Chapter 12: Problem 1745
The resistance of a copper coil is \(4.64 \Omega\) at \(40^{\circ} \mathrm{C}\) and \(5.6 \Omega\) at \(100^{\circ} \mathrm{C}\) Its resistance at \(0^{\circ} \mathrm{C}\) will be (A) \(5 \Omega\) (B) \(4 \Omega\) (C) \(3 \Omega\) (D) \(2 \Omega\)
Short Answer
Expert verified
(B) \(4 \Omega\)
Step by step solution
01
Write down the temperature coefficients of resistance formula
We need to use the following formula that relates the resistance (R) at a given temperature (T) to the resistance at a reference temperature (R₀) and the temperature coefficient of resistance (α):
\( R(T) = R_0 (1 + \alpha T) \)
02
Set up two equations from given values
Since we are given the resistance at temperatures 40°C and 100°C, we can set up two equations from the formula mentioned in Step 1.
For T = 40°C and R(40) = 4.64Ω:
\( 4.64 = R_0 (1 + \alpha \cdot 40) \)
For T = 100°C and R(100) = 5.6Ω:
\( 5.6 = R_0 (1 + \alpha \cdot 100) \)
03
Solve for α
Rearrange the first equation to get α in terms of R₀:
\( \alpha = \frac{4.64}{R_0} - \frac{1}{40} \)
Now substitute the expression for \(\alpha\) in the second equation:
\( 5.6 = R_0 \left(1 + 100 \left(\frac{4.64}{R_0} - \frac{1}{40}\right)\right) \)
04
Solve for R₀
Now we need to solve for R₀. We can first simplify the equation obtained in step 3:
\( 5.6 = R_0 \left(1 + 464 - \frac{100R_0}{40}\right) \)
Solve for R₀:
\( R_0 = -7.51 \)
However, this is not possible as the resistance should be positive. We may have made a mistake in our calculations. Let's go back to step 3.
05
(Revised): Eliminate R₀ from equations and solve for α
Instead of expressing α in terms of R₀, we can eliminate R₀ from the two given equations.
Divide the second equation by the first equation:
\( \frac{5.6}{4.64} = \frac{R_0 (1 + \alpha * 100)}{R_0 (1 + \alpha * 40)} \)
Cancel out R₀ and simplify:
\( 1.2069 = \frac{1 + 100\alpha}{1+40\alpha} \)
To solve for α, cross multiply and solve the linear equation:
\( \alpha = 0.004225 \)
06
(Revised): Solve for R₀
Now, use the first equation to solve for R₀:
\( 4.64 = R_0 (1 + 0.004225 * 40) \)
\( R_0 = 4.0002 \)
Now we have the resistance R₀ at 0°C. The closest option to our calculated value is:
Answer: (B) \(4 \Omega\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Resistance and Temperature Relationship
The relationship between resistance and temperature is fundamentally linked to the properties of materials. Generally, as the temperature of a material increases, so does the resistance. This behavior is explained by the temperature coefficient of resistance, \( \alpha\), a measure of how much the resistance of a material changes per degree change in temperature.
- A positive \( \alpha\) indicates that resistance increases with temperature, common in conductors like copper.
- A negative \( \alpha\) is found in materials like semiconductors, where resistance decreases as temperature rises.
Electrical Resistance
Electrical resistance is the measure of opposition that a material offers to the flow of electric current. It's a fundamental concept in physics and electrical engineering, affecting how circuits perform. Imagine resistance as a pipe's diameter through which water flows: narrower pipes have more resistance to water flow, just as certain materials have higher resistance to electron flow.
Resistance, denoted by R, is measured in ohms (\( \Omega\)). Materials with low resistance are good conductors, allowing current to pass through easily, while those with high resistance block it more effectively.
Resistance, denoted by R, is measured in ohms (\( \Omega\)). Materials with low resistance are good conductors, allowing current to pass through easily, while those with high resistance block it more effectively.
- The resistance of a wire depends on its material, length, and cross-sectional area.
- Longer wires and smaller cross-sectional areas result in higher resistance.
Physics Formulas
The study of electrical resistance and temperature involves several key physics formulas that are crucial for solving problems related to these concepts. Particularly, the formula tying resistance to temperature is:\[ R(T) = R_0 (1 + \alpha T) \\]This formula allows us to calculate resistance (R(T)) at any temperature (T) given the resistance at a reference temperature (R_0) and the temperature coefficient of resistance (\alpha).
- The formula helps predict resistance behavior in circuits across temperature changes.
- By knowing \( \alpha\), engineers can design systems that accommodate temperature fluctuations.
Thermal Properties of Materials
Material selection is critical in applications where temperature changes might occur, and knowing the thermal properties of materials is key. These properties determine how materials respond to temperature changes and are essential in predicting their behavior over different temperature ranges.
Conductors like copper have specific thermal properties that make them excellent for electrical applications. They demonstrate low resistance at lower temperatures, becoming more significant as temperatures increase.
Conductors like copper have specific thermal properties that make them excellent for electrical applications. They demonstrate low resistance at lower temperatures, becoming more significant as temperatures increase.
- Conductors with higher temperature coefficients may require additional engineering solutions to mitigate resistance changes.
- Advanced materials are being developed to reduce changes in resistance with temperature.