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YV
=
CI
where 
Y
is the nodal conductance matrix. However, the nodal 
conductance matrix is now asymmetric and cannot be written down easily by inspection. However, 

---

4.20
Nodal Analysis and Mesh Analysis of Memoryless Circuits
the equation confirms that all node voltages (and hence all element voltages and currents) can be 
expressed as a linear combination of independent source functions. 
Substituting the numerical values,
8
1
2
20
17
2
23
19
9
1
0
0
0
1
0
1
2
3
−
−
−
−
−




















=







v
v
v








 =










9
17
9
17
0
Solving for the voltage vector by Cramer’s rule, v
1
=
2V, v
2
=
1V and v
3
=
3V. 
Step-5: Use these node voltage values in the original circuit to obtain element voltages and 
currents.
Now, the voltage across elements and current through them can be obtained by inspection. The 
complete solution is shown in Fig. 4.5-2.
17 V
9 A
10 A
2 A
I
1
+
+
+
+
+
+
+
–
–
–
–
–
–
1 
Ω
0.5 
Ω
1 
Ω
2 V
16 V
2 V
21 
v
x
21 A
16 A
4 A
15 A
16 A
1 V
1 V
1 A
0.2 
Ω
R
2
v
x
R
3
R
4
R
5
R
6
R
1
0.5 
Ω
0.2 
Ω
–
–
V
1
2 V
0 V
1 V
3 V
3 V
Fig. 4.5-2 
Complete circuit solution in Example 4.5-1 
example: 4.5-2
Solve the circuit (a) in Fig. 4.5-3 completely.
–4 V
i
x
i
x
12 A
I
1
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
1 V
1 
Ω
0.5 
Ω
1 
Ω
0.2 
Ω
R
2
R
3
R
4
R
5
R
6
R
1
4.5
0.5 
Ω
0.2 
Ω
V
2
i
x
–
–
–
–
3
2
1
R
V
2
V
1
i
x
i
x
12 A
4 A
I
1
I
2
+
+
+
+
+
+
+
–
–
–
–
–
–
1 V
1 
Ω
0.5 
Ω
1 
Ω
0.2 
Ω
R
2
R
3
R
4
R
5
R
6
R
1
4.5
0.5 
Ω
0.2 
Ω
V
2
i
x
i
v2
v
3
v
1
v
2
–
–
–
–
3
2
1
R
V
2
(a)
(b)
Fig. 4.5-3 
(a) Circuit for Example 4.5-2 (B) Circuit after node reduction by source 
transformation

---

Nodal Analysis of Circuits Containing Dependent Current Sources 
4.21
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 4A in parallel with R
4
as shown in the circuit in Fig. 4.5-3(b). 
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.
We start at left-most node of the circuit in Fig. 4.5-3(b) and assign a node voltage variable v
1
there 
since that node is not directly constrained by a voltage source. Moving to node-2, we see that the node 
voltage at that node cannot obtained from the already assigned variable v
1
and that there is no direct 
constraint at that node. Hence, we assign a node voltage variable v
2
at that node. Now, the node voltage 
at node-3 can be obtained as v
1
+
V
2
and a node voltage variable is not needed at that node. Therefore, 
there are only two node voltage variables in this circuit.
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.
i
x
is the controlling variable in this circuit. However, i
x
=
G
5
[v
2
– (v
1
+
V
2
)]. Therefore, the current 
source function is kG
5
[v
2
– (v
1
+
V
2
)]. with k 
=
4.5.
Step-4: Prepare the node equations for the reduced circuit and solve them for node voltage 
variables. Ignore node equation at nodes where voltage sources are connected directly to reference 
node. Combine the node equations at the end nodes of voltage sources connected between two non-
reference nodes.
The node equations are listed below.
Node
G v
G v
v
G V
i
I
Node
G v
G v
v
G v
V
−
+
−
−
−
=
−
+
−
+
1
2
1 1
2
1
2
3 2
1
4 2
2
2
1
5
2
2
(
)
(
)
(
−− −
= − =
−
+
+
+
+ −
+
−
v
V
I
G V
Node
G v
V
G V
G v
V
v
kG v
1
2
2
4 1
6
1
2
3 2
5
1
2
2
5
2
3
)
(
)
(
)
(
vv
V
i
V
1
2
2
0
−
+
=
)
Combining the node equations at node-1 and node-3 to eliminate 
i
V
2
,
Node
Node
G v
G v
v
G v
V
G v
V
v
kG v
− +
−
+
−
+
+
+
+ −
+
−
1
3
1 1
2
1
2
6
1
2
5
1
2
2
5
2
(
)
(
)
(
)
(
vv
V
I
Node
G v
G v
v
G v
v
V
I
G V
1
2
1
4 2
2
2
1
5
2
1
2
2
4 1
2
−
=
−
+
−
+
− −
= − =
)
(
)
(
)
Casting these equations in matrix form,
G
G
G
k G
G
k G
G
G
G
G
G
v
v
1
2
6
5
2
5
2
5
2
4
5
1
1
1
+
+
+ −
(
)
−
+ −
−
+
+
+
(
)






(
)
(
(
) )
(
)
22
6
5
4
5
1
1
2
1 0
1
0





 =
−
+ −
















(
(
) )
G
k G
G
G
I
V
V
This equation is in the form 
YV
=
CU
where 
Y
is the nodal conductance matrix. However, the nodal 
conductance matrix is now asymmetric and cannot be written down easily by inspection. However, 
the equation confirms that all node voltages (and hence all element voltages and currents) can be 
expressed as a linear combination of independent source functions. 

