(a) Explanation Write the equation for the electric power delivered to the filament. P = ( Δ V ) 2 R (I) Here, P is the electric power delivered to the filament, Δ V is the potential difference across the ends of the wire and R is the resistance of the wire. Write the equation for resistance. R = ρ l A Here, ρ is the resistivity of the material of the length of the filament, l is the length of the filament and A is the cross sectional area. Put the above equation in equation (I). P = ( Δ V ) 2 ρ l / A = ( Δ V ) 2 A ρ l (II) Write the equation for the cross-sectional area of the wire. A = π r 2 Here, r is the radius of the filament. Put the above equation in equation (II). P = ( Δ V ) 2 π r 2 ρ l = π r 2 ( Δ V ) 2 ρ l (III) Rewrite equation (III) for l . l = π r 2 ( Δ V ) 2 P ρ (IV) The filament will radiate all power through its lateral surface area when the filament radiates in steady state. Write the equation for the power of the radiation. P = e σ T 4 A l (V) Here, e is the emissivity, σ is the Stefan’s constant, T is the temperature of the filament and A l is the lateral surface area. Write the equation for the lateral surface area. A l = 2 π r l Put the above equation in equation (V). P = e σ T 4 ( 2 π r l ) Put equation (IV) in the above equation (b) To determine The length of the filament.

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

5th Edition

ISBN: 9781305586871

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Question

Chapter 28, Problem 65P

(a)

To determine

The radius of the filament.

(b)

To determine

The length of the filament.

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Chapter 28, Problem 64P

Chapter 28, Problem 66P

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The radius of the filament.

Solution Summary: The author writes the equation for the electric power delivered to the filament and the cross-sectional area of the wire.

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The radius of the filament.

The length of the filament.

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Chapter 28 Solutions