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Chapter 26: Problem 12

12\. A Certain Wire A certain wire has a resistance \(R\). What is the resistance of a second wire, made of the same material, that is half as long and has half the diameter?

Short Answer

Expert verified
The resistance of the second wire is 8R.

Step by step solution

01

- Understanding Resistance

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

- Length of Second Wire

The second wire is half as long as the first wire. If the length of the first wire is \(L\), then the length of the second wire is \( \frac{L}{2} \).
03

- Diameter and Area of Second Wire

The second wire has half the diameter of the first wire. If the diameter of the first wire is \(d\), then the diameter of the second wire is \( \frac{d}{2} \). The cross-sectional area is given by \( A = \pi \left( \frac{d}{2} \right)^2 \). For the second wire, the new diameter is \( \frac{d}{2} \), so the cross-sectional area becomes: \( A_{new} = \pi \left( \frac{\frac{d}{2}}{2} \right)^2 = \pi \left( \frac{d}{4} \right)^2 = \frac{\pi d^2}{16} \).
04

- Resistance of the Second Wire

Using the resistance formula and substituting the new values: \( R_{new} = \rho \frac{L_{new}}{A_{new}} = \rho \frac{\frac{L}{2}}{\frac{\pi d^2}{16}} \), which simplifies to: \( R_{new} = \rho \frac{L}{2} \cdot \frac{16}{\pi d^2} = 8 \times \rho \frac{L}{\pi d^2} = 8R \). Thus, the resistance of the second wire is \(8R\).

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Key Concepts

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

Resistance Calculation
Resistance is a measure of how much a material opposes the flow of electric current. It's calculated using the formula:
\[ R = \rho \frac{L}{A} \]
Here:
  • \( R \): Resistance (measured in ohms, \( \text{惟} \))
  • \( \rho \): Resistivity (property of the material, measured in ohm-meter, \( \text{惟鈰卪} \))
  • \( L \): Length of the wire (measured in meters, \( \text{m} \))
  • \( A \): Cross-sectional area (measured in square meters, \( \text{m}^2 \))
To better understand this, think of the wire as a highway for electrons, and resistance as traffic. A longer highway or a more crowded one makes it harder for cars (or electrons) to pass through easily. In practical terms, engineers use this calculation to design circuits, ensuring that wires have the correct resistance for their particular applications.
Resistivity
Resistivity is a fundamental property of materials that defines how strongly they resist current flow. It's denoted by the symbol \( \rho \) and is a constant for a given material.

Metals like copper and aluminum have low resistivity, meaning they conduct electricity well. In contrast, materials like rubber and glass have high resistivity and act as insulators.
This is why choosing the right material for wires is crucial in electrical engineering. For instance, copper is often used in electrical wiring due to its low resistivity. Knowing the resistivity helps you predict how much a given material will resist electric current and assists in calculating \[ R = \rho \frac{L}{A} \]
Cross-Sectional Area
The cross-sectional area of a wire is another factor affecting its resistance. It's the area of the slice you would see if you cut the wire perpendicular to its length.

For a wire with a circular cross-section, the area \( A \) is given by:
  • \[ A = \frac{蟺d^2}{4} \]
  • where \( d \) is the diameter of the wire.

In our exercise, the diameter is halved so the new cross-sectional area becomes:
  • \[ A_{new} = \frac{蟺 \frac{d}{2}^2}{4} = \frac{蟺 \frac{d^2}{4}}{4} = \frac{蟺 d^2}{16} \]

It's essential to carefully calculate the cross-sectional area because it influences the overall resistance. A smaller cross-sectional area increases resistance, just like a narrower pipe makes it harder for water to flow through.

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Most popular questions from this chapter

37\. How Long How long does it take electrons to get from a car battery to the starting motor? Assume the current is \(300 \mathrm{~A}\) and the electrons travel through a copper wire with cross-sectional area \(0.21 \mathrm{~cm}^{2}\) and length \(0.85 \mathrm{~m} .\) (Hint: Assume one conduction electron per atom and take the number density of copper atoms to be \(8.5 \times\) \(10^{28}\) atoms \(\left./ \mathrm{m}^{2} .\right)\)

45\. Building a Water Heater The nickel-chromium alloy Nichrome has a resistivity of about \(10^{-6} \Omega-\mathrm{m} .\) Suppose you want to build a small heater out of a coil of Nichrome wire and a \(6 \mathrm{~V}\) battery in order to heat \(30 \mathrm{ml}\) of water from a temperature of \(20 \mathrm{C}\) to \(40 \mathrm{C}\) in \(1 \mathrm{~min}\). Assume the battery has negligible internal resistance. (a) How much heat energy (in joules) do you need to do this? (b) How much power (in watts) do you need to do it in the time indicated? (c) What resistance should your Nichrome coil have in order to produce this much power in heat? (d) Can you create a coil having these properties? (Hint: Can you find a plausible length and cross-sectional area for your wire that will give you the resistance you need?) (e) If the internal resistance of the battery were \(1 / 3 \Omega\), how would it affect your calculation? (Only explain what you would have to do; don't recalculate the size of your coil.)

18\. A Student A student kept his \(9.0 \mathrm{~V}, 7.0 \mathrm{~W}\) radio turned on at full volume from \(9: 00\) P.M. until \(2: 00\) A.M. How much charge went through it?

11\. Longer Wire A wire with a resistance of \(6.0 \Omega\) is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

5\. Electrocution A human being can be electrocuted if a current as small as \(50 \mathrm{~mA}\) passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding, one in each hand. If his resistance is \(2000 \Omega\), what might the fatal voltage be?
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