2–1
ECE2205: Circuits and Systems I
WRITING CIRCUIT EQUATIONS
• So far, we have used ad hoc methods to solve circuit equations.
• We need more systematic approaches that guarantee always coming
up with the required equations. Alternatives: Guess, crash, and burn.
• In this chapter, we study three methods (recipes):
– The exhaustive (exhausting!) method
– The simplified exhaustive method;
– Node analysis;
– We skip loop/mesh analysis.
• The goal is to write the simplest set of equations that may be solved
to find all voltages and currents in a circuit.
Planar Networks, Trees, Co-Trees, Meshes, and Such
• Before we can examine these methods, we need a few definitions.
• A network (circuit) is
planar if its circuit can
be drawn in a 2D
plane without having
a crossing path.
Planar
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
Non-planar
ECE2205, WRITING CIRCUIT EQUATIONS
2–2
• The circuits we study in this course are nearly all planar:
– Loop/mesh analysis works only for planar circuits, which is one
reason we skip it here;
– Exhaustive/node analysis works for all types of circuit.
• An example planar network is drawn to
the right.
• A tree of a circuit is a graph in which
1. Every node is in the graph;
2. One (and only one) path connects
any node with any other node;
3. No closed paths exist.
• The tree diagram is not unique.
• A co-tree is the set of branches not
included in the tree.
• There will be N − 1 edges (branches) to
connect N nodes in a tree.
• Inserting a missing branch from the co-tree forms a loop in the circuit
called a mesh. Each mesh defines a path for KVL.
The Exhaustive Method
• For every circuit with e (non-source) elements and s sources, there
are 2e + s unknowns to solve for:
– Both a voltage and current for each element;
– Either a voltage or current for each source.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–3
ECE2205, WRITING CIRCUIT EQUATIONS
• The exhaustive method finds 2e + s independent (simultaneous)
equations to solve using the following method:
1.(a) Define a voltage variable and a current variable for each
element in the circuit.
(b) Define a voltage variable for each current source and a current
variable for each voltage source.
2. Write a v-i equation for each element.
3. Write a KVL equation for each mesh.
4. Write a KCL equation for all but one node (it does not matter which
one is omitted).
5. Solve the resulting 2e + s equations.
EXAMPLE :
Solve the circuit given below.
3
+
−
is(t)
v0 (t)
+
v2 (t)
+
v1 (t)
4
−
−
+
1
v3 (t)
−
1.(a) The diagram is labeled with v1(t), v2(t), and v3(t).
The corresponding currents use the default sign convention.
(b) The current source is given a vo(t) variable.
2. By Ohm’s law,
v1(t) = i 1(t) × 4,
v2(t) = i 2(t) × 3,
v3(t) = i 3(t) × 1
3. The left mesh gives, via KVL, vo (t) + v1(t) = 0.
The right mesh gives v1(t) = v2(t) + v3(t).
4. The top-middle node gives, via KCL, i s (t) = i 1(t) + i 2(t).
The top-right node gives i 2(t) = i 3(t).
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–4
ECE2205, WRITING CIRCUIT EQUATIONS
