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

Options

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

Answer

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

### Step 1: Understand the Resistance of Each Wire
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{p \cdot L}{A} \]
where:
- \( p \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Calculate the Resistance of Each Wire
For the first wire with resistivity \( p_1 \):
\[ R_1 = \frac{p_1 \cdot L}{A} \]

For the second wire with resistivity \( p_2 \):
\[ R_2 = \frac{p_2 \cdot L}{A} \]

### Step 3: Find the 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:
\[ R = \frac{(p_1 + p_2) \cdot L}{A} \]

### Step 4: Express the Total Resistance in Terms of Equivalent Resistivity
Now, we can express the total resistance \( R \) in terms of an equivalent resistivity \( p \):
\[ R = \frac{p \cdot (2L)}{A} \]
Here, we consider the two wires as a single wire of length \( 2L \) and the same cross-sectional area \( A \).

### Step 5: Set the Two Expressions for Resistance Equal
Setting the two expressions for resistance equal gives:
\[ \frac{(p_1 + p_2) \cdot L}{A} = \frac{p \cdot (2L)}{A} \]

### Step 6: Simplify to Find Equivalent Resistivity
We can cancel \( L \) and \( A \) from both sides:
\[ p_1 + p_2 = 2p \]
Solving for \( p \) gives:
\[ p = \frac{p_1 + p_2}{2} \]

### Conclusion
The equivalent resistivity \( p \) of the two wires connected in series is:
\[ p = \frac{p_1 + p_2}{2} \]