Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

x
i
x
v
y
4 A
4 A
2 A
1 A
1 V
21 A 15 A
1 V
1 A
1 A
1 V
0.2 V
1.8 V
2 V
3 V
21
–0.9
0V
2V
1V
3V
+
+
+
+
+
+
+
–
–
–
–
–
–
–
Fig. 4.6-2 
Complete solution for 
circuit in Fig. 4.6-1 (a)

---

Nodal Analysis of Circuits Containing Dependent Voltage Sources 
4.25
adding v
x
. Hence, we do not assign a node voltage variable at that node. The node voltage at node-3 
cannot be obtained from v
1
and there is no constraining voltage source connected from that node to 
reference node. Hence, we assign a node voltage variable v
3
at that node. Therefore, there are only two 
node voltage variables in this circuit.
Step-3: Identify the controlling variables of dependent 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 for the dependent voltage source between node-1 and node-2 in the 
circuit (b) of Fig. 4.6-1. However, v
x
=
v
3
– v
1
. Therefore, the voltage source function is k(v
3
– v
1
) with 
k 
=
1.
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
k v
v
G v
v
i
G V
Node
G
v
x
−
+
−
−
−
+
−
+
=
−
1
2
1 1
2
1
1
3
1
3
1
3
1 1
4
(
(
(
)))
(
)
((
(
))
(
(
))
(
((
)
))
v
k v
v
G
k v
v
G v
k v
v
v
i
Nod
v
x
1
3
1
2
3
1
5
1
3
1
3
0
−
−
+
−
−
+
−
−
−
−
=
ee
G v
G v
v
G v
v
k v
v
I
G V
−
+
−
+
−
−
−
= +
3
6 3
3
3
1
5
3
1
3
1
1
6 2
(
)
(
(
(
)))
We eliminate the current through the dependent voltage source from the equations by adding the 
first two equations.
Node
Node
G v
G v
v
G v
k v
v
G v
k v
v
− +
−
+
−
+
−
−
+
−
−
1
2
1 1
3
1
3
4
1
3
1
5
1
3
1
(
)
(
(
))
(
((
))
))
(
)
(
(
(
)))
−
=
−
+
−
+
−
−
−
= +
v
G V
Node
G v
G v
v
G v
v
k v
v
I
G
3
1 1
6 3
3
3
1
5
3
1
3
1
1
3
66 2
V
Substituting the numerical values and casting these equations in matrix form,
16
9
6 11
5
21
1
2
−
−











 =






v
v
Solving for the voltage vector by Cramer’s rule, v
1
=
2V, v
3
=
3V. Then, v
x 
=
1V and therefore 
v
2
=
v
1
– v
x
=
1V.
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 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 voltage 
sources.
We have to find the current through the dependent voltage source by employing KCL equation at 
node-1 or node-2. Choosing node-2,
1
+ 
1(1
- 
2)
+ 
2(1
- 
3) 
=
i
v
x
⇒
i
v
x
=
-
4 A. The complete solution is marked in Fig. 4.6-4.

---

4.26
Nodal Analysis and Mesh Analysis of Memoryless Circuits
4 A
4 A
2 A
5 A
1 V
1 V
1 V
1 V
1 V
11 A
1 V
5 A
1 A
1 A
1 V
2 V
2 V
0V
2V
1V
3V
+
–
–
–
–
–
v
x
v
x
I
1
V
1
V
2
+
+
+
+
+
+
+
+
–
–
–
–
Fig. 4.6-4 
Complete solution for circuit in Fig. 4.6-3(a)
example: 4.6-3
Solve the circuit in Fig. 4.6-5 by nodal analysis.
v
2
v
x
v
x
i
x
i
x
I
1
R
1
R
2
R
3
R
4
R
5
R
6
0.2 
Ω
0.5 
Ω
1 
Ω
–17 A
R
1
2
3
1 
Ω
2
2
0.5 
Ω
0.2 
Ω
+
+
+
–
–
–
+
–
+
+
+
+
–
–
–
–
–
Fig. 4.6-5 
Circuit for nodal analysis in Example 4.6-3 
Solution
Node-1 is constrained by the dependent source to reference node and hence no node voltage variable 
can be assigned there. Node-2 is assigned a node voltage variable v
2
. Then, the node voltage variable 
at node-3 gets fixed as v
2
+
2i
x
through the dependent voltage source connected between node-2 and 
node-3. Thus, this circuit has only one node voltage variable to be solved for.
Node equation at node-1 is not needed for determining node voltages. However, it will be needed 
later for determining the current through the dependent voltage source connected at that node.
v
x 
=
v
2
+ 
2i
x
–2v
x
and i
x 
=
2v
x
-
v
2
Solving these two equations, we get v
x
=
v
2
and i
x
=
v
2
.
The KCL equations at node-2 and node-3 can be combined to form a single equation in v
2
.
This combined equation will be v
2
+
(v
2
–2v
x
) 
+ 
5(v
2
+ 
2i
x
) 
+ 
2(v
2
+ 
2i
x
- 
2v
x
) 
=
17. Substituting for 
v
x
and i
x
in terms of v
2
, we get, 17v
2
=
17 
⇒
v
2
=
1V. 
Now, v
x
=
v
2
=
1V and i
x
=
1A. Therefore, the node voltages are 2V, 1V and 3V respectively at 
node-1, node-2 and node-3. Then, the node equation at node-1 can be employed to find current into 

---

Mesh Analysis of Circuits with Resistors and Independent Voltage Sources 
4.27
the positive terminal of dependent source as –9 A. Node equation at node-3 is used to determine 
the current into the positive terminal of second dependent source as –21A. The complete solution is 
marked in Fig. 4.6-6.
i
x
i
x
I
2
–17 A
2
2
21 A
2 A
1 A
1 A
10 A
9 A
15 A
4 A
3 V
2 V
2 V
1 V
1 V
2 V
1 V
+
–
+
–
+
–
+
+
+
+
–
–
–
–
v
x
v
x
+
–
0V
2V
1V
3V
Fig. 4.6-6 
Solution for circuit in Example 4.6-3 
4.7 
 mesh AnAlysIs of cIrcuIts wIth resIstors And
IndePendent VoltAge sources
Mesh Analysis uses the KCL equations and element equations to eliminate variables, reduces the 
number of pertinent variables to (b
- 
n
+ 
1) mesh current variables and uses the second set of equations 
(KVL equations) to solve for these variables. 
Mesh analysis is applicable only to planar networks. A planar network is one that can be drawn on 
a plane without any component crossing over another component. Consider the two circuits shown in 
Fig. 4.7-1.
+
–
(a)
+
(b)
–
Fig. 4.7-1 
(a) A non-planar circuit (b) A planar circuit that appears to be non-planar
The circuit in Fig. 4.7-1(a) is non-planar since it cannot be drawn on a plane surface without 
crossovers. The circuit (b) is planar though there appears to be a crossover the way it is drawn in 
Fig. 4.7-1. However, it can be redrawn to avoid the crossover.