---

4.22
Nodal Analysis and Mesh Analysis of Memoryless Circuits
Substituting the numerical values,
4
6
3 4
1 0 2
0 1 2
12
4
1
14
2
1
2
−











 =





 −










=
−




v
v


Solving for the voltage vector by Cramer’s rule, v
1
=
2V, v
2
=
1V and v
3
=
v
1
+
V
2
=
3V.
Step-5: Use these node voltage values in the original circuit to obtain element voltages and currents 
for resistors and current sources.
The voltage across resistive elements and current sources and currents through resistive elements 
can be obtained by inspection. The currents through independent voltage sources in series with 
resistors can also be obtained at this stage.
Step-6: Use appropriate node equations to solve for 
currents through the remaining independent voltage 
sources.
The current through the independent voltage source 
V
2
has to be determined. We use the node equation at 
node-1 for this purpose.
G v
G v
v
G V
i
I
V
1 1
2
1
2
3 2
1
2
+
−
−
−
=
(
)
10 1 2
12
3
2
2
+ − −
=
⇒
= −
i
i
V
V
A
The complete solution is marked in Fig. 4.5-4.
4.6 
Nodal aNalysis of CirCuits CoNtaiNiNg depeNdeNt Voltage sourCes
A dependent voltage source can appear in three positions in a circuit. It may appear from a node to 
the reference node. Secondly, it may appear between two nodes at which more than two elements are 
incident. Thirdly, it may appear in series with some resistor. The following example shows a circuit 
that has a dependent voltage source in series with a resistor.
example: 4.6-1
Solve the circuit in Fig. 4.6-1(a).
R
1
v
1
v
1
v
2
v
2
v
3
v
3
v
y
i
x
i
x
i
x
I
1
I
1
v
y
R
2
R
3
R
4
R
5
R
6
R
6
0.2 
Ω
R
1
0.2 
Ω
0.5 
Ω
R
3
0.5 
Ω
R
5
0.5 
Ω
1 
Ω
R
4
1 
Ω
4 A
4 A
21
v
y
21
–0.9
i
x
–4.5
R
R
(a)
(b)
1
1
2
2
3
3
1 
Ω
R
2
1 
Ω
0.5 
Ω
0.2 
Ω
0.2 
Ω
+
+
+
+
+
+
+
–
+
+
+
–
–
–
–
–
–
+
–
–
–
+
–
–
+
–
Fig. 4.6-1 
(a) Circuit for Example 4.6-1 (b) Reduced circuit after source transformation
2 V
2 V
1 V
5 V
–4 V
i
x
i
x
3 V
12 A
10 A
I
1
2 V +
+
+
+
+
+
+
+
–
–
–
–
–
–
–
1 V
1 A
5 A
15 A
4.5
18 A
3 A
2 A
4 A
1 V
0 V
3 V
V
2
1 V
2 V
i
x
–
–
–
–
3 A
2 A
3 V
V
2
V
1
Fig. 4.5-4 
Complete solution for 
circuit in Example 4.5-2

---

Nodal Analysis of Circuits Containing Dependent Voltage Sources 
4.23
Solution
Step-1: Look for independent voltage sources and dependent voltage sources in series with resistors 
and apply source transformation on such combinations.
There is one such combination in this circuit. It is –0.9i
x
in series with R
1
. Applying source 
transformation on this combination results in a dependent current source of –0.9G
1
i
x
A in parallel 
with R
1
as shown in the circuit in Fig. 4.6-1(b). 
Step-2: Assign node voltage variables at those nodes where the node voltage variable is not decided 
directly by a voltage source or indirectly by already assigned node voltage variables and voltage 
source functions.
We start at left-most node of the circuit in Fig. 4.6-1(b) and assign a node voltage variable v
1
there 
since that node is not directly constrained by a voltage source. Moving to node-2, we see that the node 
voltage at that node cannot obtained from the already assigned variable v
1
and that there is no direct 
constraint at that node. Hence, we assign a node voltage variable v
2
at that node. The node voltage at 
node-3 cannot be obtained from v
1
and v
2
. Hence, we assign a node voltage variable v
3
at that node. 
Therefore, there are three node voltage variables in this circuit.
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.
i
x
is the controlling variable for the dependent current source at node-1 in the circuit (b) of Fig. 
4.6-1. However, i
x
=
G
3
[v
1
– v
3
]. Therefore, the current source function is k
1
 G
1
G
3
[v
1
– v
3
] with 
k
1
=
-
0.9.
v
y
is the controlling variable for the dependent current source at node-3. However, v
y
=
v
2
. Therefore, 
the current source function at node-3 is k
2
 v
2
 with k
2
=
21.
Step-4: Prepare the node equations for the reduced circuit and solve them for node voltage 
variables. Ignore node equation at nodes where voltage sources are connected directly to reference 
node. Combine the node equations at the end nodes of voltage sources connected between two non-
reference nodes.
The node equations are listed below.
Node
G v
G v
v
G v
v
k G G v
v
Node
G v
G
−
+
−
+
−
+
−
=
−
+
1
0
2
1 1
2
1
2
3
1
3
1 1
3
1
3
4 2
(
)
(
)
(
)
22
2
1
5
2
3
1
6 3
3
3
1
5
3
2
2 2
3
(
)
(
)
(
)
(
)
v
v
G v
v
I
Node
G v
G v
v
G v
v
k v
−
+
−
= −
−
+
−
+
−
+
==
0
Substituting the numerical values and casting these equations in matrix form,
17
1
11
1
4
2
2
23
9
0
1
0
1
2
3
−
−
−
−
−
−




















= −









v
v
v

[ ]
4
This equation is in the form