Two metal wires of identical dimesnios are connected in series. If `sigma_(1)` and `sigma_(2)` are the conducties of the metal wires respectively, the effective conductivity of the combination is

Options

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

Explanation

To find the effective conductivity of two metal wires of identical dimensions connected in series, we will follow these steps:

### Step 1: Understand the relationship between conductivity and resistivity
Conductivity (σ) is the reciprocal of resistivity (ρ). Therefore, we have:
\[ \sigma = \frac{1}{\rho} \]
This means that resistivity can be expressed as:
\[ \rho = \frac{1}{\sigma} \]

### Step 2: Write the formula for resistance
The resistance (R) of a wire can be expressed in terms of its resistivity, length (L), and cross-sectional area (A):
\[ R = \frac{\rho L}{A} \]
Substituting the expression for resistivity, we have:
\[ R = \frac{L}{\sigma A} \]

### Step 3: Calculate the individual resistances
For two wires with conductivities σ₁ and σ₂, the resistances R₁ and R₂ can be written as:
\[ R_1 = \frac{L}{\sigma_1 A} \]
\[ R_2 = \frac{L}{\sigma_2 A} \]

### Step 4: Find the effective resistance
Since the two wires are connected in series, the total or effective resistance (R_effective) is the sum of the individual resistances:
\[ R_{\text{effective}} = R_1 + R_2 = \frac{L}{\sigma_1 A} + \frac{L}{\sigma_2 A} \]

### Step 5: Simplify the effective resistance expression
Factoring out common terms, we get:
\[ R_{\text{effective}} = \frac{L}{A} \left( \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \right) \]

### Step 6: Relate effective resistance to effective conductivity
Let the effective conductivity be σ_effective. We can express the effective resistance in terms of effective conductivity:
\[ R_{\text{effective}} = \frac{L}{\sigma_{\text{effective}} A} \]
Setting the two expressions for R_effective equal to each other:
\[ \frac{L}{\sigma_{\text{effective}} A} = \frac{L}{A} \left( \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \right) \]

### Step 7: Cancel common terms and solve for effective conductivity
Canceling L and A from both sides gives:
\[ \frac{1}{\sigma_{\text{effective}}} = \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \]
Taking the reciprocal of both sides, we find:
\[ \sigma_{\text{effective}} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2} \]

### Final Result
Thus, the effective conductivity of the combination of the two metal wires is:
\[ \sigma_{\text{effective}} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2} \]

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