Understanding how the physical dimensions of a wire affect its electrical resistance is a fundamental concept in physics. Resistance is a measure of how much a material opposes the flow of electric current. For a metallic wire, resistance depends on its material, length, and cross-sectional area.

Factors Affecting Wire Resistance

The resistance ($R$) of a uniform conductor is directly proportional to its length ($L$) and inversely proportional to its cross-sectional area ($A$). The relationship is given by the formula:

\begin{equation*} R = \rho \frac{L}{A} \end{equation*}

where $\rho$ (rho) is the resistivity of the material, a property specific to the substance from which the wire is made. Resistivity depends on temperature and the material itself, but not on the dimensions of the wire.

Calculating the Increased Resistance

Let's consider the original wire with the given properties:

• Original Length = $L_{original} = L$
• Original Cross-sectional Area = $A_{original} = A$
• Original Resistance = $R_{original} = R$
• Resistivity of the material = $\rho$

Using the resistance formula, the original resistance is:

\begin{equation*} R = \rho \frac{L}{A} \quad \text{(Equation 1)} \end{equation*}

Now, the wire's dimensions are changed:

• New Length = $L_{new} = 2L$ (length is doubled)
• New Cross-sectional Area = $A_{new} = A/2$ (area is halved)
• Resistivity remains the same as the material is unchanged = $\rho$

Let the new resistance be $R_{new}$. Using the resistance formula for the new dimensions:

\begin{equation*} R_{new} = \rho \frac{L_{new}}{A_{new}} \end{equation*}

Substitute the new length and new area into the formula:

\begin{equation*} R_{new} = \rho \frac{2L}{A/2} \end{equation*}

To simplify the expression, we can rewrite it as:

\begin{equation*} R_{new} = \rho \frac{2L \times 2}{A} \end{equation*}

\begin{equation*} R_{new} = \rho \frac{4L}{A} \end{equation*}

Now, we can relate the new resistance to the original resistance by rearranging the terms:

\begin{equation*} R_{new} = 4 \times \left(\rho \frac{L}{A}\right) \end{equation*}

From Equation 1, we know that $R = \rho \frac{L}{A}$. Substitute this into the expression for $R_{new}$:

\begin{equation*} R_{new} = 4R \end{equation*}

Therefore, the new resistance of the wire is four times the original resistance.

The increase in resistance is $R_{new} - R_{original} = 4R - R = 3R$. However, the question asks for the increased resistance, which usually means the *new* resistance value under the changed conditions relative to the original R. Looking at the options, they represent the *new* resistance. Thus, the new resistance is 4R.

Step-by-Step Solution Summary

1. Recall the formula for resistance of a wire: $R = \rho \frac{L}{A}$.
2. Identify the original parameters: $L_{original} = L$, $A_{original} = A$, $R_{original} = R$.
3. Identify the new parameters: $L_{new} = 2L$, $A_{new} = A/2$.
4. Apply the formula for the new resistance: $R_{new} = \rho \frac{L_{new}}{A_{new}}$.
5. Substitute the new dimensions: $R_{new} = \rho \frac{2L}{A/2}$.
6. Simplify the expression: $R_{new} = \rho \frac{4L}{A}$.
7. Factor out the original resistance formula: $R_{new} = 4 \times \left(\rho \frac{L}{A}\right)$.
8. Substitute $R = \rho \frac{L}{A}$: $R_{new} = 4R$.

When the length of a wire is doubled and its cross-sectional area is halved, the resistance increases by a factor of 4.

Revision Table: Electrical Resistance Concepts

Additional Information: Resistivity and Conductivity

Resistivity ($\rho$) is a fundamental property of the material itself, indicating how strongly it resists electric current. Materials with low resistivity, like copper or aluminum, are good conductors. Materials with high resistivity, like rubber or glass, are poor conductors (insulators).

The inverse of resistivity is conductivity ($\sigma$). Conductivity measures how well a material conducts electric current. The relationship is $\sigma = 1/\rho$. Good conductors have high conductivity, while insulators have low conductivity.

The formula $R = \rho \frac{L}{A}$ shows that for a given material (constant $\rho$), increasing the length increases resistance (longer path for electrons) and increasing the cross-sectional area decreases resistance (wider path for electrons). This explains why changing the dimensions significantly impacts the resistance.