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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 of the wire after it's been stretched to twice its original length is four times the initial resistance \(R\).

Step by step solution

01

Identify the known quantities.

We are given that the resistivity \(\rho\) (rho) remains constant, the initial resistance \(R\), and the fact that the wire's length doubles.
02

Apply the formula for Resistance

The resistance of a wire can be calculated by the equation \(R= \rho\frac{L}{A}\), where \(\rho\) is the resistivity of the wire, \(L\) represents the length of the wire, and \(A\) represents the cross-sectional area of the wire. Since the volume of the wire (which is equal to the product of the cross-sectional area and the length of the wire) remains constant when it's stretched, we can express \(A\) as \(\frac{V}{L}\), where \(V\) represents volume.
03

Substitute \(A\) into the Resistance formula

By substituting the equation for the cross-sectional area, \(\frac{V}{L}\), into the original resistance equation we get: \(R= \rho\frac{L^{2}}{V}\). This is the new formula for resistance.
04

Understand how doubling the length affects resistance

The length of the wire is now doubled, therefore we substitute \(2L\) instead of original length \(L\) into the equation to derive the new resistance \(R'\). Doing so leads us to: \(R' = \rho\frac{(2L)^{2}}{V} = 4\rho\frac{L^{2}}{V} = 4R\).
05

Conclusion

So the new resistance \(R'\) of the wire after it's been lengthened to twice its original length is four times the initial resistance \(R\). Using the above steps, we have demonstrated how the resistance of a wire changes with changing dimensions, while keeping the resistivity constant and the volume unaltered.

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