Problem 46 - Precalculus Mathematics for Calculus

Question

Elecricity By using Kirchhoff's Laws, it can be shown that the currents \(I_{1}, I_{2},\) and \(I_{3}\) that pass through the three branches of the circuit in the figure satisfy the given linear system. Solve the system to find \(I_{1}, I_{2},\) and \(I_{3}\) $$ \left\\{\begin{aligned} I_{1}+I_{2}-I_{3} &=0 \\ 16 I_{1}-8 I_{2} &=4 \\ 8 I_{2}+4 I_{3} &=5 \end{aligned}\right. $$

Step-by-Step Solution

Answer

\(I_1 = \frac{1}{4}, I_2 = \frac{1}{6}, I_3 = \frac{5}{12}\).

Step 1: Rearrange to form an augmented matrix

First, write the system of equations in matrix form. The matrix can be written as an augmented matrix for the coefficients and constants. The system \[ \begin{cases} I_1 + I_2 - I_3 = 0 \ 16I_1 - 8I_2 = 4 \ 8I_2 + 4I_3 = 5 \end{cases} \] becomes \[ \begin{pmatrix} 1 & 1 & -1 & | & 0 \ 16 & -8 & 0 & | & 4 \ 0 & 8 & 4 & | & 5 \end{pmatrix} \].

Step 2: Use Gaussian elimination

Start by using Gaussian elimination to simplify the augmented matrix. Use the first row to eliminate the first elements of the remaining rows. Multiply the first row by 16 and subtract from the second row, resulting in \[ 16R_1 - R_2: \begin{pmatrix} 1 & 1 & -1 & | & 0 \ 0 & -24 & 16 & | & 4 \ 0 & 8 & 4 & | & 5 \end{pmatrix} \].

Step 3: Further row reduction

Continue with further row reduction to eliminate the second element of the third row. Multiply the first row by 8 and subtract from the third row: \[ 8R_1 - R_3: \begin{pmatrix} 1 & 1 & -1 & | & 0 \ 0 & -24 & 16 & | & 4 \ 0 & 8 & 4 & | & 5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 1 & -1 & | & 0 \ 0 & -24 & 16 & | & 4 \ 0 & 0 & 12 & | & 5 \end{pmatrix} \].

Step 4: Solve for one variable

From the third row \(0I_1 + 0I_2 + 12I_3 = 5 \), we solve for \(I_3\): \[ I_3 = \frac{5}{12} \].

Step 5: Back substitution to find other variables

Substitute \(I_3 = \frac{5}{12}\) into the second row equation \(0I_1 - 24I_2 + 16I_3 = 4\). This gives: \(-24I_2 + \frac{80}{12} = 4\). Simplifying, \( -24I_2 = 4 - \frac{80}{12} \), leads to \(I_2 = \frac{1}{6} \).

Step 6: Solve for the last variable

Substitute \(I_3 = \frac{5}{12}\) and \(I_2 = \frac{1}{6}\) into the first equation \(I_1 + I_2 - I_3 = 0\). This gives: \( I_1 + \frac{1}{6} - \frac{5}{12} = 0\). Solving this, \(I_1 = \frac{5}{12} - \frac{1}{6} = \frac{1}{4} \).

Key Concepts

Gaussian elimination

When solving a system of linear equations, Gaussian elimination is a powerful tool. This method transforms a given system into an equivalent one that is easier to solve. You start with converting the system into an augmented matrix, focusing on creating zeros below the leading coefficients (or pivots) of each row. The goal is to arrive at a matrix in row-echelon form, where each non-zero row has its first non-zero number (pivot) to the right of the pivot in the previous row. The steps include

• Performing row operations to eliminate the coefficients below the pivot.
• Re-ordering the equations if needed, to make pivoting easier.

Row operations should preserve the solution set of the system, which include exchanging two rows, multiplying a row by a non-zero scalar, and adding or subtracting multiples of one row to another.

Augmented Matrix

An augmented matrix is an essential concept in solving systems of equations, particularly using Gaussian elimination. It combines the coefficient matrix and the constants from the equations into one matrix. The horizontal line in the augmented matrix separates the coefficients from the constants. For instance, with the given system:

• Equation 1: \( I_1 + I_2 - I_3 = 0 \)
• Equation 2:\( 16I_1 - 8I_2 = 4 \)
• Equation 3: \( 8I_2 + 4I_3 = 5 \)

The augmented matrix representation will be \[ \begin{pmatrix} 1 & 1 & -1 & | & 0 \ 16 & -8 & 0 & | & 4 \ 0 & 8 & 4 & | & 5 \end{pmatrix} \]. This carefully organized structure provides clarity and is easy to manipulate during Gaussian elimination.

Linear Systems

Linear systems are collections of linear equations involving the same set of variables. In the context of circuit analysis, these equations can represent relationships between electrical currents in different circuit branches. A system is considered linear if each equation is a linear combination of variables with constant coefficients. The significance of understanding linear systems in circuit analysis includes:

• Predicting behaviors of complex circuits.
• Modeling real-world systems accurately.

A solution to a linear system refers to a set of values for the variables that satisfy all the equations concurrently. By applying methods such as Gaussian elimination and matrix operations, we can effectively find these solutions.

Circuit Analysis

Circuit analysis involves using mathematical methods and principles to study electrical circuits. One of the foundational techniques, used in this problem, relies on Kirchhoff's Laws, which include:

• Kirchhoff's Current Law (KCL): Sum of currents entering a junction equals the sum leaving.
• Kirchhoff's Voltage Law (KVL): Sum of voltage gains and drops around a closed loop equals zero.

These laws give rise to the linear equations that we solve using methods such as Gaussian elimination. Understanding circuit analysis is invaluable for electrical engineering, allowing us to design and evaluate circuits efficiently. Recognizing the interdependence of components in a circuit helps us predict how the system reacts to changes, ensuring proper functionality.