6. MESH ANALYSIS
INTRODUCTION
PASSIVE SIGN CONVENTION
PLANAR CIRCUITS
FORMATION OF MESHES
ANALYSIS OF A SIMPLE CIRCUIT
DETERMINANT OF A MATRIX
CRAMER’S RULE
GAUSSIAN ELIMINATION METHOD
EXAMPLES FOR MESH ANALYSIS
SUPERMESH
EXAMPLES WITH DEPENDENT SOURCES
SUMMARY
EXERCISE PROBLEMS
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6.1
INTRODUCTION
The two general analytical techniques used in network analysis are mesh or loop
analysis and nodal analysis. This chapter describes mesh analysis, whereas the
next chapter is on nodal analysis.
Mesh analysis is a systematic technique to evaluate all voltages and currents in a
circuit. It is based on Kirchoff’s Voltage Law and Ohm’s Law. This technique can
be applied to ac circuits also. The way to form meshes or loops is described first,
followed by an example. Then a set of simultaneous equations are formed, with the
number of these equations equaling the number of unknown variables to be
determined. These unknown variables are the loop currents or the mesh currents.
The simultaneous equations can be compactly described by matrix equations, and
the solution to matrix equations can be obtained using Cramer=s rule. Gaussian
elimination method can also be used to obtain the solution. Solution by hand can
be laborious if the number of unknowns exceeds three. Some examples are
presented, to illustrate the technique of mesh analysis. Since it is essential to be
familiar with passive sign convention, it is explained first.
6.2
PASSIVE SIGN CONVENTION
According to passive sign convention, the terminal by which current enters an
element should be marked positive with respect to the potential at the other
terminal of the same element. Then both the voltage and the current associated
with the element can be treated as positive values, when forming equations for the
circuit, even through their actual value may turn out to be otherwise. For the
resistor in Fig. 1, current I1 flows into the positive terminal of the resistor. Hence
the sign of voltage and the direction of current for this resistor are in accordance
with passive sign convention. When passive sign convention is followed for a
resistor, the Ohm’s law is satisfied. It can also be said that when Ohm’s law is
satisfied, passive sign convention has been followed. When passive sign
convention is followed for assigning voltage and current to a resistor, both the
voltage and the current are positive and the resistor absorbs power. Note that
power is the average of energy absorbed over time. It is energy absorbed per
second and power by definition is an average value.
Equation (6.1) expresses the Ohm’s Law for a resistor, and equation (6.2)
expresses the power absorbed by a resistor. Since the power absorbed by a resistor
varies proportionately either with the square of its voltage or with the square of its
current, the power absorbed by a resistor is always positive. When the sign for
voltage and the direction of current associated with a resistor are according to
passive sign convention, then the voltage and the current have the same sign.
For the resistor shown in the Fig. 2, current I2 flows out of the positive terminal
of the resistor. Hence the sign of voltage and the direction of current for this
resistor are not in accordance with passive sign convention. When passive sign
convention is not followed for a resistor, as shown in Fig. 2, the current and the
voltage have opposite signs. If the voltage across the resistor is positive, then the
sign of current in relation to the voltage is negative. Hence the voltage and the
current associated with the resistor are related to each other, as expressed by
equation (6.3)
As shown in Fig. 3, the passive sign convention is valid for any element or a twoterminal circuit. When both the voltage and the current I1 are positive or have the
same sign, then the power absorbed by the element is positive. When both the
voltage and the current I2 are positive or have the same sign, then the power
delivered by the element is positive. When the power is determined in accordance
with the passive sign convention, the power associated with this element is
negative.
6.3
PLANAR CIRCUITS
A planar circuit is shown in Fig. 4. It is seen that a circuit can be drawn such that
no two elements cross over each other. Let us find out what a non-planar circuit
is. Let us add one more node, marked as c, to the circuit in Fig. 4 and connect it
to all other nodes as shown in Fig. 5. It is seen that the resistor R1 connected from
node b to node c crosses over resistor R2. This circuit cannot be drawn without
cross-over of two elements and is a non-planar circuit. Mesh analysis is applicable
only to planar circuits, whereas loop analysis and nodal analysis can be applied to
non-planar circuits also.
6.4
FORMATION OF MESHES
The circuit in Fig. 6 illustrates how meshes or loops are to be formed. This circuit
has meshes within it and it is easy to see the mesh structure within the circuit. The
circuit has many windows and each window has a mesh current. It is better to
follow some rules in forming meshes.
Let the number of branches in a network be e, and let the number of nodes
including the datum node be n. Then the number of independent of mesh
equations is e - n + 1. These meshes are formed as follows:
a.
