EE100supplementarynoteweek03-2
Transcript of EE100supplementarynoteweek03-2
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Node Voltage Method
This simplest among many circuit analysis methods is
applicable only for connected circuits N made of linear 2-
terminal resistors and current sources. The only variables in the
linear equations are the n-1 node voltages e1, e2 , , en-1 for an
n-node circuit.
Step 1. Choose an arbitrary datum node and label the
remaining nodes consecutively , , , , and
let e1, e2 , , en-1 be node-to-datum voltages.
Step 2. Express the current of each resistor Rj via Ohms
law in terms of 2 node-to-datum voltages:
1 2 n-1
ji
+ -jR
je
+
je
j
j
j
jei
R
e+ = (1)
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5A2A
+
-
6v
6i
2e 3e
1e
+-3v
4v
+
-
5v+
-8 6
42+
-1v 2v
+
-
35
6i
e=
3i
11
2
2
ei
e=
12
3
4
ei
e=
24
8i e=
1
2 3
KCL at :1 31 2
( )( )52 4
e ee e
+ =1 (2)
KCL at : 1 2 2( )
22 8
e e e + = 2 (3)
KCL at : 1 3 3( ) 24 6
e e e + =3 (4)
3 1 1
1.EXAMPLE
R t E (2) (3)
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3 1 1
4 2 41 5
0
2 81 50
4 1 2
2 0d e t
3 8 4
=
1(a) To solve for , replace colum n 1 of m atrix from (5) and calculatee
1
1 15
2 45
2 08
52 0
1 2
1 1 5d e t
9 6
=
C ramer's Ru le
11
115 2023
96 384Ve = = =
2(b) To solve for , replace colum n 2 of matrix from (5) and calculatee
22
76 2015.2
96 384e V
= = =
(c) To solve for replace colum n 3 of matrix from (5) and calculatee
2
3 1
54 41
2 021 5
24 1 2
7 6d e t
9 6
=
( 6 )
( 7 )
( 8 )
( 9 )
Solving eq. (5) by Cram er's Ru le (or any other m ethod) :
C ramer's Rule
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{ }1 2 3O nce the node-to-datum voltages , , are found,e e eall resistor curren ts and voltages are found trivially via
K V L, K C L, and O hm 's law :
1 21 1
7.87.8 , 3.9
2v V i Ae e= = = =
1 32 2
4.44.4 , 1.1
4v V i Ae e= = = =
33 32 3.4 , 2v Ve e i A= = =
24 4
15.215.2 , 1.9
8v V ie A= = = =
35 518.618.6 , 3.1
6v V ie= = = =
16 623 , 5v V ie A= = =
V erification of Solution
T ellegen 's TheApp oly :rem
6
1 1 2 2 3 3 4 4
1
( ) ( ) ( ) ( )j jj
v i v i v i v i v i=
= + + +5 5 6 6( ) ( )v i v i+ +
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Modified Node Voltage Method
Circuit N +v
1e1 e2
6V2A 3
4
i4
i2
i1
i3
_v3v4
+
-
+
_
+
_v2
+1
3
2
_
11
2
4
ei
e=
12
3
ie
=
KCL at : 1 1 2( ) 23 4e e e
+ =1 (1)
KCL at : 1 23
( )0
4i
e e + =2 (2)
For each voltage source , add an equation .j jjs j s
vev e+ =
2 6e = (3)
1 2When the circuit contains " " voltages , , , , uses s sv v v
1 2their associated currents , , , when applying KCL.s s si i i
Step 1.
Step 2.
Step 3.
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Explicit matrix form of Node Voltage Equations
Assumption: Circuit N contains only linear resistors and
independent current sources which do not form cut sets.
Step 1. Delete all current sources from N and draw the
reduced digraph G of the remaining pure resistor circuit.
Assume Ghas n nodes and b branches.
