6. MESH ANALYSIS

INTRODUCTION PASSIVE SIGN CONVENTIONPLANAR CIRCUITS FORMATION OF MESHESANALYSIS OF A SIMPLE CIRCUITDETERMINANT OF A MATRIXCRAMER’S RULEGAUSSIAN ELIMINATION METHODEXAMPLES FOR MESH ANALYSISSUPERMESHEXAMPLES WITH DEPENDENT SOURCESSUMMARYEXERCISE PROBLEMS___________________________________________________________________________

6.1 INTRODUCTION

The two general analytical techniques used in network analysis are mesh or loopanalysis and nodal analysis. This chapter describes mesh analysis, whereas thenext chapter is on nodal analysis.

Mesh analysis is a systematic technique to evaluate all voltages and currents in acircuit. It is based on Kirchoff’s Voltage Law and Ohm’s Law. This technique canbe 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 thenumber of these equations equaling the number of unknown variables to bedetermined. These unknown variables are the loop currents or the mesh currents.The simultaneous equations can be compactly described by matrix equations, andthe solution to matrix equations can be obtained using Cramer=s rule. Gaussianelimination method can also be used to obtain the solution. Solution by hand canbe laborious if the number of unknowns exceeds three. Some examples arepresented, to illustrate the technique of mesh analysis. Since it is essential to befamiliar with passive sign convention, it is explained first.

6.2 PASSIVE SIGN CONVENTION

According to passive sign convention, the terminal by which current enters anelement should be marked positive with respect to the potential at the otherterminal 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 thecircuit, even through their actual value may turn out to be otherwise. For the

1resistor in Fig. 1, current I flows into the positive terminal of the resistor. Hencethe sign of voltage and the direction of current for this resistor are in accordancewith passive sign convention. When passive sign convention is followed for aresistor, the Ohm’s law is satisfied. It can also be said that when Ohm’s law issatisfied, passive sign convention has been followed. When passive signconvention is followed for assigning voltage and current to a resistor, both thevoltage and the current are positive and the resistor absorbs power. Note thatpower is the average of energy absorbed over time. It is energy absorbed persecond 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 resistorvaries proportionately either with the square of its voltage or with the square of itscurrent, the power absorbed by a resistor is always positive. When the sign forvoltage and the direction of current associated with a resistor are according topassive sign convention, then the voltage and the current have the same sign.

2For the resistor shown in the Fig. 2, current I flows out of the positive terminalof the resistor. Hence the sign of voltage and the direction of current for thisresistor are not in accordance with passive sign convention. When passive signconvention is not followed for a resistor, as shown in Fig. 2, the current and thevoltage have opposite signs. If the voltage across the resistor is positive, then thesign of current in relation to the voltage is negative. Hence the voltage and thecurrent associated with the resistor are related to each other, as expressed byequation (6.3)

As shown in Fig. 3, the passive sign convention is valid for any element or a two-

1terminal circuit. When both the voltage and the current I are positive or have thesame sign, then the power absorbed by the element is positive. When both the

2voltage and the current I are positive or have the same sign, then the powerdelivered by the element is positive. When the power is determined in accordancewith the passive sign convention, the power associated with this element isnegative.

6.3 PLANAR CIRCUITS

A planar circuit is shown in Fig. 4. It is seen that a circuit can be drawn such thatno two elements cross over each other. Let us find out what a non-planar circuitis. Let us add one more node, marked as c, to the circuit in Fig. 4 and connect it

1to all other nodes as shown in Fig. 5. It is seen that the resistor R connected from

2node b to node c crosses over resistor R . This circuit cannot be drawn withoutcross-over of two elements and is a non-planar circuit. Mesh analysis is applicableonly to planar circuits, whereas loop analysis and nodal analysis can be applied tonon-planar circuits also.

6.4 FORMATION OF MESHES

The circuit in Fig. 6 illustrates how meshes or loops are to be formed. This circuithas meshes within it and it is easy to see the mesh structure within the circuit. Thecircuit has many windows and each window has a mesh current. It is better tofollow some rules in forming meshes.

