Problem 46 - Elementary and Intermediate Algebra

Question

Solve each problem by writing a variation model. The power (in watts) lost in a resistor (in the form of heat) varies directly as the square of the current (in amperes) passing through it. The constant of proportionality is the resistance (in ohms). What power is lost in a 5 -ohm resistor carrying a 3 -ampere current?

Step-by-Step Solution

Answer

45 watts of power is lost in the 5-ohm resistor.

Step 1: Understand the problem

We need to find the power lost in a resistor. The power lost varies directly with the square of the current and is proportional to the resistance.

Step 2: Use the direct variation formula

The problem states that power, \( P \), varies directly with the square of the current, \( I^2 \), and the constant of proportionality is the resistance, \( R \). This can be expressed as the formula: \( P = R \times I^2 \).

Step 3: Substitute the known values

We are given that the resistance \( R = 5 \) ohms and the current \( I = 3 \) amperes. Substitute these values into the formula: \( P = 5 \times (3)^2 \).

Step 4: Calculate the power

First, calculate \( 3^2 = 9 \). Then multiply by the resistance: \( P = 5 \times 9 = 45 \). Therefore, the power lost is 45 watts.

Key Concepts

Understanding the Power and Resistance Relationship

In the world of electrical circuits, understanding the relationship between power and resistance is crucial. Power, in this context, refers to the energy required or lost over time in a circuit. When a current passes through a resistor, some energy is lost in the form of heat. This phenomenon is essential as it directly relates to the safety and efficiency of electrical appliances.
The relationship between power and resistance can be established using Ohm's Law. According to this principle, the power dissipation in a resistor is calculated by multiplying the square of the current by the resistance: \( P = R \times I^2 \). This relationship means that the power dissipated increases with both resistance and the square of the current.

• A higher resistance means more power is lost.
• Similarly, a higher current results in exponentially increased power loss due to the square relationship.

This concept is vital in designing circuits, as it helps assure that the components can handle the power they need to dissipate, thus preventing overheating or failure.

Exploring the Direct Variation Formula

The concept of direct variation is a straightforward yet powerful mathematical model. Understanding this assists in describing how changes in one variable affect another directly and proportionally. This is fundamental in circuit analysis, particularly when calculating power loss.
The statement that 'power varies directly with the square of the current' translates mathematically into a direct variation formula. Here, the equation \( P = R \times I^2 \) represents the power \( P \) directly varying with the square of the current \( I^2 \), whereby the resistance \( R \) acts as the constant of proportionality.
Direct variation means that if you double the current, the power increases by a factor of four, since the current is squared. This sensitivity of power to current changes imposes a need for careful control in practical applications. By understanding this formula, students can predict how different resistors or currents will affect the overall circuit performance, which is crucial for safe and efficient electrical system design.

Calculating Resistor Power Loss

Resistor power loss is an inevitable and critical factor in circuit design and analysis. When electricity flows through a resistor, it encounters opposition, converting some energy into heat. This power lost as heat can affect the performance and longevity of electronic devices.
To calculate resistor power loss, the direct variation formula \( P = R \times I^2 \) is often utilized. By knowing the resistance and current, you can easily determine the power loss. For instance, if a resistor with a resistance of 5 ohms carries a current of 3 amperes, you replace the known values in the formula: \( P = 5 \times (3)^2 \). Given \( (3)^2 \) equals 9, then multiplying by 5 ohms equals a power loss of 45 watts.
Successfully calculating the power loss helps engineers to ensure that resistors are adequately rated for their applications, preventing overheating and potential damage. It also aids in optimizing the energy efficiency of electronic devices by accounting for these losses in the overall design.