Problem 48 - Precalculus: Mathematics for Calculus

Question

Electricity 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

Currents: \(I_1 = \frac{11}{28}, I_2 = \frac{2}{7}, I_3 = \frac{19}{28}\).

Step 1: Understand the Equations

We are given a system of equations derived from Kirchhoff's Laws: 1. \( I_1 + I_2 - I_3 = 0 \) 2. \( 16I_1 - 8I_2 = 4 \) 3. \( 8I_2 + 4I_3 = 5 \) Our goal is to solve this system to find the values of currents \( I_1 \), \( I_2 \), and \( I_3 \).

Step 2: Express One Variable in Terms of Others

From the first equation \( I_1 + I_2 - I_3 = 0 \), express \( I_1 \) in terms of \( I_2 \) and \( I_3 \):\[ I_1 = I_3 - I_2 \]

Step 3: Substitute into Other Equations

Replace \( I_1 \) in the second equation \( 16I_1 - 8I_2 = 4 \) with \( I_3 - I_2 \): \[ 16(I_3 - I_2) - 8I_2 = 4 \] Simplify to: \[ 16I_3 - 24I_2 = 4 \] This simplifies to: \[ 16I_3 = 24I_2 + 4 \] Divide by 8: \[ 2I_3 = 3I_2 + \frac{1}{2} \] \[ I_3 = \frac{3}{2}I_2 + \frac{1}{4} \]

Step 4: Substitute for Third Equation

Substitute \( I_3 = \frac{3}{2}I_2 + \frac{1}{4} \) into the third equation \( 8I_2 + 4I_3 = 5 \):\[ 8I_2 + 4 \left( \frac{3}{2}I_2 + \frac{1}{4} \right) = 5 \]Simplify to:\[ 8I_2 + 6I_2 + 1 = 5 \]\[ 14I_2 = 4 \]Divide by 14:\[ I_2 = \frac{2}{7} \]

Step 5: Find Other Unknowns

Now use \( I_2 = \frac{2}{7} \) to find \( I_3 \):\[ I_3 = \frac{3}{2} \times \frac{2}{7} + \frac{1}{4} \] \[ I_3 = \frac{3}{7} + \frac{1}{4} = \frac{12}{28} + \frac{7}{28} = \frac{19}{28} \]Now find \( I_1 \):\[ I_1 = I_3 - I_2 \]\[ I_1 = \frac{19}{28} - \frac{2}{7} = \frac{19}{28} - \frac{8}{28} = \frac{11}{28} \]

Step 6: Complete the Solution

We have found the values for the currents:\[ I_1 = \frac{11}{28} \]\[ I_2 = \frac{2}{7} \]\[ I_3 = \frac{19}{28} \]These are the solutions to the system of equations.

Key Concepts

Understanding Linear Systems in Circuits

Linear systems are a powerful tool in mathematics, often used to model various physical and engineering problems. In circuit analysis, linear systems can help determine unknown electrical parameters like currents and voltages. A linear system consists of multiple linear equations, where each equation involves variables raised to the power of one. These equations may represent physical laws or relationships that must be satisfied.

When it comes to circuits, Kirchhoff's Laws are commonly expressed as linear systems. In our context, we have three equations:

• The first equation, \( I_1 + I_2 - I_3 = 0 \), reflects the conservation of current at a junction.
• The second and third equations, \( 16I_1 - 8I_2 = 4 \) and \( 8I_2 + 4I_3 = 5 \), relate to the voltage drops across various components.

To solve these systems, you can use substitution or elimination methods. By expressing one unknown in terms of another, you create simpler equations, making it easier to find each variable's exact values. This systematic approach helps ensure that all parts of the circuit interact harmoniously.

The Role of Kirchhoff's Laws in Circuit Analysis

Circuit analysis is a foundational skill in electrical engineering, where we analyze circuits to understand the behavior and magnitude of their currents and voltages. To tackle such problems, Kirchhoff's Laws are indispensable. These laws embody the fundamental principles of electric charge and energy conservation.

Kirchhoff's Current Law (KCL) says that the total current entering a junction equals the total current leaving. This is reflected in our first equation, \( I_1 + I_2 - I_3 = 0 \), ensuring all currents are accounted for within the circuit.

Kirchhoff's Voltage Law (KVL) states that the sum of electrical potential differences (voltages) around any closed circuit loop is zero. The given equations, \( 16I_1 - 8I_2 = 4 \) and \( 8I_2 + 4I_3 = 5 \), are derived from applying KVL to different loops in the circuit.

• This sum becomes a vital check as it guarantees that conduits neither create nor store net charge over time.
• Both laws, when applied judiciously, provide a structured way to predict how currents distribute across various pathways and resistances within the circuit.

Exploring Currents in Circuits

Currents in circuits are essential to comprehending how electrical systems function. The current, measured in amperes, signifies the flow of electric charge through a circuit. The exercise focuses on finding three currents, \( I_1, I_2, \) and \( I_3 \), moving through different branches.

To solve for the currents, we manipulate the system of linear equations. Here's the streamlined process:

• First, express one current in terms of the others to simplify the equations. This step effectively reduces complexity.
• Next, substitute this expression into other equations. Solving these will yield specific values for the currents.

Each current value represents the precise rate of charge flow in a specified branch. For example, the calculated solutions yield \( I_1 = \frac{11}{28} \), \( I_2 = \frac{2}{7} \), and \( I_3 = \frac{19}{28} \). Understanding these values is paramount, providing insights into the functional status of the entire circuit.