Problem 2 - Basic Engineering Mathematics

Question

The power dissipated by a resistor was measured for varying values of current flowing in the resistor and the results are as shown: \begin{tabular}{|l|cccccc|} \hline Current, \(I\) amperes & \(1.4\) & \(4.7\) & \(6.8\) & \(9.1\) & \(11.2\) & \(13.1\) \\\ Power, \(P\) watts & 49 & 552 & 1156 & 2070 & 3136 & 4290 \\ \hline \end{tabular} Prove that the law relating current and power is of the form \(P=R I^{n}\), where \(R\) and \(n\) are constants, and determine the law. Hence calculate the power when the current is 12 amperes and the current when the power is 1000 watts.

Step-by-Step Solution

Answer

The law is \( P = 35.0 I^2 \). At 12 A, power is 5040 W; at 1000 W, current is 5.34 A.

Step 1: Understand the relationship

We're given the relationship \( P = R I^n \) where \( R \) and \( n \) are constants. The goal is to find \( R \) and \( n \) such that this equation holds true for the given data set.

Step 2: Linearize the equation

Take the logarithm of both sides of the equation \( P = R I^n \), resulting in \( \log P = \log R + n \log I \). This transforms the equation into a linear form where \( \log P \) is the dependent variable and \( \log I \) is the independent variable.

Step 3: Calculate logarithms for the data

Calculate \( \log I \) and \( \log P \) for each pair \( (I, P) \) in the given data table. Use base 10 for log calculations: - \( I = 1.4, \log I \approx 0.146 \), \( P = 49, \log P \approx 1.69 \)- \( I = 4.7, \log I \approx 0.672 \), \( P = 552, \log P \approx 2.742 \)- \( I = 6.8, \log I \approx 0.833 \), \( P = 1156, \log P \approx 3.063 \)- \( I = 9.1, \log I \approx 0.959 \), \( P = 2070, \log P \approx 3.315 \)- \( I = 11.2, \log I \approx 1.049 \), \( P = 3136, \log P \approx 3.496 \)- \( I = 13.1, \log I \approx 1.117 \), \( P = 4290, \log P \approx 3.633 \)

Step 4: Determine the slope (n) and intercept (log R)

Plot \( \log P \) against \( \log I \) and perform a linear regression to find the line of best fit. This line is of the form \( y = mx + c \), corresponding to \( \log P = n \log I + \log R \), where the slope \( n \approx 2 \) and the y-intercept \( \log R \approx 1.544 \).

Step 5: Solve for R

To find \( R \), use the equation \( \log R = 1.544 \). Therefore, \( R = 10^{1.544} \approx 35.0 \). So the relationship is \( P = 35.0 I^2 \).

Step 6: Calculate the power for I = 12 A

Substitute \( I = 12 \) into \( P = 35.0 I^2 \). \[ P = 35.0 \times 12^2 = 35.0 \times 144 = 5040 \] Thus, the power is 5040 watts.

Step 7: Calculate the current for P = 1000 W

Use the equation \( 1000 = 35.0 I^2 \) and solve for \( I \).\[ I^2 = \frac{1000}{35.0} \approx 28.57 \]\[ I = \sqrt{28.57} \approx 5.34 \]Therefore, the current is approximately 5.34 amperes.

Key Concepts

Resistor Power Calculation

The power calculation in a resistor is all about understanding how electrical energy is converted into heat. This happens when current flows through the resistor, leading to the formula we're using here: \(P = R I^n\). Power \(P\) represents the energy dissipated as heat, and current \(I\) is the flow of electric charge that passes through. "\(R\)" is the resistance: a measure of how easily current can flow. The constant "\(n\)" in our equation represents the nature of the relation between the power, resistance, and current.

Resistor power calculations help us determine how much power is being used by a device, which is crucial for selecting components that can handle the generated heat without damage. Knowing the power dissipation helps in designing circuits that are both safe and efficient. It ensures that all parts, like resistors, can safely withstand the amount of current and voltage they encounter in everyday use.

Logarithmic Transformation

A logarithmic transformation is a mathematical tool that helps us tackle complex, non-linear relationships. It's particularly useful when dealing with equations where a variable is raised to the power of another, such as in \(P = R I^n\). By applying the logarithm to this equation, we can simplify it into a linear form: \(\log P = \log R + n \log I\).

The main advantage of transforming the data logarithmically is that it makes calculating and interpreting relationships more straightforward. In this scenario, the transformation converts power and current values to a log-scale, turning the original non-linear equation into a linear one. This allows us to use simple linear regression techniques to find \"\(R\)\" and \"\(n\)\".

• Logarithmic transformations can also help stabilize the variance of the data.
• They make the data conform more closely to the assumptions underlying linear regression.

This technique often results in better model fit and simpler interpretation.

Linear Regression Analysis

Linear regression analysis is a statistical method used for modeling the relationship between a dependent variable and one or more independent variables.

In our problem, we use it on log-transformed data to determine the values of \(n\) and \(R\). By plotting \(\log P\) against \(\log I\), we address it as a linear equation \(\log P = n \log I + \log R\), where:

• The slope of the line represents the exponent \(n\), dictating how current affects power.
• The y-intercept gives us \(\log R\), allowing calculation of \(R\) using the antilog.

Performing linear regression analysis helps us derive these constants with accuracy by minimizing the differences between observed values and those predicted by our linear model. This tool is a cornerstone of empirical modeling, providing valuable insights into data trends and relationships.