Let the number of branches in a network be e, and let the number of nodesincluding the datum node be n. Then the number of independent of meshequations is e - n + 1. These meshes are formed as follows:

a. Identify the meshes and mark the mesh currents with an assigned directionfor each mesh current. It is preferable to assign the direction as shown inFig. 6, all in the clockwise direction. It is better to make sure that thecurrent through any branch is either the mesh current itself or the differenceof two mesh currents. It is preferable if the current through any element isnot the sum of two mesh currents.

b. Apply KVL equation around each labeled mesh. Use the passive signconvention while forming the mesh equations.

c. Solve the resulting set of simultaneous equations for the unknown meshcurrents. For solving the problem manually, it is preferable to use theGaussian elimination method instead of Cramer’s rule.

6.5 ANALYSIS OF A SIMPLE CIRCUIT

A simple problem is presented now. The task is to find the currents and voltagespresent in the circuit in Fig. 7. A sequence of steps is followed to obtain thesolution. The first step is to identify the loop currents, as shown next. The valuesof components are shown beside the circuit.

The first step is to create adjacent, non-overlapping cells, as shown in Fig. 8. Thenmark the loop currents, as shown in Fig. 8. There are two unknowns and two loopcurrents have been identified. Let both the loop currents flow in the samedirection, preferably clockwise, as illustrated here. When the cells are adjacent,and all the loop currents have the same direction, the current through any elementis either the loop current itself, or the difference of two loop currents. The currentthrough any element is never, ever the sum of two loop currents.

1The path for loop current I is shown in red colour. We can apply KVL to getthe 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

1 3resistors, R and R , be assigned such that they are in accordance with thepassive sign convention, as shown in Fig. 9.

3The voltage across resistor R is marked in accordance with the passive sign

1convention, but this has been done with reference to current I . It can be seen,

3that two loop currents flow through resistor R in opposite directions. Hence

3whichever way the voltage across resistor R is assigned a sign, it will be inaccordance with passive sign convention for one loop current, but not so for theother loop current.

1Equation (6.4) defines the voltage across resistor R . Let current though

3 N Nresistor R be I , where I is defined by equation (6.5). Then we get the

3 3 Nvoltage across R as the product of R and I . Equation (6.6) defines the

3 1voltage across resistor R . Applying KVL to the loop containing current I , we

1 Aget equation (6.7). It is seen that current I enters source V through itsnegative terminal and hence the source voltage has a negative sign in front of itin equation (6.7).

2The path for loop current I is shown in red color in Fig. 10. The voltage across

3 1resistor R has already been assigned with reference to current I , but the voltage

2 2across resistor R can be assigned with reference to current I in accordance withthe passive sign convention. It means that both the voltage and the current of

2resistor R have the same sign.

2Equation (6.8) defines the voltage across resistor R . Applying KVL to the loop

2 2containing current I , we get equation (6.9). It is seen that current I enters resistor

3 3R through its negative terminal and hence the voltage across R has a negative

3sign in front of it in equation (6.9). The voltage across R can be expressed byequation (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 termsof 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). Theterms along the main diagonal are loop resistances, and they are positive values.The other two terms are mutual resistances. They are negative values. We takeup the solution of a matrix equation in the subsequent sections. It is possible tosolve 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

1 2 (6.17). On solving these two equations, we get that I = 2 A, and that I = 1 A.

From equation (6.15), it can be seen how mesh equations can be written just byinspection. For a resistive network, each diagonal element of the loop-impedancematrix reflects the self-resistance of the corresponding loop. The self-resistanceof a loop is the sum of resistances through which the loop current flows. Otherelements of this matrix are the mutual resistances. Since current through anyelement is either the loop current itself or the difference of two loop currents, anynon-diagonal element is either zero or a negative value. If two loop currents haveno common element, the corresponding element of the matrix would have a zeroentry. If two loop currents have a common element, the corresponding elementsof the matrix would have an entry equaling the negative of the resistance commonto those two loops.

6.6 DETERMINANT OF A MATRIX

Before we take up the application of Cramer’s rule, it is necessary to know howto 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 differencebetween the positive product and the negative product. This concept can beextended to a third-order matrix too. The determinant of a matrix can bedetermined 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 onlytechnique, but for a third order determinant, a simple technique exists and thattechnique is now explained.

Given a third-order matrix, the first step is to copy the first two columns andplace them after the third column, as shown in Fig. 12. Now we have fivecolumns and three rows, as illustrated.

We can express the value of determinant as the difference between the sum ofpositive products, and the sum of negative products. The way to get the sum ofpositive 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 formsa 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-ordermatrix. It is difficult to remember this equation, but the process, by which itcan 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 co-factors. The cofactor of an element is found out as follows. Cross out the rowand the column to which the element belongs. Then the cofactor can beidentified 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, thecofactor 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 isa column matrix with n elements. Then we obtain x as:

if the inverse of A exists. Since obtaining the inverse of a matrix is somewhatcomplex ,we can use Cramer’s rule instead.

