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4.9
mesh AnAlysIs of cIrcuIts contAInIng dePendent sources
Dependent voltage sources do not pose any problem for mesh analysis. Independent voltage sources
appear on the right-hand side of mesh equations. However, dependent voltage sources affect the
coefficients of mesh current variables in the left-hand side of mesh equations.
The controlling variable of a linear dependent current source will be a voltage or current existing
elsewhere in the circuit. But, any voltage or current variable in the circuit can be expressed in terms
of mesh current variables. Hence, the dependent current source function can be expressed in terms
of mesh current variables. Therefore, dependent voltage sources will affect the coefficients of mesh
equation, i.e., they will change the Mesh Resistance Matrix. We will see that they can destroy the
symmetry of the Mesh Resistance Matrix.
Dependent current sources place constraints on mesh current variables the same way independent
current sources do. They cause a reduction in the number of mesh currents to be determined in the circuit.
They too affect the coefficients of mesh equations and make Mesh Resistance Matrix asymmetric.
example: 4.9-1
Apply mesh analysis on the circuit in Fig. 4.9-1.
V
1
R
1
R
2
R
3
R
4
R
5
v
x
6.5
v
x
i
x
i
1
i
2
i
3
3 V
+
+
+
–
–
+
–
+
+
+
+
–
–
–
–
–
–4
i
x
2
Ω
3
Ω
1
Ω
1
Ω
4
Ω
V
2
V
3
Fig. 4.9-1
Circuit for Example 4.9-1
Solution
Step-1: Carry out mesh reduction by employing source transformation, if relevant.
There are no current sources appearing in parallel with any resistor. Hence, no mesh reduction is
possible.
4.40
Nodal Analysis and Mesh Analysis of Memoryless Circuits
Step-2: Identify meshes and assign mesh current variables.
There are three meshes and there is no current source to impose any constraints on them. Hence,
all the three meshes are assigned mesh current variables.
Step-3: Identify the controlling variable of dependent sources in terms of mesh current variables
and express dependent source functions in terms of mesh current variables.
i
x
is the controlling variable of V
2
. i
x
is the current flowing in R
4
from top to bottom and hence it is
equal to (i
2
-
i
3
). Therefore, the source function of V
2
is k
1
(i
2
-
i
3
) with k
1
=
-
4.
v
x
is the controlling variable for V
3
.v
x
is the voltage across R
1
. Therefore, v
x
=
R
1
i
1
and the source
function of V
3
is k
2
R
1
i
1
with k
2
=
6.5.
Step-4: Prepare the mesh equations and solve them.
The mesh equations are written with the dependent source functions expressed in terms of mesh
current variables.
Mesh-1
− +
−
− +
−
=
+
−
−
V
R i
R i
i
k i
i
i e
R
R i
R
k
1
1 1
2
2
1
1
2
3
1
2
1
2
0
(
)
(
)
. ., (
)
(
11
2
1 3
1
1
2
3
2
2
1
3 2
4
3
2
0
)
(
)
(
)
(
)
i
k i
V
k i
i
R i
i
R i
R i
i
−
=
−
−
+
− +
−
−
=
Mesh-2
ii e
R i
R
R
R
k i
R
k i
R i
i
R
. .,
(
)
(
)
(
)
−
+
+
+
−
−
−
=
−
+
2 1
2
3
4
1
2
4
1
3
4
3
2
0
Mesh-3
55 3
2 1 1
2 1 1
4 2
4
5
3
0
0
i
k R i
i e
k R i
R i
R
R i
−
=
−
−
+
+
=
. .,
(
)
These equations are expressed in matrix form below.
(
)
(
)
(
)
(
)
(
)
R
R
R
k
k
R
R
R
R
k
R
k
k R
R
R
R
1
2
2
1
1
2
2
3
4
1
4
1
2 1
4
4
5
+
−
−
−
−
+
+
−
−
−
−
−
+
=
[ ]
i
i
i
V
1
2
3
1
1
0
0
Note the asymmetry in Mesh Resistance Matrix.
Substituting the numerical values and solving for mesh currents by Cramer’s rule,
5
7
4
3
9
5
13
1
5
1
0
0
3
1
2
3
−
−
−
−
−
=
[ ]
i
i
i
⇒
i
1
=
1A, i
2
=
2A and i
3
=
3A
Step-5: Apply KCL at various nodes of the circuit to find all the element currents and resistor
voltages.
Consider R
4
. Applying KCL at the node formed by R
3
, R
4
and R
5
, we get the current flowing from
top to bottom in R
4
as i
2
-
i
3
. However, the reference direction that was chosen for current in R
4
is from
bottom to top. Hence, current in R
4
=
-
(i
2
-
i
3
) in the direction marked in Fig. 4.9-1. The value is 1A.
This makes the value of i
x
equal to –1A and hence the dependence voltage source V
2
source function
becomes –4
×
-
1
=
4V.
Consider R
1
. The reference direction for its current as marked in Fig. 4.9-1 is from left to right and
is in the same direction as the first mesh current. R
1
is wholly owned by first mesh and hence current
through it is i
1
itself. The value is 1A. That makes the voltage across R
1
equal to 2V. Moreover, the
value of v
x
is also 2V. Therefore, the source function of the dependent voltage V
1
is 6.5
×
2
=
13V.
The remaining voltage and current variables also may be evaluated similarly. The complete circuit
solution is shown in Fig. 4.9-2.
Mesh Analysis of Circuits Containing Dependent Sources
4.41
V
1
v
x
6.5
v
x
i
x
3 V
–1 A
–1 A
–3 A
1 A
2 A
1 A
1 A
3 A
2 V
+
+
+
–
–
+
–
+
+
+
+
–
–
–
–
–
4 V
3 V
1 V
13 V
–4
i
x
12 V
2 V
1A
2A
3A
Fig. 4.9-2
Complete mesh analysis solution for Example 4.9-1
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