4.6
Nodal Analysis and Mesh Analysis
of Memoryless Circuits
All element voltages may be related to the three node voltages in this manner by inspection.
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
R
R
R
R
R
R
I
I
1
1
2
2
1
3
3
1
4
2
5
3
2
6
3
1
2
1
2
=
=
−
= −
=
= −
=
= −
=
vv
v
v
I
3
2
3
=
−
(4.2-1)
We observe that all element voltages can be obtained either as some node voltage straightaway or
as difference between two node voltages. Hence, a set of (
n
-
1) node voltages, defined with respect
to the reference node, is a sufficient set of voltage
variables for determining b element voltages.
Note that we are using KVL equations to reduce the number of pertinent voltage variables to (
n
-
1)
from
b.
The KCL equations remain. We use them to solve for these node voltages. But KCL equations
are written in terms of currents. This is where the element equations come in. We substitute element
equations in KCL equations as and when we write KCL equations in order to substitute for element
currents in terms of element voltages. Of course, element voltages will be expressed in terms of node
voltages only. Thus, writing node equations at the
n
-
1 nodes (because only
n
-
1 KCL equations are
independent) involves two mental operations for each element – obtain element voltage in terms of
node voltages and replace current variable by voltage variable with the help of element equation. We
illustrate this for node-1 in the circuit in Fig. 4.2-1.
We write KCL at this node by equating the sum of currents
going away from the node to zero.
i
i
i
I
i e
v
R
v
R
v
R
I
i e
G v
G v
R
R
R
R
R
R
1
2
3
1
1
1
2
2
3
3
1
1 1
2
1
0
0
+
−
− =
+
−
− =
+
. .,
. .,
(
−−
+
−
=
v
G v
v
I
2
3
1
3
1
)
(
)
where
G represents the conductance value (1/
R) of corresponding resistance. The third equation,
which is the final form of node equation, can be written by inspection without writing out the first
two explicitly. With our convention of positive sign for current flowing away from the node, the node
voltage variable at the node where KCL is being written will appear with positive sign and other node
voltage variables will appear with negative sign in the equation. Moreover, the
net current delivered by
current sources to that node will appear with
positive sign on the right side of equation. The remaining
two node equations
for this circuit are
G v
v
G v
v
G v
I
I
G v
v
G v
v
G v
I
2
2
1
5
2
3
4 2
2
3
3
3
1
5
3
2
6 3
2
(
)
(
)
(
)
(
)
−
+
−
+
= − −
−
+
−
+
=
We can solve for
v
1
,
v
2
and
v
3
using the three equations listed below:
G v
G v
v
G v
v
I
G v
G v
v
G v
v
I
I
1 1
2
1
2
3
1
3
1
4 2
2
2
1
5
2
3
2
3
+
−
+
−
=
+
−
+
−
= − −
(
)
(
)
(
)
(
)
G
G v
G v
v
G v
v
I
6 3
3
3
1
5
3
2
2
+
−
+
−
=
(
)
(
)
(4.2-2)
The element voltages may be found out subsequently by using Eqn. 4.2-1 and the element currents
may be obtained by using element equations in the last step.
We followed a certain convention in writing the node equations in Eqn. 4.2-2. Adhering to such a
convention has yielded certain symmetry in these equations. Let us express
these equations in matrix
notation to see the symmetry clearly.
(
)
(
)
G
G
G v
G v
G v
I
G v
G
G
G v
G v
I
I
G
1
2
3
1
2 2
3 3
1
2 1
2
4
3
2
5 3
2
3
3
+
+
−
−
=
−
+
+
+
−
= − −
−
vv
G v
G
G
G v
I
i e
G
G
G
G
G
G
G
G
G
1
5 2
3
5
6
3
2
1
2
3
2
3
2
2
4
3
−
+
+
+
=
+
+
−
−
−
+
+
(
)
. .,
(
)
(
)
−−
−
−
+
+
=
−
−
G
G
G
G
G
G
v
v
v
5
3
5
3
5
6
1
2
3
1
0
0
0
1
1
0
1
0
(
)
=
I
I
I
i e
1
2
3
. .,
YV
CU
(4.2-3)
Y
is called the
Nodal Conductance Matrix,
V
is called the
Node Voltage Vector,
U
is called the
Input
Vector and
C
is called the
Input Matrix.