Identify the meshes and mark the mesh currents with an assigned direction
for each mesh current. It is preferable to assign the direction as shown in
Fig. 6, all in the clockwise direction. It is better to make sure that the
current through any branch is either the mesh current itself or the difference
of two mesh currents. It is preferable if the current through any element is
not the sum of two mesh currents.
b.
Apply KVL equation around each labeled mesh.
convention while forming the mesh equations.
c.
Solve the resulting set of simultaneous equations for the unknown mesh
currents. For solving the problem manually, it is preferable to use the
Gaussian elimination method instead of Cramer’s rule.
Use the passive sign
6.5
ANALYSIS OF A SIMPLE CIRCUIT
A simple problem is presented now. The task is to find the currents and voltages
present in the circuit in Fig. 7. A sequence of steps is followed to obtain the
solution. The first step is to identify the loop currents, as shown next. The values
of components are shown beside the circuit.
The first step is to create adjacent, non-overlapping cells, as shown in Fig. 8. Then
mark the loop currents, as shown in Fig. 8. There are two unknowns and two loop
currents have been identified. Let both the loop currents flow in the same
direction, preferably clockwise, as illustrated here. When the cells are adjacent,
and all the loop currents have the same direction, the current through any element
is either the loop current itself, or the difference of two loop currents. The current
through any element is never, ever the sum of two loop currents.
The path for loop current I1 is shown in red colour. We can apply KVL to get
the loop equation. Kirchoff’s voltage law states that the algebraic sum of
voltages of elements present in a loop is zero. Let the voltages across the
resistors, R1 and R3, be assigned such that they are in accordance with the
passive sign convention, as shown in Fig. 9.
The voltage across resistor R3 is marked in accordance with the passive sign
convention, but this has been done with reference to current I1. It can be seen,
that two loop currents flow through resistor R3 in opposite directions. Hence
whichever way the voltage across resistor R3 is assigned a sign, it will be in
accordance with passive sign convention for one loop current, but not so for the
other loop current.
Equation (6.4) defines the voltage across resistor R1 . Let current though
resistor R3 be IN, where IN is defined by equation (6.5). Then we get the
voltage across R3 as the product of R3 and IN . Equation (6.6) defines the
voltage across resistor R3 . Applying KVL to the loop containing current I1, we
get equation (6.7). It is seen that current I1 enters source VA through its
negative terminal and hence the source voltage has a negative sign in front of it
in equation (6.7).
The path for loop current I2 is shown in red color in Fig. 10. The voltage across
resistor R3 has already been assigned with reference to current I1 , but the voltage
across resistor R2 can be assigned with reference to current I2 in accordance with
the passive sign convention. It means that both the voltage and the current of
resistor R2 have the same sign.
Equation (6.8) defines the voltage across resistor R2 . Applying KVL to the loop
containing current I2, we get equation (6.9). It is seen that current I2 enters resistor
R3 through its negative terminal and hence the voltage across R3 has a negative
sign in front of it in equation (6.9). The voltage across R3 can be expressed by
equation (6.10).
The loop equations ( 6.7) and (6.9) can be expressed by equations (6.11) and
(6.12) by replacing the resistor voltages by the corresponding expressions in terms
of the resistor value and its current.
Equations (6.11) and (6.12) can be represented by equations by grouping terms.
Equations (6.13) and (6.14) can be represented by the matrix equation (6.15). The
terms along the main diagonal are loop resistances, and they are positive values.
The other two terms are mutual resistances. They are negative values. We take
up the solution of a matrix equation in the subsequent sections. It is possible to
solve the two simultaneous equations (6.12) and (6.13), and get the solution.
By substituting the given values of components, we get equations (6.16) and
(6.17). On solving these two equations, we get that I1 = 2 A, and that I2 = 1 A.
From equation (6.15), it can be seen how mesh equations can be written just by
inspection. For a resistive network, each diagonal element of the loop-impedance
matrix reflects the self-resistance of the corresponding loop. The self-resistance
of a loop is the sum of resistances through which the loop current flows. Other
elements of this matrix are the mutual resistances. Since current through any
element is either the loop current itself or the difference of two loop currents, any
non-diagonal element is either zero or a negative value. If two loop currents have
no common element, the corresponding element of the matrix would have a zero
entry. If two loop currents have a common element, the corresponding elements
of the matrix would have an entry equaling the negative of the resistance common
to those two loops.
6.6
DETERMINANT OF A MATRIX
Before we take up the application of Cramer’s rule, it is necessary to know how
to get the determinant of a matrix.
