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Chapter 25: Problem 30

What is the resistance of a copper wire of length \(l=\) \(10.9 \mathrm{~m}\) and diameter \(d=1.3 \mathrm{~mm} ?\) The resistivity of copper is \(1.72 \cdot 10^{-8} \Omega \mathrm{m}\)

Short Answer

Expert verified
Question: Calculate the resistance of a 10.9 meters long copper wire with a diameter of 1.3 mm and resistivity of 1.72 x 10^-8 Ωm. Answer: The resistance of the copper wire is approximately 1.414 x 10^-3 Ω.

Step by step solution

01

Convert diameter to meters

Given diameter \(d = 1.3\) mm, we need to convert it to meters. Since there are 1000mm in 1m, the conversion is: \(d = 1.3 \times 10^{-3} \mathrm{~m}\)
02

Calculate cross-sectional area

Now we need to find the cross-sectional area of the wire. The formula for the area of a circle is: \(A = \pi \cdot (\frac{d}{2})^2\) Using the diameter in meters we found in step 1: \(A = \pi \cdot (\frac{1.3 \times 10^{-3}}{2})^2\) \(A \approx 1.325 \times 10^{-6} \mathrm{~m}^2\)
03

Calculate resistance

Now that we have the area, we can use the formula for resistance: \(R = \frac{\rho \cdot L}{A}\) Plug in the given values for resistivity (\(\rho = 1.72 \times 10^{-8} \Omega \mathrm{m}\)) and length (\(L = 10.9 \mathrm{~m}\)) and the area we calculated: \(R = \frac{(1.72 \times 10^{-8} \Omega \mathrm{m}) \cdot (10.9 \mathrm{~m})}{1.325 \times 10^{-6} \mathrm{~m}^2}\) \(R \approx 1.414 \times 10^{-3} \Omega\) Thus, the resistance of the copper wire is approximately \(1.414 \times 10^{-3} \Omega\).

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Most popular questions from this chapter

A copper wire has radius \(r=0.0250 \mathrm{~cm},\) is \(3.00 \mathrm{~m}\) long, has resistivity \(\rho=1.72 \cdot 10^{-8} \Omega \mathrm{m},\) and carries a current of \(0.400 \mathrm{~A}\). The wire has density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\) a) What is the resistance, \(R,\) of the wire? b) What is the electric potential difference, \(\Delta V\), across the wire? c) What is the electric field, \(E\), in the wire?

If the current through a resistor is increased by a factor of \(2,\) how does this affect the power that is dissipated? a) It decreases by a factor of 4 . b) It increases by a factor of 2 . c) It decreases by a factor of 8 . d) It increases by a factor of 4 .

The most common material used for sandpaper, silicon carbide, is also widely used in electrical applications. One common device is a tubular resistor made of a special grade of silicon carbide called carborundum. A particular carborundum resistor (see the figure) consists of a thick-walled cylindrical shell (a pipe) of inner radius \(a=\) \(1.50 \mathrm{~cm},\) outer radius \(b=2.50 \mathrm{~cm},\) and length \(L=60.0 \mathrm{~cm} .\) The resistance of this carborundum resistor at \(20 .{ }^{\circ} \mathrm{C}\) is \(1.00 \Omega\). a) Calculate the resistivity of carborundum at room temperature. Compare this to the resistivities of the most commonly used conductors (copper, aluminum, and silver). b) Carborundum has a high temperature coefficient of resistivity: \(\alpha=2.14 \cdot 10^{-3} \mathrm{~K}^{-1} .\) If, in a particular application, the carborundum resistor heats up to \(300 .{ }^{\circ} \mathrm{C},\) what is the percentage change in its resistance between room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right)\) and this operating temperature?

Two conductors of the same length and radius are connected to the same emf device. If the resistance of one is twice that of the other, to which conductor is more power delivered?

Should light bulbs (ordinary incandescent bulbs with tungsten filaments) be considered ohmic resistors? Why or why not? How would this be determined experimentally?
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