Problem 29 - Algebra and Trigonometry with Analytic Geometry

Question

Electrical currents Shown in the figure is a schematic of an electrical circuit containing three resistors, a 6-volt battery, and a 12 -volt battery. It can be shown, using Kirchhoff's laws, that the three currents \(I_{1}, I_{2}\), and \(I_{3}\) are solutions of the following system of equations: $$ \left\\{\begin{array}{rlr} I_{1}-I_{2}+I_{3} & =0 \\ R_{1} I_{1}+R_{2} I_{2} & =6 \\ R_{2} I_{2}+R_{3} I_{3} & =12 \end{array}\right. $$ Find the three currents if (a) \(R_{1}=R_{2}=R_{3}=3\) ohms (b) \(R_{1}=4\) ohms, \(R_{2}=1 \mathrm{ohm}\), and \(R_{3}=4 \mathrm{ohms}\) Exercise 29

Step-by-Step Solution

Answer

(a) \(I_1 = 0\), \(I_2 = 2\), \(I_3 = 2\); (b) \(I_1 = 0\), \(I_2 = 6\), \(I_3 = 3\).

Step 1: Setting Up the Equations for Part (a)

For part (a), we are given that all resistances are equal, i.e., \(R_1 = R_2 = R_3 = 3\) ohms. We can substitute these values into the original system of equations:1. \( I_1 - I_2 + I_3 = 0 \)2. \( 3I_1 + 3I_2 = 6 \)3. \( 3I_2 + 3I_3 = 12 \)These are the equations we will solve for \(I_1, I_2,\) and \(I_3\).

Step 2: Solving the Equations for Part (a)

First, simplify the second and third equations by dividing through by 3:2. \( I_1 + I_2 = 2 \)3. \( I_2 + I_3 = 4 \)Now we have the simplified system:1. \( I_1 - I_2 + I_3 = 0 \)4. \( I_1 + I_2 = 2 \)5. \( I_2 + I_3 = 4 \)Solving equation 4 for \(I_1\):\( I_1 = 2 - I_2 \).Substitute \( I_1 = 2 - I_2 \) into equation 1:\( 2 - I_2 - I_2 + I_3 = 0 \), simplifying gives \( 2 - 2I_2 + I_3 = 0 \).Solving for \(I_3\):\( I_3 = 2I_2 - 2 \).Substitute \( I_3 = 2I_2 - 2 \) into equation 5:\( I_2 + 2I_2 - 2 = 4 \), which simplifies to \( 3I_2 = 6 \) so \( I_2 = 2 \).With \( I_2 = 2 \), then \( I_1 = 2 - I_2 = 0 \) and \( I_3 = 2I_2 - 2 = 2 \).

Step 3: Setting Up the Equations for Part (b)

For part (b), we have different resistances: \(R_1 = 4\) ohms, \(R_2 = 1\) ohm, and \(R_3 = 4\) ohms. Substitute these into the original equations:1. \( I_1 - I_2 + I_3 = 0 \)2. \( 4I_1 + I_2 = 6 \)3. \( I_2 + 4I_3 = 12 \)These are the equations we will solve for \(I_1, I_2,\) and \(I_3\).

Step 4: Solving the Equations for Part (b)

First, solve equation 2 for \(I_2\):\( I_2 = 6 - 4I_1 \).Substitute \( I_2 = 6 - 4I_1 \) into equation 3:\( (6 - 4I_1) + 4I_3 = 12 \), simplifying gives \( 4I_3 = 6 - 4 + 4I_1 \), which is \( 4I_3 = 6 + 4I_1 \).Solve for \(I_3\):\( I_3 = \frac{3 + 2I_1}{2} \).Substitute \( I_2 = 6 - 4I_1 \) and \( I_3 = \frac{6 + 4I_1}{4} \) into equation 1:\( I_1 - (6 - 4I_1) + \frac{3 + 2I_1}{2} = 0 \), solving this gives \( I_1 = 0 \).With \( I_1 = 0 \):\( I_2 = 6 - 4 \times 0 = 6 \) and \( I_3 = \frac{6}{4} = 3 \).

Key Concepts

Electrical Circuits

Electrical circuits are beautifully intricate systems that allow electricity to flow through a conductive path. Consider each part of a circuit, such as wires, resistors, and batteries, as a vital component that makes this flow possible. The circuit in our exercise consists of resistors connected to batteries, which serve as the power source.
What makes a circuit work is its continuity and the existence of a closed loop which allows electrons to pass from the negative end of the battery to the positive end. This journey generates a flow of electric charge known as current.
Understanding circuits requires knowledge of various components like:

• Resistors: Devices that resist the flow of current, causing an energy drop that is useful for controlling current levels.
• Batteries: Provide the voltage or electrical force needed to move charges through the circuit.
• Current Paths: Conductive paths that direct the current smoothly through the system.

The circuit analyzed using Kirchhoff's laws helps us understand how currents behave through different paths, making complex circuits manageable.

System of Equations

Solving a circuit problem often involves setting up a system of equations. This system helps us find unknown values, like currents through various paths. In our exercise, we used equations derived from Kirchhoff’s laws, which are fundamental principles for circuit analysis.
These laws are based on:

• Kirchhoff's Current Law (KCL): States that the sum of currents entering a junction equals the sum exiting. This is reflected in our first equation, where the sum of currents at a junction is set to zero.
• Kirchhoff's Voltage Law (KVL): States that the sum of voltage drops around any closed loop in a circuit equals zero. The second and third equations in the exercise are derived using this principle.

By substituting resistance values into these equations, we form a solvable set of linear equations, leading us to find the currents across various paths of the circuit. This method is powerful because it breaks down complex circuits into solvable parts.

Resistors

Resistors are crucial components that regulate the flow of current through a circuit. Imagine them as the speed bumps for electrical currents. In our exercise, different resistors were used to explore how their variations affect the total flow of currents.
Resistors are measured in ohms (Ω), and they abide by Ohm's Law, which states that the voltage across a resistor (V) equals the product of the current through it (I) and its resistance (R): \[ V = IR \].
Modeling resistors in equations helps us:

• Calculate current when voltage and resistance are known.
• Understand the voltage drop across each resistor due to the current flowing through.

For part (a) and (b) of the exercise, different resistances were considered to illustrate how they impact the circuit's behavior. Uniform resistors distribute voltage equally across paths, whereas variable resistors create more complex distributions.

Voltage

Voltage is the driving force that pushes electrical charges through a circuit. Think of it as the pressure in a water pipe that encourages water to flow. Voltage is measured in volts (V) and in our exercise, comes from two batteries of 6 volts and 12 volts.
Batteries convert chemical energy into electrical energy, providing the potential difference required for current flow. With Kirchhoff's Voltage Law in mind, we know the sum of all voltage drops in a closed loop equals the total voltage supplied by connected power sources.
Understanding voltage in circuits is essential for:

• Determining how much energy is supplied to and consumed by the components within the circuit.
• Calculating allowable currents so components like resistors do not overheat.
• Ensuring that devices connected to a circuit are functioning within safe voltage limits.

In any analysis, especially for safety and efficiency, recognizing the role of voltage helps ensure circuits operate correctly without causing damage to components or compromising function.