A heuristic argument is provided below to offer insight into how Cramer’s rulehas 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

2 3determinant, we can add x times the second column and x times the third columnto the first column without affecting the value of the left-hand side value ofequation (6.24). That is,

The first column of the determinant on the right- hand side of equation (6.25) canbe replaced by the right-hand side column matrix of equation (6.22). Thenequation (6.26) is obtained. Equation (6.26) expresses Cramer’s rule.

2 3In the same way, we can get the values of x and x .

A numerical example is presented below.

Equation (6.29) presents a matrix equation. The values of currents have to bedetermined. The value of determinant of Z in equation (6.30) can be obtained asfollows.

The first two columns are copied and are placed after the third column. Theproduct terms along the diagonals can be identified and are presented. Thedeterminant is evaluated and equation (6.31) expresses its value.

1, 2 3Using Cramer’s rule, we can obtain the values of I I , and I .

The use of Cramer’s rule tends to be tedious, whereas it is easier the Gaussianelimination method.

6.8 GAUSSIAN ELIMINATION METHOD

Compared to the use of Cramer’s rule, the Gaussian elimination method is easierto use for hand calculations. Let a set of three simultaneous equations be:

Given a set of simultaneous equations, re-arrange the equations such that

1The aim is to eliminate variable x from equations (ii) and (iii). It can be eliminatedfrom equation (ii), as shown below.

1It is seen that variable x has been eliminated. Let us express equation (v) asfollows.

1The next step is to eliminate variable x from equation (iii), as shown below.

1It is seen that variable x has been eliminated. Let us express equation (viii) asfollows.

2The next step is to eliminate x from equations (vi) and (ix).

Express equation (xi) as follows.

1 2 3The three equations that are used for obtaining the values of x , x ,and x areequations (i), (vi) and (xii), which are presented below.

3We can get the value of x from equation (xii), as shown be equation (xiii).

3 2,Knowing the value of x , we can get the value of x as shown by equation (xiv).

2 3 1Once the values of x ,and x are known, the value of x can be obtained, as shownby 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.

1Eliminate y from equation (ii). Keep equation (i).

1Eliminate y from equation (iii). Keep equation (i).

2Eliminate y from equation (vii). Keep equation (v).

3 2Now the value of y is known, evaluate y from equation (v)

1Evaluate y 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 beapplied. In this section, the circuits contain only independent sources. Exampleswith 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. Allthe three loop currents are in the clockwise direction. The cells containing theseloop currents are adjacent to each other. The current through any element is eitherthe loop current or the difference of two loop currents. Since there are three loopcurrents to be determined, three independent equations have to be formed. In thisproblem, the solution is relatively easy, because one KVL equation can be formedfor each loop, thus leading to formation of three independent equations.

1Path for loop current I is shown in red colour in Fig. 16. Write the KVL equation

1for mesh current I .

1Equation (6.31) is the KVL equation for mesh current I . From equation (6.31),we can obtain equation (6.32) by re-grouping coefficients.

2Path for loop current I is shown in red colour in Fig. 17. Write the KVL equation

2for mesh current I .

3Path for loop current I is shown in red colour in Fig. 18. Write the KVL equation

3for mesh current I .

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 fromequation (6.37)

To solve the set of simultaneous equations expressed in the matrix form, we canuse 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, CurA=[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 isa straight-forward technique. On the other hand, if a network contains a currentsource, 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 canitself be a mesh current and the other is that the current source is equal to thedifference of two mesh currents. The example below shows a case where thecurrent 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.

1The equation for mesh current I can be formed as shown in the previous example.

2 3Since the current source of 1 A is the difference of two mesh currents I and I , thesecond equation can be obtained from this equality. This current source of 1 A is

2 3present in the mesh for current I and also in the mesh for current I as well. Hencetwo expressions for the voltage can be obtained for the voltage across the 1 Acurrent source. Equating these two expressions to each other, the third independentequation can be obtained.

1The first mesh equation is formed as follows. For mesh current I , the KVLequation is:

1Equation (6.38) is the KVL equation for mesh current I . 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 andwe can get the second equation.

It is seen that two expressions for the voltage across the current source can be

Yobtained. Let it be V .