• To solve these seven equations and seven unknowns, we can use the
v-i equations to eliminate the voltages from the KVL equations.
• The KVL equations become:
vo (t) + 4i 1(t) = 0
4i 1(t) = 3i 2(t) + i 3(t).
• If we know i 1(t), we then know vo (t), so this equation is decoupled
from the rest.
• We are left with
i 1(t) + i 2(t) = i s (t)
−i 2(t) + i 3(t) = 0
−4i 1(t) + 3i 2(t) + i 3(t) = 0.
• The second equation gives i 2(t) = i 3(t), and we simplify again
i 1(t) + i 2 (t) = i s (t)
−4i 1(t) + 4i 2(t) = 0.
• We get i 1(t) = i 2(t) = i 3(t) = i s (t)/2.
• This in turn gives us
v1(t) = 4i 1(t) = 2i s (t)
3
v2(t) = 3i 2(t) = i s (t)
2
1
v3(t) = i 3(t) = i s (t)
2
vo (t) = −v1(t) = −2i s (t).
• Whew! And this is a simple example!
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–5
ECE2205, WRITING CIRCUIT EQUATIONS
The Simplified Exhaustive Method
• We can reduce the number of equations to be solved if we notice that
one equation per source can be quickly eliminated.
• To do so systematically, we define supernodes and supermeshes.
Supernodes
• Consider a section of a network that contains a voltage source that
connects two nodes.
i1
i3
vs(t)
– At node a, KCL gives i 1 + i 2 − i = 0;
i
– At node b, KCL gives i 3 + i 4 + i = 0;
a
b
i2
– Together, i 1 + i 2 + i 3 + i 4 = 0.
i4
• The last equation does not involve the source variable i, and is the
KCL for the surface drawn.
• We can use one KCL equation for the “supernode” instead of two
KCL equations for the two nodes.
• If we need to know i, then compute i = i 1 + i 2 when done.
Supermeshes
• Consider a circuit with a current source.
+
v1 (t)
• The network on the right has a current
−
source that is a member of two meshes.
−
is(t)
– KVL and the left mesh gives v1(t) + v(t) = 0.
– KVL and the right mesh gives −v(t) − v2(t) = 0.
– Together, v1(t) − v2(t) = 0.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
v(t)
+
+
v2 (t)
−
2–6
ECE2205, WRITING CIRCUIT EQUATIONS
• This is the KVL of the closed path formed (mesh formed) if the current
source is removed.
• We can use one KVL equation for the “supermesh” instead of two for
the individual meshes.
• If we need to know v(t), then compute v(t) = −v1(t).
+
• Sometimes a current source
does not separate two meshes
(an exterior current source).
is(t)
v1 (t)−
+
+
v(t)
v2 (t)
−
−
vs(t)
• In that case, we ignore the KVL equation (v2(t) − v(t) = 0) for that
source unless we need to know the current-source voltage itself.
Simplified Exhaustive Method (continued)
• The exhaustive method required 2e + s equations.
• Using supernodes and supermeshes, we only need 2e equations and
2e unknowns.
– Unknowns: e are element voltages, and e are element currents.
– Equations: e equations are element relations, and e are KVL/KCL.
– For resistors, we immediately use Ohm’s law to reduce to e
equations.
• The simplified exhaustive method is summarized below.
1. Create a supernode around each voltage source and the nodes to
which it is attached.
2. Create a supermesh from each pair of adjacent meshes that are