Given a second-order matrix, its determinant is obtained as shown by equation
(6.18). It can be seen that the determinant can be expressed as the difference
between the positive product and the negative product. This concept can be
extended to a third-order matrix too. The determinant of a matrix can be
determined in terms of the values of its elements in any of the rows or columns,
and the respective co-factors . For a higher order determinant, it is the only
technique, but for a third order determinant, a simple technique exists and that
technique is now explained.
Given a third-order matrix, the first step is to copy the first two columns and
place them after the third column, as shown in Fig. 12. Now we have five
columns and three rows, as illustrated.
We can express the value of determinant as the difference between the sum of
positive products, and the sum of negative products. The way to get the sum of
positive products and the sum of negative products has been illustrated by Fig.
13. The product of elements lying on the diagonals with a negative slope forms
a positive product term, whereas a negative product term is the product of
elements lying on the diagonals with a positive slope.
Equation (6.19) presents the expression for the determinant of a third-order
matrix. It is difficult to remember this equation, but the process, by which it
can be obtained, can be remembered easily.
Given a matrix of third order, its determinant can be determined in terms of
the values of its elements in any of the rows or columns, and the respective cofactors. The cofactor of an element is found out as follows. Cross out the row
and the column to which the element belongs. Then the cofactor can be
identified as shown above. Note that there is a negative sign for one cofactor.
When the sum of the row number and the column number is not even, the
cofactor has a negative sign.
Fig. 15: Determinant of a third-order determinant
6.7
CRAMER’S RULE
Matrix equations, are often used to solve network problems. Let
where A is a n x n square matrix, x is a column matrix with n elements, and d is
a column matrix with n elements. Then we obtain x as:
if the inverse of A exists. Since obtaining the inverse of a matrix is somewhat
complex ,we can use Cramer’s rule instead.
A heuristic argument is provided below to offer insight into how Cramer’s rule
has arisen. Let A be a non-singular 3 × 3 matrix. For example, let
The determinant of A is:
When
Since addition of a column to another column does not change the value of the
determinant, we can add x2 times the second column and x3 times the third column
to the first column without affecting the value of the left-hand side value of
equation (6.24). That is,
The first column of the determinant on the right- hand side of equation (6.25) can
be replaced by the right-hand side column matrix of equation (6.22). Then
equation (6.26) is obtained. Equation (6.26) expresses Cramer’s rule.
In the same way, we can get the values of x2 and x3 .
A numerical example is presented below.
Equation (6.29) presents a matrix equation. The values of currents have to be
determined. The value of determinant of Z in equation (6.30) can be obtained as
follows.
The first two columns are copied and are placed after the third column. The
product terms along the diagonals can be identified and are presented. The
determinant is evaluated and equation (6.31) expresses its value.
Using Cramer’s rule, we can obtain the values of I1, I2 , and I3.
The use of Cramer’s rule tends to be tedious, whereas it is easier the Gaussian
elimination method.
6.8
GAUSSIAN ELIMINATION METHOD
Compared to the use of Cramer’s rule, the Gaussian elimination method is easier
to use for hand calculations. Let a set of three simultaneous equations be:
Given a set of simultaneous equations, re-arrange the equations such that
The aim is to eliminate variable x1 from equations (ii) and (iii). It can be eliminated
from equation (ii), as shown below.
It is seen that variable x1 has been eliminated. Let us express equation (v) as
follows.
The next step is to eliminate variable x1 from equation (iii), as shown below.
It is seen that variable x1 has been eliminated. Let us express equation (viii) as
follows.
The next step is to eliminate x2 from equations (vi) and (ix).
Express equation (xi) as follows.
The three equations that are used for obtaining the values of x1 , x2 ,and x3 are
equations (i), (vi) and (xii), which are presented below.
We can get the value of x3 from equation (xii), as shown be equation (xiii).
Knowing the value of x3 , we can get the value of x2, as shown by equation (xiv).
Once the values of x2 ,and x3 are known, the value of x1 can be obtained, as shown
by equation (xv).
A numerical example for Gaussian elimination method is presented now.
Let the set of simultaneous equations be:
Rearrange the above equations, as shown below.
Eliminate y1 from equation (ii). Keep equation (i).
Eliminate y1 from equation (iii). Keep equation (i).
Eliminate y2 from equation (vii). Keep equation (v).
Now the value of y3 is known, evaluate y2 from equation (v)
Evaluate y1 from equation (i).
It can be seen that the numerical example is quite simple.
