Solution:

The resistance $R$ of a wire is given by the formula:

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

Where:

• $\rho$ is the resistivity of the material (which is constant for copper),
• $l$ is the length of the wire,
• $A$ is the cross-sectional area of the wire.

Given:

• Wire 1: Length = $l$, Area = $A$
• Wire 2: Length = $2l$, Area = $\frac{A}{2}$
• Wire 3: Length = $\frac{l}{2}$, Area = $2A$

We will calculate the resistance for each wire.

Step 1: Calculate the resistance for each wire

Wire 1: Length = $l$, Area = $A$

$$R_1 = \rho \frac{l}{A}$$

Wire 2: Length = $2l$, Area = $\frac{A}{2}$

$$R_2 = \rho \frac{2l}{\frac{A}{2}} = \rho \frac{2l \times 2}{A} = \rho \frac{4l}{A}$$

Wire 3: Length = $\frac{l}{2}$, Area = $2A$

$$R_3 = \rho \frac{\frac{l}{2}}{2A} = \rho \frac{l}{4A}$$

Step 2: Compare the resistances

• $R_1 = \rho \frac{l}{A}$
• $R_2 = \rho \frac{4l}{A}$
• $R_3 = \rho \frac{l}{4A}$

Clearly, $R_3$ is the smallest resistance because $\frac{l}{4A}$ is the smallest fraction compared to $\frac{l}{A}$ and $\frac{4l}{A}$.

Step 3: Conclusion

The wire with the minimum resistance is Wire 3, which has a cross-sectional area of $2A$.

Thus, the answer is Wire 3.