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Chapter 25: Problem 38

A potential difference of \(12.0 \mathrm{~V}\) is applied across a wire of cross-sectional area \(4.50 \mathrm{~mm}^{2}\) and length \(1000 . \mathrm{km} .\) The current passing through the wire is \(3.20 \cdot 10^{-3} \mathrm{~A}\). a) What is the resistance of the wire? b) What type of wire is this?

Short Answer

Expert verified
Answer: The wire is likely made of copper.

Step by step solution

01

Find the resistance using Ohm's Law

Ohm's Law states that \(V = IR\), where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance. We have \(V = 12.0 \mathrm{~V}\) and \(I = 3.20 \cdot 10^{-3} \mathrm{~A}\). To find the resistance, we can rearrange the formula to get \(R = V/I\). So, \(R = \frac{12.0 \mathrm{~V}}{3.20 \cdot 10^{-3} \mathrm{~A}}\).
02

Calculate the resistance

Now, we can plug in the values to find the resistance: \(R = \frac{12.0 \mathrm{~V}}{3.20 \cdot 10^{-3} \mathrm{~A}} = 3750 \Omega\).
03

Rearrange the resistance equation to find resistivity

We can use the formula \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area. We want to find the resistivity, so we can rearrange the formula to get \(\rho = R \frac{A}{L}\).
04

Convert the length and area to the proper units

The length is given in kilometers, which we need to convert to meters. We know that 1 km = 1000 m, so \(1000 \mathrm{~km} = 1000 \cdot 1000 \mathrm{~m} = 1 \cdot 10^{6} \mathrm{~m}\). The cross-sectional area is given in mm², which we need to convert to m². We know that 1 m² = 1,000,000 mm², so \(4.50 \mathrm{~mm}^{2} = \frac{4.50}{1,000,000} \mathrm{~m}^{2} = 4.50 \cdot 10^{-6} \mathrm{~m}^{2}\).
05

Calculate the resistivity

Now we can plug in the values to find the resistivity: \(\rho = 3750 \Omega \cdot \frac{4.50 \cdot 10^{-6} \mathrm{~m}^{2}}{1 \cdot 10^{6} \mathrm{~m}} = 1.69 \cdot 10^{-8} \Omega \cdot \mathrm{m}\).
06

Identify the type of wire

The resistivity value we found, \(1.69 \cdot 10^{-8} \Omega \cdot \mathrm{m}\), is close to the resistivity of copper, which is approximately \(1.68 \cdot 10^{-8} \Omega \cdot \mathrm{m}\). Therefore, the wire is likely made of copper.

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Most popular questions from this chapter

You make a parallel combination of resistors consisting of resistor A having a very large resistance and resistor B having a very small resistance. The equivalent resistance for this combination will be: a) slightly greater than the resistance of the resistor A. b) slightly less than the resistance of the resistor \(\mathrm{A}\). c) slightly greater than the resistance of the resistor B. d) slightly less than the resistance of the resistor B.

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?

Two cylindrical wires, 1 and \(2,\) made of the same material, have the same resistance. If the length of wire 2 is twice that of wire 1 , what is the ratio of their cross-sectional areas, \(A_{1}\) and \(A_{2} ?\) a) \(A_{1} / A_{2}=2\) c) \(\mathrm{A}_{1} / \mathrm{A}_{2}=0.5\) b) \(A_{1} / A_{2}=4\) d) \(A_{1} / A_{2}=0.25\)

A modern house is wired for \(115 \mathrm{~V}\), and the current is limited by circuit breakers to a maximum of \(200 .\) A. (For the purpose of this problem, treat these as DC quantities.) a) Calculate the minimum total resistance the circuitry in the house can have at any time. b) Calculate the maximum electrical power the house can consume.

Show that for resistors connected in series, it is always the highest resistance that dissipates the most power, while for resistors connected in parallel, it is always the lowest resistance that dissipates the most power.
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