Problem 11 - Basic Engineering Mathematics

Question

The following values of resistance \(R\) ohms and corresponding voltage \(V\) volts are obtained from a test on a filament lamp. $$ \begin{array}{|l|lllrr|} \hline R \text { ohms } & 30 & 48.5 & 73 & 107 & 128 \\ V \text { volts } & 16 & 29 & 52 & 76 & 94 \\ \hline \end{array} $$ Choose suitable scales and plot a graph with \(R\) representing the vertical axis and \(V\) the horizontal axis. Determine (a) the gradient of the graph, (b) the \(R\) axis intercept value, (c) the equation of the graph, (d) the value of resistance when the voltage is \(60 \mathrm{~V}\), and (e) the value of the voltage when the resistance is \(40 \mathrm{ohms}\), (f) If the graph were to continue in the same manner, what value of resistance would be obtained at \(110 \mathrm{~V}\) ?

Step-by-Step Solution

Answer

(a) Gradient: \(\frac{49}{39}\). (b) Intercept: 8 ohms. (c) Eq: \(R=\frac{49}{39}V+8\). (d) 83.4 ohms. (e) 25.8 V. (f) 146.4 ohms at 110 V.

Step 1: Plot the Graph

First, plot the given data on a graph using a suitable scale. Let the voltage \( V \) be on the horizontal axis and the resistance \( R \) on the vertical axis. Plot the points: (16, 30), (29, 48.5), (52, 73), (76, 107), and (94, 128). Draw the best fit line through these points as the data implies linearity based on the instruction.

Step 2: Determine the Gradient

To find the slope (gradient) of the line, choose two points on the line that the line passes through. Here, use the first and last points (since they are on our line): \((V_1, R_1) = (16, 30)\) and \((V_2, R_2) = (94, 128)\). The gradient \( m \) is calculated as \[ m = \frac{R_2 - R_1}{V_2 - V_1} = \frac{128 - 30}{94 - 16} = \frac{98}{78} \]. Simplifying gives \( m = \frac{49}{39} \).

Step 3: Find the R-axis Intercept

To find the \( R \)-axis intercept, use the line equation \( R = mV + c \). Use one point, say \((16, 30)\), and the slope found to solve for \( c \): \[ 30 = \frac{49}{39} \times 16 + c \]. Solving for \( c \), we get \( c \approx 8 \).

Step 4: Write the Equation of the Graph

Using the gradient and intercept, the equation of the graph can be written as: \[ R = \frac{49}{39}V + 8 \].

Step 5: Calculate Resistance for 60 V

Substitute \( V = 60 \) into the equation \( R = \frac{49}{39}V + 8 \): \[ R = \frac{49}{39} \times 60 + 8 \]. This simplifies to \( R \approx 83.4 \) ohms.

Step 6: Calculate Voltage for 40 ohms

Rearrange the graph equation to solve for \( V \): \[ V = \frac{R - 8}{\frac{49}{39}} \]. Substituting \( R = 40 \), we get \( V \approx 25.8 \) volts.

Step 7: Calculate Resistance for 110 V

Substitute \( V = 110 \) into the equation \( R = \frac{49}{39}V + 8 \): \[ R = \frac{49}{39} \times 110 + 8 \]. This gives \( R \approx 146.4 \) ohms.

Key Concepts

Linear Equations

A linear equation is an algebraic expression that represents a straight line when graphed. It is usually in the form of \(y = mx + c\), where \(m\) is the slope, \(x\) is the independent variable, \(y\) is the dependent variable, and \(c\) is the y-intercept. In our exercise, we use the equation form \(R = mV + c\) because resistance \(R\) depends linearly on voltage \(V\).
By plotting the given data points and drawing a best fit line through them, we can determine that our values indeed form a linear relationship. This means as voltage increases, resistance also increases proportionately. Understanding this simple form of equation helps us analyze various scenarios that might appear in electrical circuits and other applications.

Resistance and Voltage Relationship

The relationship between resistance \(R\) and voltage \(V\) in this context illustrates how they are directly proportional when dealing with certain electrical components like filament lamps. This means when voltage increases, resistance also increases, which can be visualized as a straight line on a graph.
In practical terms, this relationship helps in predicting how changes in one element can affect the other. For example:

• If the voltage is set to a certain value, it becomes easy to calculate the corresponding resistance using the established linear equation.
• Similarly, given a resistance, one can determine the voltage required to achieve it.

This understanding is crucial in designing and troubleshooting electrical circuits.

Slope and Intercept Calculation

Calculating the slope and intercept is essential for defining the linear relationship in a graph. The slope, or gradient, \(m\) represents how steep the line is, calculated as \[ m = \frac{R_2 - R_1}{V_2 - V_1} \]. This tells us how quickly resistance changes with respect to voltage.
For this exercise, the slope was computed using the points \((16, 30)\) and \((94, 128)\), resulting in \( \frac{49}{39} \). This fraction reveals that for every 39 volts increase, resistance goes up by 49 ohms.
The intercept \(c\) is found by substituting a known point into the equation and solving for \(c\). This tells where the line crosses the R-axis. In this case, it was approximately 8, showing the resistance starts at 8 ohms when voltage is zero. Together, these values form the line equation \(R = \frac{49}{39}V + 8\).

Data Interpretation

Interpreting the results from a graph plotting exercise involves predicting values using the derived equation. Once you find the equation of the line, it becomes a tool to estimate answers for unknown values.
For example, to find resistance at 60 volts, substitute into the line equation \(R = \frac{49}{39}V + 8\), resulting in approximately 83.4 ohms. Conversely, to find voltage at a resistance of 40 ohms, rearrange to \( V = \frac{R - 8}{\frac{49}{39}} \), which results in roughly 25.8 volts.

• Solving these lets you predict behavior not explicitly mapped in the initial data.
• It enables you to extend the line graph beyond the measured points to anticipate values outside the recorded data range.

This ability is powerful in fields ranging from physics to engineering, where such predictions are routinely employed in system design and analysis.