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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 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 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.