6.9 EXAMPLES FOR MESH ANALYSIS
Some worked examples are presented next to illustrate how mesh analysis is to be
applied. In this section, the circuits contain only independent sources. Examples
with dependent sources are presented in another section.
Worked Example 6.2
Obtain the mesh currents in the circuit in Fig. 15 using mesh analysis.
Solution:
There are two voltage sources. As can be seen, three loops can be identified. All
the three loop currents are in the clockwise direction. The cells containing these
loop currents are adjacent to each other. The current through any element is either
the loop current or the difference of two loop currents. Since there are three loop
currents to be determined, three independent equations have to be formed. In this
problem, the solution is relatively easy, because one KVL equation can be formed
for each loop, thus leading to formation of three independent equations.
Path for loop current I1 is shown in red colour in Fig. 16. Write the KVL equation
for mesh current I1.
Equation (6.31) is the KVL equation for mesh current I1. From equation (6.31),
we can obtain equation (6.32) by re-grouping coefficients.
Path for loop current I2 is shown in red colour in Fig. 17. Write the KVL equation
for mesh current I2.
Path for loop current I3 is shown in red colour in Fig. 18. Write the KVL equation
for mesh current I3.
Equations (6.32), (6.34) and (6.36) can be expressed by a matrix equation.
By substituting the values of components, we get the following equation from
equation (6.37)
To solve the set of simultaneous equations expressed in the matrix form, we can
use Cramer’s rule. On solving, we obtain that
% Matlab Solution Worked Example 1
% Circuit Matrix is A, Input Voltage Vector is V
% Currents are denoted by vector, Cur
A=[8 -3 -3;-3 10 -5;-3 -5 13];
V=[15;6;-6];
Cur=inv(A)*V
The output of the program is:
Cur =
3
2
1
When a network contains only independent voltage sources, the mesh analysis is
a straight-forward technique. On the other hand, if a network contains a current
source, the equations have to be formed with care. When a current source is
present, there are two possibilities. One possibility is that the current source can
itself be a mesh current and the other is that the current source is equal to the
difference of two mesh currents. The example below shows a case where the
current source is the difference of two mesh currents.
Worked Example 6.3:
Obtain the mesh currents of the circuit in Fig. 19 using mesh analysis.
Solution:
There are three unknowns and three independent equations have to be formed.
The equation for mesh current I1 can be formed as shown in the previous example.
Since the current source of 1 A is the difference of two mesh currents I2 and I3 , the
second equation can be obtained from this equality. This current source of 1 A is
present in the mesh for current I2 and also in the mesh for current I3 as well. Hence
two expressions for the voltage can be obtained for the voltage across the 1 A
current source. Equating these two expressions to each other, the third independent
equation can be obtained.
The first mesh equation is formed as follows. For mesh current I1 , the KVL
equation is:
Equation (6.38) is the KVL equation for mesh current I1. From equation (6.31),
we can obtain equation (6.39) by re-grouping coefficients.
It is seen that the current source of 1 A is the difference of two mesh currents and
we can get the second equation.
It is seen that two expressions for the voltage across the current source can be
obtained. Let it be VY.
Equations (6.39), (6.40) and (6.44) can be expressed by a matrix equation.
By substituting the values of components, we get the following equation from
equation (6.45)
On solving, we obtain that
Worked Example 6.4
Find the mesh currents I1 and I2 in the circuit shown in Fig. 21.
Solution:
The circuit in Fig. 21 for worked example 4 contains a current source as one of the
mesh currents. In such a case, one of the unknowns is known. The solution is
easier, since we have to solve for only two unknowns.
Write the KVL equation for mesh current I1 :
On grouping terms, we obtain that
Since I3 = 1 A, the above equation becomes:
For mesh current I2 , the loop equation is obtained as given below.
On grouping terms and substituting for I3, we obtain that
Equations (6.48) and (6.50) can be expressed by a matrix equation.
The matrix equation is:
On solving, we obtain that
Worked Example 6.5
Obtain the mesh currents of the circuit in Fig. 22 using mesh analysis.
Solution:
First write the KVL equation for loop current I2. The second equation is obtained
from expressing the source current of 2 A as the difference of two loop currents.
The voltage across the current source is present in two loops. Equate the two
expressions and get the third equation.
On solving equations (6.51), (6.52) and (6.53), we get that
It is possible to have loop currents spanning more than one cell, leading to
supermesh. We study this topic next.
6.10 SUPERMESH
When a circuit contains a current source such that it has to be expressed as the
difference of two mesh currents as shown in Worked Example 6.3, the solution can
be simplified by using a supermesh. The use of a supermesh reduces the number
of unknown mesh currents by one, if the current source is an independent source.