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 fromequation (6.45)

On solving, we obtain that

Worked Example 6.4

1 2Find the mesh currents I and I in the circuit shown in Fig. 21.

Solution:

The circuit in Fig. 21 for worked example 4 contains a current source as one of themesh currents. In such a case, one of the unknowns is known. The solution iseasier, since we have to solve for only two unknowns.

1Write the KVL equation for mesh current I :

On grouping terms, we obtain that

3Since I = 1 A, the above equation becomes:

2For mesh current I , the loop equation is obtained as given below.

3On grouping terms and substituting for I , 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:

2First write the KVL equation for loop current I . The second equation is obtainedfrom 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 twoexpressions 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 tosupermesh. We study this topic next.

6.10 SUPERMESH

When a circuit contains a current source such that it has to be expressed as thedifference of two mesh currents as shown in Worked Example 6.3, the solution canbe simplified by using a supermesh. The use of a supermesh reduces the numberof 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 expressedas the difference of two mesh currents, the use of a supermesh may simplify theprocess of obtaining one of the equations, but there may be no reduction in thenumber of equations required. The concept of supermesh is illustrated by a fewexamples.

WORKED EXAMPLE 6.6

Solution:

For the circuit in Fig. 23, the supermesh has been marked. The current for the loop

1to the left of the current source is I , whereas the current for the loop to the right

1 of the current source is (I + 2). We can form two expressions for the voltageacross the current source, and equate them as shown below.

Another example is presented next. The problem used for worked example 6.3 issolved using a supermesh.

WORKED EXAMPLE 7

1 2 3Find the mesh currents I , I and I in the circuit shown in Fig. 24.

SOLUTION:

The circuit in Figure 25 shows the super mesh. It can be seen that current in thepart in red color is different from the current in the part in blue color. The sum ofvoltages in this super mesh is zero. The relevant equations are presented next.

1 2The two unknown mesh currents are I and I . The KVL equation for mesh

1current I is presented below, after substituting values of components intoequation (6.55).

2For mesh current I , 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 avoidthe 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 usingmeshes. The output voltage or current of a dependent source may be controlled byanother voltage or another current. The presence of a dependent source introducesconstraints on the formation of equations. Mesh equations have to besupplemented 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

1first equation can be formed for mesh current I based on the KVL. The KVL

2 3equations for mesh currents I and I contain the dependent source as one of the

elements in the mesh equation. But it is seen that the dependent voltage source can

1be expressed in terms of I . Using this relationship, the KVL equations formed for

2 3mesh currents I and I can be changed as shown below and the problem can besolved.

1The KVL equation for mesh current I is expressed by equation (6.58). Ongrouping terms, it can be presented, as shown by equation (6.59).

2It is seen that the dependent source is present in two loops. For mesh current I , we

a 1get equation (6.60). Since V = 2× I , equation (6.61) is obtained after dividingeach term of equation (6.60) by 5.

3 a 1For mesh current I , we get equation (6.62). Since V = 2× I , we get equation(6.63) from equation (6.62).

From equations (6.59), (6.61) and (6.63), we get the matrix equation, shownbelow.

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

1is seen that the current source of 2 A is the difference of two mesh currents, I and

3I .

It is seen that the dependent current source is the difference of two mesh currents,

2 3I and I .

Then

a 1Since V = 2× I ,

Next a mesh equation can be formed by tracing the path along the boundary of thecircuit. Following the path along the branches on the boundary of the circuit, weget that

b 1 2Since V = 3 × (I - I ) , the above equation can be expressed as follows.

Equations (6.64), (6.65) and (6.66) can be expressed in the form of a matrixequation.

On solving, we obtain that

Worked Example 6.10:

Obtain the mesh currents in Fig. 28 using a supermesh.

Solution:

1, 2 3The three unknown mesh currents are I I and I . The KVL equation for mesh

1current I is presented below.

2For mesh current I , the mesh equation is:

The third equation is obtained from the value of the dependent current source. Itis 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 andindependent sources. Given a circuit with dependent sources, it is difficult tospecify a particular technique for forming the necessary equations. The numberof independent equations to be formed should equal the number of unknowns. Inthat case, the square matrix would have a non-zero determinant and then thesolution is unique. The next chapter describes how nodal analysis can be appliedto such circuits.

Exercise Problems:

E6.1: Obtain the loop currents in the circuit shown in Fig. 29. Verify youranswer.

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