Question

A potential difference of \(12.0 \mathrm{~V}\) is applied across a wire of cross-sectional area \(4.50 \mathrm{~mm}^{2}\) and length \(1000 . \mathrm{km} .\) The current passing through the wire is \(3.20 \cdot 10^{-3} \mathrm{~A}\). a) What is the resistance of the wire? b) What type of wire is this?

Short answer

Answer: The wire is likely made of copper.

Step-by-step solution

Find the resistance using Ohm's Law

Ohm's Law states that \(V = IR\), where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance. We have \(V = 12.0 \mathrm{~V}\) and \(I = 3.20 \cdot 10^{-3} \mathrm{~A}\). To find the resistance, we can rearrange the formula to get \(R = V/I\). So, \(R = \frac{12.0 \mathrm{~V}}{3.20 \cdot 10^{-3} \mathrm{~A}}\).

Calculate the resistance

Now, we can plug in the values to find the resistance: \(R = \frac{12.0 \mathrm{~V}}{3.20 \cdot 10^{-3} \mathrm{~A}} = 3750 \Omega\).

Rearrange the resistance equation to find resistivity

We can use the formula \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area. We want to find the resistivity, so we can rearrange the formula to get \(\rho = R \frac{A}{L}\).

Convert the length and area to the proper units

The length is given in kilometers, which we need to convert to meters. We know that 1 km = 1000 m, so \(1000 \mathrm{~km} = 1000 \cdot 1000 \mathrm{~m} = 1 \cdot 10^{6} \mathrm{~m}\). The cross-sectional area is given in mm², which we need to convert to m². We know that 1 m² = 1,000,000 mm², so \(4.50 \mathrm{~mm}^{2} = \frac{4.50}{1,000,000} \mathrm{~m}^{2} = 4.50 \cdot 10^{-6} \mathrm{~m}^{2}\).

Calculate the resistivity

Now we can plug in the values to find the resistivity: \(\rho = 3750 \Omega \cdot \frac{4.50 \cdot 10^{-6} \mathrm{~m}^{2}}{1 \cdot 10^{6} \mathrm{~m}} = 1.69 \cdot 10^{-8} \Omega \cdot \mathrm{m}\).

Identify the type of wire

The resistivity value we found, \(1.69 \cdot 10^{-8} \Omega \cdot \mathrm{m}\), is close to the resistivity of copper, which is approximately \(1.68 \cdot 10^{-8} \Omega \cdot \mathrm{m}\). Therefore, the wire is likely made of copper.

Key concepts

Resistance Calculation

When dealing with electrical circuits, understanding how to calculate resistance is crucial. Resistance, denoted as \( R \), measures how much a material opposes the flow of electric current. The basic formula to calculate resistance is derived from Ohm's Law, which states:

• \( V = IR \)

where:

• \( V \) is the voltage (potential difference) across the wire,
• \( I \) is the current flowing through the wire,
• \( R \) is the resistance.

If you know the voltage and current, you can rearrange the formula to find the resistance:

• \( R = \frac{V}{I} \)

This equation allows you to easily find the resistance of any component in a circuit when you have the voltage across it and the current flowing through it. In the given exercise, for a wire with a potential difference of \(12.0 \, \mathrm{V}\) and a current of \(3.20 \cdot 10^{-3} \, \mathrm{A}\), applying the formula gives a resistance of \(3750 \, \Omega\). This provides insight into the wire's ability to conduct electricity.

Resistivity

Resistivity is a fundamental property that gives us insight into how a material conducts electrical current. It is denoted by \( \rho \) and varies depending on the material's nature and temperature. The resistivity is calculated using the formula:

• \( R = \rho \frac{L}{A} \)

where:

• \( R \) represents resistance,
• \( L \) is the length of the wire,
• \( A \) is the cross-sectional area of the wire.

To find the resistivity, reorganize the formula to:

• \( \rho = R \frac{A}{L} \)

It is important to ensure that all measurements are in compatible units: length in meters and area in square meters. To do this, convert the wire's length and area to the appropriate units. For the cross-sectional area given in mm², convert it to m² by multiplying by \(10^{-6}\). For kilometers, convert to meters by multiplying by \(1000\). By applying these conversions, you can calculate that the resistivity of the wire in the exercise is approximately \(1.69 \times 10^{-8} \, \Omega \cdot \mathrm{m}\). This value is very close to that of copper, indicating the wire's material.

Copper Wire

Copper is a widely used material in electrical engineering because of its excellent conductivity. The resistivity of copper is approximately \( 1.68 \times 10^{-8} \, \Omega \cdot \mathrm{m} \). It is known for having one of the lowest resistivities, which makes it highly efficient in conducting electricity with minimal resistance and energy loss. Knowing copper's resistivity helps in identifying the material of a wire when performing resistivity calculations.Copper wires are:

• Efficient conductors of electricity, making them ideal for electrical components and wiring.
• Durable and robust, allowing them to withstand environmental and mechanical stresses.

In the exercise, when the calculated resistivity (\(1.69 \times 10^{-8} \, \Omega \cdot \mathrm{m}\)) was determined, it closely matched that of copper, confirming the material of the wire. As copper is a common choice for wiring, understanding its resistivity properties is essential for correctly designing and implementing electrical systems.