Two wires of the same dimensions but resistivities `rho_(1)` and `rho_(2)` are connected in series. The equivalent resistivity of the combination is

Options

A. `(rho_(1) + rho_(2))/(2)`
B. `rho_(1) + rho_(2)`
C. `2 (rho_(1) + rho_(2))`
D. `sqrt(rho_(1) rho_(2))`

Explanation

To find the equivalent resistivity of two wires of the same dimensions but different resistivities (\( \rho_1 \) and \( \rho_2 \)) connected in series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**: We have two wires of the same length \( L \) and cross-sectional area \( A \). The resistivities of the wires are \( \rho_1 \) and \( \rho_2 \).

2. **Calculate the Resistance of Each Wire**:
- The resistance \( R_1 \) of the first wire can be calculated using the formula:
\[ R_1 = \frac{\rho_1 \cdot L}{A} \]
- The resistance \( R_2 \) of the second wire is:
\[ R_2 = \frac{\rho_2 \cdot L}{A} \]

3. **Combine the Resistances**: Since the two resistances are in series, the total or equivalent resistance \( R \) is given by:
\[ R = R_1 + R_2 = \frac{\rho_1 \cdot L}{A} + \frac{\rho_2 \cdot L}{A} \]

4. **Factor Out Common Terms**: We can factor out \( \frac{L}{A} \) from the equation:
\[ R = \frac{L}{A} (\rho_1 + \rho_2) \]

5. **Determine the Equivalent Resistivity**: The equivalent resistivity \( \rho \) for the combination of the two wires can be defined as:
\[ R = \frac{\rho \cdot (2L)}{A} \]
Here, the length of the combined wire is \( 2L \) because we have two wires in series.

6. **Set the Two Expressions for Resistance Equal**:
\[ \frac{L}{A} (\rho_1 + \rho_2) = \frac{\rho \cdot (2L)}{A} \]

7. **Solve for \( \rho \)**:
- Cancel \( \frac{L}{A} \) from both sides (assuming \( L \) and \( A \) are not zero):
\[ \rho_1 + \rho_2 = 2\rho \]
- Rearranging gives:
\[ \rho = \frac{\rho_1 + \rho_2}{2} \]

### Final Answer:
The equivalent resistivity of the combination is:
\[ \rho = \frac{\rho_1 + \rho_2}{2} \]