Problem 36 - Algebra and Trigonometry

Question

Electricity 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 \(I_{3} .\) $$ I_{1}+I_{2}-I_{3}=0 $$ $$16 I_{1}-8 I_{2} \quad=4$$ $$8 I_{2}+4 I_{3}=5$$

Step-by-Step Solution

Answer

\( I_{1} = \frac{11}{28}, I_{2} = \frac{2}{7}, I_{3} = \frac{19}{28} \).

Step 1: Identify the System of Equations

We have the following system of equations: \( I_{1} + I_{2} - I_{3} = 0 \), \( 16I_{1} - 8I_{2} = 4 \), \( 8I_{2} + 4I_{3} = 5 \). Three equations with three unknowns: \( I_{1}, I_{2}, \) and \( I_{3} \).

Step 2: Solve Equation 2 for I1

From the second equation, solve for \( I_{1} \): \( 16I_{1} = 8I_{2} + 4 \). Divide through by 16 to isolate \( I_{1} \): \( I_{1} = \frac{8I_{2} + 4}{16} = \frac{1}{2}I_{2} + \frac{1}{4} \).

Step 3: Substitute I1 into Equation 1

Substitute \( I_{1} = \frac{1}{2}I_{2} + \frac{1}{4} \) into the first equation: \( \left(\frac{1}{2}I_{2} + \frac{1}{4}\right) + I_{2} - I_{3} = 0 \). Simplify to find \( I_{3} \): \( \frac{3}{2}I_{2} + \frac{1}{4} - I_{3} = 0 \). This gives \( I_{3} = \frac{3}{2}I_{2} + \frac{1}{4} \).

Step 4: Substitute I3 into Equation 3

Plug \( I_{3} = \frac{3}{2}I_{2} + \frac{1}{4} \) into the third equation: \( 8I_{2} + 4\left(\frac{3}{2}I_{2} + \frac{1}{4}\right) = 5 \). Simplify to obtain \( 8I_{2} + 6I_{2} + 1 = 5 \), leading to \( 14I_{2} = 4 \).

Step 5: Solve for I2

Solve \( 14I_{2} = 4 \) to find \( I_{2} \): \( I_{2} = \frac{4}{14} = \frac{2}{7} \).

Step 6: Find I1 and I3 using I2

Using \( I_{2} = \frac{2}{7} \), calculate \( I_{1} \) and \( I_{3} \). For \( I_{1} \), use \( I_{1} = \frac{1}{2} \times \frac{2}{7} + \frac{1}{4} = \frac{1}{7} + \frac{1}{4} = \frac{11}{28} \). For \( I_{3} \), use \( I_{3} = \frac{3}{2} \times \frac{2}{7} + \frac{1}{4} = \frac{3}{7} + \frac{1}{4} = \frac{19}{28} \).

Key Concepts

Linear System

In the realm of circuit analysis, understanding a linear system is essential. A linear system is a set of equations, where each term is either a constant or the product of a constant and a single variable. The main goal when dealing with a linear system is to find the values of the unknowns that satisfy all the equations simultaneously. The system of equations you've encountered in this exercise can be written as the matrix equation \[\begin{bmatrix} 1 & 1 & -1 \ 16 & -8 & 0 \ 0 & 8 & 4 \end{bmatrix} \begin{bmatrix} I_1 \ I_2 \ I_3 \end{bmatrix} = \begin{bmatrix} 0 \ 4 \ 5 \end{bmatrix}\] which is characteristic of linear systems.

• Each row in the matrix corresponds to one of the original equations.
• The variables are organized in a specific order, consistent across all equations.

In this specific setting, solving the system involves manipulating these equations using algebraic techniques to isolate one variable at a time. You'll encounter substitution or elimination methods as tools for reaching the solution.

Circuit Analysis

Circuit analysis is the study of how electrical components in a circuit carry electrical current. It provides a framework to calculate the current, voltage, and resistance at any point in a network. One of the fundamental techniques in this field is using Kirchhoff's Laws, which include the Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL).

• KCL states that the total current entering a junction must equal the total current leaving the junction.
• KVL states that the total sum of the voltages in any closed loop in a circuit must equal zero.

For example, in the exercise, we applied these principles to the circuit to determine the equations for currents \(I_1\), \(I_2\), and \(I_3\). This process involved identifying loop and junction equations that satisfy Kirchhoff's Laws. Understanding these laws is crucial for ensuring accurate current and voltage calculations.

Current Calculation

Once you have the linear equations set up from circuit analysis, the next step is the current calculation. This process involves solving the system of equations to find the values of the currents \(I_1\), \(I_2\), and \(I_3\).

• Start by expressing some variables in terms of others; in the exercise, this was done by solving the equations sequentially.
• For instance, calculate \(I_1\) first in terms of \(I_2\).
• Then use this value to further solve for \(I_3\) and \(I_2\).

Substituting the known values back into the equations verifies the solution ensures coherence with the original system. Ultimately, current calculation provides insights into how your circuit behaves in real-world scenarios. Each step is a logical progression from expressing one current in terms of another, leading up to solving for all currents. This methodology allows for precise control over circuit behavior, reflecting real-world electrical principles.