Note that the order of Nodal Conductance matrix is (
n
-
1)
×
(
n
-
1) and that it is
symmetric. The
diagonal element of
Y
matrix,
y
ii
, is the
sum of conductances connected at the node-i. The off-diagonal
element of
Y
matrix,
y
ij
, is the
negative of sum of all conductances connected between node-i and
node-j. There can be more than one conductance connected between two nodes. Then, they will be
in parallel and they will add in
y
ij
. That is why
y
ij
should be the negative of
sum of all conductances
connected
between node-i and node-
j. The
right-hand side product
CU
is a column vector of
net
current injected by the current sources at the corresponding nodes.
Now, we can write down this matrix equation by inspection after skipping all the intermediate
steps. The following matrix equation results in the case of the example we considered in this section.
G
S G
S G
S
G
S G
S G
1
2
3
4
5
6
1
0 2
5
1
1
1
1
0 5
2
1
1
1
1
0 5
2
=
=
=
=
=
=
=
=
=
=
.
;
;
.
;
;
.
;
Ω
Ω
Ω
Ω
Ω
==
=
−
−
−
−
−
−
= − −
1
0 2
5
8
1
2
1
4
2
2
2
9
9
21
1
2
3
.
;
(
S
v
v
v
−−
= −
17
21
9
4
21
)
(4.2-4)
Solving the matrix equation by Cramer’s rule, we get,
Nodal Analysis of Circuits Containing Resistors ...
4.7
4.8
Nodal Analysis and Mesh Analysis of Memoryless Circuits
v
1
9
1
2
4
4
2
21
2
9
8
1
2
1
4
2
2
2
9
9 36 4
1 36 42
2 8 8
=
−
−
−
−
−
÷
−
−
−
−
−
−
=
− + − +
−
−
(
)
(
)
(
44
8 36 4
1 9 4
2 2 8
446
223
2
8
9
2
1
4
2
2 21
9
8
1
1
)
(
)
(
)
(
)
− + − − −
+
=
=
=
−
−
−
−
−
÷
−
−
V
v
22
1
4
2
2
2
9
8 36 42
9 9 4
2 21 8
223
223
223
1
8
1
1
−
−
−
−
=
− +
− − − − − −
=
=
=
−
(
)
(
)
(
)
V
v
99
1
4
4
2
2 21
8
1
2
1
4
2
2
2
9
8 84 8
1 21 8
9 2 8
8 36
−
−
−
−
÷
−
−
−
−
−
−
=
− + − − +
+
−
(
)
(
)
(
)
(
44
1 9 4
2 2 8
669
223
3
)
(
)
(
)
+ − − −
+
=
=
V
The element voltages can be calculated by using Eqn. 4.2-1 or by inspection. Similarly, the element
currents may be obtained by inspection. The complete solution is marked in the circuits appearing in
Fig. 4.2-2.
I
1
I
2
I
3
R
1
0.2
Ω
R
2
R
5
R
3
R
4
R
6
1
Ω
0.5
Ω
0.5
Ω
0.2
Ω
1
Ω
9 A
2 V
21 A
–17 A
(a)
1 V
3 V
0 V
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
2 V
3 V
2 V
15 A
9 A
–2 V
2 V
10 A
1 A
2 A
1 V
1 V
–2 V
1 V
1 V
1 A
21 A
–17 A
(b)
3 V
1 V
0 V
4 A
Fig. 4.2-2
Nodal analysis solution for circuit in Fig. 4.2-1
This
section has shown that an n-node circuit containing only linear resistors and independent
current sources will have a nodal representation given by
Do'stlaringiz bilan baham: 
