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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°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°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°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°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°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°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

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100-W incandescent bulb uses only 23 W of power. The compact bulb lasts 10,000 hours, on the average, and costs \(11.00, whereas the incandescent bulb costs only \)0.75, but lasts just 750 hours. The study assumed that electricity costs $0.080 per kilowatt-hour and that the bulbs are on for 4.0 h per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100-W" fluorescent bulb? (Remember, it actually uses only 23 W of power and operates across 120 V.)

When a resistor with resistance \(R\) is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that \(R\) remains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0\(\Omega\) \(\cdot\) m. The conducting path between the hands can be represented approximately as a cylinder 1.6 m long and 0.10 m in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) What potential difference between the hands is needed for a lethal shock current of 100 mA? (Note that your result shows that small potential differences produce dangerous currents when the skin is damp.) (c) With the current in part (b), what power is dissipated in the body?

Unlike the idealized ammeter described in Section 25.4, any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_A\) is connected in series with a resistor \(R\) and a battery of emf \(\varepsilon\) and internal resistance r. The current measured by the ammeter is \(I_A\). Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of \(I_A\), \(r\), \(R_A\), and \(R\). The more "ideal" the ammeter, the smaller the difference between this current and the current IA. (b) If \(R\) = 3.80 \(\Omega\), \(\varepsilon\) = 7.50 V, and \(r\) = 0.45 \(\Omega\), find the maximum value of the ammeter resistance \(R_A\) so that \(I_A\) is within 1.0% of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a \(maximum\) value.

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce \(3.15 \times 10^4\) J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

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