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Chapter 25: Problem 11

A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0\(^\circ\)C) the ammeter reads 18.5 A, while at 92.0\(^\circ\)C it reads 17.2 A. You can ignore any thermal expansion of the rod. Find (a) the resistivity at 20.0\(^\circ\)C and (b) the temperature coefficient of resistivity at 20\(^\circ\)C for the material of the rod.

Short Answer

Expert verified
(a) Resistivity at 20.0掳C is \(1.060 \times 10^{-6} \Omega \cdot \text{m}\). (b) Temperature coefficient is \(1.042 \times 10^{-3} \text{ C}^{-1}\).

Step by step solution

01

Understand Ohm's Law

Ohm's Law relates voltage (\( V \)), current (\( I \)), and resistance (\( R \)) as \( V = IR \). For both temperatures, we can find the resistance using this equation, since voltage and current are given.
02

Calculate Resistance at 20.0掳C

Using Ohm's Law, \( R_{20} = \frac{V}{I_{20}} \), where \( V = 15.0 \text{ V} \) and \( I_{20} = 18.5 \text{ A} \).\[ R_{20} = \frac{15.0 \text{ V}}{18.5 \text{ A}} = 0.8108 \Omega \]
03

Calculate Resistance at 92.0掳C

Similarly, \( R_{92} = \frac{V}{I_{92}} \), where \( V = 15.0 \text{ V} \) and \( I_{92} = 17.2 \text{ A} \).\[ R_{92} = \frac{15.0 \text{ V}}{17.2 \text{ A}} = 0.8721 \Omega \]
04

Find the Resistivity at 20.0掳C

Use the formula for resistance \( R = \rho \frac{L}{A} \), where \( \rho \) is resistivity, \( L = 1.50 \text{ m} \), and \( A = \pi \left(\frac{d}{2}\right)^2 \), with \( d = 0.500 \text{ cm} = 0.005 \text{ m} \).First calculate area: \[ A = \pi \left(\frac{0.005}{2}\right)^2 = 1.9635 \times 10^{-5} \text{ m}^2 \]Then, using \( R_{20} = 0.8108 \Omega \):\[ 0.8108 = \rho \frac{1.50}{1.9635 \times 10^{-5}} \]Solving for resistivity \( \rho \):\[ \rho = \frac{0.8108 \times 1.9635 \times 10^{-5}}{1.50} = 1.060 \times 10^{-6} \Omega \cdot \text{m} \]
05

Calculate Temperature Coefficient of Resistivity

The temperature coefficient of resistivity \( \alpha \) can be found using the formula:\[ R_{92} = R_{20}(1 + \alpha \Delta T) \]Where \( \Delta T = 92.0 - 20.0 = 72.0 \text{ C} \).Rearrange to solve for \( \alpha \):\[ 0.8721 = 0.8108 (1 + \alpha \times 72.0) \]\[ \alpha = \frac{0.8721/0.8108 - 1}{72.0} = 1.042 \times 10^{-3} \text{ C}^{-1} \]

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in physics that describes the relationship between voltage, current, and resistance in an electrical circuit. It is often stated as the equation \( V = IR \), where:
  • \( V \) is the voltage across the component in volts (V),
  • \( I \) is the current flowing through the component in amperes (A),
  • \( R \) is the resistance of the component in ohms (\( \Omega \)).
This law implies that if you increase the voltage across a conductor, the current will increase, provided the resistance is constant. Conversely, if the resistance increases (say due to changes in temperature or material properties), the current will decrease for a given voltage.
In the given problem, we utilized Ohm's Law to find the resistance of the cylindrical rod at different temperatures using the constant voltage provided by the power supply and the changing current measured by the ammeter.
Temperature Coefficient of Resistivity
The temperature coefficient of resistivity, denoted by \( \alpha \), is a measure of how much a material's resistivity changes with temperature. For most conductors, resistivity increases with temperature. This coefficient is particularly useful for determining how a material's resistance will change under different thermal conditions.
The formula which incorporates this concept is:
  • \( R = R_{0}(1 + \alpha \Delta T) \)
where:
  • \( R \) is the resistance at temperature \( T \),
  • \( R_{0} \) is the resistance at a reference temperature (usually 20掳C),
  • \( \alpha \) is the temperature coefficient of resistivity,
  • \( \Delta T \) is the temperature change.
In this exercise, we determined \( \alpha \) by comparing the resistance of the rod at room temperature and at 92.0掳C. By understanding how resistivity changes with temperature, engineers and scientists can design circuits and materials that perform reliably across a range of conditions.
Cylindrical Rod
A cylindrical rod is a common geometric shape used in physics and engineering to simplify models of real-world objects like wires or pipelines. When analyzing the electrical properties of a rod, we focus on its length, diameter, and cross-sectional area. These factors help determine the resistance of the rod using the formula:
  • \( R = \rho \frac{L}{A} \)
where:
  • \( R \) is the resistance,
  • \( \rho \) is the resistivity of the material,
  • \( L \) is the length of the rod,
  • \( A \) is the cross-sectional area, calculated as \( A = \pi (\frac{d}{2})^2 \)
In the given exercise, the rod's diameter and length are used to find its cross-sectional area, which is crucial for calculating the resistivity. Understanding these properties allows us to predict how changes in physical dimensions and material properties will affect a rod's electrical behavior.

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Most popular questions from this chapter

A current-carrying gold wire has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4-m length of this wire?

An external resistor with resistance \(R\) is connected to a battery that has emf \(\varepsilon\) and internal resistance \(r\). Let \(P\) be the electrical power output of the source. By conservation of energy, \(P\) is equal to the power consumed by \(R\). What is the value of \(P\) in the limit that \(R\) is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when \(R = r\). What is this maximum \(P\) in terms of \(\varepsilon\) and \(r\)? (d) A battery has \(\varepsilon\) = 64.0 V and \(r =\) 4.00 \(\Omega\). What is the power output of this battery when it is connected to a resistor \(R\), for \(R =\) 2.00 \(\Omega\), \(R =\) 4.00 \(\Omega\), and \(R =\) 6.00 \(\Omega\) ? Are your results consistent with the general result that you derived in part (b)?

A cylindrical copper cable 1.50 km long is connected across a 220.0-V potential difference. (a) What should be its diameter so that it produces heat at a rate of 90.0 W? (b) What is the electric field inside the cable under these conditions?

If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area \(A\) of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread. \(A_{13}\) mm is about (a) \\(\frac{1}{10}\\) \(A_5\) mm; (b) \\(\frac{1}{_4}\\) \(A_5\) mm; (c) \\(\frac{2}{_5}\\) \(A_5\) mm; (d) the same as \(A_5\) mm.

A resistor with resistance \(R\) is connected to a battery that has emf 12.0 V and internal resistance \(r =\) 0.40\(\Omega\). For what two values of \(R\) will the power dissipated in the resistor be 80.0 W?
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