Question

What is the resistance of an aluminum wire of length \(2.0 \mathrm{~m}\), cross- sectional area \(0.12 \mathrm{~mm}^{2}\) ? The conductivity of aluminum at room temperature is \(60.7 \mathrm{MS}^{-\mathrm{m}^{-1}}\).

Short answer

The resistance of the aluminum wire is approximately \(0.274 \Omega\).

Step-by-step solution

Understand the relationship between resistance, conductivity, length, and cross-sectional area

The resistance (\(R\)) of a wire can be calculated using the formula: \(R = \frac{L}{\sigma A}\), where \(L\) is the length of the wire, \(\sigma\) is the electrical conductivity of the material, and \(A\) is the cross-sectional area of the wire.

Convert the area from mm² to m²

Initially, convert the cross-sectional area from \(\text{mm}^2\) to \(\text{m}^2\) to be consistent with the given conductivity units. \(\text{Area in} \text{m}^2 = \text{Area in} \text{mm}^2 \times 10^{-6}\). \(\text{So}, 0.12 \text{mm}^2 = 0.12 \times 10^{-6} \text{m}^2\).

Insert known values into the resistance formula

Substitute the length (\(L = 2.0 \text{m}\)), the conductivity (\(\sigma = 60.7 \times 10^6 \text{S/m}\)), and the converted cross-sectional area into the resistance formula: \(R = \frac{2.0}{60.7 \times 10^6 \times 0.12 \times 10^{-6}}\).

Calculate the resistance

Perform the calculation to find the resistance of the aluminum wire: \(R = \frac{2.0}{60.7 \times 10^6 \times 0.12 \times 10^{-6}} = \frac{2.0}{60.7 \times 0.12} \times 10^{6}\). Completing the division inside the bracket and then multiplying by \(10^6\) will give you the resistance in ohms (\(\Omega\)).

Key concepts

Electrical Conductivity

Electrical conductivity is a fundamental property that defines how well a material can conduct electricity. This concept is crucial when working with various materials in electronics and electrical engineering. Conductivity, denoted by the Greek symbol \( \sigma \), is the inverse of resistivity. When a material has high conductivity, it means that electric charges, typically electrons, can move through it more readily.

Understanding the concept of electrical conductivity involves knowing that substances like metals, particularly aluminum, copper, and silver, have high conductivities which means they make excellent conductors. Conversely, materials with low conductivity, such as rubber or glass, are considered insulators. The units of electrical conductivity are siemens per meter (S/m).

In the context of the exercise, the conductivity of aluminum is given as \(60.7 \text{ MS/m}\), which is a relatively high value, indicating aluminum's suitability as a wire material.

Cross-Sectional Area Conversion

The cross-sectional area of a wire is a significant factor influencing its electrical resistance. For accurate resistance calculations, it's essential to convert the area into consistent units. In many cases, as in our sample exercise, the area is given in square millimeters (mm²) but needs to be converted to square meters (m²) to match the units of other measurements in the formula.

To convert from mm² to m², you multiply by \(10^{-6}\), since there are one million square millimeters in a square meter. Through this conversion, \(0.12 \text{ mm}^2\) becomes \(0.12 \times 10^{-6} \text{ m}^2\). Remembering this conversion is crucial for correctly inputting values into resistance formulas and ensures that unit consistency is maintained.

Resistance Formula

The resistance formula provides the bedrock for many calculations in electrical engineering. According to Ohm's Law, resistance \(R\), measured in ohms (Ω), is calculated by the formula \(R = \frac{L}{\sigma A}\), where \(L\) represents the length of the conductor in meters (m), \(\sigma\) is the electrical conductivity in siemens per meter (S/m), and \(A\) signifies the cross-sectional area in square meters (m²).

By integrating this formula, we can analyze how various properties of the wire, such as its length, material (conductivity), and thickness (cross-sectional area), affect its ability to resist electrical flow. A key point to remember is that resistance is directly proportional to the wire's length and inversely proportional to both its conductivity and cross-sectional area. Therefore, to decrease resistance, one can either shorten the length of the conductor, increase its cross-sectional area, or use a material with higher conductivity.

In the given exercise, after converting all units appropriately and substituting the values into the resistance formula, we determine the wire's resistance. This approach is both a practical skill and an application of theoretical principles.