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Chapter 5: Problem 14

Electric resistance in copper wire changes with the temperature of the wire. If \(C(t)\) is the electric resistance at temperature \(t\), in degrees Fahrenheit, then the resistance ratio \(C(t) / C(0)\) can be measured. $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Temperature } t \\ \text { in degrees } \end{array} & \frac{C(t)}{C(0)} \text { ratio } \\ \hline 0 & 1 \\ \hline 10 & 1.0393 \\ \hline 20 & 1.0798 \\ \hline 30 & 1.1215 \\ \hline 40 & 1.1644 \\ \hline \end{array} $$ a. On the basis of the data in the table, explain why the ratio \(C(t) / C(0)\) can be reasonably modeled by a quadratic function. b. Find a quadratic formula for the ratio \(C(t) / C(0)\) as a function of temperature \(t\). c. At what temperature is the electric resistance double that at 0 degrees? d. Suppose that you have designed a household appliance to be used at room temperature ( 72 degrees) and you need to have the wire resistance inside the appliance accurate to plus or minus \(10 \%\) of the predicted resistance at 72 degrees. i. What resistance ratio do you predict at 72 degrees? (Use four decimal places.) ii. What range of resistance ratios represents plus or minus \(10 \%\) of the resistance ratio for 72 degrees? iii. What temperature range for the appliance will ensure that your appliance operates within the \(10 \%\) tolerance? Is this range reasonable for use inside a home?

Short Answer

Expert verified
a quadratic model fits the data. \[ \frac{C(t)}{C(0)} = 0.0004t^2 + 0.0039t + 1 \] The resistance doubles at 49.03掳F and at 72掳F, it is 1.2267 with a 10% tolerance ranging from 1.1040 to 1.3494, meaning 20掳F to 110掳F is acceptable.

Step by step solution

01

Analyzing the Data

The table shows that the resistance ratio \(\frac{C(t)}{C(0)}\) steadily increases as temperature \(t\) increases. We need to verify if a quadratic model fits this data. Quadratic functions exhibit a constant rate of change in their first derivative; if the second differences are roughly constant, it indicates a quadratic relation.
02

Calculating Second Differences

Given the resistance ratios for temperatures 0, 10, 20, 30, and 40, we calculate the first differences as follows: \(1.0393 - 1 = 0.0393\), \(1.0798 - 1.0393 = 0.0405\), \(1.1215 - 1.0798 = 0.0417\), \(1.1644 - 1.1215 = 0.0429\). Now, calculate second differences: \(0.0405-0.0393 = 0.0012\), \(0.0417-0.0405 = 0.0012\), \(0.0429-0.0417 = 0.0012\). The second differences are constant, supporting a quadratic model.
03

Creating the Quadratic Model

Assuming \(\frac{C(t)}{C(0)} = at^2 + bt + c\), we know: When \(t = 0\), \(\frac{C(t)}{C(0)} = 1\), hence \(c = 1\). Using the values for \(t = 10\) and \(t = 20\), we solve the system: 1.0393 = 100a + 10b + 1 1.0798 = 400a + 20b + 1. Solving these yields \(a = 0.0004\) and \(b = 0.0039\). Thus, our model is: \[\frac{C(t)}{C(0)} = 0.0004t^2 + 0.0039t + 1\]
04

Finding the Temperature for Double Resistance

We set the ratio to 2: \[\frac{C(t)}{C(0)} = 2\] \[0.0004t^2 + 0.0039t + 1 = 2\] \[0.0004t^2 + 0.0039t - 1 = 0\]. Using the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\], after calculation, \(t = 49.03\) degrees.
05

Calculating Resistance Ratio at 72 Degrees

Substituting \(t = 72\) into the model: \[\frac{C(72)}{C(0)} = 0.0004 \times 72^2 + 0.0039 \times 72 + 1\] Calculating this gives \(\frac{C(72)}{C(0)} = 1.2267\) (4 decimal places).
06

Determining the Acceptable Resistance Range

To find \(\pm 10\%\) of the resistance ratio at 72 degrees: Multiply 1.2267 by 0.10, yielding 0.1227. Hence, the range is: \[1.1040 \leq \frac{C(t)}{C(0)} \leq 1.3494\].
07

Finding Temperature Range for 10% Tolerance

Using the quadratic model, solve for \(t\) when \(\frac{C(t)}{C(0)} = 1.1040\) and \(1.3494\). First solve \(0.0004t^2 + 0.0039t + 1 = 1.1040\), yielding \(t \approx 20.5\). Then solve \(0.0004t^2 + 0.0039t + 1 = 1.3494\), yielding \(t \approx 110.5\). Thus, the appliance will function properly from approximately 20掳 to 110掳 F. This range is reasonable for home use as it includes typical room temperatures.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Resistance
Electric resistance is a fundamental property of electrical circuits, influencing how easily electricity can flow through materials. It is primarily determined by the material's properties, like its length, cross-sectional area, and temperature. For conducting materials, such as copper, resistance tends to increase with a rise in temperature. When a wire heats up, the atoms in the material vibrate more intensely, making it harder for electrons to flow through the wire without scattering. This scattering increases the electric resistance. Therefore, understanding the relationship between temperature and resistance is key in designing electrical appliances.

In our example, copper wire is used for its excellent conductivity. However, even the best conductors experience changes in resistance with temperature shifts, which must be accounted for in practical applications, such as household wiring or electronic devices, to ensure safety and efficiency.
Temperature Dependence
The dependence of a material's electric resistance on temperature is an important consideration when modeling electric circuits and components. As seen in the table from the exercise, as temperature increases, the resistance ratio \(\frac{C(t)}{C(0)}\) also increases, indicating a direct influence of temperature on resistance.

This relationship is often nonlinear, and for some materials, including metals like copper, it can be closely approximated by a quadratic function. This quadratic temperature dependence is due to the physical changes in the material's structure at an atomic level as it heats. As a result, engineers and scientists frequently model the temperature-dependent behavior of resistance using polynomial equations, particularly quadratics, which can effectively capture the gradual and consistent increase in resistance with temperature.
Modeling Data
Modeling data involves creating an equation that describes a set of observed data points. In the context of our exercise, we are looking at resistance ratios at various temperatures. To verify if our data can be modeled by a quadratic function, we observe the second differences of the data points. If these second differences are relatively constant, it suggests that a quadratic function is an appropriate model.

In our dataset, by analyzing the differences between consecutive resistance ratios, we calculated constant second differences, indicating a quadratic relation. This led us to formulate the quadratic equation \[\frac{C(t)}{C(0)} = 0.0004t^2 + 0.0039t + 1\].

Modeling data accurately helps in predicting future values. In this case, it allows us to predict resistance at temperatures not explicitly measured in the original data set.
Quadratic Equation Solutions
Quadratic equations have the general form \[ax^2 + bx + c = 0\]. Solutions to these equations can be found using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]. This formula enables us to find exact solutions for the temperature where certain resistance conditions are met.

In our example, we sought the temperature at which resistance is twice its initial value, leading us to set the ratio \[\frac{C(t)}{C(0)} = 2\]. Solving the resulting quadratic equation gave us a temperature of 49.03掳F.

Furthermore, using the quadratic model, we can also ensure appliances operate within certain tolerance levels for safety and efficacy. By solving quadratic equations, we determine the temperature range that maintains resistance within a specified tolerance, allowing for practical and safe appliance design under varying environmental conditions.

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Most popular questions from this chapter

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