There are three copper wires of length and cross-sectional area (L,A),(2L,A/2)(L/2,2A). In which case in the resistance minimum?

Options

A. it is the same in all three cases
B. Wire of cross-sectional are 2A
C. Wire of cross-sectional area A
D. Wire of cross-sectional are `(1)/(2)`A

Answer

To find out in which case the resistance is minimum among the three copper wires, we can use the formula for resistance:

\[ R = \rho \frac{L}{A} \]

where:
- \( R \) is the resistance,
- \( \rho \) 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.

### Step-by-Step Solution:

1. **Identify the parameters for each wire:**
- **Wire 1:** Length \( L \), Area \( A \)
- **Wire 2:** Length \( 2L \), Area \( \frac{A}{2} \)
- **Wire 3:** Length \( \frac{L}{2} \), Area \( 2A \)

2. **Calculate the resistance for each wire:**
- **Resistance of Wire 1:**
\[ R_1 = \rho \frac{L}{A} \]
- **Resistance of Wire 2:**
\[ R_2 = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \cdot 2}{A} = \rho \frac{4L}{A} \]
- **Resistance of Wire 3:**
\[ R_3 = \rho \frac{\frac{L}{2}}{2A} = \rho \frac{L}{4A} \]

3. **Compare the resistances:**
- From the calculations:
- \( R_1 = \rho \frac{L}{A} \)
- \( R_2 = \rho \frac{4L}{A} \)
- \( R_3 = \rho \frac{L}{4A} \)

4. **Determine which resistance is the smallest:**
- Comparing \( R_1, R_2, \) and \( R_3 \):
- \( R_1 \) is greater than \( R_3 \) because \( R_1 = \rho \frac{L}{A} \) and \( R_3 = \rho \frac{L}{4A} \).
- \( R_2 \) is the largest because \( R_2 = \rho \frac{4L}{A} \).
- Therefore, \( R_3 < R_1 < R_2 \).

5. **Conclusion:**
- The resistance is minimum in **Wire 3**.

### Final Answer:
The wire with minimum resistance is the third wire (length \( \frac{L}{2} \), area \( 2A \)).