Step 2. Pick a datum node and label the node-to-datum voltages
{ e1, e2 , , en-1 }, and derive the reduced-incidence matrix A.Define the branch admittance matrix Yb and independent
current source vector is as follow:
bY
1
2
1 1 1
2 2 2
0 0 0
0 0 0(1) , (2)
0 0 0
s
s
s
b b b s
ii Y v
ii Y v
i Y v i
=
i
i v1where , resistance of branch .j j
j
Y R jR
=
current= algebraic of all sources nodesu ,mms
i entering m
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Deriving the Node Voltage Equation
The current sources can be
deleted since they can be trivially accounted
for by representing their net contribution at
each node m by the algebraic sum of all
current sources entering node m , m = 1, 2,
, n-1. The KCL equations therefore takes
the augmented form
Substituting (1) for i in (4), we obtain
Substituting KVL
s=A i i
b s=A Y v i
(4)
(5)
( )n s=Y e i
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Writing Node-Admittance Matrix
Yn By Inspection
nY
(8)
:Node Votage Equation
1
2
1
11 12 1, 1
21 22 2, 1
1,1 1,2 1, 1
1
2
1n
sn
sn
n n n snn
iY Y Y
iY Y Y
Y eY
e
Y i
e
=
e si
Diagonal Elements of nY1
= sum of adm ittances of all resistorsm m j
j
Y Y
R
connected to node , 1, 2 , , -1m m n=
Off-Diagonal Elements of nY
1= sum of adm ittances of all resistors
jk j
j
Y YR
- ( connected across nod e and nod ej k )
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5A2A
+
-
6v
6i
2e 3e
1e
+-3v
4v
+
-
5v+
-8 6
42+
-1v 2v
+
-
35
6
ei =
3i
1 21
2
e ei
=
1 32
4
e ei
=
24
8ei =
1
2 3
KCL at :1 31 2
( )( )
5 42 4
e ee e
+ = 1 (2)
KCL at : 1 2 2( )
4 2 32 8
e e e + = 2 (3)
KCL at : 1 3 3( ) 2 14 6
e e e + = +3 (4)
Matrix Form : 3 1 1
EXAMPLE
1A
4A
3A
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Mesh Current Method
The next simplest among many circuit
analysis methods is applicable only for
connected planar circuit N (with a planardigraph) made of 2-terminal linear resistors
and voltage sources. The only variables in the
equations are l conceptual mesh currents
circulating in the l meshes in
a clockwise direction (by convention) :
1 2
, , ,lm m m
l l l
1R
10i
12i
7i4i
1i
2R
4R
5R
7R
8R
10R 12R
1i
2i3i
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ji
jR
ci
ai
b
i
di
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Mesh Current Method
The next simplest among many circuit
analysis methods is applicable only for connected
planar circuits made of 2-terminal linear resistors andvoltage sources. The only variables in the associated
mesh-current equations are l mesh current
which we define to be circulating in the l meshes in a clockwise direction (by convention).
Unlike node-to-datum voltages in the node voltage
method which are physical in the sense they can bemeasured by a volt meter, the mesh currents are
abstract variables introduced mathematically for
writing a set of equations whose solution can be usedto find each resistor current ij trivially via
2
, , lm ml l
1
,ml
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1
2
3
6 0 2 2
0 14 8 22 8 10 5
i
i
i
=
(1)
Then calculate :
1 3 1
3 10
i i i= =
1 2 3
13 8 19 , ,20 20 20
i i i= = =
2 1
13
20
i i= =
3 1 2
5 20
i i i= =
4 3 2
11 20i i i= =
5 2
8i i= =
1 1
6210
v i= =
2 2
134
5
v i= =
3 2v =
4 4
22
8 5v i= =
5 5
126v i= =
Mesh
CurrentEquations
,
,
,
,
,
(2)
(3)
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2V
+
-
6v
6i
+-3v
4v+
-
5v
+
-8 6
42
+
-
1v 2v+
-
5i3i
2i
4i
1i
3i1
i
2i
+
1 3 1 i i i=
2 1i i=
3 1 2 i i i=
4 3 2 i i i=
5 2i i=
( )1 3 1 2v i i=
2 14v i=
3 2v =
( )4 3 2 8v i i=
5 26v i=
6 5v =
,
,
,
,
,
,6 3
i i=
Loop equation around mesh 1:
1 2 3 3 1 1 0 2( ) 4 2 0v v v i i i + + = + =
1 3 6 2 2i i =
5V +-
(1)
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5V
2V
+
-
6v
6i
+-3v
4v+
-
5v
+
-8 6
42
+
-
1v 2v+
-
5i3i
2i
4i
1i
3i1
i
2i
1 3 M esh 1 : 6 2 2i i =
2 3
M esh 2 : 14 8 2i i =
1 2 3 M esh 3 : 2 8 10 5i i i + =
1 1 3
1 2 S olving from (1)3 3
i i i =
2 2 3
4 1 S olving from (2)7 7
i i i =
S ub stituting (4) and (5) into (3)
(1)
(2)
(3)
(4)
(5)
+- +
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1i
2i
4i
6i
3i5i
We can redraw this circuit so that there are
no intersecting branches.