When a circuit contains a dependent current source such that it can be expressed
as the difference of two mesh currents, the use of a supermesh may simplify the
process of obtaining one of the equations, but there may be no reduction in the
number of equations required. The concept of supermesh is illustrated by a few
examples.
WORKED EXAMPLE 6.6
Solution:
For the circuit in Fig. 23, the supermesh has been marked. The current for the loop
to the left of the current source is I1, whereas the current for the loop to the right
of the current source is (I1 + 2). We can form two expressions for the voltage
across the current source, and equate them as shown below.
Another example is presented next. The problem used for worked example 6.3 is
solved using a supermesh.
WORKED EXAMPLE 7
Find the mesh currents I1, I2 and I3 in the circuit shown in Fig. 24.
SOLUTION:
The circuit in Figure 25 shows the super mesh. It can be seen that current in the
part in red color is different from the current in the part in blue color. The sum of
voltages in this super mesh is zero. The relevant equations are presented next.
The two unknown mesh currents are I1 and I2 . The KVL equation for mesh
current I1 is presented below, after substituting values of components into
equation (6.55).
For mesh current I2, the mesh equation is:
On grouping terms, we get that
The matrix equation is:
On solving, we get that
It is a matter of choice whether to use a supermesh or not. It is preferable to avoid
the use of a supermesh and follow the technique outlined for Worked Example 6.3.
6.11 EXAMPLES WITH DEPENDENT SOURCES
This section shows how circuits with dependent sources can be analyzed using
meshes. The output voltage or current of a dependent source may be controlled by
another voltage or another current. The presence of a dependent source introduces
constraints on the formation of equations. Mesh equations have to be
supplemented by equations that satisfy the imposed constraints.
WORKED EXAMPLE 8
Obtain the mesh currents in the circuit shown in Fig. 26 using mesh analysis.
Solution:
There are three unknowns and three independent equations have to be formed. The
first equation can be formed for mesh current I1 based on the KVL. The KVL
equations for mesh currents I2 and I3 contain the dependent source as one of the
elements in the mesh equation. But it is seen that the dependent voltage source can
be expressed in terms of I1. Using this relationship, the KVL equations formed for
mesh currents I2 and I3 can be changed as shown below and the problem can be
solved.
The KVL equation for mesh current I1 is expressed by equation (6.58). On
grouping terms, it can be presented, as shown by equation (6.59).
It is seen that the dependent source is present in two loops. For mesh current I2, we
get equation (6.60). Since Va = 2× I1, equation (6.61) is obtained after dividing
each term of equation (6.60) by 5.
For mesh current I3, we get equation (6.62). Since Va = 2× I1, we get equation
(6.63) from equation (6.62).
From equations (6.59), (6.61) and (6.63), we get the matrix equation, shown
below.
On solving, we obtain that
Worked Example 9
Obtain the mesh currents for the circuit in Fig. 27 using mesh analysis.
Solution:
There are three unknowns and three independent equations have to be formed. It
is seen that the current source of 2 A is the difference of two mesh currents, I1 and
I3 .
It is seen that the dependent current source is the difference of two mesh currents,
I2 and I3.
Then
Since Va = 2× I1,
Next a mesh equation can be formed by tracing the path along the boundary of the
circuit. Following the path along the branches on the boundary of the circuit, we
get that
Since Vb = 3 × (I1 - I2) , the above equation can be expressed as follows.
Equations (6.64), (6.65) and (6.66) can be expressed in the form of a matrix
equation.
On solving, we obtain that
Worked Example 6.10:
Obtain the mesh currents in Fig. 28 using a supermesh.
Solution:
The three unknown mesh currents are I1, I2 and I3 . The KVL equation for mesh
current I1 is presented below.
For mesh current I2, the mesh equation is:
The third equation is obtained from the value of the dependent current source. It
is seen that
The matrix equation is obtained from equations (6.68), (6.69) and (6.70).
On solving, we get that
6.12 SUMMARY
This chapter has introduced mesh analysis to resistive circuits with dependent and
independent sources. Given a circuit with dependent sources, it is difficult to
specify a particular technique for forming the necessary equations. The number
of independent equations to be formed should equal the number of unknowns. In
that case, the square matrix would have a non-zero determinant and then the
solution is unique. The next chapter describes how nodal analysis can be applied
to such circuits.
Exercise Problems:
E6.1: Obtain the loop currents in the circuit shown in Fig. 29. Verify your
answer.
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