# Appendix B: Solving Simultaneous Equations

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1 Appendix B: Solving Simultaneous Equations Electric circuit analysis methods help us solve for the unknown voltages and currents in the circuits: Kirchhoff s circuit law solves for unknown branch currents Node-voltage method solves for unknown node voltages Mesh-current method solves for unknown mesh currents In order to solve for n of these unknown currents or voltages, we need n independent equations that relate known values (such as the resistors, voltage and current sources, and other elements in the circuit) to the unknown voltages or currents Once we have these n equations, we can write them in standard matrix form and solve them using either Cramer s rule (by hand) or matrix inversion (using MATLAB R, MathScript RF Module, similar software solvers, or your engineering calculator) This Appendix provides a brief overview of these approaches B- Review of Cramer s Rule Let us assume that the application of Kirchhoff s current and voltage laws to a certain circuit led to the following set of equations: 2(i + i 2 ) 0 +(3i 2 i 4 )=0 (Ba) 3(i + i 2 )+2(i + 3 )=0 (Bb) i i 2 = 0 (Bc) Our task is to solve the three independent, simultaneous, linear equations to determine the values of the three unknowns, i to [Recall that independence means that none of the three equations can be generated through a linear combination of the other two] One way to accomplish the specified task is to apply the method of elimination of variables If we solve for i in Eq (Bc), for example, and then use the expression i =(i 2 +) to replace i in Eqs (Ba and b), we end up with two new equations containing only two variables, i 2 and Repeat of the substitution procedure leads to a single equation in only one unknown, which can be solved directly Once that unknown has been determined, it is a straightforward process to solve for the values of the other two variables Such a solution method might prove effective for solving a simple set of three simultaneous equations, but what if the circuit we wish to analyze happens to contain a large number of variables? In that case, the more expeditious approach is to take advantage of Cramer s rule, whose implementation procedure is both systematic and straightforward Our review of Cramer s rule will initally use the set of three simultaneous equations given by Eq (B) to demonstrate the mechanics of the solution procedure for a system of order 3 Afterwards, we will treat the general case of a system of order n (consisting of n independent equations in n unknowns) B- System of Order 3 Step : Cast Equations in Standard Form Before we can apply Cramer s rule, we need to regularize the simultaneous equations into a standard system of the following form: a i + a 2 i 2 + a 3 = b, a 2 i + a 22 i 2 + a 23 = b 2, a 3 i + a 32 i 2 + a 33 = b 3, (B2a) (B2b) (B2c) where the a s are the coefficients of the variables, i to, and the b s are the unaffiliated constants By expanding the bracketed quantities in Eq (B) and collecting terms, we 70

2 706 December 7, 4 CHAPTER 3 FOURIER ANALYSIS TECHNIQUE can convert the equations into the standard form defined by Eq (B2) Such a process leads to: i + i 2 4 = 0, (B3a) i 3i = 0, (B3b) i i 2 = (B3c) This generates a matrix equation of the form AI = B, where I is the vector of n unknowns (currents i through in this case): i i2 = 0 0 (B4) 0 Note that a =, a 2 =, and a 33 = 0 The regularized set of three linear, simultaneous equations given by Eq (B4) is a system of order 3 Step 2: General Solution According to Cramer s rule, the solutions for i to are given by i = Δ i 2 = Δ 2 = Δ 3 (Ba) (Bb) (Bc) where Δ is the value of the characteristic determinant of the system represented by Eq (B4), andδ to Δ 3 are the affiliated determinants for variables i to The procedure for evaluating these determinants is covered in Steps 3 and 4 Before we do so, however, we should note that in view of the fact that Δ appears in the denominator in Eq (B), Cramer s rule cannot provide solutions for the unknown variables when 0 This is not surprising, because for any system of n unknowns, the condition 0 occurs when one, or more, of the equations is not independent This means that the system contains more unknowns than the available number of independent equations, in which case it has no unique solution Step 3: Evaluating the Characteristic Determinant The characteristic determinant is composed of the a-coefficients of the 3 3 system of equations: a a 2 a 3 a 2 a 22 a 23 a 3 a 32 a 33 (B6) Each element in the determinant has an address jk specified by its row number j and column number k Thus, a 2 is in the first row ( j = ) and second column (k = 2) For the system given by Eq (B4), (B7) To evaluate Δ, we expand it in terms of the elements of one of its rows For simplicity, we will always perform the expansion using the top row The expansion process converts Δ from a determinant of order 3 into the sum of 3 terms, each containing a determinant of order 2 Expanding Eq (B7) by its top row gives a C + a 2 C 2 + a 3 C 3 = C + C 2 4C 3, (B8) where C, C 2,andC 3 are the cofactors of elements a, a 2,anda 3, respectively The cofactor of any element a jk located at the intersection of row j and column k is related to the minor determinant of that element by C jk =( ) j+k M jk, (B9) and the minor determinant M jk is obtained by deleting from the parent determinant all elements contained in row j and column k Hence, M is given by Δ after removal of the top row and the left column, M = a 22 a 23 = a 32 a (B0) For a determinant of order 2, expansion by the top row gives M = a 22 M 22 a 23 M 23 = a 22 a 33 a 23 a 32, (B) which is equivalent to diagonal multiplication of the upperleft and lower-right corners to get a 22 a 33, followed with

