6. MESH ANALYSIS 6.1 INTRODUCTION

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1 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 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

2 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 I 1 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.

3 For the resistor shown in the Fig. 2, current I 2 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 I 1 are positive or have the same sign, then the power absorbed by the element is positive. When both the voltage and the current I 2 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 R 1 connected from node b to node c crosses over resistor R 2. 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.

4 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. Use the passive sign 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.

5 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. 1 The path for loop current I is shown in red colour. We can apply KVL to get the loop equation. Kirchoff s voltage law states that the algebraic sum of

6 voltages of elements present in a loop is zero. Let the voltages across the resistors, R 1 and R 3, be assigned such that they are in accordance with the passive sign convention, as shown in Fig. 9. The voltage across resistor R 3 is marked in accordance with the passive sign convention, but this has been done with reference to current I 1. It can be seen, that two loop currents flow through resistor R 3 in opposite directions. Hence whichever way the voltage across resistor R 3 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 R 1. Let current though resistor R 3 be I N, where I N is defined by equation (6.5). Then we get the voltage across R 3 as the product of R 3 and I N. Equation (6.6) defines the voltage across resistor R 3. Applying KVL to the loop containing current I 1, we get equation (6.7). It is seen that current I 1 enters source V A 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 I 2 is shown in red color in Fig. 10. The voltage across resistor R 3has already been assigned with reference to current I 1, but the voltage across resistor R 2can be assigned with reference to current I 2in accordance with the passive sign convention. It means that both the voltage and the current of resistor R have the same sign. 2

7 Equation (6.8) defines the voltage across resistor R 2. Applying KVL to the loop containing current I 2, we get equation (6.9). It is seen that current I 2enters resistor R 3 through its negative terminal and hence the voltage across R 3 has a negative sign in front of it in equation (6.9). The voltage across R 3 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.

8 By substituting the given values of components, we get equations (6.16) and (6.17). On solving these two equations, we get that I = 2 A, and that I = 1 A. 1 2 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.

9 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.

10 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

11 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 x 2times the second column and x 3times 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.

12 In the same way, we can get the values of x 2 and x 3. 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 I 1, I 2, and I 3.

13 The use of Cramer s rule tends to be tedious, whereas it is easier the Gaussian elimination method.

14 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 x 1 from equations (ii) and (iii). It can be eliminated from equation (ii), as shown below. It is seen that variable x 1 has been eliminated. Let us express equation (v) as follows. The next step is to eliminate variable x 1 from equation (iii), as shown below.

15 It is seen that variable x 1 has been eliminated. Let us express equation (viii) as follows. The next step is to eliminate x 2 from equations (vi) and (ix). Express equation (xi) as follows The three equations that are used for obtaining the values of x, x,and x are equations (i), (vi) and (xii), which are presented below.

16 We can get the value of x 3 from equation (xii), as shown be equation (xiii). Knowing the value of x 3, we can get the value of x 2, as shown by equation (xiv). Once the values of x 2,and x 3 are known, the value of x 1can 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 y 1 from equation (ii). Keep equation (i).

17 Eliminate y 1 from equation (iii). Keep equation (i). Eliminate y 2 from equation (vii). Keep equation (v). Now the value of y 3 is known, evaluate y 2 from equation (v) Evaluate y 1 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.

18 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.

19 Path for loop current I 1 is shown in red colour in Fig. 16. Write the KVL equation for mesh current I. 1 Equation (6.31) is the KVL equation for mesh current I 1. From equation (6.31), we can obtain equation (6.32) by re-grouping coefficients. Path for loop current I 2 is shown in red colour in Fig. 17. Write the KVL equation for mesh current I. 2 Path for loop current I 3 is shown in red colour in Fig. 18. Write the KVL equation for mesh current I. 3

20 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; ; ]; V=[15;6;-6]; Cur=inv(A)*V The output of the program is: Cur = 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

21 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 I 1 can be formed as shown in the previous example. Since the current source of 1 A is the difference of two mesh currents I 2and I 3, the second equation can be obtained from this equality. This current source of 1 A is present in the mesh for current I 2and also in the mesh for current I 3as 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.

22 The first mesh equation is formed as follows. For mesh current I 1, the KVL equation is: Equation (6.38) is the KVL equation for mesh current I 1. 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 V. Y 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

23 Worked Example 6.4 Find the mesh currents I 1 and I 2 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 I 1 : On grouping terms, we obtain that Since I 3 = 1 A, the above equation becomes: For mesh current I 2, the loop equation is obtained as given below. On grouping terms and substituting for I 3, we obtain that

24 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 I 2. 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

25 It is possible to have loop currents spanning more than one cell, leading to supermesh. We study this topic next 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 I 1, whereas the current for the loop to the right of the current source is (I 1 + 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.

26 WORKED EXAMPLE 7 Find the mesh currents I 1, I 2 and I 3 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.

27 The two unknown mesh currents are I 1 and I 2. The KVL equation for mesh current I 1 is presented below, after substituting values of components into equation (6.55). For mesh current I 2, 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 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 I 1 based on the KVL. The KVL equations for mesh currents I and I contain the dependent source as one of the 2 3

28 elements in the mesh equation. But it is seen that the dependent voltage source can be expressed in terms of I 1. Using this relationship, the KVL equations formed for mesh currents I 2 and I 3 can be changed as shown below and the problem can be solved. The KVL equation for mesh current I 1 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 I 2, we get equation (6.60). Since V a = 2 I 1, equation (6.61) is obtained after dividing each term of equation (6.60) by 5. For mesh current I 3, we get equation (6.62). Since V a = 2 I 1, 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.

29 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, I 1 and I. 3 It is seen that the dependent current source is the difference of two mesh currents, I and I. 2 3 Then

30 Since V a = 2 I 1, 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 V b = 3 (I 1 - I 2), 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:

31 Obtain the mesh currents in Fig. 28 using a supermesh. Solution: The three unknown mesh currents are I1, I 2 and I 3. The KVL equation for mesh current I is presented below. 1 For mesh current I 2, 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.

32 Exercise Problems: E6.1: Obtain the loop currents in the circuit shown in Fig. 29. Verify your answer.