separated by a current source.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–7
ECE2205, WRITING CIRCUIT EQUATIONS
3. Define a voltage variable and a current variable for each element
(not the sources) in the circuit.
4. Write a v-i relation for each element.
5. Write a KVL equation for each mesh or supermesh whose path
does not contain an exterior current source.
6. Write a KCL equation for all but one of the nodes or supernodes.
7. Solve the resulting set of equations for the element variables.
8. If any of the source variables are required, solve for each of them
after the element variables have been determined.
3
EXAMPLE :
Consider the circuit shown:
+
v2 (t)
+
• No voltage sources, therefore no
supernodes.
v1 (t)
is(t)
−
4
−
• Only an exterior current source—ignore KVL for that mesh.
• v-i relationships yield
v1(t) = i 1(t) × 4;
v2(t) = i 2(t) × 3;
v3(t) = i 3 (t) × 1.
• KVL for the only mesh gives v1(t) − v2(t) − v3(t) = 0.
• KCL for the top nodes gives the equations:
i s (t) − i 1(t) − i 2(t) = 0
i 2(t) − i 3(t) = 0
v1(t) v2(t)
+
4
3
v2(t) v3(t)
−
= 0.
3
1
or i s (t) =
or
• Solving,
v2(t) = 3v3(t), v1(t) = 4v3(t), i s (t) = 2v3(t) or,
v3(t) = i s (t)/2, v2(t) = 3i s (t)/2, v1(t) = 2i s (t).
and so forth for i 1 (t), i 2(t), and i 3(t).
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
+
1
v3 (t)
−
2–8
ECE2205, WRITING CIRCUIT EQUATIONS
EXAMPLE :
4Ω
i1 (t)
Consider the circuit shown:
• One voltage source makes “supernode”
i2 (t) 3 Ω
• We ignore one node using KCL—
we will ignore the supernode here.
i4 (t) 2 Ω
i3 (t)
vs(t)
2Ω
• One exterior current source i s (t)
gives mesh to ignore.
is(t)
• Middle node: i 2(t) = i 3(t) + i 4 (t).
• Right node: i 1(t) + i 4(t) = i s (t).
• Top mesh: 2i 4(t) + 3i 2(t) = 4i 1(t).
• Bottom mesh: vs (t) = 3i 2(t) + 2i 3(t).
Dependent Sources
• We need to account for dependent sources in our equations, but they
do not introduce any new variables or equations.
EXAMPLE :
Consider the circuit:
• Supernode around vs (t)
and ground is ignored
KCL equation.
i1 (t) 6 Ω
i3 (t) 2 Ω
i2 (t)
vs(t)
3Ω
i4 (t)
1
i (t)
2 2
• Supermesh around dep. source gives: 3i 2(t) = 2i 3(t) + 4i 4(t).
• KVL around left mesh gives: vs (t) = 6i 1(t) + 3i 2(t).
• KCL at top-left node gives: i 1(t) = i 2 (t) + i 3(t).
1
• KCL at middle gives: i 3(t) + i 2(t) = i 4(t).
2
• Altogether, four equations and four unknowns.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
4Ω
2–9
ECE2205, WRITING CIRCUIT EQUATIONS
Solving Circuit Equations
• We have seen enough examples to convince you that we will need to
solve numerous systems of n equations and n unknowns.
• This can be done by hand (tedious) or using M ATLAB (etc).
– Knowing some linear algebra can really make life more pleasant.
– The automatic methods rely heavily on matrix/vector operations.
• This set of notes reviews the mechanics of matrix manipulation. An
attempt is also made to aid intuition.
A Matrix Primer: Terminology and Notation
• A matrix is a rectangular array of numbers (a.k.a., scalars) written
between brackets.
EXAMPLE :


