Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 4.8-1
Apply mesh analysis procedure on the circuit in Fig. 4.8-1 (a).
1 
Ω
1 
Ω
3 
Ω
2 
Ω
1 
Ω
1 
Ω
4 
Ω
4 
Ω
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
+
–
+
+
+
+
+
+
–
–
–
–
–
+
+
+
–
–
–
–
–
–
–
6 V
6 V
(a)
(b)
2 A
2 A
5 V
2.5 A
–10 V
–10 V
3 V
3 V
R
3
R
3
I
3
I
2
I
2
I
1
R
1
2 
Ω
R
1
R
5
R
5
R
4
R
4
R
2
3 
Ω
R
2
V
1
V
1
V
2
V
2
V
3
V
4
V
3
I
1
Fig. 4.8-1 
(a) Circuit for (b) Circuit after mesh reduction by source transformation
Solution
All the circuit variables are identified following passive sign convention. The labelling of variables 
will be v or i with the name of element as subscript.
Step-1: If there are current sources that appear directly across resistors convert them into equivalent 
voltage sources in series with resistors. This is called ‘mesh reduction’.
There is one such combination in circuit of Fig. 4.8-1 (a). It is I
1 
(2.5 A) in parallel with R
1 
(2
W
). 
We replace these with a voltage source of value R
1
I
1
(5 V) in series with R
1
(2
W
) with positive polarity 
at top as shown in circuit of Fig. 4.8-1 (b).
The second current source I
2
in circuit (a) cannot be reduced this way.
Step-2: Assign mesh current variables in the reduced circuit starting with leftmost mesh. Assign 
a mesh current variable to a mesh only if its mesh current is not constrained directly by a current 
source and its mesh current is not decided by other mesh current variables already assigned along with 
current source functions.
There are three meshes in the reduced circuit. The first mesh has no current source directly 
constraining its mesh current. No other mesh was assigned mesh current variable and hence this mesh 
current cannot get decided by other already assigned mesh current variables. Therefore, we assign a 
mesh current variable to this mesh and call it i
1
.
Moving to the second mesh we observe that the independent current source I
2
directly constrains 
its mesh current to be I
2
A (2 A). Therefore, we do not assign any mesh current variable to this mesh.
There is no current source in the third mesh. Its mesh current cannot be obtained from the already 
assigned mesh current variables. Therefore, we assign a mesh current variable to this mesh and call 
it i
3
.

---

Mesh Analysis of Circuits with Independent Current Sources 
4.35
Thus, this circuit has two mesh current variables 
-
i
1
and i
3
– to be solved for.
Step-3: Prepare the mesh equations by applying KVL in meshes, starting at left bottom corner and 
traversing the mesh in clockwise direction. Ignore the meshes that are directly constrained by current 
sources. Further, if two meshes share a current source, then, add the mesh equations for those two 
meshes to generate a new equation that will be used in solution process. The number of equations at 
the end of this step will be equal to number of meshes – number of irreducible independent current
sources.
The two mesh equations are:
− +
−
− + =
+
−
=
−
V
R i
R i
i
V
i e
R
R i
R i
V
V
4
1 1
2
2
1
1
1
2
1
2 2
4
0
(
)
. ., (
)
for mesh-1
22
1 1
2
=
−
R I
V
− +
−
+
+ =
+
=
+ −
V
R i
I
R i
V
i e
R
R i
R I
V
V
2
4
3
2
5 3
3
4
5
3
4 2
2
0
(
)
. ., (
)
for mesh-3
33
These mesh equations are expressed in matrix form below.
R
R
R
R
R
i
i
R
R
I
I
V
V
1
2
2
4
5
1
3
1
3
1
2
1
0
0
0
1
0
0
1
0
1
+
−
+











 =
−
−






22
3
V
















Step-4: Solve the mesh equations by Cramer’s rule or by matrix inversion.
Substituting numerical values and simplifying,
5
3
0
5
1
15
1
3
−











 =
−






i
i
Solving for i
1
and i
3
by Cramer’s rule, i
1
=
1A and i
3
=
3A. i
2
is already known to be 2A.
Step-5: Use the mesh current values and apply KCL at various nodes in the original circuit to 
obtain element currents and voltages for all resistive elements and current through voltage sources.
Refer to circuits in Fig. 4.8-1. The resistor currents and voltages as per direction and polarity 
marked in Fig. 4.8-1 is calculated below.
Current through R
1
=
I
1
-
i
1
=
1.5A, 
\
Voltage across R
1
=
3V
Current through R
2
=
i
2
-
i
1
=
1A, 
\
Voltage across R
2
=
3V
Current through R
3
=
I
2
=
2A, 
\
Voltage across R
3
=
2V
Current through R
4
=
i
3
– I
2
=
1A, 
\
Voltage across R
4
=
1V
Current through R
5
=
i
3
=
3A, 
\
Voltage across R
5
=
12V
Now, the currents through the voltage sources are
Current through V
1
=
-
1A, current through V
2
=
-
1A and current through V
3
=
3A
Step-6: Use the element voltages calculated in the above step and apply KVL in various meshes in 
the original circuit to obtain element voltages for all current sources.
Applying KVL in the mesh formed by first current source and the resistor R
1 
in the original circuit, 
we get the voltage across I
1
as –3V.
Applying KVL in the mesh in which the second current source appears, we get the voltage across 
the current source I
2
as –1V.

---

4.36
Nodal Analysis and Mesh Analysis of Memoryless Circuits
Note that these voltages follow passive sign convention.
The complete circuit solution is shown in Fig. 4.8-2.
V
3
V
2
V
1
–10 V
–1 V
–1 A
–1 A
1 A
–3 V
3 V
3 V
6 V
1.5 A
2.5 A
3 V
1 V
2 A
1 A
2 A
3 A
3 A
12 V
2 V
1A
2A
3A
+
+
+
+
+
–
–
–
–
+
+
+
+
+
–
–
–
–
–
–
Fig. 4.8-2 
Complete mesh analysis solution for circuit in Fig. 4.8-1(a)