Understanding Wire Resistance and Diameter

The question asks what happens to the resistance of a wire if its diameter increases. This involves understanding the factors that affect the electrical resistance of a conductor.

Electrical resistance ($R$) is a measure of how much a material opposes the flow of electric current. For a conductor with uniform cross-section, the resistance depends on three main factors:

1. The material of the conductor (resistivity, $\rho$)
2. The length of the conductor ($L$)
3. The cross-sectional area of the conductor ($A$)

The relationship between resistance, resistivity, length, and cross-sectional area is given by the formula:

$$R = \rho \frac{L}{A}$$

Here,

• $R$ is the resistance.
• $\rho$ (rho) is the resistivity of the material, a constant value specific to the substance at a given temperature.
• $L$ is the length of the conductor.
• $A$ is the cross-sectional area of the conductor.

How Cross-Sectional Area Relates to Diameter

For a wire, the cross-section is typically circular. The area ($A$) of a circle is given by the formula $A = \pi r^2$, where $r$ is the radius. The diameter ($d$) is twice the radius, so $r = d/2$.

Substituting $r = d/2$ into the area formula, we get:

$$A = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4}$$

Now, let's substitute this expression for $A$ into the resistance formula:

$$R = \rho \frac{L}{\frac{\pi d^2}{4}} = \rho \frac{4L}{\pi d^2}$$

Analyzing the Relationship Between Resistance and Diameter

From the formula $R = \rho \frac{4L}{\pi d^2}$, we can see how resistance ($R$) relates to diameter ($d$). Assuming the material ($\rho$) and length ($L$) of the wire remain constant, we can observe that $R$ is inversely proportional to the square of the diameter ($d^2$).

$$R \propto \frac{1}{d^2} \quad (\text{when } \rho \text{ and } L \text{ are constant})$$

This means that if the diameter ($d$) increases, the term $d^2$ increases, and since $R$ is inversely proportional to $d^2$, the resistance ($R$) will decrease.

Think of it intuitively: a thicker wire (larger diameter) provides more space for the electrons to flow, similar to how a wider pipe allows more water to flow. More space for charge carriers means less opposition to current flow, hence lower resistance.

Evaluating the Options

Based on the relationship $R \propto \frac{1}{d^2}$, we can analyse the given options:

1. fluctuates: This is incorrect. If the diameter increases, the resistance changes in a predictable way, not randomly fluctuates (assuming other factors like temperature and length are constant).

2. increases: This is incorrect. As the diameter increases, the cross-sectional area increases, which leads to a decrease in resistance.

3. decreases: This is correct. Resistance is inversely proportional to the square of the diameter. A larger diameter means lower resistance.

4. remains the same: This is incorrect. The resistance is directly dependent on the cross-sectional area, which is determined by the diameter. Changing the diameter changes the area and thus the resistance.

Therefore, if the diameter of a wire increases, its resistance decreases.

Revision Table: Factors Affecting Resistance

Additional Information: Resistivity and Conductivity

Resistivity ($\rho$) is an intrinsic property of a material that quantifies how strongly it resists electric current. It is measured in ohm-meters ($\Omega \cdot \text{m}$). Materials with low resistivity (like copper, aluminum) are good conductors, while materials with high resistivity (like rubber, glass) are good insulators.

Conductivity ($\sigma$) is the reciprocal of resistivity, $\sigma = 1/\rho$. It measures how easily electric current flows through a material. It is measured in siemens per meter ($\text{S}/\text{m}$). Materials with high conductivity are good conductors.

The resistance of a wire depends on its geometry (length and cross-sectional area) and the material it is made of (resistivity). The formula $R = \rho L/A$ clearly shows these dependencies.