3 B- Review of Cramer s Rule December 7, multiplication of the other two corners to get a 23 a 32,andthen subtracting the latter term from the former Substituting the values of the coefficients we have M =( 3) 0 6 ( )=6 (B2a) Similarly, M 2 is obtained by removing from Δ the elements in row and the elements in column 2, M 2 = a 2 a 23 = a a 3 a 2 a 33 a 23 a 3 33 =( ) 0 6 = 6 (B2b) Finally, M 3 is obtained by removing from Δ row and column 3, M 3 = 3 =( ) ( ) ( 3) = 4 (B2c) Inserting the values of the three minor determinants in Eq (B8) gives C + C 2 4C 3 = M M 2 4M 3 = 6 ( 6) 4 4 = (B3) Step 4: Evaluating the Affiliated Determinants The affiliated determinant Δ for variable i is obtained by replacing column in the characteristic determinant Δ with a column comprised of the b s in Eq (B2)Thatis, b a 2 a 3 Δ = b 2 a 22 a 23 b 3 a 32 a 33 = (B4) Evaluation of Δ follows the same rules of expansion discussed earlier in Step 3 in connection with the evaluation of Δ Hence Δ = = 0 6 ( 30) 4 = 0 Application of Eq (Ba) gives i = Δ 0 = 7 Similarly, Δ 2 is obtained from Δ upon replacing column 2 with the b-column, and Δ 3 is obtained from Δ upon replacing column 3 with the b-column The procedure leads to Δ 2 = 0, Δ 3 = 0, i 2 = Δ 2 / 0/ = 2, and = Δ 3 / 2 B-2 System of Order n For a regularized system of linear simultaneous equations given by a i + a 2 i 2 + a a n i n = b, a 2 i + a 22 i 2 + a a 2n i n = b 2, a n i + a n2 i 2 + a n3 + + a nn i n = b n, the solution for any variable i k of the system is (Ba) (Bb) (Bn) i k = Δ k (B6) where Δ is the characteristic determinant and Δ k is the affiliated determinant for variable i k Analogous with the 3 3 system of Section 3-, Δ is composed of the a-coefficients: a a 2 a 3 a n a 2 a 22 a 23 a 2n, (B7) a n a n2 a n3 a nn and Δ k is obtained from Δ by replacing column k with the b-column For example, Δ 2 is given by a b a 3 a n a 2 b 2 a 23 a 2n Δ 2 = (B8) a n b n a n3 a nn To determine the value of a determinant of order n, we can carry out a process of successive expansion, analogous with that outlined in Step 3 of Section 3- The first step in the expansion process converts Δ from a determinant of order n into a sum of n terms, each containing a determinant of order (n ) Each of those new determinants can then be expanded into the sum of determinants of order (n 2), and the process can be contined until it reaches a determinant of order, which consists of a single element

5 B-3 MATLAB R or MathScript Solution December 7, The solution of AI = B is obtained by entering the statement I = A \ B; The MATLAB R or MathScript response would be I = More information on using MATLAB R and MathScript is available in Appendix E

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