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Ch. 25 - Electric Current and Resistance
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 25 - Electric Current and ResistanceProblem 26
Chapter 24, Problem 26

The filament of an incandescent lightbulb has a resistance of 12 Ω at 20°C and 140 Ω when hot.
(a) Calculate the temperature of the filament when it is hot, and take into account the change in length and area of the filament due to thermal expansion (assume tungsten for which the thermal expansion coefficient is ≈ 5.5 10⁻⁶ C°⁻¹ ).
(b) In this temperature range, what is the percentage change in resistance due to thermal expansion, and what is the percentage change in resistance due solely to the change in ρ? Use Eq. 25–5.

Verified step by step guidance
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Step 1: Start by understanding the relationship between resistance and temperature. The resistance of a material changes with temperature according to the formula: R=R0(1+α(T-T0)), where R0 is the resistance at the reference temperature T0, α is the temperature coefficient of resistivity, and T is the temperature of interest.
Step 2: Rearrange the formula to solve for the temperature T when the resistance is known: T=T0+R-R0R0α. Use the given values: R0=12 Ω, R=140 Ω, and the temperature coefficient of resistivity for tungsten, α=4.5×10-3 C°⁻¹.
Step 3: Account for the effects of thermal expansion on the resistance. The resistance of a wire also depends on its length and cross-sectional area, which change with temperature. The resistance can be expressed as: R=ρLA, where ρ is the resistivity, L is the length, and A is the cross-sectional area. Use the thermal expansion coefficient β=5.5×10-6 C°⁻¹ to calculate the change in length and area.
Step 4: Calculate the percentage change in resistance due to thermal expansion. The change in length ΔL is proportional to the temperature change: ΔLL=β(T-T0). The cross-sectional area A changes as the square of the length, so the resistance change due to thermal expansion can be approximated.
Step 5: Calculate the percentage change in resistance due solely to the change in resistivity ρ. Use the formula for resistivity change: Δρρ=α(T-T0). Combine the results to determine the overall percentage change in resistance and separate the contributions from thermal expansion and resistivity change.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Resistance and Temperature Relationship

The resistance of a conductor, such as tungsten, changes with temperature. This relationship is often described by the formula R(T) = R0(1 + α(T - T0)), where R0 is the resistance at a reference temperature T0, α is the temperature coefficient of resistance, and T is the new temperature. Understanding this concept is crucial for calculating the filament's temperature when it is hot.
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Thermal Expansion

Thermal expansion refers to the increase in size of a material as its temperature rises. For metals, this can be quantified using the linear expansion formula ΔL = L0αΔT, where ΔL is the change in length, L0 is the original length, α is the coefficient of linear expansion, and ΔT is the change in temperature. This concept is essential for considering how the filament's dimensions affect its resistance.
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Resistivity and Its Temperature Dependence

Resistivity (ρ) is a material property that quantifies how strongly a given material opposes the flow of electric current. For metals, resistivity increases with temperature, typically modeled by ρ(T) = ρ0(1 + β(T - T0)), where β is the temperature coefficient of resistivity. This concept is important for calculating the percentage change in resistance due to changes in resistivity as the filament heats up.
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Related Practice
Textbook Question

You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 120 V (assume its resistance does not change), how bright will it be relative to 75-W 120-V bulbs? [Hint: Assume roughly that brightness is proportional to power consumed.]

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An electric car uses a 45-kW (160-hp) motor. If the battery pack is designed for 340 V, what current would the motor need to draw from the battery? Neglect any energy losses in getting energy from the battery to the motor.

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Textbook Question

A 4.5-V battery is connected to a bulb whose resistance is 2.3 Ω. How many electrons leave the battery per minute?

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Textbook Question

A rectangular solid made of carbon has sides of lengths 1.0 cm, 2.0 cm, and 4.0 cm, lying along the x, y, and z axes, respectively (Fig. 25–36). Determine the resistance for current that passes through the solid in the y direction, (Assume the resistivity is ρ = 3.0 x 10⁻⁵ Ω•m).

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Textbook Question

Neglect the internal resistance of a battery unless the problem refers to it. Ten 7.0-W Christmas tree lights are connected in series to each other and to a 120-V source. What is the resistance of each bulb?

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Textbook Question

The heating element of an electric oven is designed to produce 3.1 kW of heat when connected to a 240-V source. What must be the resistance of the element?

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