0.1 1.2 −2.4 0.4


A =  0.5 −0.2 1.3 −2.5  .
−0.2 1.1 9.5 −1.8
• An important attribute of a matrix is its size or dimension.
• Measured in number of rows × number of columns. Above: 3 × 4.
• The entries or coefficients are the values in the array.
• The i, j entry is the value in the ith row and the jth column.
• The i, jth entry in matrix A is Ai j which is a number.
• The positive integers i and j are called the (row and column,
respectively) indices.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
ECE2205, WRITING CIRCUIT EQUATIONS
2–10
EXAMPLE :
A13 = −2.4, A31 = −0.2. The row index of the bottom row is 3,
the column index of the first column is 1.
• A matrix with only one column (i.e., size of n × 1) is called a column
vector, or just a vector.
• Sometimes, size is specified by calling it an n-vector.
• Entries are denoted with just one subscript (the other is “1”) as in v3.
• The entries are sometimes called the components of the vector.
EXAMPLE :


3


v =  0.5 
−1
is a 3-vector (or 3 × 1 matrix); its third component is v3 = −1.
• Sometimes a 1 × 1 matrix is considered to be the same as a scalar,
i.e., a number.
• Two matrices are equal if they are the same size and all the
corresponding entries (which are numbers) are equal.
Matrix Addition and Subtraction
• Two matrices of the same size can be added together to form another
matrix (of the same size) by adding the corresponding entries.
• Matrix addition is denoted by the symbol +. (Thus the symbol + is
overloaded to mean scalar addition when scalars appear on its leftand right-hand side, and matrix addition when matrices appear on its
left- and right-hand sides.)
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–11
ECE2205, WRITING CIRCUIT EQUATIONS
EXAMPLE :
"
1 2
3 4
#
+
"
5 6
7 8
#
=
"
6 8
10 12
#
.
• Matrix subtraction is similar: Replace the “+” with a “−”.
Matrix Multiplication
• It is also possible to multiply two matrices using matrix multiplication.
• You can multiply two matrices A and B provided that their dimensions
are compatible, which means that the number of columns of A equals
the number of rows of B.
EXAMPLE :
Am× p B p×n = C m×n .
• The product is defined by
p
X
Ci j =
Ai k Bk j = Ai 1 B1 j +· · ·+ Ai p B pj ,
i = 1, . . . , m,
j = 1, . . . , n.
k=1
• This looks complicated, but is not too difficult.



C 11 · · · C 1n
A11 · · · A1 p


 ..
 ..
...
... 
 B11 · · · B1 j · · · B1n
 .
 .


...  = 
...
...
 Ai 1 · · · Ai p  
Ci j
 





 ..
...
.
...
 B p1 · · · B pj · · · B pn
 ..
 .
Am1 · · · Amp
C m1 · · · C mn








• To find the i, jth entry of the product C = AB, you need to know the
ith row of A and the jth column of B.
• The summation above can be interpreted as “moving left-to-right
along the row i of A while moving top-to-bottom down column j of B.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
ECE2205, WRITING CIRCUIT EQUATIONS
2–12
As you go, keep a running sum of the product of entries: one from A
and one from B.”
• One VERY important fact is that matrix multiplication is not (in
general) commutative. We DON’T have AB = B A. In fact B A may
not even make sense (due to dimensions) and even if it does make
sense, it may have different dimension than AB so that equality in
AB = B A is meaningless.
EXAMPLE :
If A is 2 × 3 and B is 3 × 4 then AB makes sense, and is 2 × 4.
B A does not make sense.
EXAMPLE :
Even if both make sense (as in when both are square, for
example) AB 6= B A in general
#
# "
#"
#
"
# "
#"
"
23 34
1 2
5 6
19 22
5 6
1 2
.
=
,
=
31 46
3 4
7 8
43 50
7 8
3 4
• Matrix multiplication is associative; i.e., ( AB)C = A(BC). Therefore,
we write a product as ABC.
• Matrix multiplication is also associative with scalar multiplication; i.e.;
α( AB) = (α A)B.
• Matrix multiplication distributes across matrix addition:
A(B + C) = AB + AC, and ( A + B)C = AC + BC.
Matrix-Vector Product
• A very important type of matrix multiplication: matrix-vector product.
EXAMPLE :
y = Ax,
where A is an m × n matrix, x is an n-vector and y is an m-vector.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
ECE2205, WRITING CIRCUIT EQUATIONS
2–13
• We can think of matrix-vector multiplication (with an m × n matrix) as
a function that transforms n-vectors into m-vectors. The formula is:
yi = Ai 1 x1 + · · · + Ai n xn , i = 1, . . . , m
Matrix Inverse
• If A is square, and there is a matrix F such that F A = I , then we say
that A is invertible or nonsingular. We call F the inverse of A, and
denote it A−1.
• It is important to note that not all square matrices are invertible. For
example,
"
#
1 −1
−2 2
does not have an inverse.
• As an example of a matrix inverse, we have
#
"
#−1 "
2 −1
1 1
=
−1 1
1 2
(you should check this!).
• It is very useful to know the general formula for a 2 × 2 matrix inverse.
#−1
"
"
#
1
a b
d −b
=
ad − bc −c a
c d
provided ad − bc 6= 0. (If ad − bc = 0, the matrix is not invertible.)
¡ −1¢−1
= A.
• When a matrix is invertible, A
Linear Equations
• Any set of m linear equations in (scalar) variables x 1, . . . x n can be
represented by the compact matrix equation Ax = b, where x is a
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–14
ECE2205, WRITING CIRCUIT EQUATIONS
vector made from the variables, A is a m × n matrix and b is a
m-vector.
EXAMPLE :
1 + x2 − x 3 = −2x 1,
x3 = x2 − 2.
• Rewrite the equations with the variables lined up in columns, and the
constants on the right-hand side.
2x1 +x2 −x 3 = −1
0x1 −x 2 +x3 = −2.
• Now it is easy to rewrite the equations as a single matrix equation
 
