
300VV50V
i50V
321
1313
2
2
KK=+
=
=
=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
V8V
V11.75V
V62.5V
0
0.256
1.7067
V3
V2
V1
0.166710.1667
0.01180.0040.0033
0.00330.010.0383
3
2
1
==
=
=
( ) ( ) W16.56753000.235RiP
A0.235i50Vi
22
300
2
===
==
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7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(2)0i150i300i200
0)i(i200i50)i(i100
ii
0)i(i200i50)i(i100
521
12552
5
1252
KK=+
=+=
=++
KVL around loop (2):
5
543
4353
iisince
(3)0i200i400i6500i50)i(i400)i(i250
=
==+KK
(4)128i900i400
0)i(i400128i500
43
344
KK=+
=+
KVL around loop (3)
KVL around loop (4)
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(5)0i650i250i100
0)i(i100)i(i250i300
532
25355
KK=+
=++
KVL around loop (5)
=
0
128
0
0
256
i
i
i
i
i
65002501000
090040000
20040065000
15000300200
000200350
5
4
3
2
1
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
W16.5675(300)(0.235)R)(iP
22
5300
===
A0.235i
A0.24i
A0.22iA0.9775i
A1.29i
5
4
3
2
1
=
=
==
=
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example:
Use the loop analysis to find V
10Ai3 =
(3) Meshes (3) equations
i3 1
i1 i2+
-
2
5 5
V2
10 A
+
-V
57 V
7/30/2019 ELEC Chap3
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
(1)55i5i7
07520i5i7
075)i(i5)i(i2
21
21
2131
KK=
=+
=+
KVL around loop (1):
( ) ( )
(2)0i3i2
i2i2ii552i
)i(i5V
5
V2i
21
21212
21
2
KK=
=
=
=
=
Equation of dependent source:
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UAE University Department of Electrical Engineering Dr Hazem N Nounou
A10i
A15i0
55
i
i
32
57
2
1
2
1
=
=
=
V25V
V2510)(155
)i(i5V
21
=
==
=