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The resistivity of a conductor is \(\rho=1.00 \cdot 10^{-5} \Omega \mathrm{m}\). If a cylindrical wire is made of this conductor, with a crosssectional area of \(1.00 \cdot 10^{-6} \mathrm{~m}^{2},\) what should the length of the wire be for its resistance to be \(10.0 \Omega ?\)

Short Answer

Expert verified
Answer: The required length of the conductor wire is 10.0 meters.

Step by step solution

01

Recall the formula for the resistance of a wire

We will use the formula for the resistance of a wire: $$ R = \frac{\rho L}{A} $$ where \(R\) is the resistance, \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area.
02

Plug the given values

We are given the resistivity \(\rho = 1.00 \cdot 10^{-5} \Omega \mathrm{m}\), the cross-sectional area \(A = 1.00 \cdot 10^{-6} \mathrm{m}^2\), and the desired resistance \(R = 10.0 \Omega\). We plug these values into the formula: $$ 10.0 \Omega = \frac{(1.00 \times 10^{-5} \Omega \mathrm{m}) L}{(1.00 \times 10^{-6} \mathrm{m}^2)} $$
03

Solve for the length, L

Now, we will solve for the length \(L\): $$ L = \frac{10.0 \Omega \times 1.00 \times 10^{-6} \mathrm{m}^2}{1.00 \times 10^{-5} \Omega \mathrm{m}} $$
04

Simplify the expression

By dividing the numbers and canceling out the units, we get: $$ L = 10.0 \mathrm{m} $$
05

State the final answer

The length of the wire should be \(10.0\) meters for its resistance to be \(10.0 \Omega\).

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Key Concepts

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

Resistivity
Resistivity is a fundamental property of materials that quantifies how strongly a given material opposes the flow of electric current. It is denoted by the symbol \( \rho \) and measured in ohm-meters (\( \Omega \cdot \text{m} \)).
This intrinsic property varies between different materials and is determined by the type and structure of atoms, as well as the temperature. Consider resistivity like friction for electricity. Just as friction resists motion between surfaces, resistivity resists the flow of electric charges. Lower resistivity indicates that a material allows electricity to pass with ease, while higher resistivity means it's more difficult for electric current to flow through the material.To calculate the resistance \( R \) of a specific conductor using its resistivity, you can use the formula:
  • \( R = \frac{\rho L}{A} \)
where
  • \( \rho \) is the resistivity,
  • \( L \) is the length of the conductor, and
  • \( A \) is the cross-sectional area.
This equation shows how resistivity, alongside the dimensions of a conductor, determines the overall resistance to current flow.
Cylindrical Wire
A cylindrical wire is a common form used in electrical engineering due to its practical design for easy handling and uniform conduction of electricity. The geometry of the wire affects its electrical properties in significant ways. A cylinder's dimensions, such as length and cross-sectional area, are crucial factors in calculating electrical resistance.- **Length (\( L \))**: The longer the wire, the greater the distance the current must travel, leading to increased resistance.- **Cross-sectional Area (\( A \))**: A larger area allows more current to pass through, reducing resistance. This is because there's more 'room' for electrons to flow. For a wire with a circular cross-section, area is calculated as \( A = \pi r^2 \), where \( r \) is the radius.When using wires in circuits, maintaining an optimal balance between size and resistivity helps achieve the desired electrical performance. By understanding these relationships, you can determine the perfect dimensions to get the necessary resistance for your application.
Ohm's Law
Ohm's Law is a key principle in the study of electricity, defining the relationship between voltage, current, and resistance in an electrical circuit. It's usually expressed as:\( V = I \cdot R \)where
  • \( V \) is the voltage across the conductor (in volts),
  • \( I \) is the current through the conductor (in amperes), and
  • \( R \) is the resistance of the conductor (in ohms).
Ohm's Law implies that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant.This law is fundamental for designing and analyzing electrical circuits. By using this simple linear relation, you can determine any of the three variables, given the other two.
For instance, if you know the resistance and the voltage applied, you can easily find the current using:\( I = \frac{V}{R} \)Understanding Ohm's Law helps in predicting how electrical components will behave in various situations and designing circuits accordingly.
Conductors
Conductors are materials that allow the free flow of electric charge, primarily because they have electrons that are not tightly bound to their atoms. Metals are the best-known conductors, with silver, copper, and gold being at the top due to their low resistivity. - **Properties of Conductors**: - High conductivity and low resistivity - Ability to transfer electricity and thermal energy efficiently - Commonly used materials include copper in wires and aluminum in power lines In practical applications, conductors are essential for making efficient electrical connections and pathways. The choice of conductor material affects the efficiency, cost, and thermal performance of electrical systems. For example, copper is often preferred for electrical wiring due to its balance of low resistivity, durability, and affordability.
Understanding the characteristics of conductors ensures that components or devices are designed with the appropriate materials to facilitate optimal electrical flow.

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Most popular questions from this chapter

Before bendable tungsten filaments were developed, Thomas Edison used carbon filaments in his light bulbs. Though carbon has a very high melting temperature \(\left(3599^{\circ} \mathrm{C}\right)\) its sublimation rate is high at high temperatures. So carbonfilament bulbs were kept at lower temperatures, thereby rendering them dimmer than later tungsten-based bulbs. A typical carbon-filament bulb requires an average power of \(40 \mathrm{~W}\), when 110 volts is applied across it, and has a filament temperature of \(1800^{\circ} \mathrm{C}\). Carbon, unlike copper, has a negative temperature coefficient of resistivity: \(\alpha=-0.0005^{\circ} \mathrm{C}^{-1}\) Calculate the resistance at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) of this carbon filament.

When a battery is connected to a \(100 .-\Omega\) resistor, the current is \(4.00 \mathrm{~A}\). When the same battery is connected to a \(400 .-\Omega\) resistor, the current is 1.01 A. Find the emf supplied by the battery and the internal resistance of the battery.

A voltage spike causes the line voltage in a home to jump rapidly from \(110 . \mathrm{V}\) to \(150 . \mathrm{V}\). What is the percentage increase in the power output of a 100.-W tungsten-filament incandescent light bulb during this spike, assuming that the bulb's resistance remains constant?

Three resistors are connected to a power supply with \(V=110 . \mathrm{V}\) as shown in the figure a) Find the potential drop across \(R_{3}\) b) Find the current in \(R_{1}\). c) Find the rate at which thermal energy is dissipated from \(R_{2}\).

Two identical light bulbs are connected to a battery. Will the light bulbs be brighter if they are connected in series or in parallel?

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