"
#
"
# x1
−1
2 1 −1  
=
,
x
 2
−2
0 −1 1
x3
so we have two equations in three variables as Ax = b where
 
#
"
#
"
x1
−1
2 1 −1
 
.
b=
,
x =  x2  ,
A=
−2
0 −1 1
x3
Solving Linear Equations
• Suppose we have n linear equations in n variables x 1, . . . , x n , written
in the compact matrix notation Ax = b.
• A is a n × n matrix; b is an n-vector. Suppose that A−1 exists. Multiply
both sides of Ax = b by A−1.
A−1( Ax) = A−1 b
I x = A−1 b
x = A−1 b.
• We have solved the simultaneous equations.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
ECE2205, WRITING CIRCUIT EQUATIONS
2–15
• We see the importance of the matrix inverse in solving simultaneous
equations.
• We can’t always solve n simultaneous equations for n variables. One
or more of the equations may be redundant (i.e., may be obtained
from the others), or the equations may be inconsistent (i.e., x 1 = 1,
and x1 = 2).
• When these pathologies occur, A is singular (non-invertible).
Conversely, when A is non-invertible, the equations are either
redundant or inconsistent.
• From a practical point of view, either you don’t have enough equations
or you have the wrong ones. Otherwise, A−1 exists, and you can
solve x = A−1b.
Solving Linear Equations in Practice
• When we solve linear equations by computer, we don’t use x = A−1 b,
although that would work. Practical methods compute x = A−1 b
directly.
• A may be large, sparse, or poorly conditioned. There exist efficient
methods to handle each case.
• In Matlab, x=A\b;
EXAMPLE :
Suppose we wish to solve the following three simultaneous
equations.
1.4x1 + 2.3x2 + 3.7x3 = 6.5
3.3x1 + 1.6x2 + 4.3x3 = 10.3
2.5x1 + 1.9x2 + 4.1x3 = 8.8.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–16
ECE2205, WRITING CIRCUIT EQUATIONS
• Begin by re-arranging into a matrix-vector multiplication equation

  

1.4 2.3 3.7
x1
6.5

  

 3.3 1.6 4.3   x2  =  10.3  .
2.5 1.9 4.1
x3
8.8
• In M ATLAB,
A = [1.4 2.3 3.7; 3.3 1.6 4.3; 2.5 1.9 4.1];
b = [6.5; 10.3; 8.8];
x = A\b
• This gives the result:


1


x =  −1  .
2
EXAMPLE :
Let’s apply this
procedure to a specific
circuit.
i2 (t) 2 Ω i3 (t) 1 Ω
i1 (t)
i4 (t)
2Ω
is(t)
• One supermesh around the current source gives
2i 1(t) = 2i 2(t) + i 3(t) + 4i 4(t).
• The other mesh gives: 4i 4(t) = −vs (t) + 3i 5(t).
• Three KCL equations, including supernode gives
0 = i 1(t) + i 2(t)
i 2(t) + i s (t) = i 3(t)
i 3(t) = i 4(t) + i 5(t).
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
vs(t)
i5 (t)
4Ω
3Ω
2–17
ECE2205, WRITING CIRCUIT EQUATIONS
• Put these equations in matrix form Ci(t) = s(t)


 

2 −2 −1 −4 0
i 1(t)
0


 

 0 0 0 −4 3   i 2(t)   vs (t) 


 

 1 1 0 0 0   i 3(t)  =  0  .