1
2
3
4
1i
2i
3i5i
1 3
4
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1i
2i
4i
6i
3i5i
We can redraw this circuit so that there are
no intersecting branches.
1
2
3
4
1i
2i
3i5i
1 3
2
4
8
65V2V
2
5V
2V 6
4
8
+
+
+
+-
+
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All branch voltages and currents can be trivially
calculated from e1 and i3.
121 1 10 , 0
4
vv Ve i Ae= = = =
12
2 26 , 23
vv e V i A= = = =
23 36 , 0v V ie A= = =
14 46 , 2ev V i= = =
V erification of Solution T ellegen's T heorby :em
4
1 1 2 2 3 3 4 4
1
( ) ( ) ( ) ( )j jj
v i v i v i v i v i=
= + + +
(0)(0) (6)(2) (6)(0) ( 6)(2)= + + + ?
0=
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Note:
The unknown variables in the
modified node voltage method consist
of the usual n-1 node-to-datum
voltages, plus the unknown currents
associated with the voltage sources.
Hence, if there are voltage
sources, the modified node voltage
method would consist of (n-1)+
independent linear equations involving
(n-1)+ unknown variables
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Conservation of Electrical
Energy
The algebraic sum of
electrical energy flowing into all
devices in a connected circuit is
zero for all times .t>
Proof.
Tellegen's Theorem
( ) ( ) 0b
t
v t i t dt =
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1 1
0 0 0 0 0 1 1 0 1
0 0 0 0 0 1 0 1 01 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 01 0 0 1 0 0 0 0
0
00
00
0
0
0
0
0 0 0 0 1 0 0 0 0
0 1 0 0 0 4 0 0 0
0 0 1 0 0 0 3 0 0
0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0
6
0
t
2 1
de
=
A
1A
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4 4
0 0 0 0 0 1 1 0 1
0 0 0 0 0 1 0 1 01 1 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0 00 1 0 1 0 0 0 0 0
1 0 0 0 1 0 0 0 0
0 0 1 0 0 4 0 0 0
0 0 0 0 0 0 3 0 0
0 0
0
00
00
0
0
0
60 1 0 0 0 0 0
0 0 0 0 0 0 0 0 12
det
= A
4A
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9 9
0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 1 0 01 1 1 0 0 0 0 0 0
1 0 0 1 0 0 0 0 00 1 0 0 1 0 0 0 0
1 0 0 0 0 1 0 0 0
0 0 1 0 0 0 4 0 0
0 0 0 1 0 0 0 3 0
0 0 0 0 1
0
00
0
0 0 0
0
0
0
0
6 0
0 0 0
det
0 0 120 0 0
=
A
9A
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11 11 21 21 31 31 41 41 51 51
61 61 71 71 81 81 91 91 10,1 10,1
1 a A a A a A a A a Aa A a A a A a A a A
+ + + + = + + + + +
( )91 91 10,1 10,1 11 21 811 , because 0a A a A a a a= + = = =
( ) ( )10,191 91 10,1AA a a= +
11k 12k
11 126 2k k= +
11 1 12 1s sk v k i= +
11e
=
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Instantaneous Power
of a 2-terminal device
D+
-
( )v t
( )i t1
2
( ) ( ) ( )p t v t i t Under Associated Reference Convention,
1( ) 0 at p t t T > =1means ( ) Watts of power ent ersp T
1.
( )flow ats into t T=D
2( ) 0 at p t t T < =
f l
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Instantaneous Power
of ann-terminal device
1
1
( ) ( ) ( )n
j j
j
p t v t i t
=
=
1 2Energy entering from time to :T TD
D
k
n
1
2
n-1
1i
2i
ki
1ni
1v
2v kv
1nv
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Tellegens Theorem has many deep
applications. For this course, it can be used to
check whether your answers in homework
problems, midterm and final exams are
correct.