Problem 17 - Basic Engineering Mathematics

Question

The resistance \(R \Omega\) of a length of wire at \(t^{\circ} \mathrm{C}\) is given by \(R=R_{0}(1+\alpha t)\), where \(R_{0}\) is the resistance at \(0^{\circ} \mathrm{C}\) and \(\alpha\) is the temperature coefficient of resistance in \({ }^{\circ} \mathrm{C}\). Find the values of \(\alpha\) and \(R_{0}\) if \(R=30 \Omega\) at \(50^{\circ} \mathrm{C}\) and \(R=\) \(35 \Omega\) at \(100^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

\( R_0 \approx 29.83 \Omega \), \( \alpha \approx 0.00335 \).

Step 1: Understand the Formula

The resistance of a wire at a given temperature is described by the formula: \( R = R_0(1 + \alpha t) \). Here, \( R \) is the resistance at temperature \( t \), \( R_0 \) is the resistance at \( 0^{\circ} \mathrm{C} \), and \( \alpha \) is the temperature coefficient of resistance.

Step 2: Set Up System of Equations

Use the given conditions to establish two equations. For \( t = 50^{\circ} \mathrm{C} \), \( R = 30 \Omega \), so we write: \( 30 = R_0(1 + 50\alpha) \). For \( t = 100^{\circ} \mathrm{C} \), \( R = 35 \Omega \), leading to: \( 35 = R_0(1 + 100\alpha) \).

Step 3: Solve the Equations for \( R_0 \) and \( \alpha \)

Subtract the first equation from the second to eliminate \( R_0 \). \( 35 - 30 = R_0(1 + 100\alpha) - R_0(1 + 50\alpha) \). Simplifying, we get: \( 5 = 50R_0\alpha \). Thus, \( \alpha = \frac{5}{50R_0} = \frac{1}{10R_0} \).

Step 4: Substitute \( \alpha \) Back

Use \( \alpha = \frac{1}{10R_0} \) in either original equation to find \( R_0 \). Substitute into \( 30 = R_0(1 + 50 \alpha) \), replacing \( \alpha \) by \( \frac{1}{10R_0} \). We get: \( 30 = R_0 \left(1 + \frac{50}{10R_0}\right) \), which simplifies to \( 30R_0 = R_0^2 + 5 \).

Step 5: Solve for \( R_0 \)

The equation \( 30R_0 = R_0^2 + 5 \) can be rearranged to \( R_0^2 - 30R_0 + 5 = 0 \). Use the quadratic formula: \( R_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -30, c = 5 \). Calculate \( R_0 = \frac{30 \pm \sqrt{900 - 20}}{2} = \frac{30 \pm \sqrt{880}}{2} \).

Step 6: Simplify and Select \( R_0 \)

Solve \( \sqrt{880} \approx 29.66 \), then \( R_0 \approx \frac{30 + 29.66}{2} \) or \( R_0 \approx \frac{30 - 29.66}{2} \). Since \( R_0 \) must be positive, choose \( R_0 \approx 29.83 \).

Step 7: Find \( \alpha \)

With \( R_0 \approx 29.83 \), calculate \( \alpha = \frac{1}{10R_0} \approx \frac{1}{298.3} \approx 0.00335 \).

Step 8: Verify the Solution

Verify the values by substituting \( R_0 \approx 29.83 \) and \( \alpha \approx 0.00335 \) back into both conditions: \( 30 \approx 29.83(1 + 50 \times 0.00335) \) and \( 35 \approx 29.83(1 + 100 \times 0.00335) \). Both conditions hold true, confirming the solution.

Key Concepts

Temperature Coefficient of Resistance

The temperature coefficient of resistance, often represented by \( \alpha \), is a parameter that quantifies how much the resistance of a material changes with temperature. It's a crucial factor in electrical engineering as materials do not maintain constant resistance when the temperature changes. This coefficient helps predict the performance of electrical circuits across different temperatures.

This coefficient is

• Specific to the material.
• Typically provided in units of per degree Celsius \( (^{\circ} \text{C}^{-1}) \).
• Used to adjust resistance calculations from a baseline temperature, usually \( 0^{\circ} \text{C} \).

In most conductors, as the temperature increases, resistance increases because atoms vibrate more vigorously, interrupting the flow of electrons. Understanding and using \( \alpha \) allows engineers to design circuits that accommodate these changes.

Resistance Calculation

Resistance in wires varies with temperature changes and is calculated using the formula \( R = R_0(1 + \alpha t) \). This formula determines the resistance \( R \) at any temperature \( t \) based on a known resistance \( R_0 \) at a standard temperature, commonly \( 0^{\circ} \text{C} \).

The equation is structured as follows:

• \( R \): Resistance at temperature \( t \).
• \( R_0 \): Resistance at the reference temperature (typically \( 0^{\circ} \text{C} \)).
• \( \alpha \): Temperature coefficient of resistance.
• \( t \): Temperature difference from the baseline.

Calculating resistance with this formula involves substituting known values of \( \alpha \), \( t \), and \( R_0 \) to find \( R \). It's essential for designing and analyzing circuits under varying thermal conditions. This fundamental concept ensures that devices operate efficiently despite environmental temperature changes.

System of Equations

To find unknowns such as \( R_0 \) and \( \alpha \), we often use a system of equations, a set of equations with multiple variables that are solved together. In our example, we derived two equations using given temperatures and resistance values:

• For \( t = 50^{\circ} \text{C} \), the equation is \( 30 = R_0(1 + 50\alpha) \).
• For \( t = 100^{\circ} \text{C} \), the equation is \( 35 = R_0(1 + 100\alpha) \).

Solving these equations simultaneously helps in isolating and calculating the values of both \( \alpha \) and \( R_0 \).

A common technique is to subtract one equation from another to eliminate one variable. This method effectively reduces complexity and allows easier computation of the remaining variable. Once one of the variables is determined, it can be substituted back into one of the original equations to find the other value. Using this method ensures accurate and reliable results when multiple conditions need to be considered.

Quadratic Formula

The quadratic formula is a mathematical tool used to find the roots of a quadratic equation of the form \( ax^2 + bx + c = 0 \). This equation often appears in resistance problems requiring precise calculations.

In our exercise, after manipulating the equations, we arrived at the quadratic equation \( R_0^2 - 30R_0 + 5 = 0 \). Here, the quadratic formula \( R_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is applied where:

• \( a = 1 \)
• \( b = -30 \)
• \( c = 5 \)

Using the formula helps find solutions for \( R_0 \). After obtaining \( R_0 \), other parameters can be calculated by substitution.

The quadratic formula is crucial when simpler algebraic methods do not suffice, particularly when the equation cannot be factored easily. It provides an exact solution, which is critical in fields like electronics where precise measurements are necessary.