 



 

 0 −1 1 0 0   i 4(t)   i s (t) 
0 0 1 −1 −1
i 5(t)
0
|
{z
} | {z } | {z }
C
i (t)
s(t)
• To solve for i(t), we get




0
0
 
 
1
0


 
−1
−1 
−1 

i(t) = C s(t) = C  0  vs (t) + C  0 
 i s (t)
 
 
0
1
0
0


0 0


#
 1 0 "

 vs (t)
−1 
=C 0 0
 i (t) .

 s
0 1
0 0
• In M ATLAB,
C = [2 -2 -1 -4 0; 0 0 0 -4 3; ...
1 1 0 0 0; 0 -1 1 0 0; ...
0 0 1 -1 -1];
s = [0 0; 1 0; 0 0; 0 1; 0 0];
i = C\s
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–18
ECE2205, WRITING CIRCUIT EQUATIONS
• This results in

−0.0851

 0.0851

i(t) = 
 0.0851

 −0.1064
0.1915

0.4043

#
−0.4043  "
 vs (t)
0.5957 
 i (t) ,
 s
0.2553 
0.3404
or,
i 1(t) = −0.0851vs (t) + 0.4043i s (t)
i 2(t) = 0.0851vs (t) − 0.4043i s (t) etc.
Superposition of Sources
• Suppose a network is excited by more than one independent source.
• We have just seen via M ATLAB example that the effect of these
sources on the network can be separated. That is, if

  
 
s1(t)
1
0

  
 
 s2(t)   0 
0




 sn (t),
s(t) =  .  =  .  s1(t) + · · · + 
.


.
.
.
 .   . 
 . 
sn (t)
0
1
then
i(t) = C −1s(t)
 
 
0
1
 
 
 
