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Chapter 24: Problem 27

A uniform wire of resistance \(R\) is stretched until its length doubles. Assuming its density and resistivity remain constant, what's its new resistance?

Short Answer

Expert verified
The new resistance, after the wire is stretched to double its length, is 4 times the original resistance (\( 4R \)).

Step by step solution

01

Understand the properties of a wire

The resistance \( R \) of a wire can be expressed as \( R = pL/A \) where \( p \) is the resistivity, \( L \) is the length and \( A \) is the cross-sectional area. In this particular case, \( p \) remains constant because it is given that the resistivity remains constant.
02

Identify the changes in the wire

When the wire is stretched, its length doubles but its volume \( V \), which is the product of its length \( L \) and cross-sectional area \( A \), remains the same. This implies that \( A \) must decrease by half, to \( A/2 \), and \( L \) increases to \( 2L \).
03

Substitute the new values

Substitute the new values of \( L \) and \( A \) into the equation in Step 1: The new resistance \( R' \) is \( p(2L)/(A/2) = 4pL/A = 4R \).

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Most popular questions from this chapter

The National Electrical Code specifies a maximum current of \(10 \mathrm{A}\) in 16 -gauge \((1.29\) -mm-diameter) copper wire. What's the corresponding current density?

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