YV
=
CU
again. We note that the
Y
matrix is
symmetric. However, the
U
vector now has four entries – three current source values and one voltage
source value. We note carefully that the
C
matrix has suitable entries that convert the V units into A.
Thus, the matrix product
CU
is a column vector of currents
since a voltage multiplied by conductance is a current.
Moreover, we observe that, the form and entries of
Y
matrix on the left-hand side of equation cannot depend
on the particular values of I
1
, I
2
, I
3
and V
1
. The matrix
must be the same for any numerical values for these four
inputs. Therefore,
Y
matrix entries must be the same for a
case where all these inputs are zero-valued. However, an
independent current source of zero value is an open circuit
and an independent voltage source with zero value is a short-
circuit ! Therefore, it should be possible to write down that
matrix by inspection after we deactivate all the sources in the
circuit. The deactivated circuit is shown in Fig. 4.3-2.
The resistor R
6
simply goes out of picture due to the short-circuit across it. R
5
gets connected
between node-2 and reference node. R
3
gets connected between node-1 and reference node. We write
the
Y
-matrix of this circuit by using the rules we developed in the last section.
y
G
G
G
y
y
G
y
G
G
G
11
1
2
3
12
21
2
22
2
4
5
=
+
+
=
= −
=
+
+
We find that this is the same as the
Y
-matrix in Eqn. 4.3-2. Thus, we conclude that an independent
voltage source directly at a node in a circuit results in a reduction of node voltage variables by one and
in a reduction of order of nodal conductance matrix by one. The nodal conductance matrix remains
symmetric. The nodal conductance matrix may be found by using the deactivated circuit, if necessary.
However, the node equations will have to be written in order to get the right-hand side of Eqn. 4.3-2.
The resistor R
6
disappearing altogether in the analysis is an interesting aspect. We observe that the
value of conductance of this resistor does not appear anywhere in Eqn. 4.3-2. Thus, it has no effect
on the node voltages. We may even remove it when we write the equations for node voltage variables.
Why is it so? For the simple reason that the voltage source has fixed the node potential at the third
node and presence or absence of R
6
has no effect on that fact. Since node potentials decide the voltage
across all elements and currents through them, the rest of the circuit is totally unaffected by the value
of R
6
. The only element that gets affected by R
6
(apart from itself!) is the voltage source. Its current
will have a component due to R
6
. We may treat a resistor across an independent voltage source as a
trivial element.
Substituting the values for conductances and source functions in Eqn. 4.3-2 we get,
8
1
1
4
9 2 3
11 15 2 3
15
2
1
2
−
−
=
+ ×
− −
− + ×
=
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