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You apply a potential difference of 4.50 \(\mathrm{V}\) between the ends of a wire that is 2.50 \(\mathrm{m}\) in length and 0.654 \(\mathrm{mm}\) in radius. The resulting current through the wire is 17.6 \(\mathrm{A}\) . What is the resistivity of the wire?

Short Answer

Expert verified
The resistivity of the wire is approximately \( 1.376 \times 10^{-7} \, \text{\Omega} \cdot \text{m} \).

Step by step solution

01

Identify Known Values

We know the following values from the problem statement: The potential difference \( V = 4.50 \, \text{V} \), the current through the wire \( I = 17.6 \, \text{A} \), the length of the wire \( L = 2.50 \, \text{m} \), and the radius of the wire \( r = 0.654 \, \text{mm} = 0.654 \times 10^{-3} \, \text{m} \).
02

Calculate the Cross-sectional Area of the Wire

The cross-sectional area \( A \) of the wire is calculated using the formula for the area of a circle, \( A = \pi r^2 \). Substituting in the radius, we get:\[ A = \pi (0.654 \times 10^{-3})^2 \approx 1.344 \times 10^{-6} \, \text{m}^2 \]
03

Use Ohm's Law to Find Resistance

Ohm's Law states \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. We can rearrange the equation to find \( R \):\[ R = \frac{V}{I} = \frac{4.50}{17.6} \approx 0.256 \; \Omega \]
04

Use the Formula for Resistivity

The formula for resistivity \( \rho \) is \( \rho = \frac{RA}{L} \). Substitute values for \( R \), \( A \), and \( L \):\[ \rho = \frac{0.256 \times 1.344 \times 10^{-6}}{2.50} \approx 1.376 \times 10^{-7} \, \text{\Omega} \cdot \text{m} \]
05

Conclusion

By calculating step-by-step, the resistivity of the wire is determined to be \( \rho \approx 1.376 \times 10^{-7} \, \text{\Omega} \cdot \text{m} \).

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics and physics. It describes how voltage, current, and resistance interact in an electrical circuit. According to Ohm's Law, the voltage across a conductor is directly proportional to the current flowing through it, provided the temperature and other physical conditions remain constant.

This relationship is expressed through the formula:
  • \( V = IR \)
Where:
  • \( V \) is the voltage measured in volts (\( V \)),
  • \( I \) is the current measured in amperes (\( A \)), and
  • \( R \) is the resistance measured in ohms (\( \Omega \)).
Using this law, we can rearrange the equation to find an unknown quantity if the other two are known. In our exercise, rearranging the formula allows us to calculate resistance \( R \) from the known values of voltage and current. This forms the basis for solving many electrical problems and helps to understand how energy flows through a circuit.
cross-sectional area
The cross-sectional area is an important factor when calculating electrical resistance, especially for conductors like wires. The area refers to the surface area of the cut-through face if you were to take a slice across the width of the wire. This area impacts how easily current can pass through the wire.

For a wire, which is typically cylindrical, the cross-sectional area \( A \) is calculated using the formula for the area of a circle:
  • \( A = \pi r^2 \)
Where:
  • \( \pi \) is a constant approximately equal to 3.14159,
  • \( r \) is the radius of the wire. In our exercise, it is given as 0.654 mm, which we convert to meters by multiplying with \( 10^{-3} \) to match standard units (\( \text{m} \)).
Understanding how to calculate this area is crucial because a larger cross-sectional area allows more electrons to pass through the wire simultaneously, reducing overall resistance.
electrical resistance
Electrical resistance is a measure of how much an object opposes the flow of electric current. It is a key concept in Ohm's Law and is measured in ohms (\( \Omega \)). Resistance depends on several factors:
  • The material's intrinsic properties,
  • Length of the conductor,
  • Cross-sectional area, and
  • Temperature.
In our exercise, we calculate the resistance \( R \) using Ohm's Law with the formula:
  • \( R = \frac{V}{I} \)
Using known values of voltage (4.50 V) and current (17.6 A), we deduce the resistance to be around 0.256 \( \Omega \). This resistance reflects how difficult it is for current to flow through the wire, greatly influencing the final resistivity calculation.
potential difference
Potential difference, commonly referred to as voltage, is a measure of the energy difference between two points in a circuit. It is the driving force that pushes electric current through a circuit and is measured in volts (\( V \)).

In simpler terms, potential difference tells us how much work can be done by each coulomb of charge that moves between these two points. For our exercise, we are given a potential difference of 4.50 V across a wire. This potential difference causes a specific amount of current (17.6 A) to flow through the wire.

Understanding potential difference is fundamental for solving electrical circuit problems as it affects the current flow and, combined with resistance, determines the total energy interaction within a circuit. Recognizing the role of voltage can help you manage electrical devices more efficiently and safely.

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Most popular questions from this chapter

The following measurements of current and potential difference were made on a resistor constructed of Nichrome wire: $$ \begin{array}{l|llll}{\mathbf{I}(\mathbf{A})} & {0.50} & {1.00} & {2.00} & {4.00} \\ {V_{a b}(\mathbf{v})} & {1.94} & {3.88} & {7.76} & {15.52}\end{array} $$ (a) Graph \(V_{a b}\) as a function of \(I .\) (b) Does Nichrome obey Ohm's law? How can you tell? (c) What is the resistance of the resistor in ohms?

In an experiment conducted at room temperature, a current of 0.820 A flows through a wire 3.26 \(\mathrm{mm}\) in diameter. Find the magnitude of the electric field in the wire if the wire is made of (a) tungsten; and (b) aluminum.

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 \(\mathrm{A}\) . A few seconds later the current reaches the steady value of 1.23 \(\mathrm{A}\) . (a) What is the final temperature of the element? The average value of the temperature coefficient of resistivity for Nichrome over the temperature range is \(4.5 \times\) \(10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\) . (b) What is the power dissipated in the heating element initially and when the current reaches a steady value?

A lightning bolt strikes one end of a steel lightning rod, producing a \(15,000-\mathrm{A}\) current burst that lasts for 65\(\mu\) s. The rod is 20 \(\mathrm{m}\) long and 1.8 \(\mathrm{cm}\) in diameter, and its other end is connected to the ground by 35 \(\mathrm{m}\) of \(8.0-\mathrm{mm}\) -diameter copper wire (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

A current of 3.6 A flows through an automobile headlight. How many coulombs of charge flow through the beadlight in 3.0 \(\mathrm{h} ?\)

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