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Chapter 18: Problem 119

A piece of gold wire of length \(L\) has a resistance \(R_{0}\) Suppose the wire is drawn out so that its length increases by a factor of three. What is the new resistance \(R\) in terms of the original resistance?

Short Answer

Expert verified
Answer: The new resistance is 9 times the original resistance.

Step by step solution

01

Define the variables and given information

We are given: - Original length of the wire: \(L\) - Original resistance of the wire: \(R_{0}\) - New length of the wire: \(3L\) We need to find the new resistance, \(R\), when the length of the wire is increased by a factor of three.
02

Use the formula for resistance

The formula for resistance is: $$R=\rho\frac{L}{A}$$ where: - \(R\) is the resistance, - \(\rho\) is the resistivity of the material (constant for a given material), - \(L\) is the length of the wire, - \(A\) is the cross-sectional area of the wire.
03

Use the fact that volume remains constant

When the wire is drawn out, its volume remains constant. Therefore, we can write the equation for the volume of the wire before and after being drawn out as: $$A \cdot L = A' \cdot 3L$$ where \(A'\) is the new cross-sectional area of the wire. Now, we can solve for \(A'\): $$A'=\frac{A}{3}$$
04

Calculate the new resistance

Using the formula for resistance, we can find the new resistance \(R\) as follows: $$R=\rho\frac{3L}{\frac{A}{3}}=\rho\frac{9L}{A}$$
05

Relate the new resistance to the original resistance

Now let's express the new resistance in terms of the original resistance \(R_{0}\). We know that: $$R_{0}=\rho\frac{L}{A}$$ Now we can solve for \(\rho\frac{L}{A}\): $$\rho\frac{L}{A} = R_{0}$$
06

Substitute the expression for the original resistance into the new resistance formula

Now, using our expression for \(\rho\frac{L}{A}\) from Step 5, we can write the new resistance as: $$R=9(\rho\frac{L}{A})=9R_{0}$$
07

Conclusion

The new resistance \(R\) is 9 times the original resistance \(R_{0}\). So, we have found that: $$R=9R_{0}$$

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