/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 You want to precut a set of \(1.... [FREE SOLUTION] | 91Ó°ÊÓ

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You want to precut a set of \(1.00 \Omega\) strips of 14 gauge copper wire (of diameter \(1.628 \mathrm{~mm}\) ). How long should each strip be?

Short Answer

Expert verified
Each strip should be approximately 124 meters long.

Step by step solution

01

Understand the formula for resistance

The resistance \( R \) of a wire is given by the formula \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire.
02

Find the resistivity of copper

The resistivity \( \rho \) of copper is typically \( 1.68 \times 10^{-8} \Omega \cdot \mathrm{m} \).
03

Calculate the cross-sectional area

The cross-sectional area \( A \) for a wire of diameter \( d \) is given by the formula \( A = \pi \left( \frac{d}{2} \right)^2 \). For a diameter of \( 1.628 \mathrm{~mm} \), \( A = \pi \left( \frac{1.628 imes 10^{-3} \mathrm{~m}}{2} \right)^2 \approx 2.08 \times 10^{-6} \mathrm{~m^2} \).
04

Solve for the length of the wire

Rearrange the resistance formula to solve for \( L \): \( L = \frac{R \cdot A}{\rho} \). Substitute \( R = 1.00 \Omega \), \( A = 2.08 \times 10^{-6} \mathrm{~m^2} \), and \( \rho = 1.68 \times 10^{-8} \Omega \cdot \mathrm{m} \).
05

Calculate the length

Substitute the known values into the equation to find \( L \): \[ L = \frac{1.00 \Omega \cdot 2.08 \times 10^{-6} \mathrm{~m^2}}{1.68 \times 10^{-8} \Omega \cdot \mathrm{m}} \approx 124 \mathrm{~m}. \]
06

Final Step: Verification

Check the calculations ensuring correct unit conversions and correct arithmetic operations were applied. The calculated length \( L \) shows the precut length of the copper wire required.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrical Resistivity
Electrical resistivity is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. The resistivity, denoted by \( \rho \), is typically measured in ohm-meters (\( \Omega \cdot m \)). This value is intrinsic to the specific material, meaning each material has its own inherent resistivity value. For example, copper, which is widely used in electrical wiring due to its excellent conductive properties, has a low resistivity of about \( 1.68 \times 10^{-8} \Omega \cdot m \). This low resistivity makes copper very efficient in conducting electricity. Understanding resistivity is essential when calculating resistance in wires, as it significantly impacts how much wire you will need to create a specific resistance.
Cross-Sectional Area
The cross-sectional area of a wire plays a critical role in determining its resistance. It represents the size of the sliced-through section of the wire. For cylindrical wires, like typical copper wires, the cross-sectional area \( A \) is calculated using the formula \( A = \pi \left( \frac{d}{2} \right)^2 \), where \( d \) is the diameter of the wire. This formula derives from the geometry of a circle, where \( \pi \) is approximately 3.14159. If you have a wire with a known diameter, you can determine how much current it can handle, given that a larger area allows more electrons to pass through, thus reducing resistance. For the copper wire exercise, since its diameter is \( 1.628 \mathrm{~mm} \), or \( 1.628 \times 10^{-3} \mathrm{~m} \), the computed cross-sectional area is \( 2.08 \times 10^{-6} \mathrm{~m^2} \). This calculation is crucial in determining how the dimensions of the wire affect its ability to conduct electricity.
Copper Wire Properties
Copper is one of the most commonly used materials for electrical wiring, thanks to its unique properties. It is not only highly conductive but also has a balance of strength, flexibility, and cost-effectiveness. Copper’s properties include:
  • High Conductivity: Copper's conductivity is second only to silver, making it ideal for efficient power transmission.
  • Low Resistivity: With a low resistivity of \( 1.68 \times 10^{-8} \Omega \cdot m \), copper wire does not oppose the flow of electric current significantly, which means less energy is lost as heat.
  • Durability and Ductility: Copper can withstand environmental stress and can be drawn into thin wires without breaking.
These characteristics make copper wiring a superior choice for most electrical applications, including the exercise of creating \( 1.00 \Omega \) copper wire strips.
Ohm's Law Calculation
Ohm's Law is a fundamental principle used in electrical engineering and physics, illustrating the relationship between voltage, current, and resistance. However, when calculating the resistance of a material, we focus on the formula derived from geometric and material properties:\[ R = \rho \frac{L}{A} \]Here, \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. The goal is to solve for the length \( L \) that provides a specific resistance \( R \). Re-arranging the formula, we find:\[ L = \frac{R \cdot A}{\rho} \]By inputting the resistance \( 1.00 \Omega \), area \( 2.08 \times 10^{-6} \mathrm{~m^2} \), and resistivity \( 1.68 \times 10^{-8} \Omega \cdot m \), the length \( L \) of the copper wire needed is approximately 124 meters. This computation is crucial for ensuring that the wire achieves the desired resistance.

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Most popular questions from this chapter

A steel wire of length \(L\) and radius \(r_{1}\) has a resistance \(R\). A second steel wire has the same length but a radius \(r_{2}\) and a resistance of \(3 R\). Find the ratio \(r_{1} / r_{2}\)

A toaster using a Nichrome heating element operates on \(120 \mathrm{~V}\). When it is switched on at \(20^{\circ} \mathrm{C}\), the heating element carries an initial current of 1.35 A. A few seconds later, the current reaches the steady value of 1.23 A. (a) What is the final temperature of the element? The average value of the temperature coefficient of resistivity for Nichrome over the temperature range from \(20^{\circ} \mathrm{C}\) to the final temperature of the element is \(4.5 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1} .\) (b) What is the power dissipated in the heating element (i) initially; (ii) when the current reaches a steady value?

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 27 W. What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

A resistor with a \(15.0 \mathrm{~V}\) potential difference across its ends develops thermal energy at a rate of \(327 \mathrm{~W}\). (a) What is the current in the resistor? (b) What is its resistance?

Lightning strikes can involve currents as high as 25,000 A that last for about \(40 \mu\) s. If a person is struck by a bolt of lightning with these properties, the current will pass through his body. We shall assume that his mass is \(75 \mathrm{~kg},\) that he is wet (after all, he is in a rainstorm) and therefore has a resistance of \(1.0 \mathrm{k} \Omega\), and that his body is all water (which is reasonable for a rough, but plausible, approximation). (a) By how many degrees Celsius would this lightning bolt increase the temperature of 75 kg of water? (b) Given that the internal body temperature is about \(37^{\circ} \mathrm{C},\) would the person's temperature actually increase that much? Why or why not? What would happen first?

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