 
−1  0 
−1  0 
= C  .  s1(t) + · · · + C  .  sn (t).
 .. 
 .. 
1
0
• The overall current is a superposition (sum) of the currents induced
by each source.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–19
ECE2205, WRITING CIRCUIT EQUATIONS
• Note that this does not apply directly with dependent sources since
turning off one dependent source would affect the matrix C.
EXAMPLE :
Let’s analyze a circuit using
the ideas of superposition.
• First, “turn off” the current source. By
KVL,
vs (t) = R1i 1 + R2i 1
= i 1(R1 + R2 ).
vs(t)
– By KCL, i 2 + i s = i b . Combining,
0 = i 2 R1 + R2(i 2 + i s ).
• By superposition,
i(t) = i 1(t)+i 2(t) =
is(t)
R2
i1 (t) R1
• Second, “turn off” the voltage source.
– By KVL around the supermesh,
0 = i 2 R1 + i b R2 .
ia(t)
i(t) R1
vs(t)
R2
i2 (t) R1
ib(t)
is(t)
R2
R2i s (t)
vs (t)
−
.
R1 + R2 R1 + R2
This result agrees with the simplified exhaustive method.
The Nodal Analysis Method
• The exhaustive method gives us 2e + s equations and 2e + s
unknowns to simultaneously solve.
• The simplified exhaustive method gives us 2e equations and 2e
unknowns to solve.
– For resistive circuits we can immediately reduce this to e equations
and unknowns.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–20
ECE2205, WRITING CIRCUIT EQUATIONS
• Note that the “C” matrices were sparse—an indication that we might
be able to solve the problem using fewer coupled equations.
• The “node method” reduces the required equations to n − 1 equations
where n is the number of nodes/supernodes.
• The “mesh method” reduces this to l where l is the number of
meshes/supermeshes without exterior current source.
• We choose the method that gives the fewest equations to solve.
– In practice, the node method is always used (nodes are easier to
identify automatically via computer program, meshes not so much;
nodal analysis works with 3D circuits, mesh analysis only on
planar networks).
– We will not study the mesh method here, although it is discussed
in the text if you are interested.
• The node analysis method first assigns potentials to every node.
• In the figure, these potentials are ea (t), eb (t),
ec (t), and ed (t).
• Element voltages may be calculated from
node potential differences. e.g.,
v1(t) = ea (t) − eb (t).
ea(t)
v1 (t)
+
−
+
+
v4 (t)
v2 (t)
−
ed(t)
eb(t)
−
+
−
v3 (t)
ec(t)
• Using node potentials as variables, KVL is automatically satisfied:
v1(t) +v2(t) −v3(t) −v4(t) = (ea −eb ) +(eb −ec ) −(ed −ec ) −(ea −ed ) = 0.
• Note that potentials are specified with respect to an arbitrary
point (the ground node) which is assigned a zero potential.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2–21
ECE2205, WRITING CIRCUIT EQUATIONS
• The node method is then:
1. Create a supernode encircling each independent or dependent
voltage source and the two nodes to which it is attached.
2. Select one of the nodes of the circuit as the ground node.
3. Define n − 1 node potential variables at the remaining nodes or
supernodes of the circuit.
4. Set up KCL equations at n − 1 of the nodes/supernodes in the
network. The currents in these equations must be expressed in
terms of the node potentials.
5. Solve those n − 1 equations for the n − 1 node potentials.
6. Calculate the element voltages and currents of interest from the
node potentials.
EXAMPLE I :
Let’s first try an example
with only current sources.
is1 (t)
• Lowest node selected as ground.
• Three nodes defined: a, b, c.
• KCL at a gives
ea − e b ea − 0
i s1 (t) =
+
2
2
= ea − eb /2.
a
2Ω
2Ω
4Ω
eb − e c eb − 0 eb − e a
+
+
.
1
4
2
ec − e b ec − 0
• KCL at c gives: 0 = i s1 +
+
.
1
3
• KCL at b gives: i s2 (t) =
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
b
1Ω
is2 (t)
c
3Ω
2–22
ECE2205, WRITING CIRCUIT EQUATIONS
• These equations can be put in matrix form


  
1
 1 −
 ea   1 0  
0 


  

2
 i

  
  s1
 1 7
  


  = 
−1   eb   0 1  
−

 2 4
  
 i s2

  


4   
−1 0
ec
0 −1
3



.

• We have three equations versus five for the simplified method, and
twelve for the exhaustive method.
• Once we solve and know ea , eb , and ec , we can find all element
currents/voltages and all source voltages directly.
EXAMPLE II :
Consider the circuit with R = 2.
• Lowest node selected as ground.
ea − e b ea − e c
• KCL at a gives: i s =
+
.
4
1
• KCL at b gives:
eb − e a eb − e c eb − 0
+
+
.
0=
4
4
1
R/2
2R
ea
is
ec − e a ec − e b ec − 0
+
+
.
1
4
2
• Combining, we get the system of equations:
   

5
1
 ea   1 

−
−1 

   
 4
4
   

   
 1 3
1
   =   is .

−   eb   0 
−
 4 2
4    
   


1 7    
ec
0
−1 −
4 4
• KCL at c gives: 0 =
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
eb
2R
R/2
ec
R
2–23
ECE2205, WRITING CIRCUIT EQUATIONS
EXAMPLE III :
Now, try an example with voltage sources.
• Create a supernode around
vs1 (t)
each voltage source—only
two nonzero potentials are
independent. Note:
ea = vs1 (t) and ec = vs2 (t)+ed .
a
2Ω
b
4Ω
vs2 (t)
6Ω
4Ω
• Choose b and d as nodes at which to write KCL.
• KCL at b gives:
eb − e a eb − 0 eb − e c
+
+
2
4
6
eb − vs1 eb − 0 eb − (vs2 + ed )
=
+
+
.
2
4
6
0=
• KCL at d gives:
ec − e b ec − 0 ed − 0
+
+
6
4
2
vs + e d − e b vs 2 + e d − 0 e d − 0
= 2
+
+
.
6
4
2
• Put equations in matrix form
  



