   Chapter 17, Problem 55AP

Chapter
Section
Textbook Problem

A particular wire has a resistivity of 3.0 × 10−8 Ω · m and a cross-sectional area of 4.0 × 10−6 m2. A length of this wire is to be used as a resistor that will develop 48 W of power when connected across a 20.-V battery. What length of wire is required?

To determine
The length of wire.

Explanation

Given Info: The wire has resistivity 3.0×108Ωm . The cross sectional area of wire is 4.0×106m2 . The wire developed 48W when connected to 20V battery.

Explanation:

Formula to calculate the resistance of wire is,

R=(ΔV)2P

• R is the resistance of wire,
• ΔV is the voltage of battery,
• P is the power developed in the battery,

Formula to calculate the length of wire is,

L=RAρ

• L is the length of wire,
• A is the area of cross section of the wire,
• ρ is the resistivity of the wire,

Use (ΔV)2/P for R in the above equation to rewrite L.

L=((ΔV)2/P)Aρ

Substitute 20V for ΔV , 48W for P , 8

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