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Chapter 17: Problem 28

A wire \(3.00 \mathrm{~m}\) long and \(0.450 \mathrm{~mm}^{2}\) in cross-sectional area has a resistance of \(41.0 \Omega\) at \(20^{\circ} \mathrm{C}\). If its resistance increases to \(41.4 \Omega\) at \(29.0^{\circ} \mathrm{C}\), what is the temperature coefficient of resistivity?

Short Answer

Expert verified
The temperature coefficient of resistivity is \(0.0011 / ^{\circ}C\).

Step by step solution

01

Identify Given Values

We are given: Initial length = \(3.00 m\); Cross-sectional area = \(0.450 mm^{2}\); Initial resistance \(R_1 = 41.0 \Omega\) at \(T_1 = 20^{\circ}C\); Final resistance \(R_2 = 41.4 \Omega\) at \(T_2 = 29.0^{\circ}C\).
02

Calculate Change in Resistance

We first calculate the change in resistance, \(\Delta R\), using the formula: \(\Delta R = R_2 - R_1\). This yields \(\Delta R = 41.4 \Omega - 41.0 \Omega = 0.4 \Omega\).
03

Calculate Change in Temperature

Next, we calculate the change in temperature, \(\Delta T\), using the formula: \(\Delta T = T_2 - T_1\). This yields \(\Delta T = 29.0^{\circ}C - 20^{\circ}C = 9.0^{\circ}C\).
04

Find the Temperature Coefficient of Resistivity

The temperature coefficient of resistivity, \(\alpha\), can be found using the formula: \(\alpha = \frac{\Delta R}{R_1 \cdot \Delta T}\). Substituting the found values, we get: \(\alpha = \frac{0.4 \Omega}{41.0 \Omega \cdot 9.0^{\circ}C} = 0.0011 / ^{\circ}C\).

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Key Concepts

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

Resistivity
Resistivity is a property of materials that describes how strongly they oppose the flow of electric current. In simple terms, it's a measure of how much a material resists the passage of electricity. Different materials have different resistivities.
The resistivity of a material is affected by its temperature and physical properties. For example, metals have low resistivities, meaning they conduct electricity well, while insulators like rubber have high resistivities.
Resistivity is represented by the Greek letter \( \rho \) and is typically measured in ohm-meters \( (\Omega \cdot m) \).
  • Low resistivity means high conductivity.
  • Resistivity changes with temperature.
  • In calculations, resistivity helps determine the resistance of a wire.
Ohm's Law
Ohm's Law is a fundamental principle in electronics and electrical engineering. It relates the current flowing through a conductor to the voltage across it and the resistance it encounters. In equation form, Ohm's Law is stated as:
\( V = I \cdot R \)
where:
  • \( V \) is the voltage (in volts)
  • \( I \) is the current (in amperes)
  • \( R \) is the resistance (in ohms)
Ohm's Law shows how voltage, current, and resistance are interdependent. If you know two of these three values, you can calculate the third. This law is pivotal for understanding circuits and designing electronic devices.
Electric Resistance
Electric resistance is the quantity that measures how a device or material reduces the electric current flow through it. It is the result of collisions, which cause energy loss as heat in the material.
Resistance is dependent on the material's resistivity, as well as its length and cross-sectional area:
\( R = \frac{\rho \cdot L}{A} \)
where \( L \) is the length of the conductor and \( A \) is the cross-sectional area.
Key points to remember:
  • Longer wires have more resistance.
  • Thicker wires have less resistance.
  • Resistance changes with temperature due to changes in resistivity.
Thermal Effects on Resistance
Thermal effects on resistance refer to the change in resistance that occurs due to temperature variations. For many conductors, as temperature increases, resistance also increases.
This is because heating a conductor causes the atoms to vibrate more vigorously, which makes it harder for electrons to flow through.
The temperature coefficient of resistivity, \( \alpha \), quantifies this change in resistance with temperature:
\( \Delta R = \alpha \cdot R_1 \cdot \Delta T \)
where \( \Delta R \) is the change in resistance, \( R_1 \) is the original resistance, and \( \Delta T \) is the temperature change.
Considerations:
  • Positive \( \alpha \) means resistance increases with temperature.
  • For precise electronics, temperature effects on resistance must be managed.
  • Temperature coefficients vary with materials.

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Most popular questions from this chapter

When a straight wire is heated, its resistance changes according to the equation $$ R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right] $$ (Eq. 17.7), where \(\alpha\) is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the length and area of a wire change when it is heated, is $$ R=\frac{R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\left[1+\alpha^{\prime}\left(T-T_{0}\right)\right]}{\left[1+2 \alpha^{\prime}\left(T-T_{0}\right)\right]} $$ where \(\alpha^{\prime}\) is the coefficient of linear expansion. (See Chapter \(10 .\) (b) Compare the two results for a \(2.00\) -m-long copper wire of radius \(0.100 \mathrm{~mm}\), starting at \(20.0^{\circ} \mathrm{C}\) and heated to \(100.0^{\circ} \mathrm{C}\).

What is the required resistance of an immersion heater that will increase the temperature of \(1.50 \mathrm{~kg}\) of water from \(10.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) in \(10.0 \mathrm{~min}\) while operating at \(120 \mathrm{~V} ?\)

An 11-W energy-efficient fluorescent lamp is designed to produce the same illumination as a conventional \(40-\mathrm{W}\) lamp. How much does the energy- efficient lamp save during 100 hours of use? Assume a cost of \(\$ 0.080 / \mathrm{kWh}\) for electrical energy.

The cost of electricity varies widely throughout the United States; \( 0.120 / \mathrm{kWh}\) is a typical value. At this unit price, calculate the cost of (a) leaving a \(40.0-\mathrm{W}\) porch light on for 2 weeks while you are on vacation, (b) making a piece of dark toast in \(3.00 \mathrm{~min}\) with a \(970-\mathrm{W}\) toaster, and (c) drying a load of clothes in \(40.0 \mathrm{~min}\) in a \(5200-\mathrm{W}\) dryer.

If electrical energy costs 12 cents, or \( 0.12\), per kilowatthour, how much does it cost to (a) burn a \(100-\mathrm{W}\) lightbulb for \(24 \mathrm{~h}\) ? (b) Operate an electric oven for \(5.0 \mathrm{~h}\) if it carries a current of \(20.0 \mathrm{~A}\) at \(220 \mathrm{~V}\) ?
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