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Chapter 10: Problem 17

Wire is sold in a series of standard diameters, called "gauges." The difference in diameter between one gauge and the next in the series is about \(20 \%\). How would the resistance of a given length of wire compare with the resistance of the same length of wire in the next gauge in the series? (answer check available at lightandmatter.com)

Short Answer

Expert verified
The resistance of the next gauge wire is about 69% of the current gauge wire.

Step by step solution

01

Understand the Problem

We need to compare the resistance of wires of the same material and length but differing in diameter due to different gauge sizes. The key is understanding how diameter affects resistance, given that each gauge size varies by about 20% in diameter from the next.
02

Relationship between Resistance and Diameter

The resistance of a wire is given by the formula \( R = \frac{\rho L}{A} \), where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. For a circular wire, \( A = \pi (\frac{d}{2})^2 \), meaning that resistance is inversely proportional to the square of the diameter \( d \).
03

Calculate Change in Diameter

Given that each gauge changes the diameter by about 20%, we can express the new diameter of the next gauge as \( d_2 = 1.2d_1 \), where \( d_1 \) is the diameter of the smaller gauge and \( d_2 \) is the diameter of the next larger gauge.
04

Calculate New Cross-Sectional Area

With the new diameter \( d_2 = 1.2d_1 \), the new area \( A_2 \) becomes \( \pi (\frac{1.2d_1}{2})^2 = \pi (0.6d_1)^2 = 0.36\pi d_1^2 \), which is 1.44 times the original area \( A_1 = \pi (\frac{d_1}{2})^2 \).
05

Compare Resistances

Since resistance is inversely proportional to area, the resistance of the wire with diameter \( d_2 \), \( R_2 \), is \( \frac{R_1}{1.44} \). Hence, the resistance of the next gauge is about \( \frac{R_1}{1.44} \), or approximately 69% of the original resistance \( R_1 \).
06

Conclude the Comparison

Thus, the resistance of a given length of wire decreases when the diameter increases by a factor of 20% due to gauge change. The resistance of the larger gauge (next one) is approximately 69% of the resistance of the smaller (current) gauge.

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Key Concepts

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

Wire Gauge
Wire gauge is a system used to denote the diameter of a wire. It is commonly utilized in various applications from electrical wiring to string instruments. Understanding gauge helps us manage the resistance of the wire, as the diameter plays a crucial role in determining resistance. Generally, wire gauge numbers decrease as diameter increases, meaning a lower gauge number corresponds to a thicker wire. This system is standardized so that each change in gauge represents a consistent percentage change in diameter 鈥� typically about 20% between consecutive gauges. This stepped change ensures predictability in electrical applications, where resistance is a key factor in performance. Learning how to navigate the gauge system helps one select the appropriate wire for their specific needs, balancing diameter, resistance, and conductivity.
Diameter and Resistance
The diameter of a wire greatly impacts its electrical resistance. Resistance is a measure of how much a material opposes the flow of electric current through it. A larger diameter means more space for electrons to travel without bumping into the atoms of the wire, which results in lower resistance. The equation to find resistance is given as: \( R = \frac{\rho L}{A} \) where \( R \) is resistance, \( \rho \) is the resistivity of the wire material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. For circular wires, the area \( A \) can be calculated using the diameter by the formula: \( A = \pi \left( \frac{d}{2} \right)^2 \)indicating that resistance is inversely proportional to the square of the diameter. Therefore, increasing the diameter by approximately 20% results in a reduction of resistance by a factor of 1.44, drastically changing the wire's performance and allowing for greater current flow with less energy loss.
Cross-Sectional Area of Wire
The cross-sectional area of a wire is a crucial factor in determining its resistance. It is essentially the surface area of the "slice" through the wire and affects how easily current can flow through the wire. For a cylindrical wire, calculating this area is straightforward using the formula: \( A = \pi \left( \frac{d}{2} \right)^2 \)Here, \( d \) represents the diameter of the wire. As the diameter increases, the cross-sectional area increases exponentially (since it depends on the square of the diameter), allowing more current to pass through for the same length of wire. This results in a lower resistance. When moving to the next larger gauge, if the diameter increases by 20%, the cross-sectional area increases by a factor of 1.44, thus reducing the resistance by the same factor. This illustrates the critical relationship between diameter, area, and electrical resistance, highlighting the importance of the cross-sectional area in the design and application of wires.

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Most popular questions from this chapter

You are given a battery, a flashlight bulb, and a single piece of wire. Draw at least two configurations of these items that would result in lighting up the bulb, and at least two that would not light it. (Don't draw schematics.) If you're not sure what's going on, borrow the materials from your instructor and try it. Note that the bulb has two electrical contacts: one is the threaded metal jacket, and the other is the tip (at the bottom in the figure). [Problem by Arnold Arons.]

If a typical light bulb draws about \(900 \mathrm{~mA}\) from a 110-V household circuit, what is its resistance? (Don't worry about the fact that it's alternating current.) (answer check available at lightandmatter.com)

(a) Express the power dissipated by a resistor in terms of \(R\) and \(\Delta V\) only, eliminating \(I\). (answer check available at lightandmatter.com) (b) Electrical receptacles in your home are mostly \(110 \mathrm{~V}\), but circuits for electric stoves, air conditioners, and washers and driers are usually \(220 \mathrm{~V}\). The two types of circuits have differently shaped receptacles. Suppose you rewire the plug of a drier so that it can be plugged in to a \(110 \mathrm{~V}\) receptacle. The resistor that forms the heating element of the drier would normally draw 200 W. How much power does it actually draw now? (answer check available at lightandmatter.com)

In AM (amplitude-modulated) radio, an audio signal \(f(t)\) is multiplied by a sine wave \(\sin \omega t\) in the megahertz frequency range. For simplicity, let's imagine that the transmitting antenna is a whip, and that charge goes back and forth between the top and bottom. Suppose that, during a certain time interval, the audio signal varies linearly with time, giving a charge \(q=(a+b t) \sin \omega t\) at the top of the whip and \(-q\) at the bottom. Find the current as a function of time. (answer check available at lightandmatter.com)

We have referred to resistors dissipating heat, i.e., we have assumed that \(P=I \Delta V\) is always greater than zero. Could \(I \Delta V\) come out to be negative for a resistor? If so, could one make a refrigerator by hooking up a resistor in such a way that it absorbed heat instead of dissipating it?
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