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4.4.2 
 Applying source transformation in nodal 
Analysis of circuits
The application of this theorem in nodal analysis of circuits 
containing independent voltage sources is illustrated now. 
Consider the circuit (a) in Fig. 4.4-3. The circuit has five 
nodes including the reference node. All the nodes are 
identified. We observe that the voltage source V
3
is in 
series with resistor R
6
. This combination may be replaced 
by a current sourceI
2 
=
V
3
/ R
6
in parallel with R
6
as shown 
in (b) of Fig. 4.4-3. This results in elimination of node-4. 
The circuit in (b) is a 4-node circuit with two independent 
voltage sources (V
1
and V
2
) constraining its node voltage 
variables. Thus, there is only one node voltage variable.
We start assigning node voltage variables from the leftmost node. The first node in circuit (b) is 
unconstrained and therefore we assign the node voltage variable v
1
to that node. That fixes the node 
voltage at node-3 as v
1
+
 V
2
. Hence, a new node voltage variable is not needed at node-3. Node-2 
potential is directly constrained by the source V
1
and hence a node voltage variable is not needed 
there.
R
S
R
S
R
S
i
(
t
)
i
(
t
)
v
(
t
)
v
(
t
)
v
S
(
t
)
v
S
(
t
)
+
+
+
–
–
R
S
i
(
t
)
v
(
t
)
R
S
i
S
(
t
)
+
+
–
–
–
R
S
i
(
t
)
v
(
t
)
i
S
(
t
)
+
–
Fig. 4.4-2 
Source equivalence 
between voltage and 
current sources

---

Source Transformation Theorem and its Use in Nodal Analysis 
4.17
V
1
V
2
V
3
R
(a)
4
3
2
1
R
2
R
1
v
1
I
1
R
4
0.2 
Ω
1 
Ω
11 A
1 V
1 V
3.8 V
1 
Ω
R
6
0.2 
Ω
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
R
3
0.5 
Ω
R
5
0.5 
Ω
V
1
V
2
R
(b)
19 A
3
1
R
2
R
1
v
1
v
3
= 
v
1 
+ 1
v
2
= 1 V 
i
v1
i
v2
I
1
I
2
R
4
0.2 
Ω
1 
Ω
11 A
1 V
1 V
1 
Ω
R
6
0.2 
Ω
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
R
3
0.5 
Ω
R
5
0.5 
Ω
Fig. 4.4-3 
(a) Nodal analysis example circuit (b) Circuit after node reduction by source 
transformation
We need to write the node equations at node-1 and node-3 and combine them to get an equation in 
the single variable v
1
. The node equation at node-2 is not needed for solving node potentials since it 
is a directly-constrained node.
KCL at Node 1
KCL at Node 3
− →
+
−
−
−
=
− →
+
G v
G v
V
G V
i
I
G v
V
V
1 1
2
1
1
3 2
1
6
1
2
2
(
)
(
))
(
)
+
+
+ −
+
=
G V
G v
V
V
i
I
V
3 2
5
1
2
1
2
2
Adding these two equations, we get,
G v
G v
V
G v
V
G v
V
V
I
I
I
G V
1 1
2
1
1
6
1
2
5
1
2
1
1
2
1
6 3
+
−
+
+
+
+ −
= +
= +
(
)
(
)
(
)
Expressiing this in matrix form,
G
G
G
G
v
G
G
1
2
5
6
1
2
5
1
+
+
+
(
)


[ ]
=
+
(
))
(
)
−
+
[
]












G
G
G
I
V
V
V
5
6
6
1
1
2
3
We note that the trivial element R
3
 has no role in deciding the node voltages. Further, we note that 
the node equation has the format 
YV
=
CI
 
where the 
Y
-matrix is of 1 
× 
1 and is the nodal conductance 
matrix of the deactivated circuit. On deactivating the circuit in Fig. 4.4-3(a) by replacing voltage 
sources with short-circuits and current sources with open circuits, a simple circuit containing four 
resistors – R
1
, R
2
, R
5
and R
6
– will be the result.
Substituting the numerical values and solving for v
1
, we get, 13
26
2
1
1
v
v
=
⇒ =
V. Therefore v
3
=
3V.
Now, we fit these values of node voltages into the circuit in Fig. 4.4-3(a) and obtain the voltage 
across the resistors and current sources and currents through resistors by inspection.
i
V
1
is obtained by applying KCL at node-2 of the original circuit.
G v
v
G v
G v
v
i
i e
i
V
V
2
2
1
4 2
5
2
3
1
1
0
1 1 2
1 1
2 1 3
(
)
(
)
. ., (
)
( )
(
)
−
+
+
−
+ =
− +
+
− + =
00
4
1
⇒ =
i
V
A

