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

Options

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

Answer

To find the equivalent resistivity of two wires of the same dimension but different resistivities \( p_1 \) and \( p_2 \) connected in series, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Resistances**:
- Let the resistances of the two wires be \( R_1 \) and \( R_2 \).
- The resistivity of the first wire is \( p_1 \) and the second wire is \( p_2 \).
- Both wires have the same length \( L \) and the same cross-sectional area \( A \).

2. **Calculate Individual Resistances**:
- The resistance of the first wire can be calculated using the formula:
\[ R_1 = \frac{p_1 \cdot L}{A} \]
- Similarly, the resistance of the second wire is:
\[ R_2 = \frac{p_2 \cdot L}{A} \]

3. **Calculate Total Resistance in Series**:
- When resistors are connected in series, the total resistance \( R \) is the sum of the individual resistances:
\[ R = R_1 + R_2 = \frac{p_1 \cdot L}{A} + \frac{p_2 \cdot L}{A} \]
- This can be simplified to:
\[ R = \frac{(p_1 + p_2) \cdot L}{A} \]

4. **Express Total Resistance in Terms of Equivalent Resistivity**:
- We can express the total resistance \( R \) in terms of an equivalent resistivity \( P \):
\[ R = \frac{P \cdot (2L)}{A} \]
- Here, \( 2L \) is the combined length of the two wires when treated as a single wire.

5. **Equate the Two Expressions for Resistance**:
- Now, we can set the two expressions for \( R \) equal to each other:
\[ \frac{(p_1 + p_2) \cdot L}{A} = \frac{P \cdot (2L)}{A} \]

6. **Cancel Common Terms**:
- Since \( L \) and \( A \) are common in both sides, we can cancel them out:
\[ p_1 + p_2 = 2P \]

7. **Solve for Equivalent Resistivity \( P \)**:
- Rearranging the equation gives us:
\[ P = \frac{p_1 + p_2}{2} \]

### Final Answer:
The equivalent resistivity of the combination is:
\[ P = \frac{p_1 + p_2}{2} \]