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.