---

4.18
Nodal Analysis and Mesh Analysis of Memoryless Circuits
i
V
2
is obtained by applying KCL either at node-1 or at node-3 of the original circuit. Choosing node-1,
G v
v
G v
G v
v
i
I
i e
i
V
V
2
1
2
1 1
3
1
3
1
2
2
1 2 1
5 2
2 2 3
(
)
(
)
. ., (
)
( )
(
)
−
+
+
−
−
=
− +
+
− −
==
⇒ = −
9
2
2
i
V
A
The current delivered by V
3
is the same as the current in R
6
. Hence, 
i
V
3
4
= −
A.
The complete solution is marked in Fig. 4.4-4.
V
1
V
3
1 
Ω
11 A
10 A
1 A
4 A
4 A
0.8 V
3.8 V
4 A
2 A
2 A
1 A
2 V
2 V
3 V
0 V
1 V
1 V
1 V
1 V
2 V
1 V
1 V
I
1
R
4
+
+
+
+
+
–
–
–
–
–
+
–
+
+
+
–
–
–
V
2
Fig. 4.4-4 
Complete solution for circuit in Fig. 4.4-3(a)
4.5 
 nodAl AnAlysIs of cIrcuIts contAInIng dePendent
current sources
Dependent current sources do not pose any problem for nodal analysis. Independent current sources 
appear in the right-hand side of node equations. However, dependent current sources affect the 
coefficients of node voltage variables in the left-hand side of node equations.
The controlling variable of a linear dependent current source will be a voltage or current existing 
elsewhere in the circuit. However, any voltage or current variable in the circuit can be expressed in 
terms of node voltage variables. Hence, the dependent current source function can be expressed in 
terms of node voltage variables. Therefore, dependent current sources will affect the coefficients of 
node equation, i.e., they will change the nodal conductance matrix. We will see that they can destroy 
the symmetry of the nodal conductance matrix.
We develop the nodal analysis procedure for this kind of circuits through two examples. The first 
example has node voltages that are not constrained by independent voltage sources and the second one 
has node voltage variables constrained by independent voltage source.
example: 4.5-1 
Solve the circuit in Fig. 4.5-1(a) completely.

---

Nodal Analysis of Circuits Containing Dependent Current Sources 
4.19
(a)
(b)
17 V
9 A
I
1
+
+
+
+
+
+
+
–
–
–
–
–
–
1 
Ω
0.5 
Ω
1 
Ω
21 
v
x
0.2 
Ω
R
2
v
x
v
1
v
2
v
3
R
3
R
4
R
5
R
6
R
1
0.5 
Ω
0.2 
Ω
–
–
V
1
R
2
1
3
17 A
9 A
I
1
I
2
+
+
+
+
+
+
–
–
–
–
–
1 
Ω
0.5 
Ω
1 
Ω
21 
v
x
0.2 
Ω
R
2
v
x
v
1
v
2
v
3
R
3
R
4
R
5
R
6
R
1
0.5 
Ω
0.2 
Ω
–
–
R
2
1
3
Fig. 4.5-1 
(a) Circuit for Example 4.5-1 (b) Circuit after node reduction by source 
transformation
Solution
Step-1: Look for independent voltage sources in series with resistors and apply source transformation 
on such combinations.
There is one such combination in this circuit. It is V
1
in series with R
4
. Applying source 
transformation on this combination results in an independent current source of 17A in parallel with R
4
as shown in circuit (b) of Fig. 4.5-1. 
Step-2: Assign node voltage variables at those nodes where the node voltage variable is not decided 
directly by an independent voltage source or indirectly by already assigned node voltage variables and 
independent voltage source functions.
Now, all the three non-reference nodes in circuit (b) are unconstrained nodes and hence we assign 
three node voltage variables v
1, 
v
2
and v
3
as shown in the figure.
Step-3: Identify the controlling variables of dependent current sources in terms of the node voltage 
variables assigned in the last step and rewrite the source functions of dependent sources in terms of 
node voltage variables.
v
x
is the controlling variable in this circuit. However, v
x
is the voltage across R
2
and 
=
v
1
-
v
2
. 
Therefore, the current source function is k(v
1
-
v
2
) with k 
=
21.
Step-4: Prepare the node equations for the reduced circuit and solve them for node voltage variables.
The node equations are listed below.
Node
G v
G v
v
G v
v
I
Node
G v
G v
v
G v
−
+
−
+
−
=
−
+
−
+
1
2
1 1
2
1
2
3
1
3
1
4 2
2
2
1
5
(
)
(
)
(
)
(
22
3
1
2
2
6 3
3
3
1
5
3
2
1
2
3
−
+
−
=
−
+
−
+
−
−
−
=
v
k v
v
I
Node
G v
G v
v
G v
v
k v
v
)
(
)
(
)
(
)
(
) 00
Casting these equations in matrix form,
(
)
(
)
(
)
G
G
G
G
G
G
k
G
G
G
k
G
G
k
G
k
G
G
G
1
2
3
2
3
2
2
4
5
5
3
5
3
5
6
+
+
−
−
− +
+
+
−
−
− −
− +
+
+




















=
















v
v
v
G
I
V
1
2
3
4
1
1
1
0
0
0
0
(4.5-1) 
Eqn. 4.5-1 is in the form