A d.c. circuit comprises three closed loops. Applying Kirchhoff's laws to the closed loops gives

step 1 This is the problem number 8. A DC circuit comprises three closed loops applying to Chorps law t

A d.c. circuit comprises three closed loops. Applying...

Key Concepts

D.C. Circuit Analysis

D.C. circuit analysis refers to the process of analyzing circuits that have direct current power sources. This analysis relies on Ohm’s law and Kirchhoff’s laws to determine quantities such as current, voltage, and resistance in different parts of the circuit.

Kirchhoff's Laws

Kirchhoff's laws, which include the current law (KCL) and the voltage law (KVL), are fundamental principles in circuit analysis. They assert that the algebraic sum of currents entering a node is zero and that the sum of voltage differences around any closed loop is zero, providing the foundation for setting up equations in circuit analysis.

System of Linear Equations

A system of linear equations consists of multiple equations, each representing linear relationships among a set of variables. In the context of circuit analysis, applying Kirchhoff's laws typically results in a system of linear equations that represents the behavior of the circuit in terms of currents and voltages.

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations. It involves using elementary row operations to transform the system's augmented matrix into an upper triangular or row-echelon form, which then allows solution for the variables through back-substitution.

Transcript

00:01 This is the problem number 8.

00:04 A dc circuit comprises three closed loops applying to chorps law to the closed loops gives the following equation for current flow in millie -amperts and the equations are 2 i1 plus 3 i2 minus 4 i3 equal to 26 and second equation i 1 minus 5 i 2 minus 3 i 3 equal to minus 87 this is 7 i 1 plus 2 i 2 plus 6 i 3 equal to 12 let's assume this is equation number 1 this one is equation number 2 this is equation number three now we have to find out the value of this current i1 i2 and i3 using gaussian elimination method so first you have to know about goss elimination method and this is a method to solve the linear equation and it is also known as a row reduction method and it is a algorithm to solve linear equations.

02:44 In this method, we can switch, we can do the following operation, that is, the row rows can be swept, we can multiply by, we can multiply, we can multiply, a row by a non -zero number we can add or subtract the rows and by following these method in these operations we will get a upper triangular matrix which is known as equivalent form of the matrix matrix let's do this all of the given equation the given equation can be written as in the matrix form as the first equation is 2 i1 and plus 3 i2 3 i2 and minus 4 i3 3 i3 equal to 26 so these are the coefficient of i 1 i 2 i 3 so we have to multiply this matrix by this matrix there is this row by column that is 2 i1 plus 3 i 2 i 2 minus 4 i 3 is equal to 26 similarly this is 1 minus 5 i 1 minus 5 i 2 minus 3 i 3 is equal to minus safety 7 and third equation can be 29 that is minus 7 i1 plus 2 i 2 i 2 and 6 i3 is equal to 12.

06:43 Now using the gauss elimination method, we will write this matrix as 2, 3, minus 4, 26, 1, minus 5, minus 3, minus 87, minus 7, 2, 6, and 6, and 6, and 6, and 6, and 7, and 6, and 6, and, and, and, 7, and, and, now when we change or sweep the rows there will be no sign in the matrix so first we will not swapping any rows so let's first do the operation that is r2 can written as r1 minus 2 r2 and r3 can written as 7 r2 plus r3 let's apply these operations on this matrix and this matrix converts to it's right here this first row will be same that is 2 3 minus 4 26 and the second row will be here we are subtracting this second r1 minus 2 r2 there is 2 minus 2 is 0 here 3 and this will be 10 and 3 minus 10 that is already minus 10 that is it will be 3 plus 10 so it will be 13 similarly this minus 4 plus 6 that is 2 and here 26 plus 2 into 87 and this will be 200 now r 3 will be 7 r2 plus r3 that is 7 minus 7 that is 0 that is minus 35 plus 2 that is minus 33 and this is minus 21 plus 6 and it will be minus 15 and similarly minus 87 plus 12 and then you will get minus 5 97 now again apply 1 operation which is that is r2 is 33 we are multiplying this equation by 33 and we are multiplying by 13 that is row third row is multiplied by the 13 and row 2 is multiplied by the 33 so here r3 is multiplied by 13 r3 now writes this operation here this will be 2 3 minus 4 26 and this will be 0 and if we are multiplied by 13 then it will be 429 and this will be 33 to 2 that is 66 and this will be 66 0 0 that is 6600 and this value will be minus 33 into 13 that is minus 4 to 9 and this value will be minus 195 and this value will be minus 777 6 now as you can see that if we add this row then you will get 0 here again applying another operation that is r3 is r2 plus r3 in this operation in this operation is 2 3 minus 4 26 0 429 66 6 600 and this this is 0 and minus 195 plus 66 then you will get minus 129 and similarly 776 6600 plus 600 then you will get minus 129 and similarly 7761 plus 600 then you will get minus 11161.

13:04 Now as you can see this is the upper triangular form in this matrix this is upper triangle and this is known in the equivalent form of the matrix, z matrix and z matrix that is coefficient matrix...

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