11
1
1 1
 v 

−   eb  
 12
    2 6   s1 
6

  = 

.
 1 11    



5
−
0 −
ed
vs 2
6 12
12
0=
• Solve for eb and ed , then find any other value of interest.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
c
d
2Ω
2–24
ECE2205, WRITING CIRCUIT EQUATIONS
5V
EXAMPLE I v:
Top supernode gives
e1 = e3 + 5.
1Ä
e1
• Right supernode gives e3 = −(−7) = 7
and e1 = 12.
e2
6A
6Ä
e3
−7 V
2A
• KCL at 2 gives
e2 − e 1 e2 − e 3
+
1
6
e2 − 12 e2 − 7
+
1
6
7e2
e2
=2
=2
= 6(14) + 7
= 13.
• Note: The 6A current source does not affect any equation!
v: Linear dependent
source. Find v R5 in terms of v{s}.
All resistances have value 1 Ä.
EXAMPLE
ea
R1
eb
ec R4
R2
ed
ib
• Each voltage source gets a
supernode so that ea = vs1 and
e b = vs 2 .
vs1
vs2
R3
βi b
• KCL at c gives:
ec − e a ec − e b ec − e d ec − 0
+
+
+
= 0
1
1
1
1
(ec − vs1 ) + (ec − vs2 ) + (ec − ed ) + ec =
4ec − ed = vs1 + vs2 .
• KCL at d gives:
β
ed − e c ed − 0
ec − e d
=
+
1
1
1
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
R5
2–25
ECE2205, WRITING CIRCUIT EQUATIONS
0 = βec − βed + ec − 2ed
= (1 + β)ec − (2 + β)ed .
• Combining,
"
4
−1
1 + β −(2 + β)
#"
ec
ed
#
=
"
1 1
0 0
#"
vs 1
vs 2
#
.
• Then, v R5 = ed .
Conservation of Power
• We conclude this chapter with a discussion of power.
– Recall that power dissipated by a resistor = v × i.
– Recall that power supplied by a source = v × i.
• By our default sign convention, we
automatically get that dissipated power is
nonnegative and generated power is
nonpositive.
• As an example of power calculation,
• Nodal analysis gives
10 − 0
+ is = 0
10, 000
i s = −1 mA.
• Note that the power dissipated by the source is
Ps (t) = i s (t)vs (t)
= −1 mA × 10 V
= −10 mW.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
i
i
v
v
is
10 V
10 kÄ
2–26
ECE2205, WRITING CIRCUIT EQUATIONS
• Because power dissipated is less than zero, then power supplied is
10 mW, which is greater than zero.
• Note that resistor current = 1 mA by Ohm’s law and
Presistor = 1 mA × 10 V = 10 mW.
• The signs of power match our expectations.
X
• Another important observation is that
p(t) = 0 in the network.
• This is a general fact that is not unique to this network, and is given
the name “Conservation of Power Principle”.
• In any circuit,
p=
e+s
X
vk i k = 0.
k=1
• Net power absorbed in any circuit is zero.
The Next Step
• In this chapter, we have studied methods guaranteed to result in
equations that may be solved to find all voltages and currents in a
circuit.
• We saw that the nodal analysis method resulted in far fewer
equations to solve simultaneously than either variant of all the
exhaustive methods.
• But, what if we don’t want all voltages/currents?
• Next chapter looks at methods of circuit simplification that result in a
desired subset of voltages/currents only.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
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